diff --git a/APMO/md/en-apmo2025_sol.md b/APMO/md/en-apmo2025_sol.md
index 49a6a55ade87f90d80cb66cfe55afff85d3b5654..1174d96b48d5a4640a11ad627dd5d0cc3fe25674 100644
--- a/APMO/md/en-apmo2025_sol.md
+++ b/APMO/md/en-apmo2025_sol.md
@@ -9,7 +9,7 @@ Answer: No.
First notice that, since angles \(\angle A A_{1}B_{1}\) and \(\angle A A_{1}C_{1}\) are both right, the points \(B_{1}\) and \(C_{1}\) lie on the circle with \(A A_{1}\) as a diameter. Therefore, \(A C_{1} = A A_{1}\sin \angle A A_{1}C_{1} = A A_{1}\sin (90^{\circ}-\) \(\angle A_{1}A C) = A A_{1}\sin \angle C\) , similarly \(A B_{1} = A A_{1}\sin \angle B\) , and \(B_{1}C_{1} = A A_{1}\sin \angle A\) . Hence; triangles \(A C_{1}B_{1}\) and \(A B C\) are similar.
-
+
Let \(O\) be the circumcenter of \(A B C\) and \(A D\) be one of its diameters. Since \(\angle O A C = \frac{1}{2} (180^{\circ}-\) \(\angle A O C) = 90^{\circ} - \angle B = 90^{\circ} - \angle A C_{1}B_{1}\) , it follows that \(A D\) is perpendicular to \(B_{1}C_{1}\) . Let \(A D = 2R\) ; recall that, from the law of sines, \(\frac{B C}{\sin\angle A} = 2R\iff B C = 2R\sin \angle A\) . The area of quadrilateral \(A B_{1}D C_{1}\) is
diff --git a/APMO/segmented/en-apmo2025_sol.jsonl b/APMO/segmented/en-apmo2025_sol.jsonl
index 4fa29fba02d68e16c375c01d691ef8fc8bbf761f..412eeaf3c7e51364e24eab2e8f3842fe2069450f 100644
--- a/APMO/segmented/en-apmo2025_sol.jsonl
+++ b/APMO/segmented/en-apmo2025_sol.jsonl
@@ -1,4 +1,4 @@
-{"year": "2025", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "APMO", "problem": "Let \\(A B C\\) be an acute triangle inscribed in a circle \\(\\Gamma\\) . Let \\(A_{1}\\) be the orthogonal projection of \\(A\\) onto \\(B C\\) so that \\(A A_{1}\\) is an altitude. Let \\(B_{1}\\) and \\(C_{1}\\) be the orthogonal projections of \\(A_{1}\\) onto \\(A B\\) and \\(A C\\) , respectively. Point \\(P\\) is such that quadrilateral \\(A B_{1}P C_{1}\\) is convex and has the same area as triangle \\(A B C\\) . Is it possible that \\(P\\) strictly lies in the interior of circle \\(\\Gamma\\) ? Justify your answer. \n\nAnswer: No.", "solution": "First notice that, since angles \\(\\angle A A_{1}B_{1}\\) and \\(\\angle A A_{1}C_{1}\\) are both right, the points \\(B_{1}\\) and \\(C_{1}\\) lie on the circle with \\(A A_{1}\\) as a diameter. Therefore, \\(A C_{1} = A A_{1}\\sin \\angle A A_{1}C_{1} = A A_{1}\\sin (90^{\\circ}-\\) \\(\\angle A_{1}A C) = A A_{1}\\sin \\angle C\\) , similarly \\(A B_{1} = A A_{1}\\sin \\angle B\\) , and \\(B_{1}C_{1} = A A_{1}\\sin \\angle A\\) . Hence; triangles \\(A C_{1}B_{1}\\) and \\(A B C\\) are similar. \n\n\n \n\nLet \\(O\\) be the circumcenter of \\(A B C\\) and \\(A D\\) be one of its diameters. Since \\(\\angle O A C = \\frac{1}{2} (180^{\\circ}-\\) \\(\\angle A O C) = 90^{\\circ} - \\angle B = 90^{\\circ} - \\angle A C_{1}B_{1}\\) , it follows that \\(A D\\) is perpendicular to \\(B_{1}C_{1}\\) . Let \\(A D = 2R\\) ; recall that, from the law of sines, \\(\\frac{B C}{\\sin\\angle A} = 2R\\iff B C = 2R\\sin \\angle A\\) . The area of quadrilateral \\(A B_{1}D C_{1}\\) is \n\n\\[\\frac{B_{1}C_{1}\\cdot A D}{2} = \\frac{A A_{1}\\sin\\angle A\\cdot 2R}{2} = \\frac{A A_{1}\\cdot B C}{2},\\] \n\nwhich is indeed the area of \\(A B C\\) . \n\nSince \\(B_{1}\\) and \\(C_{1}\\) are fixed points, the loci of the points \\(P\\) such that \\(A B_{1}P C_{1}\\) is a convex quadrilateral with the same area as \\(A B C\\) is a line parallel to \\(B_{1}C_{1}\\) . That is, perpendicular to \\(A D\\) . Since the area of \\(A B_{1}D C_{1}\\) is the same as the area of \\(A B C\\) , this locus is the line perpendicular to \\(A D\\) through \\(D\\) , which is tangent to the circumcircle of \\(A B C\\) . Therefore, it is not possible that the point \\(P\\) lies inside the circumcircle of \\(A B C\\) .", "metadata": {"resource_path": "APMO/segmented/en-apmo2025_sol.jsonl", "problem_match": "# Problem 1 ", "solution_match": "# Solution \n\n"}}
+{"year": "2025", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "APMO", "problem": "Let \\(A B C\\) be an acute triangle inscribed in a circle \\(\\Gamma\\) . Let \\(A_{1}\\) be the orthogonal projection of \\(A\\) onto \\(B C\\) so that \\(A A_{1}\\) is an altitude. Let \\(B_{1}\\) and \\(C_{1}\\) be the orthogonal projections of \\(A_{1}\\) onto \\(A B\\) and \\(A C\\) , respectively. Point \\(P\\) is such that quadrilateral \\(A B_{1}P C_{1}\\) is convex and has the same area as triangle \\(A B C\\) . Is it possible that \\(P\\) strictly lies in the interior of circle \\(\\Gamma\\) ? Justify your answer. \n\nAnswer: No.", "solution": "First notice that, since angles \\(\\angle A A_{1}B_{1}\\) and \\(\\angle A A_{1}C_{1}\\) are both right, the points \\(B_{1}\\) and \\(C_{1}\\) lie on the circle with \\(A A_{1}\\) as a diameter. Therefore, \\(A C_{1} = A A_{1}\\sin \\angle A A_{1}C_{1} = A A_{1}\\sin (90^{\\circ}-\\) \\(\\angle A_{1}A C) = A A_{1}\\sin \\angle C\\) , similarly \\(A B_{1} = A A_{1}\\sin \\angle B\\) , and \\(B_{1}C_{1} = A A_{1}\\sin \\angle A\\) . Hence; triangles \\(A C_{1}B_{1}\\) and \\(A B C\\) are similar. \n\n\n \n\nLet \\(O\\) be the circumcenter of \\(A B C\\) and \\(A D\\) be one of its diameters. Since \\(\\angle O A C = \\frac{1}{2} (180^{\\circ}-\\) \\(\\angle A O C) = 90^{\\circ} - \\angle B = 90^{\\circ} - \\angle A C_{1}B_{1}\\) , it follows that \\(A D\\) is perpendicular to \\(B_{1}C_{1}\\) . Let \\(A D = 2R\\) ; recall that, from the law of sines, \\(\\frac{B C}{\\sin\\angle A} = 2R\\iff B C = 2R\\sin \\angle A\\) . The area of quadrilateral \\(A B_{1}D C_{1}\\) is \n\n\\[\\frac{B_{1}C_{1}\\cdot A D}{2} = \\frac{A A_{1}\\sin\\angle A\\cdot 2R}{2} = \\frac{A A_{1}\\cdot B C}{2},\\] \n\nwhich is indeed the area of \\(A B C\\) . \n\nSince \\(B_{1}\\) and \\(C_{1}\\) are fixed points, the loci of the points \\(P\\) such that \\(A B_{1}P C_{1}\\) is a convex quadrilateral with the same area as \\(A B C\\) is a line parallel to \\(B_{1}C_{1}\\) . That is, perpendicular to \\(A D\\) . Since the area of \\(A B_{1}D C_{1}\\) is the same as the area of \\(A B C\\) , this locus is the line perpendicular to \\(A D\\) through \\(D\\) , which is tangent to the circumcircle of \\(A B C\\) . Therefore, it is not possible that the point \\(P\\) lies inside the circumcircle of \\(A B C\\) .", "metadata": {"resource_path": "APMO/segmented/en-apmo2025_sol.jsonl", "problem_match": "# Problem 1 ", "solution_match": "# Solution \n\n"}}
{"year": "2025", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "APMO", "problem": "Let \\(\\alpha\\) and \\(\\beta\\) be positive real numbers. Emerald makes a trip in the coordinate plane, starting off from the origin \\((0,0)\\) . Each minute she moves one unit up or one unit to the right, restricting herself to the region \\(|x - y|< 2025\\) , in the coordinate plane. By the time she visits a point \\((x,y)\\) she writes down the integer \\(\\lfloor x\\alpha +y\\beta \\rfloor\\) on it. It turns out that Emerald wrote each non- negative integer exactly once. Find all the possible pairs \\((\\alpha ,\\beta)\\) for which such a trip would be possible. \n\nAnswer: \\((\\alpha ,\\beta)\\) such that \\(\\alpha +\\beta = 2\\)", "solution": "Let \\((x_{n},y_{n})\\) be the point that Emerald visits after \\(n\\) minutes. Then \\((x_{n + 1},y_{n + 1})\\in \\{(x_{n}+\\) \\(1,y_{n}),(x_{n},y_{n} + 1)\\}\\) . Either way, \\(x_{n + 1} + y_{n + 1} = x_{n} + y_{n} + 1\\) , and since \\(x_{0} + y_{0} = 0 + 0 = 0\\) \\(x_{n} + y_{n} = n\\) \n\nThe \\(n\\) - th number would be then \n\n\\[z_{n} = \\lfloor x_{n}\\alpha +(n - x_{n})\\beta \\rfloor \\Longrightarrow n\\beta +x_{n}(\\alpha -\\beta) - 1< z_{n}< n\\beta +x_{n}(\\alpha -\\beta),\\] \n\nin which \n\n\\[-2025< x_{n} - y_{n}< 2025\\iff \\frac{n - 2025}{2} < x_{n}< \\frac{n + 2025}{2}.\\] \n\nSuppose without loss of generality that \\(\\alpha \\geq \\beta\\) . Then \n\n\\[n\\beta +\\frac{n - 2025}{2} (\\alpha -\\beta) - 1< z_{n}< n\\beta +\\frac{n + 2025}{2} (\\alpha -\\beta),\\] \n\nwhich reduces to \n\n\\[\\left|z_{n} - \\frac{\\alpha + \\beta}{2} n\\right|< \\frac{2025}{2} (\\alpha -\\beta) + 1.\\] \n\nOn the other hand, \\(z_{n + 1} = \\lfloor x_{n + 1}\\alpha +y_{n + 1}\\beta \\rfloor \\in \\{\\lfloor x_{n}\\alpha +y_{n}\\beta +\\alpha \\rfloor ,\\lfloor x_{n}\\alpha +y_{n}\\beta +\\beta \\rfloor \\}\\) , which implies \\(z_{n + 1}\\geq z_{n}\\) . Since every non- negative integer appears exactly once, in increasing order, it follows that \\(z_{n} = n\\) . \n\nTherefore, for all positive integers \\(n\\) \n\n\\[\\left|n - \\frac{\\alpha + \\beta}{2} n\\right|< \\frac{2025}{2} (\\alpha -\\beta) + 1,\\] \n\nwhich can only be possible if \\(\\alpha +\\beta = 2\\) ; otherwise, the left hand side would be unbounded. If \\(\\alpha +\\beta = 2\\) , consider \\(x_{n} = \\left\\lfloor \\frac{n}{2}\\right\\rfloor\\) and \\(y_{n} = \\left\\lfloor \\frac{n}{2}\\right\\rfloor\\) . If \\(n\\) is even, \n\n\\[z_{n} = \\left\\lfloor \\frac{n}{2}\\alpha +\\frac{n}{2}\\beta \\right\\rfloor = n;\\] \n\nif \\(n\\) is odd, \n\n\\[z_{n} = \\left\\lfloor \\frac{n + 1}{2}\\alpha +\\frac{n - 1}{2}\\beta \\right\\rfloor = n + \\left\\lfloor \\frac{\\alpha - \\beta}{2}\\right\\rfloor ,\\] \n\nwhich equals \\(n\\) because \\(0< \\beta \\leq \\alpha < \\alpha +\\beta = 2\\Longrightarrow 0\\leq \\alpha -\\beta < 2\\)", "metadata": {"resource_path": "APMO/segmented/en-apmo2025_sol.jsonl", "problem_match": "# Problem 2 ", "solution_match": "# Solution \n\n"}}
{"year": "2025", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "APMO", "problem": "Let \\(P(x)\\) be a non- constant polynomial with integer coefficients such that \\(P(0) \\neq 0\\) . Let \\(a_{1}, a_{2}, a_{3}, \\ldots\\) be an infinite sequence of integers such that \\(P(i - j)\\) divides \\(a_{i} - a_{j}\\) for all distinct positive integers \\(i, j\\) . Prove that the sequence \\(a_{1}, a_{2}, a_{3}, \\ldots\\) must be constant, that is, \\(a_{n}\\) equals a constant \\(c\\) for all \\(n\\) positive integer.", "solution": "Let \\(a_{0} = P(0) \\neq 0\\) be the independent coefficient, i.e., the constant term of \\(P(x)\\) . Then there are infinitely many primes \\(p\\) such that \\(p\\) divides \\(P(k)\\) but \\(p\\) does not divide \\(k\\) . In fact, since \\(P(k) - a_{0}\\) is a multiple of \\(k\\) , \\(\\gcd (P(k), k) = \\gcd (k, a_{0}) \\leq a_{0}\\) is bounded, so pick, say, \\(k\\) with prime factors each larger than \\(a_{0}\\) . \n\nSince \\(P(k)\\) divides \\(a_{i + k} - a_{i}\\) , \\(p\\) divides \\(a_{i + k} - a_{i}\\) . Moreover, since \\(P(k + p) \\equiv P(k) \\equiv 0\\) (mod \\(p\\) ), \\(p\\) also divides \\(a_{i + k + p} - a_{i}\\) . Therefore, \\(a_{i}\\) mod \\(p\\) is periodic with periods \\(k + p\\) and \\(k\\) . By Bezout's theorem, \\(\\gcd (k + p, k) = 1\\) is also a period, that is, \\(p\\) divides \\(a_{i + 1} - a_{i}\\) for all \\(i\\) and \\(p\\) such that \\(p \\mid P(k)\\) and \\(p \\nmid k\\) for some \\(k\\) . Since there are infinitely many such primes \\(p\\) , \\(a_{i + 1} - a_{i}\\) is divisible by infinitely many primes, which implies \\(a_{i + 1} = a_{i}\\) , that is, the sequence is constant.", "metadata": {"resource_path": "APMO/segmented/en-apmo2025_sol.jsonl", "problem_match": "# Problem 3 ", "solution_match": "# Solution \n\n"}}
{"year": "2025", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "APMO", "problem": "Let \\(n \\geq 3\\) be an integer. There are \\(n\\) cells on a circle, and each cell is assigned either 0 or 1. There is a rooster on one of these cells, and it repeats the following operations: \n\nIf the rooster is on a cell assigned 0, it changes the assigned number to 1 and moves to the next cell counterclockwise. If the rooster is on a cell assigned 1, it changes the assigned number to 0 and moves to the cell after the next cell counterclockwise. \n\nProve that the following statement holds true after sufficiently many operations: \n\nIf the rooster is on a cell \\(C\\) , then the rooster would go around the circle exactly three times, stopping again at \\(C\\) . Moreover, every cell would be assigned the same number as it was assigned right before the rooster went around the circle 3 times.", "solution": "Reformulate the problem as a \\(n\\) - string of numbers in \\(\\{0,1\\}\\) and a position at which the action described in the problem is performed, and add 1 or 2 modulo \\(n\\) to the position according to the action. Say that a lap is complete for each time the position resets to 0 or 1. We will prove that the statement claim holds after at most two laps, after which the \\(n\\) - tuple cycles every three laps. \n\nSay the rooster stops at a position in a certain lap if it performs an action at that position on that lap; otherwise, the rooster bypasses that position. We start with some immediate claims: \n\nThe rooster has to stop at at least one of each two consecutive positions. The rooster stops at every position preceded by a 0. Indeed, if the numbers preceding that position are 00 then the rooster will definitely stop at the second zero, and it the numbers preceding that position are 10 then the rooster will either stop at 1 and go directly to the position or bypass 1 and stop at the second zero, and then stop at the position. Therefore, if the rooster bypasses a position, then it is preceded by a 1, and that 1 must be changed to a 0. This means that the rooster never bypasses a position in two consecutive laps. The rooster bypasses every position preceded by 01. Indeed, the rooster stops at either 1 or at 0, after which it will move to 1; at any rate, it stops at 1 and bypasses the position. \n\nOur goal is to prove that, eventually, for every three consecutive laps, each position is bypassed exactly once. Then each position changes states exactly twice, so it gets back to its initial state after three laps. The following two lemmata achieve this goal: \n\nLemma 1. If the rooster stops at a certain position in two laps in a row, it bypasses it on the next lap, except for the \\(n\\) - string \\(1010\\ldots 10\\) , for which the problem statement holds. \n\nProof. If the rooster stopped at a position in lap \\(t\\) , then it is preceded by either (A) a 0 that was changed to 1, (B) a 11 that was changed to 01, or (C) a 10 in which the rooster stopped at 1. In case (A), the position must be preceded by 11 in the lap \\(t + 1\\) , which becomes 01, so the rooster will bypass the position in the lap \\(t + 2\\) . In case (B), the position will be bypassed in lap \\(t + 1\\) . \n\nNow we deal with case (C): suppose that the position was preceded by \\(m\\) occurrences of 10, that is, \\((10)^{m}\\) , on lap \\(t\\) and take \\(m \\leq \\frac{n}{2}\\) maximal. The rooster stopped at the 1 from each occurrence of 10, except possibly the first one.\n\n\n\nFirst, suppose that \\(n \\geq 2m + 2\\) . After lap \\(t\\) , \\((10)^{m}\\) becomes either \\((00)^{m}\\) or \\(11(00)^{m - 1}\\) in lap \\(t + 1\\) . In the latter case, initially we had \\(1(10)^{m}\\) , which became \\(011(00)^{m - 1}\\) . It will then become \\(a01(11)^{m - 1}\\) in lap \\(t + 2\\) , in which the rooster will bypass the position. In the former case, \\((10)^{m}\\) becomes \\((00)^{m}\\) , so it was either \\(00(10)^{m}\\) or \\(11(10)^{m}\\) in lap \\(t\\) , which becomes respectively \\(a1(00)^{m}\\) and \\(01(00)^{m}\\) in lap \\(t + 1\\) , respectively. In the second sub- case, it becomes \\(b001(11)^{m - 1}\\) in lap \\(t + 2\\) , and the position will be bypassed. In the first sub- case, it must be \\(11(00)^{m}\\) after which it becomes either \\(01(11)^{m}\\) or \\(1001(11)^{m - 1}\\) . In any case, the position will be bypassed in lap \\(t + 2\\) . If \\(n = 2m + 1\\) , the possible configurations are \n\n\\[(10)^{m}0\\rightarrow (00)^{m}1\\rightarrow (11)^{m}0\\rightarrow (10)^{m}0,\\] \n\nthe rooster stops at the 1 from the first 10 because it was preceded by a 0. \n\n\\[(10)^{m}1\\rightarrow (00)^{m}0\\rightarrow 0(11)^{m}\\rightarrow 1(01)^{m},\\] \n\nor \n\n\\[(10)^{m}1\\rightarrow 11(00)^{m - 1}0\\rightarrow 01(11)^{m - 1}1\\rightarrow (10)^{m}1,\\] \n\nIn any case, the position is bypassed in lap \\(t + 2\\) . \n\nIf \\(n = 2m\\) , the entire configuration is \\((10)^{m}\\) , \\(m \\geq 2\\) . If the rooster did not stop at the first 1, it becomes \\(11(00)^{m - 1}\\) in the lap \\(t + 1\\) , then \\(01(11)^{m - 1}\\) in the lap \\(t + 2\\) , so the position is bypassed in this last lap. If the rooster stopped at the first 1, it becomes \\((00)^{m}\\) , then \\((11)^{m}\\) , then \\((01)^{m}\\) , then \\(10(00)^{m - 1}\\) , then \\((11)^{m}\\) , and then it cycles between \\((11)^{m}\\) , \\((01)^{m}\\) and \\(10(00)^{m - 1}\\) . \n\nSo, apart from this specific string, the rooster will stop at most two laps in a row at each position. \n\nLemma 2. If the rooster bypasses one position on a lap, then it stops at that position on the next two laps, with the same exception as lemma 1. \n\nProof. The position must be preceded by 1 in lap \\(t\\) . If it is preceded by 11, it changes to 10 in lap \\(t + 1\\) . Then it becomes 00 because the 1 was already skipped in the previous lap, and the rooster will stop at the position in the lap \\(t + 2\\) . \n\nNow suppose that the position was preceded by \\((01)^{m}\\) on lap \\(t\\) and take \\(m \\leq \\frac{n}{2}\\) maximal. It becomes \\(10(00)^{m - 1}\\) or \\((00)^{m}\\) in lap \\(t + 1\\) . In the former case, in lap \\(t + 2\\) it becomes either \\(00(11)^{m - 1}\\) , after which the rooster stops at the position again, or \\((11)^{m}\\) , which we'll study later. In the former case, \\((00)^{m}\\) becomes \\((11)^{m}\\) or \\(01(11)^{m - 1}\\) . In the latter case, the 0 was bypassed, so it must be sopped in the next lap, becoming \\((10)^{m}\\) . In the \\((11)^{m}\\) case, in order to bypass the position in lap \\(t + 2\\) , it must become \\((10)^{m}\\) . All in all, the preceding terms are \\((01)^{m}\\) . Then, either \\(10(00)^{m - 1}\\) or \\((00)^{m}\\) , then either \\((11)^{m}\\) or \\(01(11)^{m - 1}\\) , then \\((10)^{m}\\) . Then the second term in \\((01)^{m}\\) is 1, then 0, then 1, and then 0, that is, it changed three times. So we fall under the exception to lemma 1. \\(\\square\\) \n\nThe result then immediately follows from lemmata 1 and 2.", "metadata": {"resource_path": "APMO/segmented/en-apmo2025_sol.jsonl", "problem_match": "# Problem 4 ", "solution_match": "# Solution 1"}}
diff --git a/Balkan_MO/md/en-2025-BMO-type1.md b/Balkan_MO/md/en-2025-BMO-type1.md
index 26a670cd7a177da994ee035842e070cfc1a4d591..dac1bdd998df196a791347909ddc465e3bf9792d 100644
--- a/Balkan_MO/md/en-2025-BMO-type1.md
+++ b/Balkan_MO/md/en-2025-BMO-type1.md
@@ -105,7 +105,7 @@ Problem 2. Let \(\triangle ABC\) be an acute- angled triangle with orthocentre \
## Solution 1
-
+
We have \(\angle PCD = \angle ECD = \angle EAD\) and \(\angle PBD = \angle FBD = \angle FAD\) , therefore
@@ -133,7 +133,7 @@ As in Solution 1, we introduce the point \(Y\) and prove that the quadrilateral
Since \(\angle HZD = \angle HZX = \angle HCX = \angle HCP = \angle HYP = \angle HYD\) , we get the points \(H, D, Z, Y\) are concyclic. We also know that \(\angle DZC = \angle XZC = \angle XHC = \angle BYC =\)
-
+
\(\angle L C B\) , where the last equality holds because \(L C\) is tangent to \((B H P C)\) . This implies that the circles \((B Y C)\) and \((Z D C)\) share a common tangent at the point \(C\) . Finally, applying the radical axis theorem to the circles \((Z D C)\) , \((H D Z Y)\) and \((B H P C)\) , we conclude that the line \(X D\) passes through the point \(L\) , finishing the proof.
@@ -151,7 +151,7 @@ If we denote the intersection of \(L D\) and \(B Z\) as \(X^{\prime}\) , project
Combining the above, we have \((B,X;H,Z) = (B,D;H_{A},C) = (B,X^{\prime};H,Z)\) . It is well- known that, with 3 fixed points and a fixed cross- ratio, the fourth point is uniquely determined. This implies \(X\equiv X^{\prime}\) and we are done.
-
+
## Solution 4
@@ -302,7 +302,7 @@ Next, we show that if we have an edge between \(V_{i}\) and \(V_{j}\) , then we
\[V_{i + 1}V_{i + 2}\ldots V_{j}V_{i}V_{i - 1}\ldots V_{j + 1}\]
-
+
of length \(n\) . As before, we conclude that there is an edge between \(V_{i + 1}\) and \(V_{j + 1}\) . Repeating this, we get that if we have an edge between two vertices at distance \(r\) , then we have edges between any two vertices at distance \(r\) .
@@ -313,7 +313,7 @@ For positive integers \(a\) and \(b\) with \(a + b\leqslant n - 1\) , consider t
\[V_{1},V_{a + b},V_{a + b - 1},\ldots ,V_{a + 1},V_{a + b + 1},V_{a + b + 2},\ldots ,V_{n},V_{a},V_{a - 1},\ldots ,V_{1}.\]
-
+
The distance between two consecutive vertices in this ordering is \(1,a,b\) or \(a + b - 1\) . This implies that if two numbers from the multiset \(\{a,b,a + b - 1\}\) belong to \(S\) , so does the third one. Now, if \(2\in S\) , we take \(b = 2\) and easily get that that \(S\) contains any number from 1 to \(n - 1\) . This gives us the solution \(K_{n}\) .
@@ -328,7 +328,7 @@ If \(S = \{1, n - 1\}\) , we get the solution \(C_{n}\) . Otherwise, there exist
of length \(n\) .
-
+
Same as before, we get that there is an edge between \(V_{t}\) and \(V_{t + 3}\) . Therefore, we have \(3 \in S\) . Now, taking \(b = 3\) , we get that any odd number smaller than or equal to \(n - 1\) lies in \(S\) . Since we assumed \(S\) doesn’t contain consecutive integers, we get that \(n\) is even and \(S = \{1 \leqslant i \leqslant n - 1 | i \text{odd}\}\) . This gives us the solution \(K_{\frac{n}{2}, \frac{n}{2}}\) .
diff --git a/Balkan_MO/segmented/en-2025-BMO-type1.jsonl b/Balkan_MO/segmented/en-2025-BMO-type1.jsonl
index 6741d368723a8ff43787758927b57e08457219a7..f2a5b38129623bf199aaba59dcf8a79692d1cc0b 100644
--- a/Balkan_MO/segmented/en-2025-BMO-type1.jsonl
+++ b/Balkan_MO/segmented/en-2025-BMO-type1.jsonl
@@ -1,11 +1,11 @@
{"year": "2025", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "Balkan_MO", "problem": "An integer \\(n > 1\\) is called good if there exists a permutation \\(a_{1},a_{2},a_{3},\\ldots ,a_{n}\\) of the numbers \\(1,2,3,\\ldots ,n\\) , such that: \n\n- \\(a_{i}\\) and \\(a_{i + 1}\\) have different parities for every \\(1 \\leqslant i \\leqslant n - 1\\) ; \n\n- the sum \\(a_{1} + a_{2} + \\dots + a_{k}\\) is a quadratic residue modulo \\(n\\) for every \\(1 \\leqslant k \\leqslant n\\) . \n\nProve that there exist infinitely many good numbers, as well as infinitely many positive integers which are not good. \n\nRemark: Here an integer \\(x\\) is considered a quadratic residue modulo \\(n\\) if there exists an integer \\(y\\) such that \\(x \\equiv y^{2} \\pmod {n}\\) .", "solution": "We will split the problem into two parts - the first one proving there are infinitely many numbers that are not good, and the second part proving there are infinitely many good numbers. \n\n## Infinitely many numbers are not good \n\n## Proof #1 \n\nWe will show that all numbers \\(n = 4^{m}\\) with \\(m \\in \\mathbb{Z}^{+}\\) are not good. Indeed, consider the last sum in the given condition \n\n\\[a_{1} + a_{2} + \\dots +a_{n} = 1 + 2 + \\dots +n = \\frac{4^{m}(4^{m} + 1)}{2}\\] \n\nSuppose that there exists \\(x \\in \\mathbb{Z}\\) such that \n\n\\[\\frac{4^{m}(4^{m} + 1)}{2} \\equiv x^{2} \\pmod{4^{m}} \\Longleftrightarrow 4^{m} \\equiv 2x^{2} \\pmod{2 \\cdot 4^{m}} \\Longleftrightarrow 2 \\cdot 4^{m} \\mid 4^{m} - 2x^{2}\\] \n\nThis means that \\(4^{m} \\mid 2x^{2}\\) , that is \\(2^{2m - 1} \\mid x^{2}\\) , so \\(2^{m} \\mid x\\) . Let \\(x = c \\cdot 2^{m}\\) with \\(c \\in \\mathbb{Z}\\) . Thus \n\n\\[4^{m} \\equiv 2(2^{m}c)^{2} \\equiv 2c^{2} \\cdot 4^{m} \\equiv 0 \\pmod{2 \\cdot 4^{m}}\\] \n\nthis implies that \\(4^{m} \\equiv 0 \\pmod{2 \\cdot 4^{m}}\\) , which is not true. This proves the second part of the problem, i.e. that there are infinitely many numbers that are not good.\n\n\n\n## Proof #2.1 \n\nWe will show that all numbers \\(n = 4m\\) with \\(m\\in \\mathbb{Z}^{+}\\) are not good. Assume otherwise and let \\(a_{k} = 2\\) for some \\(1\\leqslant k\\leqslant n\\) \n\nLet \\(S_{i} = a_{1} + a_{2} + \\cdot \\cdot \\cdot +a_{i}\\) (if \\(i< 1\\) \\(S_{i}\\) is the empty sum). Let \\(S_{k - 1}\\equiv x^{2}\\) (mod \\(4m\\) ) and \\(S_{k}\\equiv y^{2}\\) (mod \\(4m\\) ). Thus \\(x^{2} + 2\\equiv y^{2}\\) (mod \\(4m\\) ), which means that \\(x\\) and \\(y\\) have the same parity. Now \\(4m\\mid (x - y)(x + y) + 2\\) , but since \\(4\\mid (x - y)(x + y)\\) , we get \\(4\\mid 2\\) , a contradiction. \n\n## Proof #2.2 \n\nWe will show that all numbers \\(n = 2^{m}\\) with \\(m\\in \\mathbb{Z}^{+}\\) , \\(m > 3\\) are not good. For the sake of contradiction, assume that \\(n\\) is good. \n\nLemma. Let \\(n = 2^{m}\\) with \\(m\\in \\mathbb{Z}^{+}\\) , \\(m > 3\\) and let \\(r\\) be an odd quadratic residue modulo \\(n\\) . Then \\(r\\equiv 1\\) (mod 8). \n\nProof. Since \\(r\\) is a quadratic residue, we know that \\(r\\equiv t^{2}\\) (mod \\(2^{m}\\) ) for some odd integer \\(t\\) . Then we have that \\(2^{m}\\mid r - t^{2}\\) , and because \\(m > 3\\) , we have that \\(8\\mid r - t^{2}\\) . Since \\(t\\) is odd, \\(t^{2}\\equiv 1\\) (mod 8), so we get that \\(r\\equiv 1\\) (mod 8). \n\nClaim. Let \\(n = 2^{m}\\) with \\(m\\in \\mathbb{Z}^{+}\\) , \\(m > 3\\) and let \\(r\\) be a quadratic residue modulo \\(n\\) . If \\(v_{2}(r)\\leqslant m - 3\\) then \\(r = 4^{a}\\cdot (8b + 1)\\) for some nonnegative integers \\(a\\) and \\(b\\) . \n\nProof. If \\(r\\) is odd, from the previous lemma we have that \\(r = 8b + 1\\) ( \\(a = 0\\) ) for some integer \\(b\\) . If \\(r = 2^{c}r_{1}\\) for some \\(1\\leqslant c\\leqslant m - 3\\) and odd \\(r_{1}\\) , we get that \\(2^{m}\\mid r - k^{2}\\) for some integer \\(k\\) . That is, we have \\(2^{m}\\mid 2^{c}r_{1} - k^{2}\\) . Let \\(k^{2} = 2^{2t}k_{1}^{2}\\) for some nonnegative integer \\(t\\) and odd integer \\(k_{1}\\) . Since \\(2^{c}\\mid k^{2}\\) , we get \\(2t\\geqslant c\\) . If \\(2t > c\\) , it follows that \\(v_{2}(2^{c}r_{1} - k^{2}) = c< m\\) , a contradiction. Therefore \\(c = 2t\\) and so \\(v_{2}(r)\\) is even. Now we have that \\(2^{m}\\mid 2^{c}r_{1} - 2^{c}k_{1}^{2}\\) , thus \\(2^{m - c}\\mid r_{1} - k_{1}^{2}\\) . Since \\(m - c\\geqslant 3\\) , we have that \\(8\\mid r_{1} - k_{1}^{2}\\) and because \\(k_{1}\\) is odd we get \\(r_{1}\\equiv 1\\) (mod 8). \n\nAssume \\(2^{m}\\) with \\(m > 3\\) is good with some permutation \\(a_{1},a_{2},\\ldots ,a_{2^{m}}\\) and let \\(a_{i} = 2\\) , for some \\(i > 1\\) (from the claim we know that 2 is not a quadratic residue modulo \\(2^{m}\\) ). Consider the following cases: \n\nCase 1. If \\(2^{m - 2}\\mid a_{1} + a_{2} + \\cdot \\cdot \\cdot +a_{i - 1}\\) . Let \\(a_{1} + a_{2} + \\cdot \\cdot \\cdot +a_{i- 1} = 2^{m- 2}c\\) for some integer \\(c\\) . Then \\(v_{2}(a_{1} + a_{2} + \\cdot \\cdot \\cdot +a_{i}) = v_{2}(2^{m- 2}c + 2) = 1\\) and from the claim this is not a quadratic residue, a contradiction. \n\nCase 2. If \\(2^{m - 2}\\mid a_{1} + a_{2} + \\cdot \\cdot \\cdot +a_{i}\\) . Let \\(a_{1} + a_{2} + \\cdot \\cdot \\cdot +a_{i} = 2^{m - 2}c\\) for some integer \\(c\\) . Then \\(v_{2}(a_{1} + a_{2} + \\cdot \\cdot \\cdot +a_{i - 1}) = v_{2}(2^{m - 2}c - 2) = 1\\) and from the claim this is not a quadratic residue, a contradiction. \n\nCase 3. Otherwise, the claim implies \\(a_{1} + a_{2} + \\cdot \\cdot \\cdot +a_{i - 1} = 4^{k_{1}}(8l_{1} + 1)\\) and \\(a_{1} + a_{2} + \\cdot \\cdot \\cdot +a_{i} =\\) \\(4^{k_{2}}(8l_{2} + 1)\\) for some nonnegative integers \\(k_{1},k_{2},l_{1},l_{2}\\) . Then we have \\(4^{k_{1}}(8l_{1} + 1) + 2 =\\) \\(4^{k_{2}}(8l_{2} + 1)\\) . Looking at the equation modulo 4, we get that at least one of \\(k_{1},k_{2}\\) is 0. If exactly one of \\(k_{1},k_{2}\\) is equal to 0 we get a contradiction modulo 2. Therefore \\(k_{1} = k_{2} = 0\\) and thus \\(8l_{1} + 3 = 8l_{2} + 1\\) , which is impossible.\n\n\n\n## Infinitely many numbers are good \n\n## Proof #1 \n\nNow let \\(n = p\\) be a prime number of the form \\(4k + 3,k\\in \\mathbb{Z}\\) . Consider the numbers \n\n\\[1^{2},2^{2},\\ldots ,\\left(\\frac{p - 1}{2}\\right)^{2},p - 1^{2},p - 2^{2},\\ldots ,p - \\left(\\frac{p - 1}{2}\\right)^{2}.\\] \n\nClearly, in this sequence, no two numbers are congruent modulo \\(p\\) . Indeed, suppose that there is \\(i^{2}\\equiv j^{2}\\) (mod \\(p\\) ) with \\(1\\leqslant i< j\\leqslant{\\frac{p - 1}{2}}\\) then \\(p\\mid (j - i)(j + i)\\) . But \\(0< j + i< p\\) \\(0< j - i< p\\) , a contradiction. From there, it follows that the first \\(\\textstyle{\\frac{p - 1}{2}}\\) numbers have distinct remainders in modulo \\(p\\) . We reason similarly for the last \\(\\textstyle{\\frac{p - 1}{2}}\\) numbers. Next suppose that there is \\(i^{2}\\equiv p - j^{2}\\) (mod \\(p\\) ) with \\(1\\leqslant i,j\\leqslant{\\frac{p - 1}{2}}\\) then \\(p\\mid i^{2} + j^{2}\\) . According to the well known properties of quadratic residues modulo a prime \\(p = 4k + 3\\) , we conclude that \\(p\\mid i\\) and \\(p\\mid j\\) which is also a contradiction. Thus, the claim is proved. \n\nNotice that for \\(1\\leqslant i\\leqslant{\\frac{p - 1}{2}}\\) , two numbers \\(i^{2}\\) and \\(p - i^{2}\\) have different parity remainders in modulo \\(p\\) (since the sum of the two remainders is \\(p\\) , which is odd). Consider the remainder of \\(1^{2},2^{2},\\ldots ,\\left(\\frac{p - 1}{2}\\right)^{2}\\) when divided by \\(p\\) . We denote by \\(a_{1},a_{2},\\ldots ,a_{m}\\) the odd remainders and by \\(b_{1},b_{2},\\ldots ,b_{n}\\) the even remainders; note that \\(m + n = \\textstyle {\\frac{p - 1}{2}}\\) . Finally, consider the following permutation: \n\n\\[a_{1},p - a_{1},a_{2},p - a_{2},\\ldots ,a_{m},p - a_{m},p,b_{1},p - b_{1},b_{2},p - b_{2},\\ldots ,b_{n},p - b_{n}\\] \n\nObviously, according to the above arguments, two consecutive numbers in the above permutation have different parity, and the sum of any first \\(i\\) numbers in the permutation is either congruent to 0 or congruent to some number in \\(\\{1^{2},2^{2},\\ldots ,\\left(\\frac{p - 1}{2}\\right)^{2}\\}\\) , which is clearly a quadratic residue modulo \\(p\\) . Thus, the constructed permutation as above satisfies the given conditions. Since there are infinitely many primes of the form \\(p = 4k + 3\\) , we have proved that there are infinitely many good numbers as well. \n\n## Proof #2 \n\nLet \\(n\\) be an odd integer. We will prove that the number \\(2n\\) is good. Consider the numbers: \n\n\\[1,3 + n,5,7 + n,9,11 + n,\\ldots ,4n - 1 + n.\\] \n\nIt can be easily proven that no two numbers in this sequence are congruent modulo \\(2n\\) . Since there are \\(2n\\) numbers in the sequence, they form a complete residue system modulo \\(2n\\) . Also, note that the sum of the first \\(k\\) ( \\(1\\leqslant k\\leqslant 2n\\) ) numbers in the sequence is a quadratic residue modulo \\(2n\\) (it is a quadratic residue modulo \\(n\\) as it is congruent to \\(1 + 3 + \\ldots +2k - 1 = k^{2}\\) modulo \\(n\\) and since \\(x^{2} + n\\equiv (x + n)^{2}\\) (mod \\(2n\\) ) for all integers \\(x\\) , it is also a quadratic residue modulo \\(2n\\) ). The parity condition is also satisfied (even after reduction by modulo \\(2n\\) , since \\(2n\\) is even). Finally, taking the numbers modulo \\(2n\\) gives the desired permutation.\n\n\n\n## Proof #3 \n\nLet \\(p > 2\\) be a prime number of the form \\(p = 3k + 2, k \\in \\mathbb{Z}\\) . We will prove that the number \\(n = 2p\\) is good. Consider the numbers: \n\n\\[1^{3},2^{3},3^{3},\\ldots ,(2p)^{3}.\\] \n\nIt is well known that if \\(p \\equiv 2\\) (mod 3) then the above numbers form a complete residue system modulo \\(p\\) . It can easily be proven that they also form a complete residue system modulo \\(2p\\) (by also taking parity into account). Now, since \\(1^{3} + 2^{3} + \\ldots + k^{3} = (1 + 2 + \\ldots + k)^{2}\\) , we get that the sum of the first \\(k\\) ( \\(1 \\leqslant k \\leqslant 2p\\) ) numbers in the sequence modulo \\(p\\) is a quadratic residue modulo \\(p\\) . The parity condition is also satisfied (even after reduction modulo \\(n = 2p\\) , since \\(2p\\) is even). Finally, taking the numbers modulo \\(n\\) gives the desired permutation.", "metadata": {"resource_path": "Balkan_MO/segmented/en-2025-BMO-type1.jsonl", "problem_match": "\nProblem 1.", "solution_match": "# Solution "}}
-{"year": "2025", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "Balkan_MO", "problem": "Let \\(\\triangle ABC\\) be an acute- angled triangle with orthocentre \\(H\\) and let \\(D\\) be an arbitrary interior point on side \\(BC\\) . Suppose \\(E\\) and \\(F\\) are points on the segments \\(AB\\) and \\(AC\\) respectively such that the quadrilaterals \\(ABDF\\) and \\(ACDE\\) are cyclic, and let \\(BF\\) and \\(CE\\) intersect at \\(P\\) . Let \\(L\\) be the point of line \\(HA\\) such that \\(LC\\) is tangent to the circumcircle of triangle \\(\\triangle PBC\\) at point \\(C\\) . Let lines \\(BH\\) and \\(CP\\) intersect at \\(X\\) . Prove that \\(D, L\\) and \\(X\\) are collinear.", "solution": "\n \n\nWe have \\(\\angle PCD = \\angle ECD = \\angle EAD\\) and \\(\\angle PBD = \\angle FBD = \\angle FAD\\) , therefore \n\n\\[\\angle BPC = 180^{\\circ} - \\angle EAD - \\angle FAD = 180^{\\circ} - \\angle BAC = \\angle BHC,\\] \n\nmeaning that \\(BHPC\\) is cyclic. \n\nWe also have \\(\\angle PFD = \\angle BFD = \\angle BAD = \\angle PCD\\) showing that \\(DPFC\\) is cyclic. Then, using that \\(BAFD\\) is also cyclic, we have \n\n\\[\\angle DPC = \\angle DFC = \\angle ABC.\\] \n\nLet \\(Y\\) be the point of intersection of \\(AH\\) with the circumcircle of \\(BHPC\\) . Then \n\n\\[\\angle YPC = \\angle YHC = \\angle ABC = \\angle DPC\\] \n\nshowing that \\(D\\) belongs on \\(YP\\) . \n\nFinally, applying Pascal's theorem on the hexagon \\(BHYPCC\\) , we get that \\(X = BH \\cap PC, L = HY \\cap CC\\) and \\(D = YP \\cap CB\\) are collinear, as required.", "metadata": {"resource_path": "Balkan_MO/segmented/en-2025-BMO-type1.jsonl", "problem_match": "\nProblem 2.", "solution_match": "# Solution 1"}}
-{"year": "2025", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "Balkan_MO", "problem": "Let \\(\\triangle ABC\\) be an acute- angled triangle with orthocentre \\(H\\) and let \\(D\\) be an arbitrary interior point on side \\(BC\\) . Suppose \\(E\\) and \\(F\\) are points on the segments \\(AB\\) and \\(AC\\) respectively such that the quadrilaterals \\(ABDF\\) and \\(ACDE\\) are cyclic, and let \\(BF\\) and \\(CE\\) intersect at \\(P\\) . Let \\(L\\) be the point of line \\(HA\\) such that \\(LC\\) is tangent to the circumcircle of triangle \\(\\triangle PBC\\) at point \\(C\\) . Let lines \\(BH\\) and \\(CP\\) intersect at \\(X\\) . Prove that \\(D, L\\) and \\(X\\) are collinear.", "solution": "As in Solution 1, we introduce the point \\(Y\\) and prove that the quadrilateral \\(BHPC\\) is cyclic and that points \\(Y, D, P\\) are collinear. Define point \\(Z\\) as the second intersection of \\((CXH)\\) with the line \\(DX\\) . \n\nSince \\(\\angle HZD = \\angle HZX = \\angle HCX = \\angle HCP = \\angle HYP = \\angle HYD\\) , we get the points \\(H, D, Z, Y\\) are concyclic. We also know that \\(\\angle DZC = \\angle XZC = \\angle XHC = \\angle BYC =\\)\n\n\n\n \n\n\\(\\angle L C B\\) , where the last equality holds because \\(L C\\) is tangent to \\((B H P C)\\) . This implies that the circles \\((B Y C)\\) and \\((Z D C)\\) share a common tangent at the point \\(C\\) . Finally, applying the radical axis theorem to the circles \\((Z D C)\\) , \\((H D Z Y)\\) and \\((B H P C)\\) , we conclude that the line \\(X D\\) passes through the point \\(L\\) , finishing the proof.", "metadata": {"resource_path": "Balkan_MO/segmented/en-2025-BMO-type1.jsonl", "problem_match": "\nProblem 2.", "solution_match": "# Solution 2"}}
-{"year": "2025", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "Balkan_MO", "problem": "Let \\(\\triangle ABC\\) be an acute- angled triangle with orthocentre \\(H\\) and let \\(D\\) be an arbitrary interior point on side \\(BC\\) . Suppose \\(E\\) and \\(F\\) are points on the segments \\(AB\\) and \\(AC\\) respectively such that the quadrilaterals \\(ABDF\\) and \\(ACDE\\) are cyclic, and let \\(BF\\) and \\(CE\\) intersect at \\(P\\) . Let \\(L\\) be the point of line \\(HA\\) such that \\(LC\\) is tangent to the circumcircle of triangle \\(\\triangle PBC\\) at point \\(C\\) . Let lines \\(BH\\) and \\(CP\\) intersect at \\(X\\) . Prove that \\(D, L\\) and \\(X\\) are collinear.", "solution": "Let \\(H_{A}\\) and \\(H_{C}\\) be the feet of the altitudes from \\(A\\) and \\(C\\) , respectively. Also, let \\(B A\\) and \\(B H\\) meet \\(C L\\) at \\(Z\\) and \\(T\\) respectively. \n\nWe prove that \\(B C P H\\) is cyclic as in Solution 1. \\(C H_{A}H C_{A}\\) and \\(C D E A\\) are cyclic, so \\(\\angle B H_{A}H_{C} = \\angle B D E = \\angle B A C\\) . Since \\(C L\\) is tangent to \\((B H P C)\\) , we have \\(\\angle L C B = 180^{\\circ}-\\) \\(\\angle B H C = \\angle B A C\\) . We obtained \\(\\angle B H_{A}H_{C} = \\angle B D E = \\angle B C L\\) , so \\(H_{C}H_{A}\\parallel E D\\parallel L C\\) \n\nProjecting from \\(C\\) , we obtain \\((B,X;H,Z) = (B,E;H_{C},T)\\) . From \\(H_{C}H_{A}\\parallel E D\\parallel T C\\) we have \\((B,E;H_{C},T) = (B,D;H_{A},C)\\) , which can be seen by applying Thales' theorem or by projecting from infinity. \n\nIf we denote the intersection of \\(L D\\) and \\(B Z\\) as \\(X^{\\prime}\\) , projecting from \\(L\\) we get \\((B,X^{\\prime};H,Z) =\\) \\((B,D;H_{A},C)\\) . \n\nCombining the above, we have \\((B,X;H,Z) = (B,D;H_{A},C) = (B,X^{\\prime};H,Z)\\) . It is well- known that, with 3 fixed points and a fixed cross- ratio, the fourth point is uniquely determined. This implies \\(X\\equiv X^{\\prime}\\) and we are done.\n\n\n", "metadata": {"resource_path": "Balkan_MO/segmented/en-2025-BMO-type1.jsonl", "problem_match": "\nProblem 2.", "solution_match": "# Solution 3"}}
+{"year": "2025", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "Balkan_MO", "problem": "Let \\(\\triangle ABC\\) be an acute- angled triangle with orthocentre \\(H\\) and let \\(D\\) be an arbitrary interior point on side \\(BC\\) . Suppose \\(E\\) and \\(F\\) are points on the segments \\(AB\\) and \\(AC\\) respectively such that the quadrilaterals \\(ABDF\\) and \\(ACDE\\) are cyclic, and let \\(BF\\) and \\(CE\\) intersect at \\(P\\) . Let \\(L\\) be the point of line \\(HA\\) such that \\(LC\\) is tangent to the circumcircle of triangle \\(\\triangle PBC\\) at point \\(C\\) . Let lines \\(BH\\) and \\(CP\\) intersect at \\(X\\) . Prove that \\(D, L\\) and \\(X\\) are collinear.", "solution": "\n \n\nWe have \\(\\angle PCD = \\angle ECD = \\angle EAD\\) and \\(\\angle PBD = \\angle FBD = \\angle FAD\\) , therefore \n\n\\[\\angle BPC = 180^{\\circ} - \\angle EAD - \\angle FAD = 180^{\\circ} - \\angle BAC = \\angle BHC,\\] \n\nmeaning that \\(BHPC\\) is cyclic. \n\nWe also have \\(\\angle PFD = \\angle BFD = \\angle BAD = \\angle PCD\\) showing that \\(DPFC\\) is cyclic. Then, using that \\(BAFD\\) is also cyclic, we have \n\n\\[\\angle DPC = \\angle DFC = \\angle ABC.\\] \n\nLet \\(Y\\) be the point of intersection of \\(AH\\) with the circumcircle of \\(BHPC\\) . Then \n\n\\[\\angle YPC = \\angle YHC = \\angle ABC = \\angle DPC\\] \n\nshowing that \\(D\\) belongs on \\(YP\\) . \n\nFinally, applying Pascal's theorem on the hexagon \\(BHYPCC\\) , we get that \\(X = BH \\cap PC, L = HY \\cap CC\\) and \\(D = YP \\cap CB\\) are collinear, as required.", "metadata": {"resource_path": "Balkan_MO/segmented/en-2025-BMO-type1.jsonl", "problem_match": "\nProblem 2.", "solution_match": "# Solution 1"}}
+{"year": "2025", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "Balkan_MO", "problem": "Let \\(\\triangle ABC\\) be an acute- angled triangle with orthocentre \\(H\\) and let \\(D\\) be an arbitrary interior point on side \\(BC\\) . Suppose \\(E\\) and \\(F\\) are points on the segments \\(AB\\) and \\(AC\\) respectively such that the quadrilaterals \\(ABDF\\) and \\(ACDE\\) are cyclic, and let \\(BF\\) and \\(CE\\) intersect at \\(P\\) . Let \\(L\\) be the point of line \\(HA\\) such that \\(LC\\) is tangent to the circumcircle of triangle \\(\\triangle PBC\\) at point \\(C\\) . Let lines \\(BH\\) and \\(CP\\) intersect at \\(X\\) . Prove that \\(D, L\\) and \\(X\\) are collinear.", "solution": "As in Solution 1, we introduce the point \\(Y\\) and prove that the quadrilateral \\(BHPC\\) is cyclic and that points \\(Y, D, P\\) are collinear. Define point \\(Z\\) as the second intersection of \\((CXH)\\) with the line \\(DX\\) . \n\nSince \\(\\angle HZD = \\angle HZX = \\angle HCX = \\angle HCP = \\angle HYP = \\angle HYD\\) , we get the points \\(H, D, Z, Y\\) are concyclic. We also know that \\(\\angle DZC = \\angle XZC = \\angle XHC = \\angle BYC =\\)\n\n\n\n \n\n\\(\\angle L C B\\) , where the last equality holds because \\(L C\\) is tangent to \\((B H P C)\\) . This implies that the circles \\((B Y C)\\) and \\((Z D C)\\) share a common tangent at the point \\(C\\) . Finally, applying the radical axis theorem to the circles \\((Z D C)\\) , \\((H D Z Y)\\) and \\((B H P C)\\) , we conclude that the line \\(X D\\) passes through the point \\(L\\) , finishing the proof.", "metadata": {"resource_path": "Balkan_MO/segmented/en-2025-BMO-type1.jsonl", "problem_match": "\nProblem 2.", "solution_match": "# Solution 2"}}
+{"year": "2025", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "Balkan_MO", "problem": "Let \\(\\triangle ABC\\) be an acute- angled triangle with orthocentre \\(H\\) and let \\(D\\) be an arbitrary interior point on side \\(BC\\) . Suppose \\(E\\) and \\(F\\) are points on the segments \\(AB\\) and \\(AC\\) respectively such that the quadrilaterals \\(ABDF\\) and \\(ACDE\\) are cyclic, and let \\(BF\\) and \\(CE\\) intersect at \\(P\\) . Let \\(L\\) be the point of line \\(HA\\) such that \\(LC\\) is tangent to the circumcircle of triangle \\(\\triangle PBC\\) at point \\(C\\) . Let lines \\(BH\\) and \\(CP\\) intersect at \\(X\\) . Prove that \\(D, L\\) and \\(X\\) are collinear.", "solution": "Let \\(H_{A}\\) and \\(H_{C}\\) be the feet of the altitudes from \\(A\\) and \\(C\\) , respectively. Also, let \\(B A\\) and \\(B H\\) meet \\(C L\\) at \\(Z\\) and \\(T\\) respectively. \n\nWe prove that \\(B C P H\\) is cyclic as in Solution 1. \\(C H_{A}H C_{A}\\) and \\(C D E A\\) are cyclic, so \\(\\angle B H_{A}H_{C} = \\angle B D E = \\angle B A C\\) . Since \\(C L\\) is tangent to \\((B H P C)\\) , we have \\(\\angle L C B = 180^{\\circ}-\\) \\(\\angle B H C = \\angle B A C\\) . We obtained \\(\\angle B H_{A}H_{C} = \\angle B D E = \\angle B C L\\) , so \\(H_{C}H_{A}\\parallel E D\\parallel L C\\) \n\nProjecting from \\(C\\) , we obtain \\((B,X;H,Z) = (B,E;H_{C},T)\\) . From \\(H_{C}H_{A}\\parallel E D\\parallel T C\\) we have \\((B,E;H_{C},T) = (B,D;H_{A},C)\\) , which can be seen by applying Thales' theorem or by projecting from infinity. \n\nIf we denote the intersection of \\(L D\\) and \\(B Z\\) as \\(X^{\\prime}\\) , projecting from \\(L\\) we get \\((B,X^{\\prime};H,Z) =\\) \\((B,D;H_{A},C)\\) . \n\nCombining the above, we have \\((B,X;H,Z) = (B,D;H_{A},C) = (B,X^{\\prime};H,Z)\\) . It is well- known that, with 3 fixed points and a fixed cross- ratio, the fourth point is uniquely determined. This implies \\(X\\equiv X^{\\prime}\\) and we are done.\n\n\n", "metadata": {"resource_path": "Balkan_MO/segmented/en-2025-BMO-type1.jsonl", "problem_match": "\nProblem 2.", "solution_match": "# Solution 3"}}
{"year": "2025", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "Balkan_MO", "problem": "Let \\(\\triangle ABC\\) be an acute- angled triangle with orthocentre \\(H\\) and let \\(D\\) be an arbitrary interior point on side \\(BC\\) . Suppose \\(E\\) and \\(F\\) are points on the segments \\(AB\\) and \\(AC\\) respectively such that the quadrilaterals \\(ABDF\\) and \\(ACDE\\) are cyclic, and let \\(BF\\) and \\(CE\\) intersect at \\(P\\) . Let \\(L\\) be the point of line \\(HA\\) such that \\(LC\\) is tangent to the circumcircle of triangle \\(\\triangle PBC\\) at point \\(C\\) . Let lines \\(BH\\) and \\(CP\\) intersect at \\(X\\) . Prove that \\(D, L\\) and \\(X\\) are collinear.", "solution": "Same as in Solution 3, we prove that \\(H_{C}H_{A}\\parallel E D\\parallel l\\) , where \\(l\\) is the tangent to ( \\(B H P C\\) ) at \\(C\\) . Now define \\(L\\) as the intersection of \\(A H\\) and \\(D X\\) . Apply Desargues's theorem on triangles \\(\\triangle B H_{A}H_{C}\\) and \\(\\triangle X L C\\) . Since \\(L H_{A}\\) , \\(C H_{C}\\) and \\(B X\\) are concurrent at \\(H\\) , we obtain that the intersection of \\(B H_{C}\\) and \\(C X\\) which is \\(E\\) , the intersection of \\(B H_{A}\\) and \\(L X\\) which is \\(D\\) and the intersection of \\(L C\\) and \\(H_{A}H_{C}\\) are collinear. However, since \\(D E\\parallel H_{A}H_{C}\\) , \\(L C\\) is also parallel to these lines, therefore it coincides with the tangent and we are done.", "metadata": {"resource_path": "Balkan_MO/segmented/en-2025-BMO-type1.jsonl", "problem_match": "\nProblem 2.", "solution_match": "# Solution 4"}}
{"year": "2025", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "Balkan_MO", "problem": "Let \\(\\triangle ABC\\) be an acute- angled triangle with orthocentre \\(H\\) and let \\(D\\) be an arbitrary interior point on side \\(BC\\) . Suppose \\(E\\) and \\(F\\) are points on the segments \\(AB\\) and \\(AC\\) respectively such that the quadrilaterals \\(ABDF\\) and \\(ACDE\\) are cyclic, and let \\(BF\\) and \\(CE\\) intersect at \\(P\\) . Let \\(L\\) be the point of line \\(HA\\) such that \\(LC\\) is tangent to the circumcircle of triangle \\(\\triangle PBC\\) at point \\(C\\) . Let lines \\(BH\\) and \\(CP\\) intersect at \\(X\\) . Prove that \\(D, L\\) and \\(X\\) are collinear.", "solution": "Let \\(H_{A},H_{B},H_{C}\\) to be the feet of the altitudes. We again obtain that \\(B H C P\\) is cyclic and that \\(D E\\parallel L C\\) . We want to prove \\(L C,A H\\) and \\(D X\\) are concurrent. Applying Trigonometric Ceva's theorem to \\(\\triangle A E D\\) , we need to prove: \n\n\\[\\frac{\\sin\\angle A B H}{\\sin\\angle H B C}\\cdot \\frac{\\sin\\angle L D B}{\\sin\\angle L D E}\\cdot \\frac{\\sin\\angle C E D}{\\sin\\angle C E B} = \\frac{\\cos\\angle B A C}{\\cos\\angle A C B}\\cdot \\frac{\\sin\\angle L D C}{\\sin\\angle D L C}\\cdot \\frac{\\sin\\angle C A D}{\\sin\\angle A D B}.\\] \n\nFrom law of sines in \\(\\triangle A D C\\) and \\(\\triangle A D B\\) , the last expression is equal to: \n\n\\[\\frac{\\cos\\angle B A C}{\\cos\\angle A C B}\\cdot \\frac{L C}{C D}\\cdot \\frac{C D\\cdot\\sin\\angle A C B\\cdot A D}{A B\\cdot\\sin\\angle A B C\\cdot A D}.\\] \n\nHowever, \\(L S = \\frac{H_{A}C}{\\cos\\angle B A C} = \\frac{A C\\cdot\\cos\\angle A C B}{\\cos\\angle B A C}\\) and we get: \n\n\\[\\frac{\\cos B A C}{\\cos A C B}\\cdot \\frac{A C\\cdot\\cos A C B}{\\cos B A C}\\cdot \\frac{\\sin A C B}{A B\\cdot\\sin A B C} = 1.\\]", "metadata": {"resource_path": "Balkan_MO/segmented/en-2025-BMO-type1.jsonl", "problem_match": "\nProblem 2.", "solution_match": "# Solution 5"}}
{"year": "2025", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "Balkan_MO", "problem": "Find all functions \\(f: \\mathbb{R} \\to \\mathbb{R}\\) such that, for all real numbers \\(x\\) and \\(y\\), \n\n\\[f(x + y f(x)) + y = x y + f(x + y).\\]", "solution": ". \n\nLet \\(P(x,y)\\) denote the given relation. If there is an \\(a\\in \\mathbb{R}\\) such that \\(f(a) = 0\\) , then \\(P(a,y)\\) gives that \\(y = a y + f(a + y)\\) , and so \\(f\\) must be linear. Then we can easily check and get that the only linear solutions are \\(f(x) = x\\) and \\(f(x) = 2 - x\\) ( \\(x\\in \\mathbb{R}\\) ). \n\nNow suppose that \\(f(x)\\neq 0\\) for all real numbers \\(x\\) . From \\(P(x - y,y)\\) we get that: \n\n\\[f(x - y + y f(x - y)) = -y^{2} + y(x - 1) + f(x).\\] \n\nSince \\(f(t)\\neq 0\\) for all real numbers \\(t\\) , it follows that \\(- y^{2} + y(x - 1) + f(x)\\neq 0\\) for all real numbers \\(x,y\\) , and so, its discriminant (as a polynomial in \\(y\\) ) must be negative. That is, \\((x - 1)^{2} + 4f(x)< 0\\) , which gives us \n\n\\[f(x)< -\\frac{(x - 1)^{2}}{4}\\leqslant 0\\] \n\nfor all real numbers \\(x\\) . Since \\((x + 1)^{2}\\geqslant 0\\) implies that \\(- \\frac{(x - 1)^{2}}{4}\\leqslant x\\) , we see that \n\n\\[f(x)< -\\frac{(x - 1)^{2}}{4}\\leqslant x\\] \n\nfor all real numbers \\(x\\) . Now from \\(P(x,y)\\) for \\(y > 0\\) and \\(x\\in \\mathbb{R}\\) , we get that \n\n\\[x y - y + f(x + y) = f(x + y f(x))< x + y f(x)< x - y\\frac{(x - 1)^{2}}{4}\\] \n\nand so \n\n\\[f(x + y)< x + y - y(x + \\frac{(x - 1)^{2}}{4}) = x + y - y\\frac{(x + 1)^{2}}{4}.\\] \n\nSetting \\(x = - y\\) above, we get that: \n\n\\[f(0)< -y\\frac{(-y + 1)^{2}}{4}.\\] \n\nfor all positive real numbers \\(y\\) . Letting \\(y\\to +\\infty\\) above, we reach a contradiction. Hence, the only solutions in this functional equation are \\(f(x) = x\\) and \\(f(x) = 2 - x\\)", "metadata": {"resource_path": "Balkan_MO/segmented/en-2025-BMO-type1.jsonl", "problem_match": "\nProblem 3.", "solution_match": "# Solution 1"}}
{"year": "2025", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "Balkan_MO", "problem": "Find all functions \\(f: \\mathbb{R} \\to \\mathbb{R}\\) such that, for all real numbers \\(x\\) and \\(y\\), \n\n\\[f(x + y f(x)) + y = x y + f(x + y).\\]", "solution": ". \n\nLet \\(P(x,y)\\) denote the given relation. Similarly to the first solution, if a root exists ( \\(f(a) = 0\\) for any \\(a\\) ), we get that the function is linear and that the two solutions are \\(f(x) = x\\) and \\(f(x) = 2 - x\\) . Assertion \\(P(x,c - x)\\) gives us the following relation: \n\n\\[f(x + (c - x)f(x)) = (c - x)(x - 1) + f(c) = -x^{2} + (c + 1)x + (f(c) - c)\\] \n\nThe right hand side of the expression is a quadratic equation in \\(x\\) with the discriminant \\(\\Delta = \\Delta (c) = (c + 1)^{2} + 4(f(c) - c) = (c - 1)^{2} + 4f(c)\\) . Therefore, if there exists a \\(c\\) such that \\((c - 1)^{2} + 4f(c)\\geq 0\\) , the quadratic equation has a real solution which implies the existence of a root, in which case we are done. \n\nIf \\(f(1) = 0\\) , then we found a root and are done. If \\(f(1) = 1\\) , then by taking \\(c = 1\\) we obtain that \\(\\Delta (1) = 4\\) , implying the existence of a root. We now check the case when \\(f(1) = - 1\\) . From the assertion \\(P(1 - x,x)\\) , we obtain: \n\n\\[f(1 - x + x f(1 - x)) = -x^{2} - 1\\] \n\nPlugging in \\(x = 1\\) , in the above assertion, we obtain that \\(f(f(0)) = - 2\\) . Now plugging in \\(x = 1 - f(0)\\) in the above assertion we get that \\(f(f(0) + (1 - f(0))f(f(0))) = -(1 - f(0))^{2} - 1\\) , simplifying and utilizing \\(f(f(0)) = - 2\\) we obtain \\(f(3f(0) - 2) = - f(0)^{2} + 2f(0) - 2\\) . Note that if \\(f(0)\\geq 0\\) , we have that \\(\\Delta (0) = 1 + 4f(0) > 0\\) , implying the existence of a root, so assume that \\(f(0)< 0\\) . Now using \\(c = 3f(0) - 2\\) for our discriminant value, we obtain \\(\\Delta (3f(0) - 2) = (3f(0) - 3)^{2} + 4f(3f(0) - 2) = 9(f(0) - 1)^{2} + 4(- f(0)^{2} + 2f(0) - 2) =\\) \\(5f(0)^{2} - 10f(0) + 1 > 0\\) , implying the existence of a root, and resolving the case when \\(f(1) = - 1\\) \n\nNow assume that \\(f(1)\\notin \\{0,1, - 1\\}\\) . From \\(P(1,y)\\) , we obtain the relation that \\(f(1 + yf(1)) =\\) \\(f(1 + y)\\) . As \\(f(1)\\neq 0\\) , we can inductively show that \\(f(1 + yf(1)^{k}) = f(1 + y)\\) for all \\(k\\in \\mathbb{Z}\\) Since \\(f(1)\\notin \\{1, - 1\\}\\) , there exists an unbounded sequence \\(a_{n}\\) such that \\(f(a_{n})\\) is constant. Namely, one can take \\(a_{n} = 1 + f(1)^{2n}\\) if \\(|f(1)| > 1\\) , and \\(a_{n} = 1 + f(1)^{- 2n}\\) if \\(|f(1)|< 1\\) both times it holds that \\(f(a_{n}) = f(2)\\) . The value of the discriminant along this sequence is \\(\\Delta (a_{n}) = (a_{n} - 1)^{2} + 4f(a_{n}) = (a_{n} - 1)^{2} + 4f(2)\\) , and since \\(a_{n}\\) is unbounded this there exists \\(n\\) where the value of the discriminant is positive, yielding our root. This finishes the problem.", "metadata": {"resource_path": "Balkan_MO/segmented/en-2025-BMO-type1.jsonl", "problem_match": "\nProblem 3.", "solution_match": "# Solution 2"}}
{"year": "2025", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "Balkan_MO", "problem": "Find all functions \\(f: \\mathbb{R} \\to \\mathbb{R}\\) such that, for all real numbers \\(x\\) and \\(y\\), \n\n\\[f(x + y f(x)) + y = x y + f(x + y).\\]", "solution": ". \n\nLet \\(P(x,y)\\) denote the given relation. As in the previous solutions, if a root exists, then we are done. From \\(P(1,y)\\) we obtain \\(f(1 + yc) = f(1 + y)\\) , where we have put \\(c = f(1)\\) . From the substitution \\(P(1 + x,y)\\) , we get: \n\n\\[f(1 + x + yf(1 + x)) + y = (1 + x)y + f(1 + x + y) \\quad (1)\\]\n\n\n\nSubstituting \\(P(1 + cx, cy)\\) instead, we obtain \n\n\\[f(1 + cx + cyf(1 + cx)) + cy = (1 + cx)cy + f(1 + cx + cy) \\quad (2)\\] \n\nNote that \n\n\\[f(1 + cx + cyf(1 + cx)) = f(1 + cx + cyf(1 + x)) = f(1 + c(x + yf(1 + x))) = f(1 + x + yf(1 + x))\\] \n\nand that \\(f(1 + cx + cy) = f(1 + c(x + y)) = f(1 + x + y)\\) . By subtracting (1) and (2) we obtain \\(c^{2}xy = xy\\) for all \\(x,y\\) , concluding that \\(c^{2} = 1\\) . From here, one can proceed in numerous ways (some of which have been highlighted in the previous solutions) to finish the problem.", "metadata": {"resource_path": "Balkan_MO/segmented/en-2025-BMO-type1.jsonl", "problem_match": "\nProblem 3.", "solution_match": "# Solution 3"}}
{"year": "2025", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "Balkan_MO", "problem": "Find all functions \\(f: \\mathbb{R} \\to \\mathbb{R}\\) such that, for all real numbers \\(x\\) and \\(y\\), \n\n\\[f(x + y f(x)) + y = x y + f(x + y).\\]", "solution": ". (by Stefan Šebez) \n\nLet \\(P(x,y)\\) denote the given relation. Putting \\(P(0,x + y)\\) gives us: \n\n\\[f((x + y)f(0)) + x + y = 0 + f(x + y)\\] \n\nSubtracting this identity from the relation \\(P(x,y)\\) yields: \n\n\\[P(x + yf(x)) - P((x + y)f(0)) = xy + x\\] \n\nSuppose that \\(f(x)\\neq f(0)\\) for some \\(x\\in \\mathbf{R}\\) (thus in particular \\(x\\neq 0\\) ). Then letting \\(y\\) equal \\(x(f(0) - 1) / (f(x) - f(0))\\) makes the left- hand side vanish, so that \\(x(y + 1) = 0\\) and \\(y = - 1\\) We conclude that, for an arbitrary \\(x\\in \\mathbf{R}\\) , either \\(f(x) = f(0)\\) or \\(f(x) = x(1 - f(0)) + f(0)\\) Consider the values \\(f(x)\\) and \\(f(xf(0))\\) . They are related by \\(P(0,x)\\) .. \n\n\\[f(xf(0)) = f(x) - x\\] \n\nFix some \\(x\\neq 0\\) . Then, for this \\(x\\) , (at least) one of four possible cases holds: \n\nCase \\(f(x) = f(0)\\) and \\(f(xf(0)) = f(0)\\) \n\nCase \\(f(x) = f(0)\\) and \\(f(xf(0)) = xf(0)(1 - f(0)) + f(0)\\) \n\nCase \\(f(x) = x(1 - f(0)) + f(0)\\) and \\(f(xf(0)) = f(0)\\) \n\nCase \\(f(x) = x(1 - f(0)) + f(0)\\) and \\(f(xf(0)) = xf(0)(1 - f(0)) + f(0)\\) \n\nThe first case implies that \\(x = 0\\) , a contradiction. The second gives \\(f(0)^{2} - f(0) - 1 = 0\\) . The third gives \\(f(0) = 0\\) and the fourth \\(f(0)\\in \\{0,2\\}\\) . \n\nIt is now clear that \\(f(0) = 0\\) implies \\(f(x) = x\\) for all \\(x\\) , and that \\(f(0) = 2\\) implies \\(f(x) = 2 - x\\) for all \\(x\\) . We check that these two functions indeed satisfy the starting equation. If, on the other hand, \\(f(0)\\notin \\{0,2\\}\\) , then the second case holds for all \\(x\\neq 0\\) and hence \\(f(x) = f(0)\\) for all \\(x\\) . However, this is a contradiction with \\(P(0,x)\\) . Thus there are no more solutions.", "metadata": {"resource_path": "Balkan_MO/segmented/en-2025-BMO-type1.jsonl", "problem_match": "\nProblem 3.", "solution_match": "# Solution 4"}}
-{"year": "2025", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "Balkan_MO", "problem": "There are \\(n\\) cities in a country, where \\(n \\geqslant 100\\) is an integer. Some pairs of cities are connected by direct (two- way) flights. For two cities \\(A\\) and \\(B\\) we define: \n\n- a path between \\(A\\) and \\(B\\) as a sequence of distinct cities \\(A = C_{0}, C_{1}, \\ldots , C_{k}, C_{k + 1} = B\\) , \\(k \\geqslant 0\\) , such that there are direct flights between \\(C_{i}\\) and \\(C_{i + 1}\\) for every \\(0 \\leqslant i \\leqslant k\\) ;- a long path between \\(A\\) and \\(B\\) as a path between \\(A\\) and \\(B\\) such that no other path between \\(A\\) and \\(B\\) has more cities;- a short path between \\(A\\) and \\(B\\) as a path between \\(A\\) and \\(B\\) such that no other path between \\(A\\) and \\(B\\) has fewer cities. \n\nAssume that for any pair of cities \\(A\\) and \\(B\\) in the country, there exist a long path and a short path between them that have no cities in common (except \\(A\\) and \\(B\\) ). Let \\(F\\) be the total number of pairs of cities in the country that are connected by direct flights. In terms of \\(n\\) , find all possible values of \\(F\\) .", "solution": "Use the obvious graph interpretation. We show that any such graph is one of the following: the full graph \\(K_{n}\\) , the circular graph \\(C_{n}\\) , and for \\(n\\) even, the bipartite graph \\(K_{\\frac{n}{2}, \\frac{n}{2}}\\) . First, we show that these graphs satisfy the condition. \n\n- For \\(K_{n}\\) , we can choose any long path and the short path is the edge.- For \\(C_{n}\\) , we have exactly two paths between any two vertices, and one of them has at most as many vertices as the other.- For \\(K_{\\frac{n}{2}, \\frac{n}{2}}\\) , if the vertices are on different sides, the short path is the edge. Otherwise, take any long path. We observe that it alternates between the sides and begins and ends on one side. Therefore, there is a vertex on the other side that doesn't appear in the long path. Additionally, there is a short path that passes through this vertex. \n\nNext, we show that only these graphs work for \\(n\\) large enough. \n\nThe graph is clearly connected, as any two vertices belong to a path. Consider a longest path in the graph. Let \\(p\\) be its length and denote the vertices in the path by \\(V_{1}, V_{2}, \\ldots , V_{p}\\) in the corresponding order. We can assume that this path is the long path between \\(V_{1}\\) and \\(V_{p}\\) that has a corresponding short path through other vertices. We show that the edge \\(V_{1} V_{p}\\) belongs to the graph. If the edge doesn't exist, the short path has length at least two, implying that there is a vertex \\(X\\) different from \\(V_{i}, i \\in \\{1, \\ldots , p\\}\\) such that there exists an edge from \\(V_{1}\\) to \\(X\\) . Then the path \\(X V_{1} V_{2} \\ldots V_{p}\\) has length \\(p + 1\\) , which gives a contradiction. \n\nNext we show that \\(p = n\\) , i.e. that the cycle \\(V_{1} \\ldots V_{p}\\) contains all the vertices. If there exists another vertex \\(A\\) connected with an edge to a vertex \\(V_{i}\\) , then the path \\(A V_{i} V_{i + 1} \\ldots V_{i - 1}\\) has length \\(p + 1\\) , which gives a contradiction. Since the graph is connected, the cycle contains all vertices.\n\n\n\nFor two vertices of the graph, we say that they have distance \\(r\\) if there are exactly \\(r - 1\\) vertices between them on a side of the cycle. Observe that they also have distance \\(n - r\\) . \n\nIf we relabel the vertices by \\(A_{1},A_{2},\\ldots ,A_{n}\\) in such a way that we know the graph has \\(n - 1\\) of the edges \\(A_{i}A_{i + 1},i\\in \\{1,\\ldots ,n\\}\\) (where \\(A_{n + 1} = A_{1}\\) ), then it also has the last one. This is shown same as before. \n\nNext, we show that if we have an edge between \\(V_{i}\\) and \\(V_{j}\\) , then we also have an edge between \\(V_{i + 1}\\) and \\(V_{j + 1}\\) . Assume \\(i< j\\) . Consider the path \n\n\\[V_{i + 1}V_{i + 2}\\ldots V_{j}V_{i}V_{i - 1}\\ldots V_{j + 1}\\] \n\n\n \n\nof length \\(n\\) . As before, we conclude that there is an edge between \\(V_{i + 1}\\) and \\(V_{j + 1}\\) . Repeating this, we get that if we have an edge between two vertices at distance \\(r\\) , then we have edges between any two vertices at distance \\(r\\) . \n\nDefine \\(S\\) as the set of numbers \\(1\\leqslant r\\leqslant n - 1\\) such that the graph has the edges of distance \\(r\\) . Note that \\(1,n - 1\\in S\\) . \n\nFor positive integers \\(a\\) and \\(b\\) with \\(a + b\\leqslant n - 1\\) , consider the ordering \n\n\\[V_{1},V_{a + b},V_{a + b - 1},\\ldots ,V_{a + 1},V_{a + b + 1},V_{a + b + 2},\\ldots ,V_{n},V_{a},V_{a - 1},\\ldots ,V_{1}.\\] \n\n\n \n\nThe distance between two consecutive vertices in this ordering is \\(1,a,b\\) or \\(a + b - 1\\) . This implies that if two numbers from the multiset \\(\\{a,b,a + b - 1\\}\\) belong to \\(S\\) , so does the third one. Now, if \\(2\\in S\\) , we take \\(b = 2\\) and easily get that that \\(S\\) contains any number from 1 to \\(n - 1\\) . This gives us the solution \\(K_{n}\\) .\n\n\n\nAssume now \\(2 \\notin S\\) . This implies that we do not have two consecutive numbers smaller than \\(n - 2\\) in \\(S\\) . But as \\(2 \\notin S\\) , we also have \\(n - 2 \\notin S\\) , so \\(S\\) doesn’t contain two consecutive integers. \n\nIf \\(S = \\{1, n - 1\\}\\) , we get the solution \\(C_{n}\\) . Otherwise, there exists \\(t \\in S\\) such that \\(3 \\leqslant t \\leqslant n - 3\\) . Consider the path \n\n\\[V_{t}V_{t - 1}\\ldots V_{2}V_{t + 2}V_{t + 1}V_{1}V_{n}\\ldots V_{t + 3}\\] \n\nof length \\(n\\) . \n\n\n \n\nSame as before, we get that there is an edge between \\(V_{t}\\) and \\(V_{t + 3}\\) . Therefore, we have \\(3 \\in S\\) . Now, taking \\(b = 3\\) , we get that any odd number smaller than or equal to \\(n - 1\\) lies in \\(S\\) . Since we assumed \\(S\\) doesn’t contain consecutive integers, we get that \\(n\\) is even and \\(S = \\{1 \\leqslant i \\leqslant n - 1 | i \\text{odd}\\}\\) . This gives us the solution \\(K_{\\frac{n}{2}, \\frac{n}{2}}\\) . \n\nFinally, the number of edges can be \\(n\\) , \\(\\frac{n(n - 1)}{2}\\) , and if \\(n\\) is even it can also be \\(\\frac{n^{2}}{4}\\) . \n\nRemark: Even if \\(n\\) is not big enough, we still characterize all such graphs similarly. The condition was added as at some point we choose a number \\(t\\) between 3 and \\(n - 3\\) , and this wouldn’t make sense for small \\(n\\) and we would need to quickly discuss why those cases also have the same graphs.", "metadata": {"resource_path": "Balkan_MO/segmented/en-2025-BMO-type1.jsonl", "problem_match": "\nProblem 4.", "solution_match": "# Solution "}}
+{"year": "2025", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "Balkan_MO", "problem": "There are \\(n\\) cities in a country, where \\(n \\geqslant 100\\) is an integer. Some pairs of cities are connected by direct (two- way) flights. For two cities \\(A\\) and \\(B\\) we define: \n\n- a path between \\(A\\) and \\(B\\) as a sequence of distinct cities \\(A = C_{0}, C_{1}, \\ldots , C_{k}, C_{k + 1} = B\\) , \\(k \\geqslant 0\\) , such that there are direct flights between \\(C_{i}\\) and \\(C_{i + 1}\\) for every \\(0 \\leqslant i \\leqslant k\\) ;- a long path between \\(A\\) and \\(B\\) as a path between \\(A\\) and \\(B\\) such that no other path between \\(A\\) and \\(B\\) has more cities;- a short path between \\(A\\) and \\(B\\) as a path between \\(A\\) and \\(B\\) such that no other path between \\(A\\) and \\(B\\) has fewer cities. \n\nAssume that for any pair of cities \\(A\\) and \\(B\\) in the country, there exist a long path and a short path between them that have no cities in common (except \\(A\\) and \\(B\\) ). Let \\(F\\) be the total number of pairs of cities in the country that are connected by direct flights. In terms of \\(n\\) , find all possible values of \\(F\\) .", "solution": "Use the obvious graph interpretation. We show that any such graph is one of the following: the full graph \\(K_{n}\\) , the circular graph \\(C_{n}\\) , and for \\(n\\) even, the bipartite graph \\(K_{\\frac{n}{2}, \\frac{n}{2}}\\) . First, we show that these graphs satisfy the condition. \n\n- For \\(K_{n}\\) , we can choose any long path and the short path is the edge.- For \\(C_{n}\\) , we have exactly two paths between any two vertices, and one of them has at most as many vertices as the other.- For \\(K_{\\frac{n}{2}, \\frac{n}{2}}\\) , if the vertices are on different sides, the short path is the edge. Otherwise, take any long path. We observe that it alternates between the sides and begins and ends on one side. Therefore, there is a vertex on the other side that doesn't appear in the long path. Additionally, there is a short path that passes through this vertex. \n\nNext, we show that only these graphs work for \\(n\\) large enough. \n\nThe graph is clearly connected, as any two vertices belong to a path. Consider a longest path in the graph. Let \\(p\\) be its length and denote the vertices in the path by \\(V_{1}, V_{2}, \\ldots , V_{p}\\) in the corresponding order. We can assume that this path is the long path between \\(V_{1}\\) and \\(V_{p}\\) that has a corresponding short path through other vertices. We show that the edge \\(V_{1} V_{p}\\) belongs to the graph. If the edge doesn't exist, the short path has length at least two, implying that there is a vertex \\(X\\) different from \\(V_{i}, i \\in \\{1, \\ldots , p\\}\\) such that there exists an edge from \\(V_{1}\\) to \\(X\\) . Then the path \\(X V_{1} V_{2} \\ldots V_{p}\\) has length \\(p + 1\\) , which gives a contradiction. \n\nNext we show that \\(p = n\\) , i.e. that the cycle \\(V_{1} \\ldots V_{p}\\) contains all the vertices. If there exists another vertex \\(A\\) connected with an edge to a vertex \\(V_{i}\\) , then the path \\(A V_{i} V_{i + 1} \\ldots V_{i - 1}\\) has length \\(p + 1\\) , which gives a contradiction. Since the graph is connected, the cycle contains all vertices.\n\n\n\nFor two vertices of the graph, we say that they have distance \\(r\\) if there are exactly \\(r - 1\\) vertices between them on a side of the cycle. Observe that they also have distance \\(n - r\\) . \n\nIf we relabel the vertices by \\(A_{1},A_{2},\\ldots ,A_{n}\\) in such a way that we know the graph has \\(n - 1\\) of the edges \\(A_{i}A_{i + 1},i\\in \\{1,\\ldots ,n\\}\\) (where \\(A_{n + 1} = A_{1}\\) ), then it also has the last one. This is shown same as before. \n\nNext, we show that if we have an edge between \\(V_{i}\\) and \\(V_{j}\\) , then we also have an edge between \\(V_{i + 1}\\) and \\(V_{j + 1}\\) . Assume \\(i< j\\) . Consider the path \n\n\\[V_{i + 1}V_{i + 2}\\ldots V_{j}V_{i}V_{i - 1}\\ldots V_{j + 1}\\] \n\n\n \n\nof length \\(n\\) . As before, we conclude that there is an edge between \\(V_{i + 1}\\) and \\(V_{j + 1}\\) . Repeating this, we get that if we have an edge between two vertices at distance \\(r\\) , then we have edges between any two vertices at distance \\(r\\) . \n\nDefine \\(S\\) as the set of numbers \\(1\\leqslant r\\leqslant n - 1\\) such that the graph has the edges of distance \\(r\\) . Note that \\(1,n - 1\\in S\\) . \n\nFor positive integers \\(a\\) and \\(b\\) with \\(a + b\\leqslant n - 1\\) , consider the ordering \n\n\\[V_{1},V_{a + b},V_{a + b - 1},\\ldots ,V_{a + 1},V_{a + b + 1},V_{a + b + 2},\\ldots ,V_{n},V_{a},V_{a - 1},\\ldots ,V_{1}.\\] \n\n\n \n\nThe distance between two consecutive vertices in this ordering is \\(1,a,b\\) or \\(a + b - 1\\) . This implies that if two numbers from the multiset \\(\\{a,b,a + b - 1\\}\\) belong to \\(S\\) , so does the third one. Now, if \\(2\\in S\\) , we take \\(b = 2\\) and easily get that that \\(S\\) contains any number from 1 to \\(n - 1\\) . This gives us the solution \\(K_{n}\\) .\n\n\n\nAssume now \\(2 \\notin S\\) . This implies that we do not have two consecutive numbers smaller than \\(n - 2\\) in \\(S\\) . But as \\(2 \\notin S\\) , we also have \\(n - 2 \\notin S\\) , so \\(S\\) doesn’t contain two consecutive integers. \n\nIf \\(S = \\{1, n - 1\\}\\) , we get the solution \\(C_{n}\\) . Otherwise, there exists \\(t \\in S\\) such that \\(3 \\leqslant t \\leqslant n - 3\\) . Consider the path \n\n\\[V_{t}V_{t - 1}\\ldots V_{2}V_{t + 2}V_{t + 1}V_{1}V_{n}\\ldots V_{t + 3}\\] \n\nof length \\(n\\) . \n\n\n \n\nSame as before, we get that there is an edge between \\(V_{t}\\) and \\(V_{t + 3}\\) . Therefore, we have \\(3 \\in S\\) . Now, taking \\(b = 3\\) , we get that any odd number smaller than or equal to \\(n - 1\\) lies in \\(S\\) . Since we assumed \\(S\\) doesn’t contain consecutive integers, we get that \\(n\\) is even and \\(S = \\{1 \\leqslant i \\leqslant n - 1 | i \\text{odd}\\}\\) . This gives us the solution \\(K_{\\frac{n}{2}, \\frac{n}{2}}\\) . \n\nFinally, the number of edges can be \\(n\\) , \\(\\frac{n(n - 1)}{2}\\) , and if \\(n\\) is even it can also be \\(\\frac{n^{2}}{4}\\) . \n\nRemark: Even if \\(n\\) is not big enough, we still characterize all such graphs similarly. The condition was added as at some point we choose a number \\(t\\) between 3 and \\(n - 3\\) , and this wouldn’t make sense for small \\(n\\) and we would need to quickly discuss why those cases also have the same graphs.", "metadata": {"resource_path": "Balkan_MO/segmented/en-2025-BMO-type1.jsonl", "problem_match": "\nProblem 4.", "solution_match": "# Solution "}}
diff --git a/Benelux_MO/md/Benelux_en-olympiad_en-bxmo-problems-2025-zz.md b/Benelux_MO/md/Benelux_en-olympiad_en-bxmo-problems-2025-zz.md
index 73503acaded7c64f01c11a266ca92b47afee6df7..df9fd7c55af7996e1e2bc75db598ac82b0ced693 100644
--- a/Benelux_MO/md/Benelux_en-olympiad_en-bxmo-problems-2025-zz.md
+++ b/Benelux_MO/md/Benelux_en-olympiad_en-bxmo-problems-2025-zz.md
@@ -41,7 +41,7 @@ Problem 3. Let \(ABC\) be a triangle with incentre \(I\) and circumcircle \(\Ome
Solution.
-
+
By definition of \(D\) , \(E\) and \(F\) , we know that \(ID, IE\) and \(IF\) are the angle bisectors of \(ABC\) . Using angles in \(\Omega\) , we can compute
@@ -55,7 +55,7 @@ proving that \(D'E\parallel FI\) . Moreover, we can also compute
proving that \(EI\parallel D'F\) . Therefore, \(ED'FI\) is a parallelogram and its diagonals intersect in their midpoints, proving that \(D'I\) contains \(M\) .
-
+
Alternative Solution. By definition of \(D, E\) and \(F\) , we know that \(ID, IE\) and \(IF\) are the angle bisectors of \(ABC\) . If \(D' = A\) , this means that \(\overrightarrow{OAB} = 90^\circ - \overrightarrow{ACB} = \frac{\overrightarrow{CAB}}{2}\) and \(ABC\) is isosceles in \(A\) . In that case, the symmetry with respect to \(AI\) sends \(E\) to \(F\) and so \(M \in AI\) . We can thus suppose that \(D' \neq A\) .
diff --git a/Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2025-zz.jsonl b/Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2025-zz.jsonl
index eb2f997d848f2056fbacb0c10452b8a05afcca32..c23747c8aada9e822b38ae8e9770d5b5046903da 100644
--- a/Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2025-zz.jsonl
+++ b/Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2025-zz.jsonl
@@ -1,7 +1,7 @@
{"year": "2025", "tier": "T3", "problem_label": "1", "problem_type": null, "exam": "Benelux_MO", "problem": "Does there exist a function \\(f\\colon \\mathbb{R}\\to \\mathbb{R}\\) such that \n\n\\[f\\big(x^{2} + f(y)\\big) = f(x)^{2} - y\\] \n\nfor all \\(x,y\\in \\mathbb{R}?\\)", "solution": "There does not exist such a function. Let us suppose by contradiction it does. By substituting \\(x\\gets 0\\) , we get \n\n\\[f(f(y)) = f(0)^{2} - y\\] \n\nfor all \\(y\\in \\mathbb{R}\\) . Since the right- hand side is bijective, this implies that \\(f\\) is also bijective. Taking \\(y\\gets 0\\) we get \n\n\\[f\\big(x^{2} + f(0)\\big) = f(x)^{2}\\] \n\nfor all \\(x\\in \\mathbb{R}\\) and so \\(f(- x)^{2} = f(x^{2} + f(0)) = f(x)^{2}\\) . Since \\(f\\) is injective, we get \\(f(- x) = - f(x)\\) for all \\(x\\neq 0\\) . Since \\(f\\) is a surjection, there exists \\(r\\in \\mathbb{R}\\) such that \\(f(r) = 0\\) . If \\(r\\neq 0\\) , \\(f(- r) = - f(r) = 0 = f(r)\\) contradicting the fact that \\(f\\) is injective. So \\(r = 0\\) and \\(f(0) = 0\\) . Substituting \\((x,y)\\gets (1,0)\\) yields \\(f(1) = f(1)^{2}\\) and so \\(f(1) = 1\\) since \\(f(0) = 0\\) and \\(f\\) is injective. Taking \\((x,y)\\gets (0,1)\\) , we finally get \\(1 = f(f(1)) = - 1\\) which is the desired contradiction.", "metadata": {"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2025-zz.jsonl", "problem_match": "\nProblem 1.", "solution_match": "\nSolution."}}
{"year": "2025", "tier": "T3", "problem_label": "1", "problem_type": null, "exam": "Benelux_MO", "problem": "Does there exist a function \\(f\\colon \\mathbb{R}\\to \\mathbb{R}\\) such that \n\n\\[f\\big(x^{2} + f(y)\\big) = f(x)^{2} - y\\] \n\nfor all \\(x,y\\in \\mathbb{R}?\\)", "solution": "We observe that, for all \\(y,z\\in \\mathbb{R}\\) , we have \n\n\\[z\\geqslant f(y)\\Longrightarrow f(z)\\geq -y.\\] \n\nIndeed, we can take \\(x = \\sqrt{z - f(y)}\\) and get \\(f(z) = f(x)^{2} - y\\geq - y\\) . We deduce that \\(\\lim_{z\\to +\\infty}f(z) =\\) \\(+\\infty\\) . Indeed, for any \\(K\\in \\mathbb{R}\\) , we have \\(z\\geqslant f(- K)\\Longrightarrow f(z)\\geqslant K\\) \n\nLet us now fix \\(x\\) , and let \\(y\\to +\\infty\\) . We have \\(x^{2} + f(y)\\to +\\infty\\) , hence \\(f(x^{2} + f(y))\\to +\\infty\\) . On the other side, we have \\(f(x)^{2} - y\\to -\\infty\\) , a contradiction.", "metadata": {"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2025-zz.jsonl", "problem_match": "\nProblem 1.", "solution_match": "\nAlternative Solution."}}
{"year": "2025", "tier": "T3", "problem_label": "2", "problem_type": null, "exam": "Benelux_MO", "problem": "Let \\(N \\geq 2\\) be a natural number. At a mathematical olympiad training camp, the same \\(N\\) courses are organised every day. Each student takes exactly one of the \\(N\\) courses each day. At the end of the camp, every student has taken each course exactly once, and any two students took the same course on at least one day, but took different courses on at least one other day. What is, in terms of \\(N\\) , the largest possible number of students at the camp?", "solution": "The largest number of students at the camp is \\((N - 1)!\\) . Since each student takes exactly one course each day and, at the end, has taken each course exactly once, the schedule of a student can be represented by a permutation of the set of the \\(N\\) courses. To show that \\((N - 1)!\\) is possible, we can e.g. assign to each of the \\((N - 1)!\\) students a unique permutation of the \\(N - 1\\) first courses and making them all take the \\(N\\) - th course on the last day. It is easy to observe that such a construction satisfies the properties of the statement. \n\nTo prove that, for a set \\(S\\) of students, one has \\(|S| \\leq (N - 1)!\\) , one first subdivides the set of permutations into disjoint subsets of size \\(N\\) . Two permutations are said to be in the same subset if and only if one can be obtained from the other by cyclically permute the order of the course. Clearly, this is a well- defined subdivision since cyclically permuting twice the order of the courses can be obtained by cyclically permute them only once. Moreover, each of these subsets contains exactly \\(N\\) permutations since there are \\(N\\) cycles of length \\(N\\) . There are thus \\(\\frac{N!}{N} = (N - 1)!\\) such subsets. If \\(|S| > (N - 1)!\\) , two students will have their associated permutations in the same subset. However, they cannot be the same permutation (otherwise the two students took the same course every day), and they cannot be obtained by a non- trivial cyclic permutation from each other (otherwise the two students never took the same course).", "metadata": {"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2025-zz.jsonl", "problem_match": "\nProblem 2.", "solution_match": "\nSolution."}}
-{"year": "2025", "tier": "T3", "problem_label": "3", "problem_type": null, "exam": "Benelux_MO", "problem": "Let \\(ABC\\) be a triangle with incentre \\(I\\) and circumcircle \\(\\Omega\\) . Let \\(D, E, F\\) be the midpoints of the arcs \\(\\overline{BC}, \\overline{CA}, \\overline{AB}\\) of \\(\\Omega\\) not containing \\(A, B, C\\) , respectively. Let \\(D'\\) be the point of \\(\\Omega\\) diametrically opposite to \\(D\\) . Show that \\(I, D'\\) , and the midpoint \\(M\\) of \\([EF]\\) lie on a line.", "solution": "\n \n\nBy definition of \\(D\\) , \\(E\\) and \\(F\\) , we know that \\(ID, IE\\) and \\(IF\\) are the angle bisectors of \\(ABC\\) . Using angles in \\(\\Omega\\) , we can compute \n\n\\[\\overline{EFI} = \\overline{EFC} = \\overline{EBC} = \\frac{\\overline{ABC}}{2} = 90^{\\circ} - \\frac{\\overline{ACB}}{2} - \\frac{\\overline{BAC}}{2}\\] \\[\\qquad = 90^{\\circ} - \\overline{FCB} - \\overline{BAD} = 90^{\\circ} - \\overline{FEB} - \\overline{BED} = 90^{\\circ} - \\overline{FED} = \\overline{D'EF}\\] \n\nproving that \\(D'E\\parallel FI\\) . Moreover, we can also compute \n\n\\[\\overline{FEI} = \\overline{FEB} = \\overline{FCB} = \\frac{\\overline{ACB}}{2} = 90^{\\circ} - \\frac{\\overline{ABC}}{2} - \\frac{\\overline{CAB}}{2}\\] \\[\\qquad = 90^{\\circ} - \\overline{EB} - \\overline{CAD} = 90^{\\circ} - \\overline{EFC} - \\overline{CFD} = 90^{\\circ} - \\overline{EFD} = \\overline{D'FE}\\] \n\nproving that \\(EI\\parallel D'F\\) . Therefore, \\(ED'FI\\) is a parallelogram and its diagonals intersect in their midpoints, proving that \\(D'I\\) contains \\(M\\) .\n\n\n", "metadata": {"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2025-zz.jsonl", "problem_match": "\nProblem 3.", "solution_match": "\nSolution."}}
+{"year": "2025", "tier": "T3", "problem_label": "3", "problem_type": null, "exam": "Benelux_MO", "problem": "Let \\(ABC\\) be a triangle with incentre \\(I\\) and circumcircle \\(\\Omega\\) . Let \\(D, E, F\\) be the midpoints of the arcs \\(\\overline{BC}, \\overline{CA}, \\overline{AB}\\) of \\(\\Omega\\) not containing \\(A, B, C\\) , respectively. Let \\(D'\\) be the point of \\(\\Omega\\) diametrically opposite to \\(D\\) . Show that \\(I, D'\\) , and the midpoint \\(M\\) of \\([EF]\\) lie on a line.", "solution": "\n \n\nBy definition of \\(D\\) , \\(E\\) and \\(F\\) , we know that \\(ID, IE\\) and \\(IF\\) are the angle bisectors of \\(ABC\\) . Using angles in \\(\\Omega\\) , we can compute \n\n\\[\\overline{EFI} = \\overline{EFC} = \\overline{EBC} = \\frac{\\overline{ABC}}{2} = 90^{\\circ} - \\frac{\\overline{ACB}}{2} - \\frac{\\overline{BAC}}{2}\\] \\[\\qquad = 90^{\\circ} - \\overline{FCB} - \\overline{BAD} = 90^{\\circ} - \\overline{FEB} - \\overline{BED} = 90^{\\circ} - \\overline{FED} = \\overline{D'EF}\\] \n\nproving that \\(D'E\\parallel FI\\) . Moreover, we can also compute \n\n\\[\\overline{FEI} = \\overline{FEB} = \\overline{FCB} = \\frac{\\overline{ACB}}{2} = 90^{\\circ} - \\frac{\\overline{ABC}}{2} - \\frac{\\overline{CAB}}{2}\\] \\[\\qquad = 90^{\\circ} - \\overline{EB} - \\overline{CAD} = 90^{\\circ} - \\overline{EFC} - \\overline{CFD} = 90^{\\circ} - \\overline{EFD} = \\overline{D'FE}\\] \n\nproving that \\(EI\\parallel D'F\\) . Therefore, \\(ED'FI\\) is a parallelogram and its diagonals intersect in their midpoints, proving that \\(D'I\\) contains \\(M\\) .\n\n\n", "metadata": {"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2025-zz.jsonl", "problem_match": "\nProblem 3.", "solution_match": "\nSolution."}}
{"year": "2025", "tier": "T3", "problem_label": "3", "problem_type": null, "exam": "Benelux_MO", "problem": "Let \\(ABC\\) be a triangle with incentre \\(I\\) and circumcircle \\(\\Omega\\) . Let \\(D, E, F\\) be the midpoints of the arcs \\(\\overline{BC}, \\overline{CA}, \\overline{AB}\\) of \\(\\Omega\\) not containing \\(A, B, C\\) , respectively. Let \\(D'\\) be the point of \\(\\Omega\\) diametrically opposite to \\(D\\) . Show that \\(I, D'\\) , and the midpoint \\(M\\) of \\([EF]\\) lie on a line.", "solution": "By definition of \\(D, E\\) and \\(F\\) , we know that \\(ID, IE\\) and \\(IF\\) are the angle bisectors of \\(ABC\\) . If \\(D' = A\\) , this means that \\(\\overrightarrow{OAB} = 90^\\circ - \\overrightarrow{ACB} = \\frac{\\overrightarrow{CAB}}{2}\\) and \\(ABC\\) is isosceles in \\(A\\) . In that case, the symmetry with respect to \\(AI\\) sends \\(E\\) to \\(F\\) and so \\(M \\in AI\\) . We can thus suppose that \\(D' \\neq A\\) . \n\nLet us consider \\(I_A, I_B\\) and \\(I_C\\) the excenters of \\(ABC\\) . The lines \\(I_B AI_C, I_C BI_A\\) and \\(I_A CI_B\\) are thus the exterior bisectors of the triangle. Hence, \\(I_B IC \\perp AI_A\\) and, since \\(\\overrightarrow{D'AD} = 90^\\circ\\) , one has \\(D' \\in I_B IC\\) . In the triangle \\(I_A I_B IC\\) , the points \\(A, B\\) and \\(C\\) are the feet of the heights and \\(\\Omega\\) is thus the Euler circle of \\(I_A I_B IC\\) . Since \\(D' \\in [I_B IC] \\cap \\Omega\\) and \\(D' \\neq A\\) , it is the midpoint of \\([I_B IC]\\) . Moreover, \\(E\\) and \\(F\\) , being on the Euler circle and the heights, there are the midpoints of \\([I_B I]\\) and \\([IC I]\\) respectively. The homothety of centre \\(I\\) and ratio \\(\\frac{1}{2}\\) sends \\(I_B\\) on \\(E\\) and \\(IC\\) on \\(F\\) . It thus also sends \\(D'\\) on \\(M\\) , proving that \\(D', M\\) and \\(I\\) are collinear.", "metadata": {"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2025-zz.jsonl", "problem_match": "\nProblem 3.", "solution_match": "\nAlternative Solution."}}
{"year": "2025", "tier": "T3", "problem_label": "4", "problem_type": null, "exam": "Benelux_MO", "problem": "Let \\(a_{0},a_{1},\\ldots ,a_{10}\\) be integers such that, for each \\(i\\in \\{0,1,\\ldots ,2047\\}\\) , there exists a subset \\(S\\subseteq \\{0,1,\\ldots ,10\\}\\) with \n\n\\[\\sum_{j\\in S}a_{j}\\equiv i\\pmod {2048}.\\] \n\nShow that for each \\(i\\in \\{0,1,\\ldots ,10\\}\\) , there is exactly one \\(j\\in \\{0,1,\\ldots ,10\\}\\) such that \\(a_{j}\\) is divisible by \\(2^{i}\\) but not by \\(2^{i + 1}\\) . \n\nNote: \\(\\sum_{j\\in S}a_{j}\\) is the summation notation, for instance, \\(\\sum_{j\\in \\{2,5\\}}a_{j} = a_{2} + a_{5}\\) , while, for the empty set \\(\\varnothing\\) , one defines \\(\\sum_{j\\in \\varnothing}a_{j} = 0\\) .", "solution": "We denote by \\(\\nu_{2}(a)\\) the valuation 2- adic of the integer \\(a\\) . Let us prove by induction the more general statement that, for \\(n\\in \\mathbb{N}_{>0}\\) , if \\(a_{0},a_{1},\\ldots ,a_{n - 1}\\) are integers such that, for each \\(i\\in \\{0,1,\\ldots ,2^{n} - 1\\}\\) , there exists a subset \\(S\\subseteq \\{0,1,\\ldots ,n - 1\\}\\) with \\(\\sum_{j\\in S}a_{j}\\equiv i\\) (mod \\(2^{n}\\) ), then, for each \\(i\\in \\{0,1,\\ldots ,n - 1\\}\\) , there is exactly one \\(j\\in \\{0,1,\\ldots ,n - 1\\}\\) such that \\(\\nu_{2}(a_{j}) = i\\) . The result then follows by setting \\(n = 11\\) . The case \\(n = 1\\) is trivial. \n\nWe suppose by induction that the result is true for \\(n - 1\\) for prove it for \\(n\\geq 2\\) . Let us first notice that, since there are \\(2^{n}\\) elements in \\(\\{0,1,\\ldots ,2^{n} - 1\\}\\) and \\(2^{n}\\) subsets of \\(\\{0,1,\\ldots ,n - 1\\}\\) , for each \\(i\\in\\) \\(\\{0,1,\\ldots ,2^{n} - 1\\}\\) , there exists exactly one \\(S_{i}\\subseteq \\{0,1,\\ldots ,n - 1\\}\\) such that \\(\\sum_{j\\in S_{i}}a_{j}\\equiv i\\) (mod \\(2^{n}\\) ). By summing all these sums for all \\(i\\) , we obtain \n\n\\[\\sum_{i = 0}^{2^{n} - 1}\\sum_{j\\in S_{i}}a_{j}\\equiv \\sum_{i = 0}^{2^{n} - 1}i = \\frac{2^{n}\\cdot(2^{n} - 1)}{2} = 2^{n - 1}\\cdot (2^{n} - 1)\\pmod {2^{n}}.\\] \n\nFor a fixed \\(j\\in \\{0,1,\\ldots ,n - 1\\}\\) , \\(a_{j}\\) appears in exactly \\(2^{n - 1}\\) of these sums (since each of the \\(n - 1\\) remaining indices may or may not be in \\(S_{i}\\) ). Therefore, \n\n\\[2^{n - 1}\\cdot \\sum_{j = 0}^{n - 1}a_{j} = \\sum_{i = 0}^{2^{n} - 1}\\sum_{j\\in S_{i}}a_{j}\\equiv 2^{n - 1}\\cdot (2^{n} - 1)\\pmod {2^{n}}\\] \n\nwhich is equivalent to \n\n\\[\\sum_{j = 0}^{n - 1}a_{j}\\equiv 2^{n} - 1\\equiv 1\\pmod {2}.\\] \n\nSince \\(\\sum_{j = 0}^{n - 1}a_{j}\\) is odd, at least one of the \\(a_{j}\\) 's is odd. \n\nLet us suppose by contradiction that at least two of them are odd. We now sum all the \\(\\sum_{j\\in S_{i}}a_{j}\\) for which \\(\\sum_{j\\in S_{i}}a_{j}\\) (or, equivalently, \\(i\\) ) is even: \n\n\\[\\sum_{i\\in \\{0,\\ldots ,2^{n} - 1\\}}\\sum_{j\\in S_{i}}a_{j}\\equiv \\sum_{i\\in \\{0,\\ldots ,2^{n} - 1\\}}i = 2^{n - 1}\\cdot (2^{n - 1} - 1)\\pmod {2^{n}}\\] \n\nFor a fixed \\(j\\in \\{0,1,\\ldots ,n - 1\\}\\) , \\(a_{j}\\) appears in exactly \\(2^{n - 2}\\) of these sums. Indeed, since there is at least one \\(k\\in \\{0,1,\\ldots ,n - 1\\} \\backslash \\{j\\}\\) such that \\(a_{k}\\) is odd, one can consider any \\(S\\subseteq \\{0,1,\\ldots ,n - 1\\} \\backslash \\{j,k\\}\\) , add \\(j\\) to it, and potentially also \\(k\\) in order to make the partial sum even (exactly one possibility for each such \\(S\\) ). Therefore, \n\n\\[2^{n - 2}\\cdot \\sum_{j = 0}^{n - 1}a_{j} = \\sum_{i\\in \\{0,\\ldots ,2^{n} - 1\\}}\\sum_{j\\in S_{i}}a_{j}\\equiv 2^{n - 1}\\cdot (2^{n - 1} - 1)\\pmod {2^{n}}\\]\n\n\n\nor equivalently \n\n\\[\\sum_{j = 0}^{n - 1}a_{j}\\equiv 2\\cdot (2^{n - 1} - 1)\\pmod {4}.\\] \n\nHence \\(\\sum_{j = 0}^{n - 1}a_{j}\\) is even, which is a contradiction. Therefore, there is exactly one \\(k\\in \\{0,1,\\ldots ,n - 1\\}\\) such that \\(a_{k}\\) is odd. \n\nA sum \\(\\sum_{j\\in S}a_{j}\\) is odd if and only if \\(k\\in S\\) . Moreover, the sums \\(\\sum_{j\\in S}a_{j}\\) for which \\(k\\notin S\\) cover all the even numbers modulo \\(2^{n}\\) . Thus, omitting \\(a_{k}\\) , the \\(n - 1\\) integers \\(\\frac{a_{0}}{2},\\ldots ,\\frac{a_{n - 1}}{2}\\) satisfy the condition of the statement. By the induction hypothesis, for each \\(i\\in \\{0,1,\\ldots ,n - 2\\}\\) , there is exactly one \\(j\\in \\{0,1,\\ldots ,n - 1\\} \\backslash \\{k\\}\\) such that \\(\\nu_{2}\\left(\\frac{a_{j}}{2}\\right) = i\\) , i.e., \\(\\nu_{2}(a_{j}) = i + 1\\) . Since \\(\\nu_{2}(a_{k}) = 0\\) , this concludes the proof. \n\nRemark: If one considers the sum of all \\(\\sum_{j\\in S_{i}}a_{j}\\) such that \\(i\\) is odd, one obtains, in the case where at least two \\(a_{i}\\) 's are odd, that \\(2^{n - 2}\\cdot \\sum_{j = 0}^{n - 1}a_{j}\\equiv 0\\) (mod \\(2^{n}\\) ) also reaching a contradiction.", "metadata": {"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2025-zz.jsonl", "problem_match": "\nProblem 4.", "solution_match": "\nSolution."}}
{"year": "2025", "tier": "T3", "problem_label": "4", "problem_type": null, "exam": "Benelux_MO", "problem": "Let \\(a_{0},a_{1},\\ldots ,a_{10}\\) be integers such that, for each \\(i\\in \\{0,1,\\ldots ,2047\\}\\) , there exists a subset \\(S\\subseteq \\{0,1,\\ldots ,10\\}\\) with \n\n\\[\\sum_{j\\in S}a_{j}\\equiv i\\pmod {2048}.\\] \n\nShow that for each \\(i\\in \\{0,1,\\ldots ,10\\}\\) , there is exactly one \\(j\\in \\{0,1,\\ldots ,10\\}\\) such that \\(a_{j}\\) is divisible by \\(2^{i}\\) but not by \\(2^{i + 1}\\) . \n\nNote: \\(\\sum_{j\\in S}a_{j}\\) is the summation notation, for instance, \\(\\sum_{j\\in \\{2,5\\}}a_{j} = a_{2} + a_{5}\\) , while, for the empty set \\(\\varnothing\\) , one defines \\(\\sum_{j\\in \\varnothing}a_{j} = 0\\) .", "solution": "We do the same induction as in the main Solution. If all \\(a_{j}\\) are even, then all the sums are even, contradicting the hypothesis. Therefore, without loss of generality, we can assume that \\(a_{0}\\) is odd. For each \\(i\\in \\{0,1,2,\\ldots ,2^{n} - 1\\}\\) , let \\(S_{i}^{\\prime}\\subseteq \\{0,1,\\ldots ,n - 1\\}\\) be the subset such that \\(\\sum_{j\\in S_{i}^{\\prime}}a_{j}\\equiv i\\cdot a_{0}\\) (mod \\(2^{n}\\) ) (this subset exists by the hypothesis and is unique since there are exactly \\(2^{n}\\) subsets of \\(\\{0,1,\\ldots ,n - 1\\}\\) ). By definition, one has \\(S_{0}^{\\prime} = \\emptyset\\) and \\(S_{1}^{\\prime} = \\{0\\}\\) . \n\nLet us prove by induction that, for each \\(i\\in \\{0,1,\\ldots ,2^{n - 1} - 1\\}\\) , one has \\(0\\notin S_{2i}^{\\prime}\\) but \\(0\\in S_{2i + 1}^{\\prime}\\) . The case \\(i = 0\\) is trivial. For \\(i > 0\\) , if we suppose that \\(0\\in S_{2i - 1}^{\\prime}\\) , we prove that \\(0\\notin S_{2i}^{\\prime}\\) . Indeed, if \\(0\\in S_{2i}^{\\prime}\\) , then \\(\\sum_{j\\in S_{2i}^{\\prime}\\backslash \\{0\\}}a_{j}\\equiv 2i a_{0} - a_{0} = (2i - 1)a_{0}\\equiv \\sum_{j\\in S_{2i - 1}^{\\prime}}a_{j}\\) (mod \\(2^{n}\\) ) and so \\(S_{2i}^{\\prime}\\backslash \\{0\\} = S_{2i - 1}^{\\prime}\\) by uniqueness. But \\(0\\in S_{2i - 1}^{\\prime}\\) so this is a contradiction, proving that \\(0\\notin S_{2i}^{\\prime}\\) . Then, \\(\\sum_{j\\in S_{2i}^{\\prime}\\cup \\{0\\}}a_{j}\\equiv 2i a_{0} + a_{0} = (2i + 1)a_{0}\\equiv \\sum_{j\\in S_{2i + 1}^{\\prime}}a_{j}\\) (mod \\(2^{n}\\) ). By uniqueness, \\(S_{2i}^{\\prime}\\cup \\{0\\} = S_{2i + 1}^{\\prime}\\) and so \\(0\\in S_{2i + 1}^{\\prime}\\) . \n\nSince \\(a_{0}\\) is odd, it is invertible modulo \\(2^{n}\\) and so \\(0, a_{0}, 2a_{0}, \\ldots , (2^{n} - 1)a_{0}\\) is a permutation of 0, 1, 2, ..., \\(2^{n} - 1\\) modulo \\(2^{n}\\) . This shows that the subsets \\(S_{0}^{\\prime}, S_{2}^{\\prime}, S_{4}^{\\prime}, \\ldots , S_{2^{n} - 2}^{\\prime}\\) are all distinct. There are thus \\(2^{n - 1}\\) subsets \\(S_{2i}^{\\prime}\\) and they are a precisely the subsets of \\(\\{1,2,\\ldots ,n - 1\\}\\) (since there are also \\(2^{n - 1}\\) such subsets). Moreover, their corresponding sums are all even. Since this list of subsets contains \\(\\{1\\} , \\{2\\} , \\ldots , \\{n - 1\\}\\) , this shows that \\(a_{1}, a_{2}, \\ldots , a_{n - 1}\\) are all even. \n\nWe then conclude the induction as in the main Solution.", "metadata": {"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2025-zz.jsonl", "problem_match": "\nProblem 4.", "solution_match": "\nAlternative Solution."}}
diff --git a/CANADA_MO/md/en-CMO2025-solutions.md b/CANADA_MO/md/en-CMO2025-solutions.md
index fca0fe0341d814cab1b46c548da65332bb17c18f..92fd4806ca78975b3e058f1094a8665636464620 100644
--- a/CANADA_MO/md/en-CMO2025-solutions.md
+++ b/CANADA_MO/md/en-CMO2025-solutions.md
@@ -157,7 +157,7 @@ Consider the following version of the problem:
A rectangle \(R\) is divided into a set \(S\) of finitely many smaller rectangles such that no three rectangles in \(S\) share a common corner. For each \(r \in S\) , draw two non- intersecting arcs inside \(r\) , connecting the pairs of adjacent corners of \(r\) (there are two ways to do this, by connecting either the horizontally or vertically adjacent corners). Prove that there does not exist a path from the bottom- left corner of \(R\) to the top- right corner of \(R\) by walking only along these arcs.
-
+
Figure 1: A possible diagram of all the arcs.
@@ -175,7 +175,7 @@ Suppose that no rectangle was chosen in at least two operations. In particular,
At any point in the process, consider whether the last move by the ant was horizontal or vertical, and whether the most recently chosen rectangle was to the left or the right of the ant's path. In the first move, either the ant moved horizontally and the rectangle was to the left, or the ant moved vertically and the rectangle was to the right. We claim that this invariant is preserved throughout the entire process (see Figure 2 for a sample path). Assuming this claim, the final move to the top right corner must select the top right rectangle. If it is vertical, this is to the left of the path, and if it is horizontal, it is to the right of the path, both of which are impossible, providing a contradiction.
-
+
Figure 2: A possible path by the ant. The red arrows are all vertical, with the corresponding rectangle to the right. The blue arrows are all horizontal, with the corresponding rectangle to the left.
@@ -184,7 +184,7 @@ It remains to show that the invariant is preserved. Since four rectangles cannot
First, assume the ant moves up, hence the chosen rectangle \(r\) is on the right. The possible configurations are depicted in the first two diagrams in Figure 3.
-
+
Figure 3: Possible ant moves going up or right.
diff --git a/CANADA_MO/segmented/en-CMO2025-solutions.jsonl b/CANADA_MO/segmented/en-CMO2025-solutions.jsonl
index efe9903999ba06f14cd220fb3a1de8305a2c2836..30b8c53aa16517415fdeb2ab4cf8941bed08c643 100644
--- a/CANADA_MO/segmented/en-CMO2025-solutions.jsonl
+++ b/CANADA_MO/segmented/en-CMO2025-solutions.jsonl
@@ -4,6 +4,6 @@
{"year": "2025", "tier": "T2", "problem_label": "3", "problem_type": null, "exam": "Canada_MO", "problem": "A polynomial \\(c_{d}x^{d} + c_{d - 1}x^{d - 1} + \\dots + c_{1}x + c_{0}\\) with degree \\(d\\) is reflexive if there is an integer \\(n \\geq d\\) such that \\(c_{i} = c_{n - i}\\) for every \\(0 \\leq i \\leq n\\) , where \\(c_{i} = 0\\) for \\(i > d\\) . Let \\(\\ell \\geq 2\\) be an integer and \\(p(x)\\) be a polynomial with integer coefficients. Prove that there exist reflexive polynomials \\(q(x), r(x)\\) with integer coefficients such that \n\n\\[(1 + x + x^{2} + \\dots + x^{\\ell -1})p(x) = q(x) + x^{\\ell}r(x).\\]", "solution": "Let \\(d\\) be the degree of \\(p\\) and let \\(k\\) be any non- negative integer. We will choose \n\n\\[q(x) = \\frac{x^{d + k + \\ell}p\\left(\\frac{1}{x}\\right) - p(x)}{x - 1},\\] \\[r(x) = \\frac{p(x) - x^{d + k}p\\left(\\frac{1}{x}\\right)}{x - 1}.\\] \n\nFirst, we must show that both \\(q\\) and \\(r\\) are integer polynomials. Consider the numerator in \\(q\\) 's definition, \\(x^{d + k + \\ell}p\\left(\\frac{1}{x}\\right) - p(x)\\) . This is clearly an integer polynomial. As it is equal to 0 when evaluated at \\(x - 1\\) , \\(x - 1\\) divides it. Furthermore, as \\(x - 1\\) is monic, the quotient has integer coefficients. The argument for \\(r\\) is similar. \n\nNext, we will show that this choice of \\(q\\) and \\(r\\) satisfies the desired equation. Plugging them into the RHS of the equation gives \n\n\\[q(x) + x^{\\ell}r(x) = \\frac{x^{d + k + \\ell}p\\left(\\frac{1}{x}\\right) - p(x)}{x - 1} +x^{\\ell}\\left(\\frac{p(x) - x^{d + k}p\\left(\\frac{1}{x}\\right)}{x - 1}\\right)\\] \\[\\qquad = \\frac{x^{d + k + \\ell}p\\left(\\frac{1}{x}\\right) - p(x) + x^{\\ell}p(x) - x^{d + k + \\ell}p\\left(\\frac{1}{x}\\right)}{x - 1}\\] \\[\\qquad = \\left(\\frac{x^{\\ell} - 1}{x - 1}\\right)p(x)\\] \\[\\qquad = (1 + x + \\cdot \\cdot \\cdot +x^{\\ell -1})p(x)\\] \n\nas desired. \n\nFinally, we will show that \\(q\\) and \\(r\\) are indeed reflexive. We can re- interpret the reflexive condition as such: \n\nPolynomial \\(a(x)\\) is reflexive iff there is an integer \\(n \\geq \\deg (a)\\) for which \n\n\\[a(x) = x^{n}a\\left(\\frac{1}{x}\\right).\\]\n\n\n\nWe have \n\n\\[q(x) = \\frac{x^{d + k + \\ell}p\\left(\\frac{1}{x}\\right) - p(x)}{x - 1}\\] \\[\\qquad = x^{d + k + \\ell -1}\\cdot \\frac{p\\left(\\frac{1}{x}\\right) - x^{-(d + k + \\ell)}p(x)}{\\frac{x - 1}{x}}\\] \\[\\qquad = x^{d + k + \\ell -1}\\cdot \\frac{x^{-(d + k + \\ell)}p(x) - p\\left(\\frac{1}{x}\\right)}{\\frac{1}{x} - 1}\\] \\[\\qquad = x^{d + k + \\ell -1}q\\left(\\frac{1}{x}\\right)\\] \n\nas desired. Similarly, \n\n\\[r(x) = \\frac{p(x) - x^{d + k}p\\left(\\frac{1}{x}\\right)}{x - 1}\\] \\[\\qquad = x^{d + k - 1}\\cdot \\frac{x^{-(d + k)}p(x) - p\\left(\\frac{1}{x}\\right)}{\\frac{x - 1}{x}}\\] \\[\\qquad = x^{d + k - 1}\\cdot \\frac{p\\left(\\frac{1}{x}\\right) - x^{-(d + k)}p(x)}{\\frac{1}{x} - 1}\\] \\[\\qquad = x^{d + k - 1}r\\left(\\frac{1}{x}\\right).\\]", "metadata": {"resource_path": "CANADA_MO/segmented/en-CMO2025-solutions.jsonl", "problem_match": "\nP3.", "solution_match": "\n## Solution 1"}}
{"year": "2025", "tier": "T2", "problem_label": "3", "problem_type": null, "exam": "Canada_MO", "problem": "A polynomial \\(c_{d}x^{d} + c_{d - 1}x^{d - 1} + \\dots + c_{1}x + c_{0}\\) with degree \\(d\\) is reflexive if there is an integer \\(n \\geq d\\) such that \\(c_{i} = c_{n - i}\\) for every \\(0 \\leq i \\leq n\\) , where \\(c_{i} = 0\\) for \\(i > d\\) . Let \\(\\ell \\geq 2\\) be an integer and \\(p(x)\\) be a polynomial with integer coefficients. Prove that there exist reflexive polynomials \\(q(x), r(x)\\) with integer coefficients such that \n\n\\[(1 + x + x^{2} + \\dots + x^{\\ell -1})p(x) = q(x) + x^{\\ell}r(x).\\]", "solution": "We write degree \\(n\\) polynomial \\(p\\) as \n\n\\[p(x):= \\sum_{i = 0}^{n}p_{i}x^{i}.\\] \n\nDefine vector \\(P\\in \\mathbb{Z}^{n + 1}\\) as \n\n\\[P:= \\left(p_{0} p_{1} \\dots p_{n}\\right)^{T}.\\] \n\nWe also denote \\(X\\in \\mathbb{Z}[x]^{N}\\) for \\(N\\) some sufficiently high degree (e.g. \\(N > 2n + \\ell\\) ) as the vector of powers of \\(x\\) , i.e. \n\n\\[X:= \\left(1 x x^{2}\\dots x^{N - 1}\\right)^{T}.\\] \n\nFor a matrix \\(M\\in \\mathbb{Z}^{(n + 1)\\times N}\\) , \\(P^{T}M X\\) is an integer polynomial of degree \\(< N\\) . Note that if the non- zero entries of matrix \\(M\\) are horizontally symmetric, then the resulting polynomial must be reflexive.\n\n\n\nThen their total is the matrix whose entries are 1 at the parallelogram formed by \n\n\\[(0,0),(0,\\ell -1),(n,n + \\ell -1),(n,n),\\] \n\nwhich is precisely \\(A\\) .", "metadata": {"resource_path": "CANADA_MO/segmented/en-CMO2025-solutions.jsonl", "problem_match": "\nP3.", "solution_match": "\n## Solution 2"}}
{"year": "2025", "tier": "T2", "problem_label": "4", "problem_type": null, "exam": "Canada_MO", "problem": "Let \\(ABC\\) be a triangle with circumcircle \\(\\Gamma\\) and \\(AB \\neq AC\\) . Let \\(D\\) and \\(E\\) lie on the arc \\(BC\\) of \\(\\Gamma\\) not containing \\(A\\) such that \\(\\angle BAE = \\angle DAC\\) . Let the incenters of \\(BAE\\) and \\(CAD\\) be \\(X\\) and \\(Y\\) respectively, and let the external tangents of the incircles of \\(BAE\\) and \\(CAD\\) intersect at \\(Z\\) . Prove that \\(Z\\) lies on the common chord of \\(\\Gamma\\) and the circumcircle of \\(AXY\\) .", "solution": "Let \\(AX\\) and \\(AY\\) intersect \\((ABC)\\) again at \\(P\\) and \\(Q\\) , let the inradii of \\(ABE\\) and \\(ACD\\) be \\(r_B\\) and \\(r_C\\) , and let \\((AXY)\\) intersect \\((ABC)\\) again at \\(N\\) . \n\nFirst note that \\(\\angle BAP = \\frac{1}{2}\\angle BAE = \\frac{1}{2}\\angle CAD = \\angle QAC\\) , so \\(XP = BP = CQ = CY\\) . This thus implies that, since \\(NXP\\) and \\(NYQ\\) are spirally similar, \\(NX = NY\\) and \\(NP = NQ\\) (in particular, \\(N\\) is the midpoint of arc \\(XAY\\) on \\((AXY)\\) ). Now, note that \n\n\\[\\frac{ZX}{ZY} = \\frac{r_b}{r_c\\] \\[\\qquad = \\frac{AX\\sin(\\angle PAE)}{AY\\sin(\\angle QAD)\\] \\[\\qquad = \\frac{AX}{AY}.\\] \n\nSince \\(Z\\) lies on the ray \\(YX\\) , we have that \\(Z\\) lies on the external angle bisector of \\(\\angle XAY\\) . But so does \\(N\\) , hence \\(Z\\) , \\(A\\) , \\(N\\) are collinear, as desired.", "metadata": {"resource_path": "CANADA_MO/segmented/en-CMO2025-solutions.jsonl", "problem_match": "\nP4.", "solution_match": "\n## Solution \n\n"}}
-{"year": "2025", "tier": "T2", "problem_label": "5", "problem_type": null, "exam": "Canada_MO", "problem": "A rectangle \\(R\\) is divided into a set \\(S\\) of finitely many smaller rectangles with sides parallel to the sides of \\(R\\) such that no three rectangles in \\(S\\) share a common corner. An ant is initially located at the bottom-left corner of \\(R\\) . In one operation, we can choose a rectangle \\(r \\in S\\) such that the ant is currently located at one of the corners of \\(r\\) , say \\(c\\) , and move the ant to one of the two corners of \\(r\\) adjacent to \\(c\\) . \n\nSuppose that after a finite number of operations, the ant ends up at the top-right corner of \\(R\\) . Prove that some rectangle \\(r \\in S\\) was chosen in at least two operations.", "solution": "Consider the following version of the problem: \n\nA rectangle \\(R\\) is divided into a set \\(S\\) of finitely many smaller rectangles such that no three rectangles in \\(S\\) share a common corner. For each \\(r \\in S\\) , draw two non- intersecting arcs inside \\(r\\) , connecting the pairs of adjacent corners of \\(r\\) (there are two ways to do this, by connecting either the horizontally or vertically adjacent corners). Prove that there does not exist a path from the bottom- left corner of \\(R\\) to the top- right corner of \\(R\\) by walking only along these arcs. \n\n\n\nFigure 1: A possible diagram of all the arcs. \n\nIn this problem, consider an undirected graph where nodes correspond to corners of rectangles in \\(S\\) and edges correspond to the arcs, connecting the two nodes that the arc connects. The degrees of the nodes corresponding to the corners of \\(R\\) are all exactly 1. Since no three rectangles in \\(S\\) share a common corner, all intersection points have a pattern like \\(\\vdash\\) , \\(\\neg\\) , \\(\\perp\\) , or \\(\\top\\) , so the degree of all other nodes is exactly 2. Therefore, this graph can be decomposed into several paths and cycles. The only possible endpoints of paths are the degree 1 nodes, which are the corners of \\(R\\) . It follows that if there exists a path from the bottom- left corner to the top- right corner of \\(R\\) , then there also exists a path from the bottom- right corner to the top- left corner of\n\n\n\n\\(R\\) . However, this is impossible because these two paths (viewed as planar curves inside \\(R\\) ) must intersect, which cannot occur. Therefore, this claim is proved. \n\nReturning to the original problem, suppose some operations were performed while choosing each rectangle at most once. Draw two arcs inside every rectangle, either both horizontal if the ant used this rectangle to move horizontally or both vertical if the ant used this rectangle to move vertically (or pick one arbitrarily if this rectangle was not used). By the new version of the problem, there does not exist a path from the bottom- left corner to the top- right corner of \\(R\\) , and it follows that it is impossible for the ant to have reached the top- right corner of \\(R\\) , finishing the proof.", "metadata": {"resource_path": "CANADA_MO/segmented/en-CMO2025-solutions.jsonl", "problem_match": "\nP5.", "solution_match": "\n## Solution 1"}}
-{"year": "2025", "tier": "T2", "problem_label": "5", "problem_type": null, "exam": "Canada_MO", "problem": "A rectangle \\(R\\) is divided into a set \\(S\\) of finitely many smaller rectangles with sides parallel to the sides of \\(R\\) such that no three rectangles in \\(S\\) share a common corner. An ant is initially located at the bottom-left corner of \\(R\\) . In one operation, we can choose a rectangle \\(r \\in S\\) such that the ant is currently located at one of the corners of \\(r\\) , say \\(c\\) , and move the ant to one of the two corners of \\(r\\) adjacent to \\(c\\) . \n\nSuppose that after a finite number of operations, the ant ends up at the top-right corner of \\(R\\) . Prove that some rectangle \\(r \\in S\\) was chosen in at least two operations.", "solution": "Suppose that no rectangle was chosen in at least two operations. In particular, a rectangle cannot be selected in two consecutive operations. \n\nAt any point in the process, consider whether the last move by the ant was horizontal or vertical, and whether the most recently chosen rectangle was to the left or the right of the ant's path. In the first move, either the ant moved horizontally and the rectangle was to the left, or the ant moved vertically and the rectangle was to the right. We claim that this invariant is preserved throughout the entire process (see Figure 2 for a sample path). Assuming this claim, the final move to the top right corner must select the top right rectangle. If it is vertical, this is to the left of the path, and if it is horizontal, it is to the right of the path, both of which are impossible, providing a contradiction. \n\n\n\nFigure 2: A possible path by the ant. The red arrows are all vertical, with the corresponding rectangle to the right. The blue arrows are all horizontal, with the corresponding rectangle to the left. \n\nIt remains to show that the invariant is preserved. Since four rectangles cannot intersect at a corner, each intersection has a pattern like \\(\\vdash\\) , \\(\\neg\\) , \\(\\perp\\) , or \\(\\top\\) . \n\nFirst, assume the ant moves up, hence the chosen rectangle \\(r\\) is on the right. The possible configurations are depicted in the first two diagrams in Figure 3.\n\n\n\n\nFigure 3: Possible ant moves going up or right. \n\nIn each case, the ant must choose rectangle \\(s\\) next (to avoid repeating \\(r\\) twice), and we see that both side choices preserve a horizontal move with \\(s\\) left, or a vertical move with \\(s\\) right. By rotating the picture by \\(180^{\\circ}\\) , we cover the two possibilities for the ant moving downward. \n\nIf the ant moves right, then \\(r\\) must occur on the left, and the possible configurations are the last two diagrams of Figure 3. Once again, rectangle \\(s\\) must be chosen next, and the invariant is similarly preserved. The case of the ant moving left is again handled by a \\(180^{\\circ}\\) rotation, completing the proof.", "metadata": {"resource_path": "CANADA_MO/segmented/en-CMO2025-solutions.jsonl", "problem_match": "\nP5.", "solution_match": "\n## Solution 2"}}
+{"year": "2025", "tier": "T2", "problem_label": "5", "problem_type": null, "exam": "Canada_MO", "problem": "A rectangle \\(R\\) is divided into a set \\(S\\) of finitely many smaller rectangles with sides parallel to the sides of \\(R\\) such that no three rectangles in \\(S\\) share a common corner. An ant is initially located at the bottom-left corner of \\(R\\) . In one operation, we can choose a rectangle \\(r \\in S\\) such that the ant is currently located at one of the corners of \\(r\\) , say \\(c\\) , and move the ant to one of the two corners of \\(r\\) adjacent to \\(c\\) . \n\nSuppose that after a finite number of operations, the ant ends up at the top-right corner of \\(R\\) . Prove that some rectangle \\(r \\in S\\) was chosen in at least two operations.", "solution": "Consider the following version of the problem: \n\nA rectangle \\(R\\) is divided into a set \\(S\\) of finitely many smaller rectangles such that no three rectangles in \\(S\\) share a common corner. For each \\(r \\in S\\) , draw two non- intersecting arcs inside \\(r\\) , connecting the pairs of adjacent corners of \\(r\\) (there are two ways to do this, by connecting either the horizontally or vertically adjacent corners). Prove that there does not exist a path from the bottom- left corner of \\(R\\) to the top- right corner of \\(R\\) by walking only along these arcs. \n\n\n\nFigure 1: A possible diagram of all the arcs. \n\nIn this problem, consider an undirected graph where nodes correspond to corners of rectangles in \\(S\\) and edges correspond to the arcs, connecting the two nodes that the arc connects. The degrees of the nodes corresponding to the corners of \\(R\\) are all exactly 1. Since no three rectangles in \\(S\\) share a common corner, all intersection points have a pattern like \\(\\vdash\\) , \\(\\neg\\) , \\(\\perp\\) , or \\(\\top\\) , so the degree of all other nodes is exactly 2. Therefore, this graph can be decomposed into several paths and cycles. The only possible endpoints of paths are the degree 1 nodes, which are the corners of \\(R\\) . It follows that if there exists a path from the bottom- left corner to the top- right corner of \\(R\\) , then there also exists a path from the bottom- right corner to the top- left corner of\n\n\n\n\\(R\\) . However, this is impossible because these two paths (viewed as planar curves inside \\(R\\) ) must intersect, which cannot occur. Therefore, this claim is proved. \n\nReturning to the original problem, suppose some operations were performed while choosing each rectangle at most once. Draw two arcs inside every rectangle, either both horizontal if the ant used this rectangle to move horizontally or both vertical if the ant used this rectangle to move vertically (or pick one arbitrarily if this rectangle was not used). By the new version of the problem, there does not exist a path from the bottom- left corner to the top- right corner of \\(R\\) , and it follows that it is impossible for the ant to have reached the top- right corner of \\(R\\) , finishing the proof.", "metadata": {"resource_path": "CANADA_MO/segmented/en-CMO2025-solutions.jsonl", "problem_match": "\nP5.", "solution_match": "\n## Solution 1"}}
+{"year": "2025", "tier": "T2", "problem_label": "5", "problem_type": null, "exam": "Canada_MO", "problem": "A rectangle \\(R\\) is divided into a set \\(S\\) of finitely many smaller rectangles with sides parallel to the sides of \\(R\\) such that no three rectangles in \\(S\\) share a common corner. An ant is initially located at the bottom-left corner of \\(R\\) . In one operation, we can choose a rectangle \\(r \\in S\\) such that the ant is currently located at one of the corners of \\(r\\) , say \\(c\\) , and move the ant to one of the two corners of \\(r\\) adjacent to \\(c\\) . \n\nSuppose that after a finite number of operations, the ant ends up at the top-right corner of \\(R\\) . Prove that some rectangle \\(r \\in S\\) was chosen in at least two operations.", "solution": "Suppose that no rectangle was chosen in at least two operations. In particular, a rectangle cannot be selected in two consecutive operations. \n\nAt any point in the process, consider whether the last move by the ant was horizontal or vertical, and whether the most recently chosen rectangle was to the left or the right of the ant's path. In the first move, either the ant moved horizontally and the rectangle was to the left, or the ant moved vertically and the rectangle was to the right. We claim that this invariant is preserved throughout the entire process (see Figure 2 for a sample path). Assuming this claim, the final move to the top right corner must select the top right rectangle. If it is vertical, this is to the left of the path, and if it is horizontal, it is to the right of the path, both of which are impossible, providing a contradiction. \n\n\n\nFigure 2: A possible path by the ant. The red arrows are all vertical, with the corresponding rectangle to the right. The blue arrows are all horizontal, with the corresponding rectangle to the left. \n\nIt remains to show that the invariant is preserved. Since four rectangles cannot intersect at a corner, each intersection has a pattern like \\(\\vdash\\) , \\(\\neg\\) , \\(\\perp\\) , or \\(\\top\\) . \n\nFirst, assume the ant moves up, hence the chosen rectangle \\(r\\) is on the right. The possible configurations are depicted in the first two diagrams in Figure 3.\n\n\n\n\nFigure 3: Possible ant moves going up or right. \n\nIn each case, the ant must choose rectangle \\(s\\) next (to avoid repeating \\(r\\) twice), and we see that both side choices preserve a horizontal move with \\(s\\) left, or a vertical move with \\(s\\) right. By rotating the picture by \\(180^{\\circ}\\) , we cover the two possibilities for the ant moving downward. \n\nIf the ant moves right, then \\(r\\) must occur on the left, and the possible configurations are the last two diagrams of Figure 3. Once again, rectangle \\(s\\) must be chosen next, and the invariant is similarly preserved. The case of the ant moving left is again handled by a \\(180^{\\circ}\\) rotation, completing the proof.", "metadata": {"resource_path": "CANADA_MO/segmented/en-CMO2025-solutions.jsonl", "problem_match": "\nP5.", "solution_match": "\n## Solution 2"}}
{"year": "2025", "tier": "T2", "problem_label": "5", "problem_type": null, "exam": "Canada_MO", "problem": "A rectangle \\(R\\) is divided into a set \\(S\\) of finitely many smaller rectangles with sides parallel to the sides of \\(R\\) such that no three rectangles in \\(S\\) share a common corner. An ant is initially located at the bottom-left corner of \\(R\\) . In one operation, we can choose a rectangle \\(r \\in S\\) such that the ant is currently located at one of the corners of \\(r\\) , say \\(c\\) , and move the ant to one of the two corners of \\(r\\) adjacent to \\(c\\) . \n\nSuppose that after a finite number of operations, the ant ends up at the top-right corner of \\(R\\) . Prove that some rectangle \\(r \\in S\\) was chosen in at least two operations.", "solution": "This is a variant of Solution 2. As in that proof, assume that no rectangle was chosen in two consecutive operations. We claim that for every move, the ant is in either the bottom- left or top- right corner of the chosen rectangle, and moves to the bottom- right or top- left corner. \n\nThis is clearly true of the first move. If the ant starts at the bottom- left or top- right corner on a move (choosing rectangle \\(r\\) ), it is clear that they must move to the bottom- right or top- left of \\(r\\) . Assume they moved to the bottom- right corner of \\(r\\) , and chose rectangle \\(s\\) in the next move. If they are at the bottom- right corner of \\(s\\) , then \\(r\\) and \\(s\\) are either equal or overlap, a contradiction. If they are at the top- left of \\(s\\) , then \\(r\\) and \\(s\\) intersect at a corner and no sides, and we must have 4 rectangles intersecting at a corner, again a contradiction. \n\nTherefore they must be at the bottom- left or top- right corner of \\(s\\) , as desired. The case where the ant is at the top- left corner of \\(r\\) is analogous. \n\nIf the ant is able to make it to the top right corner, their final move must select \\(r\\) as the top- right rectangle, and they move to the top- right corner, which is therefore impossible.", "metadata": {"resource_path": "CANADA_MO/segmented/en-CMO2025-solutions.jsonl", "problem_match": "\nP5.", "solution_match": "\n## Solution 3"}}
diff --git a/Dutch_TST/md/nl-2025-B2025_uitwerkingen.md b/Dutch_TST/md/nl-2025-B2025_uitwerkingen.md
index ef924c01ae12c394b99a3b6199499320e0d3cc15..32c3435610e868aca0b7b23f5cbce5623b588449 100644
--- a/Dutch_TST/md/nl-2025-B2025_uitwerkingen.md
+++ b/Dutch_TST/md/nl-2025-B2025_uitwerkingen.md
@@ -107,7 +107,7 @@ Oplossing I. Voor de tegenstandercomputers van \(a_i\) bekijken we tegen welke \
Merk op dat de computers \(a_{i-2}\), \(a_{i-1}\), \(a_{i+1}\) en \(a_{i+2}\) elk twee gemeenschappelijk tegenstandercomputers hebben met \(a_i\). Verder hebben \(a_{i-4}\), \(a_{i-3}\), \(a_{i+3}\) en \(a_{i+4}\) elk één gemeenschappelijke tegenstandercomputer met \(a_i\). Alleen voor de eerste twee, \(a_{i-4}\) en \(a_{i-3}\), is de gemeenschappelijk tegenstander met \(a_i\) hetzelfde, namelijk \(b_i\).
-
+
@@ -127,7 +127,7 @@ Oplossing II. We noemen twee computers *superconnected* als ze twee tegenstander
- Er is één computer superconnected met zowel \(a_i\) als \(a_{i+2}\), namelijk de computer \(a_{i+1}\).
-
+
@@ -145,7 +145,7 @@ Opgave 4. Gegeven is \(\triangle ABC\) met omgeschreven cirkel \(\Gamma\). Zij \
Bewijs dat \(ND\), \(MF\) en \(PE\) concurrent zijn.
-
+
diff --git a/Dutch_TST/md/nl-2025-C2025_uitwerkingen.md b/Dutch_TST/md/nl-2025-C2025_uitwerkingen.md
index 5e46bf269f7dfe3c0d7aadebd7d404708dd02657..b74d32612889fa5f0995eff893438b7b3767338f 100644
--- a/Dutch_TST/md/nl-2025-C2025_uitwerkingen.md
+++ b/Dutch_TST/md/nl-2025-C2025_uitwerkingen.md
@@ -100,7 +100,7 @@ Oplossing. Als \(m, n \ge 2\) en minstens één van de twee is even, dan is het
We tekenen eerst een voorbeeld voor \(n = 1\) en \(m \ge 1\) met twee bijzondere driehoeken, een voorbeeld voor \(n = 2\) en \(m \ge 3\) met nul bijzondere driehoeken, en het speciale geval \(n = 2\), \(m = 2\) met wederom nul bijzondere driehoeken.
-
+
@@ -111,7 +111,7 @@ Als minstens één van \(m\) en \(n\) even is, kunnen we de rechthoek opknippen
Voor elke partitionering maken we nu het volgende plaatje. We zetten een rood punt in het midden van elke schuine zijde van elke driehoek, en als er twee schuine zijden zijn dan verbinden we de twee rode punten met een rode lijn. Merk op dat elke driehoek hoogstens twee schuine zijden heeft, en alleen bijzondere driehoeken hebben er maar één.
-
+
In elk rood punt komen nu hoogstens twee rode lijnen samen. Dus deze rode lijnen vormen gesloten cykels of open paden, en de bijzondere driehoeken zijn precies de eindes van de de open paden. Zo hebben we hierboven een open pad van lengte nul, en een gesloten cykel gevormd door 6 rode segmenten. Een rood pad kan alleen van richting veranderen op een schuine zijde die in beide richtingen “hoogte” 1 heeft (oftewel die hoort bij twee driehoeken die in verschillende richtingen hoogte 1 hebben). Zo’n schuine zijde moet dus precies de diagonaal zijn van een vakje. In het bijzonder verandert een rood pad alleen van richting in het midden van vakjes. (Alternatief kun je zeggen dat de rode zijden nooit over de roosterlijnen lopen, want de hoogte van elke driehoek is 1 en de rode lijnen lopen op halve hoogte. Dus je kan nooit van richting veranderen op de roosterlijnen.)
@@ -123,7 +123,7 @@ Opgave 3. Zij \(\triangle ABC\) een scherphoekige driehoek zo dat \(|AB| + |BC|
Een mier begint zijn reis in \(X\) en gaat vanaf daar naar een punt op \(AC\), dan een punt op \(\ell\), dan weer terug naar een (eventueel ander) punt op \(AC\) en uiteindelijk naar \(Y\). Bewijs dat de lengte van de kortst mogelijke route van de mier gelijk is aan \(4|XY|\).
-
+
diff --git a/Dutch_TST/md/nl-2025-D2025_uitwerkingen.md b/Dutch_TST/md/nl-2025-D2025_uitwerkingen.md
index 9929803f3e6dd2e03b02251f2df607c393d2b73d..455699e1c8a8120d9b24ec35a2b856bbdd3c6e0f 100644
--- a/Dutch_TST/md/nl-2025-D2025_uitwerkingen.md
+++ b/Dutch_TST/md/nl-2025-D2025_uitwerkingen.md
@@ -21,7 +21,7 @@ Opgave 2. Zij \(\triangle ABC\) een scherphoekige driehoek met \(|AB| > |AC|\),
Bewijs dat de lijn \(KT\) door \(O\) gaat.
-
+
diff --git a/Dutch_TST/md/nl-2025-E2025_uitwerkingen.md b/Dutch_TST/md/nl-2025-E2025_uitwerkingen.md
index 10d6f1d3f77ad1873b2d6e7e7dc387f5a0a231b8..b60698cdffb4379b16b50a11d37190167f82fef3 100644
--- a/Dutch_TST/md/nl-2025-E2025_uitwerkingen.md
+++ b/Dutch_TST/md/nl-2025-E2025_uitwerkingen.md
@@ -9,7 +9,7 @@ Opgave 1. Zij \(ABCD\) een parallellogram en zij \(M\) het snijpunt van de diago
Bewijs dat \(\angle ACB = \angle DCF\).
-
+
diff --git a/Dutch_TST/segmented/nl-2025-B2025_uitwerkingen.jsonl b/Dutch_TST/segmented/nl-2025-B2025_uitwerkingen.jsonl
index a81d9af3bfde657848710a0b19c75e8247b4ce4c..e0d6d305c7d7f92b2e99cd64ee6a6e4e82e42030 100644
--- a/Dutch_TST/segmented/nl-2025-B2025_uitwerkingen.jsonl
+++ b/Dutch_TST/segmented/nl-2025-B2025_uitwerkingen.jsonl
@@ -3,9 +3,9 @@
{"year": "2025", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "Dutch_TST", "problem": "Zij \\(n \\ge 2\\) een geheel getal, en laat \\(z_1, \\dots, z_n\\) positieve gehele getallen zijn die voldoen aan: \n\n- \\(z_j \\le j\\) voor \\(j = 1, \\dots, n\\); \n\n- \\(z_1 + \\dots + z_n\\) is even. \n\nBewijs dat er \\(s_1, \\dots, s_n \\in \\{-1, 1\\}\\) bestaan zodat: \n\n\\[s_1 z_1 + s_2 z_2 + \\dots + s_n z_n = 0.\\]", "solution": "We bewijzen dit met tweestapsinductie naar \\(n\\). Voor \\(n = 2\\) volgt uit de voorwaarden dat \\(z_1 = z_2 = 1\\), en dan geldt dus dat \\(z_1 - z_2 = 0\\). Voor \\(n = 3\\) volgt uit de voorwaarde dat \\((z_1, z_2, z_3)\\) gelijk is aan \\((1, 1, 2)\\), \\((1, 2, 1)\\) of \\((1, 2, 3)\\). Daarvoor kiezen we respectievelijk \\((s_1, s_2, s_3)\\) gelijk aan \\((1, 1, -1)\\), \\((1, -1, 1)\\) en \\((1, 1, -1)\\). \n\nStel nu dat we het gevraagde bewezen hebben voor \\(n = k\\) en \\(n = k-1\\), waarbij \\(k \\ge 2\\). Laat \\(z_1, \\dots, z_{k+1}\\) getallen zijn die voldoen aan de voorwaarden in de opgave. We onderscheiden twee gevallen. Stel eerst dat \\(z_k = z_{k+1}\\). Door de inductiehypothese voor \\(n = k-1\\) toe te passen op \\(z_1, \\dots, z_{k-1}\\) vinden we \\(s_1, \\dots, s_{k-1}\\) zodat \\(s_1 z_1 + \\dots + s_{k-1} z_{k-1} = 0\\). Nu hebben we ook dat \n\n\\[s_1 z_1 + \\dots + s_{k-1} z_{k-1} + z_k - z_{k+1} = 0,\\]\n\ndus volgt het gevraagde. Stel nu dat \\(z_k \\neq z_{k+1}\\). Dan geldt dat \\(|z_k - z_{k+1}| \\ge 1\\). Daarnaast weten we ook dat \\(|z_k - z_{k+1}| \\le k\\) omdat \\(1 \\le z_k, z_{k+1} \\le k+1\\). Omdat \\(|z_k - z_{k+1}| \\equiv z_k + z_{k+1} \\mod 2\\), geldt ook dat \\(z_1 + \\dots + z_{k-1} + |z_k - z_{k+1}|\\) even is. We kunnen dus de inductiehypothese voor \\(n = k\\) toepassen op \\(z_1, \\dots, z_{k-1}, |z_k - z_{k+1}|\\). Hierdoor vinden we \\(s_1, \\dots, s_k\\) zodanig dat: \n\n\\[s_1 z_1 + \\dots + s_{k-1} z_{k-1} + s_k |z_k - z_{k+1}| = 0.\\]\n\nOmdat \\(|z_k - z_{k+1}| = \\pm (z_k - z_{k+1})\\), volgt ook nu het gevraagde.", "metadata": {"resource_path": "Dutch_TST/segmented/nl-2025-B2025_uitwerkingen.jsonl", "problem_match": "\nOpgave 2.", "solution_match": "\nOplossing I."}}
{"year": "2025", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "Dutch_TST", "problem": "Zij \\(n \\ge 2\\) een geheel getal, en laat \\(z_1, \\dots, z_n\\) positieve gehele getallen zijn die voldoen aan: \n\n- \\(z_j \\le j\\) voor \\(j = 1, \\dots, n\\); \n\n- \\(z_1 + \\dots + z_n\\) is even. \n\nBewijs dat er \\(s_1, \\dots, s_n \\in \\{-1, 1\\}\\) bestaan zodat: \n\n\\[s_1 z_1 + s_2 z_2 + \\dots + s_n z_n = 0.\\]", "solution": "We construeren recursief \\(s_n, s_{n-1}, \\dots, s_1\\), op de volgende manier. Om te beginnen kiezen we \\(s_n = 1\\). Bekijk vervolgens een \\(k \\in \\{2, \\dots, n\\}\\) en stel dat we \\(s_k, \\dots, s_n\\) gekozen hebben. Definieer \\(b_k = s_k z_k + \\dots + s_n z_n\\), en kies: \n\n\\[s_{k-1} = \\begin{cases} -1 & \\text{als } b_k \\ge 0; \\\\ 1 & \\text{als } b_k < 0. \\end{cases}\\]\n\n\n\nDefinieer ten slotte \\(b_1 = s_1z_1 + \\dots + s_nz_n\\). \n\nWe beweren dat \\(|b_k| \\le k\\) voor \\(k = 1, \\dots, n\\). We bewijzen dit met neerwaartse inductie naar \\(k\\). Voor \\(k = n\\) geldt inderdaad dat \\(|b_n| = |s_nz_n| = |z_n| \\le n\\). Stel nu dat we bewezen hebben dat \\(|b_k| \\le k\\) voor een zekere \\(k \\in \\{2, \\dots, n\\}\\). Als \\(b_k \\ge 0\\), dan geldt dat \\(b_{k-1} = b_k - z_{k-1}\\). We hebben \\(-(k-1) \\le -z_{k-1} \\le b_k - z_{k-1} \\le b_k - 1 \\le k-1\\), dus \\(|b_{k-1}| = |b_k - z_{k-1}| \\le k-1\\). Als \\(b_k < 0\\), dan geldt dat \\(b_{k-1} = b_k + z_{k-1}\\). In dit geval hebben we \\(-(k-1) \\le b_k + 1 \\le b_k + z_{k-1} < z_{k-1} \\le k-1\\), dus volgt ook dat \\(|b_{k-1}| = |b_k + z_{k-1}| \\le k-1\\). Dit voltooit de inductie. \n\nIn het bijzonder geldt dat \\(|b_1| \\le 1\\). Daarnaast weten we dat \\(b_1 \\equiv z_1 + \\dots + z_n \\equiv 0 \\mod 2\\), dus moet \\(b_1 = 0\\), zoals we wilden. \\(\\square\\)", "metadata": {"resource_path": "Dutch_TST/segmented/nl-2025-B2025_uitwerkingen.jsonl", "problem_match": "\nOpgave 2.", "solution_match": "\nOplossing II."}}
{"year": "2025", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "Dutch_TST", "problem": "Zij \\(n \\ge 2\\) een geheel getal, en laat \\(z_1, \\dots, z_n\\) positieve gehele getallen zijn die voldoen aan: \n\n- \\(z_j \\le j\\) voor \\(j = 1, \\dots, n\\); \n\n- \\(z_1 + \\dots + z_n\\) is even. \n\nBewijs dat er \\(s_1, \\dots, s_n \\in \\{-1, 1\\}\\) bestaan zodat: \n\n\\[s_1 z_1 + s_2 z_2 + \\dots + s_n z_n = 0.\\]", "solution": "We bewijzen de volgende sterkere claim, waaruit de opgave volgt door voor \\(m = \\frac{1}{2}S\\) te kiezen. \n\n**Claim.** Zij \\(n\\) een positief geheel getal, en laat \\(z_1, \\dots, z_n\\) positieve gehele getallen zo dat \\(z_j \\le j\\) voor \\(j = 1, \\dots, n\\). Zij \\(S = z_1 + \\dots + z_n\\) (niet-noodzakelijk even) en \\(0 \\le m \\le S\\). Dan bestaat er een deelverzameling van \\(\\{z_1, \\dots, z_n\\}\\) met som \\(m\\). \n\nWe bewijzen dit met inductie naar \\(n\\). Voor \\(n = 1\\), heeft de lege deelverzameling som 0 en de hele verzameling \\(\\{z_1\\}\\) som 1. \n\nStel nu dat de claim waar is voor \\(n = k\\) en beschouw getallen \\(z_1, \\dots, z_{k+1}\\) die aan de voorwaarde voldoen. We definiëren \\(Z = \\max(z_1, \\dots, z_{k+1})\\) en zij \\(j\\) maximaal zo dat \\(z_j = Z\\). Dat betekent dat \\(z_i < Z \\le j\\) voor alle \\(i > j\\). \n\nAls we \\(z_j\\) weglaten krijgen we een verzameling van \\(k\\) elementen, gedefinieerd als \n\n\\[ \\text{voor } i < j: \\quad \\tilde{z}_i = z_i \\le i; \\\\ \\text{voor } i \\ge j: \\quad \\tilde{z}_i = z_{i+1} < Z \\le j \\le i, \\]\n\nmet som \\(\\tilde{S} = \\sum_{i=1}^n \\tilde{z}_i = S-Z\\). Merk op dat \\(Z \\le j \\le k+1\\) en \\(S-Z = \\sum_{i=1; i \\ne j}^{k+1} z_i \\ge k \\cdot 1 = k\\), oftewel \\(Z \\le S-k\\). Dat betekent dat \\(2Z \\le (k+1) + (S-k) = S+1\\). Dus \\(Z\\) is hoogstens de helft van de som \\(S\\) naar boven afgerond, \\(Z \\le \\lceil \\frac{1}{2}S \\rceil\\), en \\(\\tilde{S}\\) is minstens de helft van \\(S\\) naar beneden afgerond, \\(\\tilde{S} \\ge \\lfloor \\frac{1}{2}S \\rfloor\\). Hiermee kunnen we nu de inductiehypothese toepassen. \n\n- Voor \\(0 \\le m \\le \\lfloor \\frac{1}{2}S \\rfloor\\), geldt dus \\(0 \\le m \\le \\tilde{S}\\). Dan is er een deelverzameling van de \\(\\tilde{z}_i\\) met som \\(m\\), wat meteen een deelverzameling van de \\(z_i\\) geeft met dezelfde som. \n\n- Voor \\(\\lceil \\frac{1}{2}S \\rceil \\le m \\le S\\), geldt dat \\(0 \\le m-Z \\le S-Z = \\tilde{S}\\). Dus er is een deelverzameling van de \\(\\tilde{z}_i\\) met som \\(m-Z\\). Als we hier \\(z_j\\) aan toevoegen krijgen we een deelverzameling van de \\(z_i\\) met som \\(m\\).\n\n\n\nHiermee hebben we voor \\(n = k + 1\\) voor elke \\(0 \\le m \\le S\\) een deelverzameling gevonden met som \\(m\\), en is de inductie voltooid. \n\nAls de som \\(S\\) even is, dan kunnen we \\(m = \\frac{1}{2}S\\) kiezen. Dat geeft een deelverzameling met som \\(\\frac{1}{2}S\\), en de som van de overige getallen is ook gelijk aan aan \\(S - \\frac{1}{2}S = \\frac{1}{2}S\\). Neem nu de bijbehorende \\(s_i = 1\\) voor de \\(z_i\\) in deze deelverzameling, en alle overige \\(s_i = -1\\). Dan geldt dus \\(s_1z_1 + s_2z_2 + \\dots + s_nz_n = \\frac{1}{2}S - \\frac{1}{2}S = 0\\). \\(\\square\\) \n\nOpmerking. Met de substitutie \\(s_i = 2t_i - 1\\) wordt de vraag equivalent aan een reeks \\(t_1, \\dots, t_n\\) met \\(t_i \\in \\{0, 1\\}\\) zo dat \n\n\\[t_1z_1 + t_2z_2 + \\dots + t_nz_n = \\frac{1}{2}S.\\]\n\nDan is de omschreven strategie van Oplossing 2 om de \\(t_i\\) te kiezen beginnende vanaf \\(i = n\\) zodanig dat de som \\(t_i z_i + \\dots + t_n z_n\\) niet groter is dan \\(\\frac{1}{2}S\\). En de deelverzameling van Oplossing 3 wordt gegeven door de \\(i\\) waarvoor \\(t_i = 1\\).", "metadata": {"resource_path": "Dutch_TST/segmented/nl-2025-B2025_uitwerkingen.jsonl", "problem_match": "\nOpgave 2.", "solution_match": "\nOplossing III."}}
-{"year": "2025", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "Dutch_TST", "problem": "Een groep van 4050 vrienden speelt een videospel-toernooi. Daarvoor staan er 2025 computers in een zaal gelabeld \\(a_1, \\dots, a_{2025}\\) en 2025 computers in een andere zaal gelabeld \\(b_1, \\dots, b_{2025}\\). De speler op computer \\(a_i\\) speelt altijd tegen de spelers \\(b_i\\), \\(b_{i+2}\\), \\(b_{i+3}\\) en \\(b_{i+4}\\) (in het bijzonder dus juist niet tegen \\(b_{i+1}\\)), waarbij we de computers cyclisch doornummeren. Na de eerste ronde kiezen alle spelers een computer binnen hun zaal voor de tweede ronde. Daarna merken ze op dat iedereen in de tweede ronde dezelfde tegenstanders heeft als in de eerste ronde. \n\nBewijs dat als er iemand dezelfde computer heeft gekozen in beide rondes, dan iedereen dezelfde computer heeft gekozen in beide rondes.", "solution": "Voor de tegenstandercomputers van \\(a_i\\) bekijken we tegen welke \\(a_j\\) zij spelen in de volgende tabel. \n\n| \\(b_i\\) : | \\(a_{i-4}\\) | \\(a_{i-3}\\) | \\(a_{i-2}\\) | \\(a_i\\) |
\n\n| \\(b_{i+2}\\) : | \\(a_{i-2}\\) | \\(a_{i-1}\\) | \\(a_i\\) | \\(a_{i+2}\\) |
| \\(b_{i+3}\\) : | | \\(a_{i-1}\\) | \\(a_i\\) | \\(a_{i+1}\\) | \\(a_{i+3}\\) |
| \\(b_{i+4}\\) : | | | \\(a_i\\) | \\(a_{i+1}\\) | \\(a_{i+2}\\) | \\(a_{i+4}\\) |
\n\nMerk op dat de computers \\(a_{i-2}\\), \\(a_{i-1}\\), \\(a_{i+1}\\) en \\(a_{i+2}\\) elk twee gemeenschappelijk tegenstandercomputers hebben met \\(a_i\\). Verder hebben \\(a_{i-4}\\), \\(a_{i-3}\\), \\(a_{i+3}\\) en \\(a_{i+4}\\) elk één gemeenschappelijke tegenstandercomputer met \\(a_i\\). Alleen voor de eerste twee, \\(a_{i-4}\\) en \\(a_{i-3}\\), is de gemeenschappelijk tegenstander met \\(a_i\\) hetzelfde, namelijk \\(b_i\\). \n\n\n\n \n\nStel nu dat er een speler is blijven zitten, zeg de speler op \\(a_{2025}\\). We gaan nu met inductie bewijzen dat alle spelers op \\(a_i\\) en \\(b_i\\) zijn blijven zitten. \n\nStel per inductiehypothese dat de speler op een zekere computer \\(a_i\\) in beide rondes hetzelfde is. Aangezien iedereen dezelfde tegenstanders heeft in beide rondes, hebben de spelers die op \\(a_{i-4}\\) en \\(a_{i-3}\\) zaten wederom elk één dezelfde gemeenschappelijke speler met\n\n\n\n\\(a_i\\), namelijk de speler die op \\(b_i\\) zat. Dat betekent dat de speler op \\(b_i\\) ook is blijven zitten. Nu merken we op dat alleen de speler die op \\(a_{i-2}\\) zat naast \\(b_i\\) nog een tweede gemeenschappelijke tegenstander met \\(a_i\\) heeft. Dus ook op \\(a_{i-2}\\) is dezelfde speler blijven zitten. \n\nOmdat 2025 oneven is, volgt nu met inductie eenvoudig dat iedereen op dezelfde computer is blijven zitten, als er één iemand is blijven zitten. □", "metadata": {"resource_path": "Dutch_TST/segmented/nl-2025-B2025_uitwerkingen.jsonl", "problem_match": "\nOpgave 3.", "solution_match": "\nOplossing I."}}
-{"year": "2025", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "Dutch_TST", "problem": "Een groep van 4050 vrienden speelt een videospel-toernooi. Daarvoor staan er 2025 computers in een zaal gelabeld \\(a_1, \\dots, a_{2025}\\) en 2025 computers in een andere zaal gelabeld \\(b_1, \\dots, b_{2025}\\). De speler op computer \\(a_i\\) speelt altijd tegen de spelers \\(b_i\\), \\(b_{i+2}\\), \\(b_{i+3}\\) en \\(b_{i+4}\\) (in het bijzonder dus juist niet tegen \\(b_{i+1}\\)), waarbij we de computers cyclisch doornummeren. Na de eerste ronde kiezen alle spelers een computer binnen hun zaal voor de tweede ronde. Daarna merken ze op dat iedereen in de tweede ronde dezelfde tegenstanders heeft als in de eerste ronde. \n\nBewijs dat als er iemand dezelfde computer heeft gekozen in beide rondes, dan iedereen dezelfde computer heeft gekozen in beide rondes.", "solution": "We noemen twee computers *superconnected* als ze twee tegenstandercomputers gemeen hebben. We merken op dat de enige computers superconnected met \\(a_i\\) de computers \\(a_{i-2}\\), \\(a_{i-1}\\), \\(a_{i+1}\\) en \\(a_{i+2}\\) zijn. Nu gaan we kijken naar drietallen die superconnected zijn. \n\n- Er zijn twee computers superconnected met zowel \\(a_i\\) als \\(a_{i+1}\\), namelijk de computers \\(a_{i-1}\\) en \\(a_{i+2}\\). \n\n- Er is één computer superconnected met zowel \\(a_i\\) als \\(a_{i+2}\\), namelijk de computer \\(a_{i+1}\\). \n\n\n\n \n\nDus als een computer zelf superconnected is met \\(a_i\\) en twee gemeenschappelijke superconnected computers met \\(a_i\\) heeft, dan is deze computer \\(a_{i-1}\\) of \\(a_{i+1}\\). \n\nAangezien iedereen dezelfde tegenstanders heeft in beide rondes, geldt dat als twee vrienden in de eerste ronde op superconnected computers speelden, ze in de tweede ronde ook op superconnected computers spelen. Uit het bovenstaande volgt dan dat, als twee vrienden in de eerste ronde naast elkaar zaten, ze dat in de tweede ronde ook doen. \n\nStel nu dat de speler op \\(a_1\\) beide rondes hetzelfde is. We nemen verder uit het ongerijmde aan dat de speler die de eerste ronde op \\(a_2\\) zat, in de tweede ronde op \\(a_{2025}\\) zat. Dan moeten inductief de spelers op \\(a_3\\), \\(a_4\\) en \\(a_5\\) verhuisd zijn naar \\(a_{2024}\\), \\(a_{2023}\\) en \\(a_{2022}\\). Dus de speler die in de eerste ronde op \\(b_5\\) zat (en tegen \\(a_1\\), \\(a_2\\), \\(a_3\\), \\(a_5\\) speelde), moet in de tweede ronde tegen de spelers op computers \\(a_{2022}\\), \\(a_{2024}\\), \\(a_{2025}\\), \\(a_1\\). Maar er is geen computer met die tegenstanders. We concluderen dus dat de speler op \\(a_2\\) wel moet zijn blijven zitten. Dit laat inductief zien dat iedereen in die zaal op dezelfde computer is blijven zitten. \n\nAangezien de spelers in de tweede zaal dezelfde tegenstanders hebben in beide rondes, weten we nu dat ze ook tegen dezelfde computers spelen. Dit legt de computers in de tweede zaal vast, dus iedereen in de tweede zaal heeft ook voor dezelfde computer gekozen in beide rondes. □", "metadata": {"resource_path": "Dutch_TST/segmented/nl-2025-B2025_uitwerkingen.jsonl", "problem_match": "\nOpgave 3.", "solution_match": "\nOplossing II."}}
-{"year": "2025", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "Dutch_TST", "problem": "Gegeven is \\(\\triangle ABC\\) met omgeschreven cirkel \\(\\Gamma\\). Zij \\(M\\) het midden van de boog \\(BC\\) van \\(\\Gamma\\) waar \\(A\\) niet op ligt. Het punt \\(N\\) op \\(\\Gamma\\) is de antipode van \\(A\\). De lijn door \\(B\\) loodrecht op \\(AM\\) snijdt \\(AM\\) in het punt \\(D\\) en snijdt \\(\\Gamma\\) een tweede keer in het punt \\(P \\neq B\\). De lijn door \\(D\\) loodrecht op \\(AC\\) snijdt \\(AC\\) in het punt \\(E\\) en snijdt \\(BC\\) in het punt \\(F\\). \n\nBewijs dat \\(ND\\), \\(MF\\) en \\(PE\\) concurrent zijn. \n\n", "solution": "We noteren de helft van de hoek bij \\(A\\) als \\(\\alpha = \\frac{1}{2} \\angle BAC = \\angle BAM = \\angle MAC\\), want \\(M\\) is het midden van boog \\(BC\\). Dan merken we op dat \\(\\angle ABP = \\angle ABD = 90^\\circ - \\angle DAB = 90^\\circ - \\alpha\\) en dat \\(\\angle EDA = 90^\\circ - \\angle DAE = 90^\\circ - \\alpha\\). Ook rekenen we uit dat, wegens de gestrekte hoek \\(\\angle ADM\\), geldt dat \\(\\angle PDE = 180^\\circ - \\angle EDA - \\angle MDP = 180^\\circ - (90^\\circ - \\alpha) - 90^\\circ = \\alpha\\). \n\nNu definiëren we \\(K\\) als het tweede snijpunt van de omgeschreven cirkel van \\(\\triangle ADE\\) met \\(\\Gamma\\) (naast \\(A\\)). Dan merken we op dat \\(\\angle AKE = \\angle ADE = 90^\\circ - \\alpha = \\angle ABP = \\angle AKP\\), dus \\(K\\), \\(E\\) en \\(P\\) zijn collinear. Evenzo geldt dat \\(\\angle AKD = 180^\\circ - \\angle AED = 90^\\circ = \\angle AKN\\), wegens Thales omdat \\(A\\) en \\(N\\) antipoden zijn. Dus \\(K\\), \\(D\\) en \\(N\\) zijn collinear. \n\nVoor de laatste lijn claimen we dat \\(BFDK\\) ook een koordenvierhoek is. Inderdaad, \\(\\angle KBF = \\angle KBC = 180^\\circ - \\angle KAC = 180^\\circ - \\angle KAE = \\angle KDE = 180^\\circ - \\angle KDF\\).\n\n\n\n(Merk op dat dit in feite de stelling van Miquel is in \\(\\triangle CEF\\) en koordenvierhoeken \\(AKBC\\) en \\(DEAK\\).) Uit de omtrekhoekstelling in de koordenvierhoek \\(BFDK\\) volgt dat \\(\\angle BKF = \\angle BDF = \\angle PDE = \\alpha = \\angle BAM = \\angle BKM\\), dus \\(K\\), \\(F\\) en \\(M\\) zijn collinear. We concluderen dat \\(ND\\), \\(MF\\) en \\(PE\\) concurrent zijn in het punt \\(K\\). \\(\\square\\)", "metadata": {"resource_path": "Dutch_TST/segmented/nl-2025-B2025_uitwerkingen.jsonl", "problem_match": "\nOpgave 4.", "solution_match": "\nOplossing I."}}
-{"year": "2025", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "Dutch_TST", "problem": "Gegeven is \\(\\triangle ABC\\) met omgeschreven cirkel \\(\\Gamma\\). Zij \\(M\\) het midden van de boog \\(BC\\) van \\(\\Gamma\\) waar \\(A\\) niet op ligt. Het punt \\(N\\) op \\(\\Gamma\\) is de antipode van \\(A\\). De lijn door \\(B\\) loodrecht op \\(AM\\) snijdt \\(AM\\) in het punt \\(D\\) en snijdt \\(\\Gamma\\) een tweede keer in het punt \\(P \\neq B\\). De lijn door \\(D\\) loodrecht op \\(AC\\) snijdt \\(AC\\) in het punt \\(E\\) en snijdt \\(BC\\) in het punt \\(F\\). \n\nBewijs dat \\(ND\\), \\(MF\\) en \\(PE\\) concurrent zijn. \n\n", "solution": "Net als in Oplossing I schrijven we \\(\\alpha = \\frac{1}{2}\\angle BAC\\) en merken we op dat \\(\\angle ANP = \\angle ABP = 90^\\circ - \\alpha = \\angle ADE\\). Omdat ook \\(\\angle APN = 90^\\circ = \\angle AED\\) zien we dat \\(\\triangle ANP \\sim \\triangle ADE\\). Bovendien zijn ze gelijk georiënteerd. Uit de theorie van draaiver- menigvuldigingen volgt nu dat \\(\\triangle AND \\sim \\triangle APE\\) en dat \\(ND\\) en \\(PE\\) elkaar snijden in het tweede snijpunt van de omgeschreven cirkels van \\(\\triangle ANP\\) en \\(\\triangle ADE\\), oftewel in \\(K\\). Nu gaan we verder zoals in Oplossing I om te laten zien dat \\(MF\\) door \\(K\\) gaat. \\(\\square\\)", "metadata": {"resource_path": "Dutch_TST/segmented/nl-2025-B2025_uitwerkingen.jsonl", "problem_match": "\nOpgave 4.", "solution_match": "\nOplossing II."}}
+{"year": "2025", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "Dutch_TST", "problem": "Een groep van 4050 vrienden speelt een videospel-toernooi. Daarvoor staan er 2025 computers in een zaal gelabeld \\(a_1, \\dots, a_{2025}\\) en 2025 computers in een andere zaal gelabeld \\(b_1, \\dots, b_{2025}\\). De speler op computer \\(a_i\\) speelt altijd tegen de spelers \\(b_i\\), \\(b_{i+2}\\), \\(b_{i+3}\\) en \\(b_{i+4}\\) (in het bijzonder dus juist niet tegen \\(b_{i+1}\\)), waarbij we de computers cyclisch doornummeren. Na de eerste ronde kiezen alle spelers een computer binnen hun zaal voor de tweede ronde. Daarna merken ze op dat iedereen in de tweede ronde dezelfde tegenstanders heeft als in de eerste ronde. \n\nBewijs dat als er iemand dezelfde computer heeft gekozen in beide rondes, dan iedereen dezelfde computer heeft gekozen in beide rondes.", "solution": "Voor de tegenstandercomputers van \\(a_i\\) bekijken we tegen welke \\(a_j\\) zij spelen in de volgende tabel. \n\n| \\(b_i\\) : | \\(a_{i-4}\\) | \\(a_{i-3}\\) | \\(a_{i-2}\\) | \\(a_i\\) |
\n\n| \\(b_{i+2}\\) : | \\(a_{i-2}\\) | \\(a_{i-1}\\) | \\(a_i\\) | \\(a_{i+2}\\) |
| \\(b_{i+3}\\) : | | \\(a_{i-1}\\) | \\(a_i\\) | \\(a_{i+1}\\) | \\(a_{i+3}\\) |
| \\(b_{i+4}\\) : | | | \\(a_i\\) | \\(a_{i+1}\\) | \\(a_{i+2}\\) | \\(a_{i+4}\\) |
\n\nMerk op dat de computers \\(a_{i-2}\\), \\(a_{i-1}\\), \\(a_{i+1}\\) en \\(a_{i+2}\\) elk twee gemeenschappelijk tegenstandercomputers hebben met \\(a_i\\). Verder hebben \\(a_{i-4}\\), \\(a_{i-3}\\), \\(a_{i+3}\\) en \\(a_{i+4}\\) elk één gemeenschappelijke tegenstandercomputer met \\(a_i\\). Alleen voor de eerste twee, \\(a_{i-4}\\) en \\(a_{i-3}\\), is de gemeenschappelijk tegenstander met \\(a_i\\) hetzelfde, namelijk \\(b_i\\). \n\n\n\n \n\nStel nu dat er een speler is blijven zitten, zeg de speler op \\(a_{2025}\\). We gaan nu met inductie bewijzen dat alle spelers op \\(a_i\\) en \\(b_i\\) zijn blijven zitten. \n\nStel per inductiehypothese dat de speler op een zekere computer \\(a_i\\) in beide rondes hetzelfde is. Aangezien iedereen dezelfde tegenstanders heeft in beide rondes, hebben de spelers die op \\(a_{i-4}\\) en \\(a_{i-3}\\) zaten wederom elk één dezelfde gemeenschappelijke speler met\n\n\n\n\\(a_i\\), namelijk de speler die op \\(b_i\\) zat. Dat betekent dat de speler op \\(b_i\\) ook is blijven zitten. Nu merken we op dat alleen de speler die op \\(a_{i-2}\\) zat naast \\(b_i\\) nog een tweede gemeenschappelijke tegenstander met \\(a_i\\) heeft. Dus ook op \\(a_{i-2}\\) is dezelfde speler blijven zitten. \n\nOmdat 2025 oneven is, volgt nu met inductie eenvoudig dat iedereen op dezelfde computer is blijven zitten, als er één iemand is blijven zitten. □", "metadata": {"resource_path": "Dutch_TST/segmented/nl-2025-B2025_uitwerkingen.jsonl", "problem_match": "\nOpgave 3.", "solution_match": "\nOplossing I."}}
+{"year": "2025", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "Dutch_TST", "problem": "Een groep van 4050 vrienden speelt een videospel-toernooi. Daarvoor staan er 2025 computers in een zaal gelabeld \\(a_1, \\dots, a_{2025}\\) en 2025 computers in een andere zaal gelabeld \\(b_1, \\dots, b_{2025}\\). De speler op computer \\(a_i\\) speelt altijd tegen de spelers \\(b_i\\), \\(b_{i+2}\\), \\(b_{i+3}\\) en \\(b_{i+4}\\) (in het bijzonder dus juist niet tegen \\(b_{i+1}\\)), waarbij we de computers cyclisch doornummeren. Na de eerste ronde kiezen alle spelers een computer binnen hun zaal voor de tweede ronde. Daarna merken ze op dat iedereen in de tweede ronde dezelfde tegenstanders heeft als in de eerste ronde. \n\nBewijs dat als er iemand dezelfde computer heeft gekozen in beide rondes, dan iedereen dezelfde computer heeft gekozen in beide rondes.", "solution": "We noemen twee computers *superconnected* als ze twee tegenstandercomputers gemeen hebben. We merken op dat de enige computers superconnected met \\(a_i\\) de computers \\(a_{i-2}\\), \\(a_{i-1}\\), \\(a_{i+1}\\) en \\(a_{i+2}\\) zijn. Nu gaan we kijken naar drietallen die superconnected zijn. \n\n- Er zijn twee computers superconnected met zowel \\(a_i\\) als \\(a_{i+1}\\), namelijk de computers \\(a_{i-1}\\) en \\(a_{i+2}\\). \n\n- Er is één computer superconnected met zowel \\(a_i\\) als \\(a_{i+2}\\), namelijk de computer \\(a_{i+1}\\). \n\n\n\n \n\nDus als een computer zelf superconnected is met \\(a_i\\) en twee gemeenschappelijke superconnected computers met \\(a_i\\) heeft, dan is deze computer \\(a_{i-1}\\) of \\(a_{i+1}\\). \n\nAangezien iedereen dezelfde tegenstanders heeft in beide rondes, geldt dat als twee vrienden in de eerste ronde op superconnected computers speelden, ze in de tweede ronde ook op superconnected computers spelen. Uit het bovenstaande volgt dan dat, als twee vrienden in de eerste ronde naast elkaar zaten, ze dat in de tweede ronde ook doen. \n\nStel nu dat de speler op \\(a_1\\) beide rondes hetzelfde is. We nemen verder uit het ongerijmde aan dat de speler die de eerste ronde op \\(a_2\\) zat, in de tweede ronde op \\(a_{2025}\\) zat. Dan moeten inductief de spelers op \\(a_3\\), \\(a_4\\) en \\(a_5\\) verhuisd zijn naar \\(a_{2024}\\), \\(a_{2023}\\) en \\(a_{2022}\\). Dus de speler die in de eerste ronde op \\(b_5\\) zat (en tegen \\(a_1\\), \\(a_2\\), \\(a_3\\), \\(a_5\\) speelde), moet in de tweede ronde tegen de spelers op computers \\(a_{2022}\\), \\(a_{2024}\\), \\(a_{2025}\\), \\(a_1\\). Maar er is geen computer met die tegenstanders. We concluderen dus dat de speler op \\(a_2\\) wel moet zijn blijven zitten. Dit laat inductief zien dat iedereen in die zaal op dezelfde computer is blijven zitten. \n\nAangezien de spelers in de tweede zaal dezelfde tegenstanders hebben in beide rondes, weten we nu dat ze ook tegen dezelfde computers spelen. Dit legt de computers in de tweede zaal vast, dus iedereen in de tweede zaal heeft ook voor dezelfde computer gekozen in beide rondes. □", "metadata": {"resource_path": "Dutch_TST/segmented/nl-2025-B2025_uitwerkingen.jsonl", "problem_match": "\nOpgave 3.", "solution_match": "\nOplossing II."}}
+{"year": "2025", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "Dutch_TST", "problem": "Gegeven is \\(\\triangle ABC\\) met omgeschreven cirkel \\(\\Gamma\\). Zij \\(M\\) het midden van de boog \\(BC\\) van \\(\\Gamma\\) waar \\(A\\) niet op ligt. Het punt \\(N\\) op \\(\\Gamma\\) is de antipode van \\(A\\). De lijn door \\(B\\) loodrecht op \\(AM\\) snijdt \\(AM\\) in het punt \\(D\\) en snijdt \\(\\Gamma\\) een tweede keer in het punt \\(P \\neq B\\). De lijn door \\(D\\) loodrecht op \\(AC\\) snijdt \\(AC\\) in het punt \\(E\\) en snijdt \\(BC\\) in het punt \\(F\\). \n\nBewijs dat \\(ND\\), \\(MF\\) en \\(PE\\) concurrent zijn. \n\n", "solution": "We noteren de helft van de hoek bij \\(A\\) als \\(\\alpha = \\frac{1}{2} \\angle BAC = \\angle BAM = \\angle MAC\\), want \\(M\\) is het midden van boog \\(BC\\). Dan merken we op dat \\(\\angle ABP = \\angle ABD = 90^\\circ - \\angle DAB = 90^\\circ - \\alpha\\) en dat \\(\\angle EDA = 90^\\circ - \\angle DAE = 90^\\circ - \\alpha\\). Ook rekenen we uit dat, wegens de gestrekte hoek \\(\\angle ADM\\), geldt dat \\(\\angle PDE = 180^\\circ - \\angle EDA - \\angle MDP = 180^\\circ - (90^\\circ - \\alpha) - 90^\\circ = \\alpha\\). \n\nNu definiëren we \\(K\\) als het tweede snijpunt van de omgeschreven cirkel van \\(\\triangle ADE\\) met \\(\\Gamma\\) (naast \\(A\\)). Dan merken we op dat \\(\\angle AKE = \\angle ADE = 90^\\circ - \\alpha = \\angle ABP = \\angle AKP\\), dus \\(K\\), \\(E\\) en \\(P\\) zijn collinear. Evenzo geldt dat \\(\\angle AKD = 180^\\circ - \\angle AED = 90^\\circ = \\angle AKN\\), wegens Thales omdat \\(A\\) en \\(N\\) antipoden zijn. Dus \\(K\\), \\(D\\) en \\(N\\) zijn collinear. \n\nVoor de laatste lijn claimen we dat \\(BFDK\\) ook een koordenvierhoek is. Inderdaad, \\(\\angle KBF = \\angle KBC = 180^\\circ - \\angle KAC = 180^\\circ - \\angle KAE = \\angle KDE = 180^\\circ - \\angle KDF\\).\n\n\n\n(Merk op dat dit in feite de stelling van Miquel is in \\(\\triangle CEF\\) en koordenvierhoeken \\(AKBC\\) en \\(DEAK\\).) Uit de omtrekhoekstelling in de koordenvierhoek \\(BFDK\\) volgt dat \\(\\angle BKF = \\angle BDF = \\angle PDE = \\alpha = \\angle BAM = \\angle BKM\\), dus \\(K\\), \\(F\\) en \\(M\\) zijn collinear. We concluderen dat \\(ND\\), \\(MF\\) en \\(PE\\) concurrent zijn in het punt \\(K\\). \\(\\square\\)", "metadata": {"resource_path": "Dutch_TST/segmented/nl-2025-B2025_uitwerkingen.jsonl", "problem_match": "\nOpgave 4.", "solution_match": "\nOplossing I."}}
+{"year": "2025", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "Dutch_TST", "problem": "Gegeven is \\(\\triangle ABC\\) met omgeschreven cirkel \\(\\Gamma\\). Zij \\(M\\) het midden van de boog \\(BC\\) van \\(\\Gamma\\) waar \\(A\\) niet op ligt. Het punt \\(N\\) op \\(\\Gamma\\) is de antipode van \\(A\\). De lijn door \\(B\\) loodrecht op \\(AM\\) snijdt \\(AM\\) in het punt \\(D\\) en snijdt \\(\\Gamma\\) een tweede keer in het punt \\(P \\neq B\\). De lijn door \\(D\\) loodrecht op \\(AC\\) snijdt \\(AC\\) in het punt \\(E\\) en snijdt \\(BC\\) in het punt \\(F\\). \n\nBewijs dat \\(ND\\), \\(MF\\) en \\(PE\\) concurrent zijn. \n\n", "solution": "Net als in Oplossing I schrijven we \\(\\alpha = \\frac{1}{2}\\angle BAC\\) en merken we op dat \\(\\angle ANP = \\angle ABP = 90^\\circ - \\alpha = \\angle ADE\\). Omdat ook \\(\\angle APN = 90^\\circ = \\angle AED\\) zien we dat \\(\\triangle ANP \\sim \\triangle ADE\\). Bovendien zijn ze gelijk georiënteerd. Uit de theorie van draaiver- menigvuldigingen volgt nu dat \\(\\triangle AND \\sim \\triangle APE\\) en dat \\(ND\\) en \\(PE\\) elkaar snijden in het tweede snijpunt van de omgeschreven cirkels van \\(\\triangle ANP\\) en \\(\\triangle ADE\\), oftewel in \\(K\\). Nu gaan we verder zoals in Oplossing I om te laten zien dat \\(MF\\) door \\(K\\) gaat. \\(\\square\\)", "metadata": {"resource_path": "Dutch_TST/segmented/nl-2025-B2025_uitwerkingen.jsonl", "problem_match": "\nOpgave 4.", "solution_match": "\nOplossing II."}}
{"year": "2025", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "Dutch_TST", "problem": "Bepaal alle natuurlijke getallen \\(n\\) waarvoor geldt dat alle priemfactoren van \\(2^n - 1\\) hoogstens 7 zijn.", "solution": "We merken op dat \\(2 \\nmid 2^n - 1\\) voor alle \\(n > 1\\), dus we zijn op zoek naar alle \\(n\\) zodanig dat 3, 5 en 7 de enige delers zijn, oftewel \\(2^n - 1 = 3^a 5^b 7^c\\). De antwoorden zijn \\(n = 1, 2, 3, 4, 6\\), die we gemakkelijk controleren met uitkomsten 1, 3, 7, 3 · 5, 3² · 7. \n\nWe rekenen uit dat \\(2^2 \\equiv 1 \\mod 3\\), \\(2^4 \\equiv 1 \\mod 5\\) en \\(2^3 \\equiv 1 \\mod 7\\), én dat dit de kleinste positieve exponenten zijn waarvoor \\(2^n\\) congruent is aan 1. Dat betekent dat \\(3 \\mid 2^n - 1\\) dan en slechts dan als \\(2 \\mid n\\), dat \\(5 \\mid 2^n - 1\\) dan en slechts dan als \\(4 \\mid n\\) en dat \\(7 \\mid 2^n - 1\\) dan en slechts dan als \\(3 \\mid n\\). \n\nAls \\(n\\) oneven is dan volgt uit het bovenstaande dat \\(2^n - 1\\) geen factoren 3 of 5 heeft. We onderscheiden twee gevallen. Als \\(3 \\nmid n\\) en \\(n > 1\\), geldt dat ook 7 geen deler is, terwijl \\(2^n - 1 > 1\\). Dus \\(2^n - 1\\) moet wel een grotere priemdeler dan 7 hebben, en deze \\(n\\) voldoen niet. De andere optie voor oneven \\(n\\) is dat \\(3 \\mid n\\). Om aan de voorwaarde te voldoen moet dan gelden dat \\(2^n - 1 = 7^c\\). Als \\(n \\ge 4\\) vinden we dus de relatie \\(7^c \\equiv -1 \\mod 16\\). Dat is echter in tegenspraak met het feit dat \\(7^c \\equiv 7, 1 \\mod 16\\), want \\(7^2 = 49 = 3 \\cdot 16 + 1\\). Dan is de enige mogelijkheid dus \\(n = 3\\). \n\nWe geven nu voor even \\(n \\ge 8\\) een bewijs dat \\(2^n - 1\\) een priemfactor heeft groter dan 7 met volledige inductie naar \\(n\\). Als inductiebasis rekenen we uit dat \n\n\\[ \n\\begin{align*} \n2^8 - 1 &= 3 \\cdot 5 \\cdot 17, \\\\\n2^{10} - 1 &= 3 \\cdot 11 \\cdot 31, \\\\\n2^{12} - 1 &= 3^2 \\cdot 5 \\cdot 7 \\cdot 13. \n\\end{align*} \n\\]\n\nStel nu dat \\(n = 2m\\) met \\(m \\ge 7\\), en stel dat voor alle even \\(k\\) met \\(8 \\le k < n\\) geldt dat \\(2^k - 1\\) een priemfactor groter dan 7 heeft. Beschouw de ontbinding \\(2^n - 1 = 2^{2m} - 1 = (2^m - 1)(2^m + 1)\\). Als \\(m\\) oneven is, dan heeft de eerst factor een priemfactor groter dan 7 aangezien \\(m \\ge 7 > 3\\). Als \\(m\\) even is, dan geldt hetzelfde op basis van de inductiehypothese aangezien nu geldt dat \\(8 \\le m < 2m = n\\). Dit voltooit de inductie. \\(\\square\\)", "metadata": {"resource_path": "Dutch_TST/segmented/nl-2025-B2025_uitwerkingen.jsonl", "problem_match": "\nOpgave 5.", "solution_match": "\nOplossing I."}}
{"year": "2025", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "Dutch_TST", "problem": "Bepaal alle natuurlijke getallen \\(n\\) waarvoor geldt dat alle priemfactoren van \\(2^n - 1\\) hoogstens 7 zijn.", "solution": "Het bewijs dat \\(2^n - 1\\) een priemfactor groter dan 7 heeft voor oneven \\(n > 3\\) gaat hetzelfde als hierboven. We geven een alternatief bewijs dat de enige even \\(n\\) die voldoen 2, 4 en 6 zijn door drie gevallen te onderscheiden. \n\nStel dat \\(4 \\mid n\\). Dan schrijven we \\(n = 4k\\) en ontbinden we \\(2^n - 1\\) als \\(2^n - 1 = (2^{2k})^2 - 1 = (4^k - 1)(4^k + 1)\\). Merk op dat \\(3 \\nmid 4^k + 1\\) want \\(4^k + 1 \\equiv 2 \\mod 3\\) voor alle \\(k\\), en merk op dat \\(7 \\nmid 4^k + 1\\) want de restklasses van \\(4^k\\) modulo 7 zijn 1, 4 en 2. Aangezien \\(2^n - 1\\)\n\n\n\ngeen priemfactor mag hebben groter dan 7, moet nu gelden dat \\(4^k + 1 = 5^d\\), met \\(d > 0\\) want \\(k > 0\\). Als \\(k \\ge 2\\), dan krijgen we dus dat 16 een deler is van \\(5^d - 1\\). Hieruit volgt dan weer dat \\(b\\) een viervoud is, dus we schrijven \\(b = 4d'\\), met \\(d' > 0\\). In dat geval vinden we dus dat \\(2^{2k} = 4^k = 5^{4d'} - 1 = (5^{2d'} - 1)(5^{2d'} + 1) = (25^{d'} - 1)(25^{d'} + 1)\\). Aangezien \\(\\text{ggd}(25^{d'} - 1, 25^{d'} + 1) = 2\\) betekent dit dan minstens één van de termen \\((25^{d'} - 1\\) of \\(25^{d'} + 1)\\) kleiner of gelijk is aan 2. Dat is een tegenspraak met \\(d' > 0\\). Dus de enige mogelijkheid in dit geval is \\(k = 1\\), oftewel \\(n = 4\\). \n\nStel dat 6 | \\(n\\), maar 4 \\(\\nmid\\) \\(n\\). Dan schrijven we \\(n = 6k\\) met \\(k\\) oneven, en ontbinden we \\(2^n - 1 = (2^{3k})^2 - 1 = (8^k - 1)(8^k + 1)\\). Merk op dat \\(5 \\nmid 8^k + 1\\) want \\(8^k \\equiv 3, -3 \\mod 5\\) voor oneven \\(k\\), en merk op dat \\(7 \\nmid 8^k + 1\\) want \\(8^k + 1 \\equiv 2 \\mod 7\\). Nu moet dus gelden dat \\(8^k + 1 = 3^a\\). Aangezien nu \\(8 \\mid 3^a - 1\\), is \\(a\\) even wat we schrijven als \\(a = 2a'\\), met \\(a' > 0\\). Dan krijgen we \\(8^k = (3^{a'} - 1)(3^{a'} + 1)\\). Omdat \\(\\text{ggd}(3^{a'} - 1, 3^{a'} + 1) = 2\\) volgt hieruit dat één van de termen \\((3^{a'} - 1\\) of \\(3^{a'} + 1)\\) kleiner of gelijk is aan 2. Dat kan alleen als \\(a' = 1\\), oftewel \\(a = 2\\), \\(k = 1\\) en \\(n = 6\\). \n\nHet laatste geval is dat 2 | \\(n\\), maar 4 \\(\\nmid\\) \\(n\\) en 3 \\(\\nmid\\) \\(n\\). Dan schrijven we \\(n = 2k\\) en vinden we meteen dat \\(3^a = 2^n - 1 = (2^k - 1)(2^k + 1)\\). Omdat \\(\\text{ggd}(2^k - 1, 2^k + 1) = 1 < 3\\), is de enige mogelijkheid dat \\(2^k - 1 = 1\\). Oftewel \\(k = 1\\) en \\(n = 2\\). \\(\\square\\)", "metadata": {"resource_path": "Dutch_TST/segmented/nl-2025-B2025_uitwerkingen.jsonl", "problem_match": "\nOpgave 5.", "solution_match": "\nOplossing II."}}
diff --git a/Dutch_TST/segmented/nl-2025-C2025_uitwerkingen.jsonl b/Dutch_TST/segmented/nl-2025-C2025_uitwerkingen.jsonl
index 794484759c3eff1fce61cc09564833afc0499eea..d6b50873ff8cd76586e26905ad96b4e77bc61351 100644
--- a/Dutch_TST/segmented/nl-2025-C2025_uitwerkingen.jsonl
+++ b/Dutch_TST/segmented/nl-2025-C2025_uitwerkingen.jsonl
@@ -1,9 +1,9 @@
{"year": "2025", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "Dutch_TST", "problem": "Beschouw de rij \\(y_0, y_1, \\ldots\\) met \\(y_0 = -\\frac{1}{4}\\) en \\(y_1 = 0\\) en die verder voldoet aan \n\n\\[y_{n+1} + y_{n-1} = 4y_n + 1\\]\n\nvoor alle \\(n \\ge 1\\). Bewijs dat voor alle \\(n \\ge 0\\) de uitdrukking \\(2y_{2n} + \\frac{3}{2}\\) \n\na) een positief geheel getal is en \n\nb) het kwadraat van een geheel getal is.", "solution": "We maken de substitutie \\(x_n = 4y_n + 2\\). Dan wordt de vergelijking homogeen: \n\n\\[x_{n+1} + x_{n-1} = 4y_{n+1} + 2 + 4y_{n-1} + 2 = 4(4y_n + 1) + 4 = 16y_n + 8 = 4x_n,\\]\n\nmet beginvoorwaarden \\(x_0 = 4(-\\frac{1}{4}) + 2 = 1\\) en \\(x_1 = 4 \\cdot 0 + 2 = 2\\). Alle getallen in de rij \\((x_i)\\) zijn dus geheel en we zien dat \\(x_{n+1}\\) en \\(x_{n-1}\\) altijd dezelfde pariteit hebben. In het bijzonder is \\(x_{2n}\\) altijd oneven. Dus \\(2y_{2n} + \\frac{3}{2} = \\frac{4y_{2n}+3}{2} = \\frac{x_{2n}+1}{2}\\) is altijd geheel. Om te laten zien dat ze ook positief zijn, bewijzen we met inductie dat \\(x_n\\) een stijgende rij positieve getallen is. Dit geldt inderdaad voor \\(x_1 > x_0 > 0\\). Stel nu als inductiehypothese dat \\(x_n > x_{n-1} > 0\\). Dan vinden we ook dat \\(x_{n+1} - x_n = 3x_n - x_{n-1} > x_n - x_{n-1} > 0\\). Hiermee concluderen we het bewijs van onderdeel (a). \n\nVoor onderdeel (b) merken we op dat de karakteristieke vergelijking voor het homogene deel \\(y_{n+1} + y_{n-1} = 4y_n\\) wordt gegeven door \\(x^2 + 1 = 4x\\). Hiervan zijn de oplossingen \\(x = 2 - \\sqrt{3}\\) en \\(x = 2 + \\sqrt{3}\\). Nu kiezen we een oplossing voor de inhomogene vergelijking, zeg \\(y_n = -\\frac{1}{2}\\). Dan is de algemene oplossing dus \n\n\\[y_n = A(2 - \\sqrt{3})^n + B(2 + \\sqrt{3})^n - \\frac{1}{2}.\\]\n\nAls we dit oplossen met behulp van \\(n = 0\\) en \\(n = 1\\), dan vinden we \\(-\\frac{1}{4} = A + B - \\frac{1}{2}\\) en \\(0 = A(2 - \\sqrt{3}) + B(2 + \\sqrt{3}) - \\frac{1}{2} = 2(A + B) + \\sqrt{3}(B - A) - \\frac{1}{2}\\). Dat betekent dat \\(A + B = \\frac{1}{4}\\) en \\(B - A = 0\\), oftewel \\(A = B = \\frac{1}{8}\\). Dus \n\n\\[y_n = \\frac{1}{8}(2 - \\sqrt{3})^n + \\frac{1}{8}(2 + \\sqrt{3})^n - \\frac{1}{2}.\\]\n\n\n\nAangezien \\((2 - \\sqrt{3})(2 + \\sqrt{3}) = 4 - 3 = 1\\) controleren we eenvoudig dat \n\n\\[ \n\\begin{align*} \n(4y_n + 2)^2 &= \\left(\\frac{1}{2}(2 - \\sqrt{3})^n + \\frac{1}{2}(2 + \\sqrt{3})^n\\right)^2 \\\\ \n&= \\frac{1}{4}(2 - \\sqrt{3})^{2n} + \\frac{1}{4}(2 + \\sqrt{3})^{2n} + \\frac{1}{2}(2 - \\sqrt{3})^n(2 + \\sqrt{3})^n \\\\ \n&= \\frac{1}{4}(2 - \\sqrt{3})^{2n} + \\frac{1}{4}(2 + \\sqrt{3})^{2n} + \\frac{1}{2} \\\\ \n&= 2y_{2n} + \\frac{3}{2}. \n\\end{align*} \n\\]\n\nDit bewijst onderdeel (b), omdat \\(4y_n + 2 = x_n\\) geheel is. \\(\\square\\)", "metadata": {"resource_path": "Dutch_TST/segmented/nl-2025-C2025_uitwerkingen.jsonl", "problem_match": "\nOpgave 1.", "solution_match": "\nOplossing I."}}
{"year": "2025", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "Dutch_TST", "problem": "Beschouw de rij \\(y_0, y_1, \\ldots\\) met \\(y_0 = -\\frac{1}{4}\\) en \\(y_1 = 0\\) en die verder voldoet aan \n\n\\[y_{n+1} + y_{n-1} = 4y_n + 1\\]\n\nvoor alle \\(n \\ge 1\\). Bewijs dat voor alle \\(n \\ge 0\\) de uitdrukking \\(2y_{2n} + \\frac{3}{2}\\) \n\na) een positief geheel getal is en \n\nb) het kwadraat van een geheel getal is.", "solution": "Voor een alternatieve aanpak van onderdeel (b) rekenen we uit dat \n\n\\[ x_{n+1}^2 - x_{n-1}^2 = (x_{n+1} + x_{n-1})(x_{n+1} - x_{n-1}) = 4x_n(x_{n+1} - x_{n-1}). \\]\n\nDit betekent dat \\(x_{n+1}^2 - 4x_{n+1}x_n + x_n^2 = x_n^2 - 4x_nx_{n-1} + x_{n-1}^2\\). Met inductie naar \\(n\\) betekent dit dat \\(x_{n+1}^2 - 4x_{n+1}x_n + x_n^2 = -3\\), want voor \\(n = 0\\) rekenen we uit dat \\(2^2 - 4 \\cdot 2 \\cdot 1 + 1^2 = -3\\). \n\nNu bewijzen we met inductie dat \\(x_{2n} = 2x_n^2 - 1\\) voor \\(n \\ge 0\\). Voor \\(n = 0\\) is dit waar want \\(1 = 2 \\cdot 1^2 - 1\\). Ook voor \\(n = 1\\) rekenen we uit dat \\(x_2 = 4x_1 - x_0 = 7 = 2x_1^2 - 1\\). Omdat we inductie gaan doen op oneven getallen merken we alvast op dat \\(16x_n = 4x_{n-1} + 4x_{n+1} = x_{n-2} + 2x_n + x_{n+2}\\), dus \\(14x_n = x_{n-2} + x_{n+2}\\). Stel nu dat de bewering waar is voor \\(n = k\\) en \\(n = k-1\\). Dan rekenen we met behulp van het voorgaande uit dat \n\n\\[ \n\\begin{align*} \nx_{2k+2} &= 14x_{2k} - x_{2k-2} \\\\ \n&\\overset{\\text{(IH)}}{=} 14(2x_k^2 - 1) - (2x_{k-1}^2 - 1) \\\\ \n&= 2(14x_k^2 - x_{k-1}^2 - 6) - 1 \\\\ \n&= 2(14x_k^2 - (4x_k - x_{k+1})^2 - 6) - 1 \\\\ \n&= 2(-2x_k^2 + 8x_kx_{k+1} - x_{k+1}^2 - 6) - 1 \\\\ \n&= 2x_{k+1}^2 - 1. \n\\end{align*} \n\\]\n\nNu hebben we dus met inductie laten zien dat \\(x_{2n} = 2x_n^2 - 1\\), oftewel \n\n\\[ 2y_{2n} + \\frac{3}{2} = \\frac{x_{2n} + 1}{2} = x_n^2. \\]", "metadata": {"resource_path": "Dutch_TST/segmented/nl-2025-C2025_uitwerkingen.jsonl", "problem_match": "\nOpgave 1.", "solution_match": "\nOplossing II."}}
{"year": "2025", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "Dutch_TST", "problem": "Beschouw de rij \\(y_0, y_1, \\ldots\\) met \\(y_0 = -\\frac{1}{4}\\) en \\(y_1 = 0\\) en die verder voldoet aan \n\n\\[y_{n+1} + y_{n-1} = 4y_n + 1\\]\n\nvoor alle \\(n \\ge 1\\). Bewijs dat voor alle \\(n \\ge 0\\) de uitdrukking \\(2y_{2n} + \\frac{3}{2}\\) \n\na) een positief geheel getal is en \n\nb) het kwadraat van een geheel getal is.", "solution": "Als alternatief voor onderdeel (b) claimen we met inductie naar \\(m\\) dat \n\n\\[ x_{n-m} + x_{n+m} = 2x_m x_n, \\]\n\n\n\nvoor alle getallen \\(m\\) en \\(n\\) met \\(0 \\le m \\le n\\). Voor \\(m = 0\\) en \\(m = 1\\), voldoen de vergelijkingen \\(x_n + x_n = 2x_n\\) en \\(x_{n-1} + x_{n+1} = 4x_n\\) inderdaad. Stel nu dat de bewering waar is voor \\(m = k\\) en \\(m = k-1\\). Dan vinden we voor \\(m = k+1\\) dat \n\n\\[ \n\\begin{align*} \nx_{n-k-1} + x_{n+k+1} &= (4x_{n-k} - x_{n-k+1}) + (4x_{n+k} - x_{n+k-1}) \\\\ \n&= 4(x_{n-k} + x_{n+k}) - (x_{n-k+1} + x_{n+k-1}) \\\\ \n&\\overset{\\text{(IH)}}{=} 8x_k x_n - 2x_{k-1} x_n \\\\ \n&= 2(4x_k - x_{k-1}) x_n \\\\ \n&= 2x_{k+1} x_n. \n\\end{align*} \n\\]\n\nHiermee is de inductie voltooid. Als we nu \\(m = n\\) invullen, dan vinden we \\(1 + x_{2n} = 2x_n^2\\).\nDus we concluderen dat \\(2y_{2n} + \\frac{3}{2} = \\frac{x_{2n+1}}{2} = x_n^2\\). \\(\\square\\) \n\nOpmerking. Voor de oneven indices geldt \\(x_{2n+1} = 3\\left(\\frac{x_{n+1}+x_n}{3}\\right)^2 - 1\\).", "metadata": {"resource_path": "Dutch_TST/segmented/nl-2025-C2025_uitwerkingen.jsonl", "problem_match": "\nOpgave 1.", "solution_match": "\nOplossing III."}}
-{"year": "2025", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "Dutch_TST", "problem": "Beschouw een rechthoekig bord van \\(m \\times n\\) vakjes met \\(m, n \\ge 1\\). De hoekpunten van de vakjes vormen een \\((m+1) \\times (n+1)\\)-grid. We noemen een driehoek met hoekpunten punten van het grid *laag* als die minstens één zijde heeft die parallel is met een zijde van het bord zo dat de hoogte van de driehoek op die zijde 1 is. We noemen een lage driehoek *bijzonder* als die twee zijden heeft die parallel zijn met een zijde van het bord. We partitioneren het bord in lage driehoeken. \n\nBepaal het minimale aantal bijzondere driehoeken over alle mogelijke partitioneringen van het \\(m \\times n\\)-bord.", "solution": "Als \\(m, n \\ge 2\\) en minstens één van de twee is even, dan is het antwoord 0. Anders (minstens één van de twee is 1, of ze zijn beide oneven) is het antwoord 2. \n\nWe tekenen eerst een voorbeeld voor \\(n = 1\\) en \\(m \\ge 1\\) met twee bijzondere driehoeken, een voorbeeld voor \\(n = 2\\) en \\(m \\ge 3\\) met nul bijzondere driehoeken, en het speciale geval \\(n = 2\\), \\(m = 2\\) met wederom nul bijzondere driehoeken. \n\n\n\n \n\nAls \\(m = 1\\) of \\(n = 1\\), dan is er dus een partitionering met twee bijzondere driehoeken. Stel nu dat \\(m, n \\ge 2\\). \n\nAls minstens één van \\(m\\) en \\(n\\) even is, kunnen we de rechthoek opknippen in stroken van breedte 2, dus dan kunnen we een partitionering maken met nul bijzondere driehoeken. Als ze beide oneven zijn, dan knippen we de rechthoek op in stroken van breedte 2 plus één strook van breedte 1. Zo maken we een partitionering met twee bijzondere driehoeken. Nu moeten we nog laten zien dat deze bovengrens het beste is wat we kunnen doen. \n\nVoor elke partitionering maken we nu het volgende plaatje. We zetten een rood punt in het midden van elke schuine zijde van elke driehoek, en als er twee schuine zijden zijn dan verbinden we de twee rode punten met een rode lijn. Merk op dat elke driehoek hoogstens twee schuine zijden heeft, en alleen bijzondere driehoeken hebben er maar één.\n\n\n\n\nIn elk rood punt komen nu hoogstens twee rode lijnen samen. Dus deze rode lijnen vormen gesloten cykels of open paden, en de bijzondere driehoeken zijn precies de eindes van de de open paden. Zo hebben we hierboven een open pad van lengte nul, en een gesloten cykel gevormd door 6 rode segmenten. Een rood pad kan alleen van richting veranderen op een schuine zijde die in beide richtingen “hoogte” 1 heeft (oftewel die hoort bij twee driehoeken die in verschillende richtingen hoogte 1 hebben). Zo’n schuine zijde moet dus precies de diagonaal zijn van een vakje. In het bijzonder verandert een rood pad alleen van richting in het midden van vakjes. (Alternatief kun je zeggen dat de rode zijden nooit over de roosterlijnen lopen, want de hoogte van elke driehoek is 1 en de rode lijnen lopen op halve hoogte. Dus je kan nooit van richting veranderen op de roosterlijnen.) \n\nNu hebben we dus open en gesloten paden van orthogonale rode lijnen die alleen van richt- ing veranderen in het midden van de vakjes. Als we de vakjes in een schaakbordpatroon kleuren, dan zijn de vakjes waardoor het pad loopt om-en-om zwart en wit. Een gesloten cykel gaat dus altijd door een even aantal vakjes. Als het bord een oneven aantal vak- jes heeft, dan is er dus minstens één open pad. En dit open pad heeft twee bijzondere driehoeken aan de uiteindes. □", "metadata": {"resource_path": "Dutch_TST/segmented/nl-2025-C2025_uitwerkingen.jsonl", "problem_match": "\nOpgave 2.", "solution_match": "\nOplossing."}}
-{"year": "2025", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "Dutch_TST", "problem": "Zij \\(\\triangle ABC\\) een scherphoekige driehoek zo dat \\(|AB| + |BC| = 4|AC|\\) en \\(|AB| < |BC|\\). Zij \\(D\\) het snijpunt van de bissectrice van \\(\\angle ABC\\) met de zijde \\(AC\\). Punten \\(P\\) en \\(Q\\) liggen op het lijnstuk \\(BD\\) zo dat \\(|BP| = 2|DQ|\\). Zij \\(\\ell\\) de lijn door \\(P\\) parallel aan \\(AC\\). De lijn door \\(Q\\) loodrecht op \\(BD\\) snijdt de lijnstukken \\(AB\\) en \\(BC\\) in respectievelijk \\(X\\) en \\(Y\\). Stel dat \\(X\\) en \\(Y\\) aan de andere kant liggen van \\(\\ell\\) dan \\(B\\). \n\nEen mier begint zijn reis in \\(X\\) en gaat vanaf daar naar een punt op \\(AC\\), dan een punt op \\(\\ell\\), dan weer terug naar een (eventueel ander) punt op \\(AC\\) en uiteindelijk naar \\(Y\\). Bewijs dat de lengte van de kortst mogelijke route van de mier gelijk is aan \\(4|XY|\\). \n\n", "solution": "Laat \\(R\\), \\(S\\) en \\(T\\) respectievelijk de punten zijn waar de mier de eerste keer op \\(AC\\) is, op \\(\\ell\\) is en de tweede keer op \\(AC\\) is. Laat \\(\\ell'\\) en \\(S'\\) de spiegelingen van \\(\\ell\\) en \\(S\\) in \\(AC\\) zijn, en zij \\(Y'\\) de spiegeling van \\(Y\\) in \\(\\ell'\\). Dan geldt wegens de driehoekongelijkheid voor de totale lengte van het pad van de mier dat \n\n\\[ \n\\begin{align*}\n|XR| + |RS| + |ST| + |TY| &= |XR| + |RS'| + |S'T| + |TY| \\\\\n&\\ge |XS'| + |S'Y| \\\\\n&= |XS'| + |S'Y'| \\\\\n&\\ge |XY'|,\n\\end{align*}\n\\]\n\nmet gelijkheid als \\(S'\\) op \\(XY'\\) ligt, \\(R\\) op \\(XS'\\) en \\(T\\) op \\(S'Y\\). Dus nu rest ons te bewijzen dat \\(|XY'| = 4|XY|\\). \n\nNu herdefiniëren we \\(S'\\) als het snijpunt van \\(XY'\\) en \\(\\ell'\\). Dan is \\(\\ell'\\) per definitie de buiten-bissectrice van \\(\\angle XS'Y\\). We definiëren ook \\(B'\\) als het snijpunt van \\(BQ\\) met \\(\\ell'\\). Dan ligt \\(B'\\) ook op de middelloodlijn van \\(XY\\), dus \\(B'\\) is het tweede BOM-punt van \\(\\triangle XYS'\\). (Merk op dat \\(BQ\\) niet parallel is aan \\(\\ell'\\) of er loodrecht op staat, want \\(\\ell' \\parallel AC\\) en \\(\\triangle BAC\\) is niet gelijkbenig. Dus dit BOM-punt is goed gedefinieerd en ongelijk aan \\(S'\\).) In het bijzonder is \\(XYB'S'\\) een koordenvierhoek. Wegens onze definitie van \\(\\ell'\\) is \\(B'\\) ook de spiegeling van\n\n\n\n\\(P\\) in \\(D\\). Dus samen met de voorwaarde dat \\(2|QD| = |BP|\\) rekenen we uit dat \n\n\\[ \n\\begin{align*} \n|QB'| &= |QD| + |DB'| \\\\ \n&= |QD| + |PD| \\\\ \n&= |QD| + |PQ| + |QD| \\\\ \n&= |BP| + |PQ| \\\\ \n&= |BQ|. \n\\end{align*} \n\\]\n\nOmdat \\(Q\\) ook het midden is van \\(XY\\) en de diagonalen loodrecht op elkaar staan, concludeer en we dat \\(BXB'Y\\) een ruit is. Nu gaan we hoekenjagen met deze koordenvierhoek, deze ruit en F-hoeken: \n\n\\[ \n\\begin{align*} \n\\angle S'XY &= \\angle S'B'Y \\\\ \n&= \\angle S'B'B - \\angle YB'B \\\\ \n&= \\angle CDB - \\angle YBB' \\\\ \n&= (180^\\circ - \\frac{1}{2}\\angle CBA - \\angle ACB) - \\frac{1}{2}\\angle CBA \\\\ \n&= \\angle BAC. \n\\end{align*} \n\\]\n\nAnalog geldt dat \\(\\angle S'YX = \\angle BCA\\). Dus \\(\\triangle XYS' \\sim \\triangle ACB\\), waardoor de voorwaarde \\(|AB| + |BC| = 4|AC|\\) ook in \\(\\triangle XYS'\\) geldt. We concluderen dat \\(|XY'| = |XS'| + |S'Y| = 4|XY'|\\), wat precies is wat we zochten. \\(\\square\\)", "metadata": {"resource_path": "Dutch_TST/segmented/nl-2025-C2025_uitwerkingen.jsonl", "problem_match": "\nOpgave 3.", "solution_match": "\nOplossing I."}}
-{"year": "2025", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "Dutch_TST", "problem": "Zij \\(\\triangle ABC\\) een scherphoekige driehoek zo dat \\(|AB| + |BC| = 4|AC|\\) en \\(|AB| < |BC|\\). Zij \\(D\\) het snijpunt van de bissectrice van \\(\\angle ABC\\) met de zijde \\(AC\\). Punten \\(P\\) en \\(Q\\) liggen op het lijnstuk \\(BD\\) zo dat \\(|BP| = 2|DQ|\\). Zij \\(\\ell\\) de lijn door \\(P\\) parallel aan \\(AC\\). De lijn door \\(Q\\) loodrecht op \\(BD\\) snijdt de lijnstukken \\(AB\\) en \\(BC\\) in respectievelijk \\(X\\) en \\(Y\\). Stel dat \\(X\\) en \\(Y\\) aan de andere kant liggen van \\(\\ell\\) dan \\(B\\). \n\nEen mier begint zijn reis in \\(X\\) en gaat vanaf daar naar een punt op \\(AC\\), dan een punt op \\(\\ell\\), dan weer terug naar een (eventueel ander) punt op \\(AC\\) en uiteindelijk naar \\(Y\\). Bewijs dat de lengte van de kortst mogelijke route van de mier gelijk is aan \\(4|XY|\\). \n\n", "solution": "We geven een alternatief voor het (subtiele gebruik van het) BOM-lemma. Definieer \\(B'\\) weer als het snijpunt van \\(BQ\\) met \\(\\ell'\\). Omdat \\(BXB'Y\\) een ruit is en \\(Y'\\) de spiegeling van \\(Y\\), geldt dat \\(|B'X| = |B'Y| = |B'Y'|\\). Dus \\(B'\\) is het middelpunt van de omgeschreven cirkel van \\(\\triangle XYY'\\). Daaruit volgt dat \\(\\angle XYY' = \\frac{1}{2}\\angle XB'Y\\). Dus met de buitenhoekstelling geldt nu dat \\(\\angle XS'Y = \\angle XYY' + \\angle Y'YS' = 2\\angle XYY' = \\angle XB'Y\\). Dus \\(XYB'S'\\) is een koordenvierhoek. Vanaf hier gaan we weer verder als Oplossing I. \\(\\square\\)", "metadata": {"resource_path": "Dutch_TST/segmented/nl-2025-C2025_uitwerkingen.jsonl", "problem_match": "\nOpgave 3.", "solution_match": "\nOplossing II."}}
+{"year": "2025", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "Dutch_TST", "problem": "Beschouw een rechthoekig bord van \\(m \\times n\\) vakjes met \\(m, n \\ge 1\\). De hoekpunten van de vakjes vormen een \\((m+1) \\times (n+1)\\)-grid. We noemen een driehoek met hoekpunten punten van het grid *laag* als die minstens één zijde heeft die parallel is met een zijde van het bord zo dat de hoogte van de driehoek op die zijde 1 is. We noemen een lage driehoek *bijzonder* als die twee zijden heeft die parallel zijn met een zijde van het bord. We partitioneren het bord in lage driehoeken. \n\nBepaal het minimale aantal bijzondere driehoeken over alle mogelijke partitioneringen van het \\(m \\times n\\)-bord.", "solution": "Als \\(m, n \\ge 2\\) en minstens één van de twee is even, dan is het antwoord 0. Anders (minstens één van de twee is 1, of ze zijn beide oneven) is het antwoord 2. \n\nWe tekenen eerst een voorbeeld voor \\(n = 1\\) en \\(m \\ge 1\\) met twee bijzondere driehoeken, een voorbeeld voor \\(n = 2\\) en \\(m \\ge 3\\) met nul bijzondere driehoeken, en het speciale geval \\(n = 2\\), \\(m = 2\\) met wederom nul bijzondere driehoeken. \n\n\n\n \n\nAls \\(m = 1\\) of \\(n = 1\\), dan is er dus een partitionering met twee bijzondere driehoeken. Stel nu dat \\(m, n \\ge 2\\). \n\nAls minstens één van \\(m\\) en \\(n\\) even is, kunnen we de rechthoek opknippen in stroken van breedte 2, dus dan kunnen we een partitionering maken met nul bijzondere driehoeken. Als ze beide oneven zijn, dan knippen we de rechthoek op in stroken van breedte 2 plus één strook van breedte 1. Zo maken we een partitionering met twee bijzondere driehoeken. Nu moeten we nog laten zien dat deze bovengrens het beste is wat we kunnen doen. \n\nVoor elke partitionering maken we nu het volgende plaatje. We zetten een rood punt in het midden van elke schuine zijde van elke driehoek, en als er twee schuine zijden zijn dan verbinden we de twee rode punten met een rode lijn. Merk op dat elke driehoek hoogstens twee schuine zijden heeft, en alleen bijzondere driehoeken hebben er maar één.\n\n\n\n\nIn elk rood punt komen nu hoogstens twee rode lijnen samen. Dus deze rode lijnen vormen gesloten cykels of open paden, en de bijzondere driehoeken zijn precies de eindes van de de open paden. Zo hebben we hierboven een open pad van lengte nul, en een gesloten cykel gevormd door 6 rode segmenten. Een rood pad kan alleen van richting veranderen op een schuine zijde die in beide richtingen “hoogte” 1 heeft (oftewel die hoort bij twee driehoeken die in verschillende richtingen hoogte 1 hebben). Zo’n schuine zijde moet dus precies de diagonaal zijn van een vakje. In het bijzonder verandert een rood pad alleen van richting in het midden van vakjes. (Alternatief kun je zeggen dat de rode zijden nooit over de roosterlijnen lopen, want de hoogte van elke driehoek is 1 en de rode lijnen lopen op halve hoogte. Dus je kan nooit van richting veranderen op de roosterlijnen.) \n\nNu hebben we dus open en gesloten paden van orthogonale rode lijnen die alleen van richt- ing veranderen in het midden van de vakjes. Als we de vakjes in een schaakbordpatroon kleuren, dan zijn de vakjes waardoor het pad loopt om-en-om zwart en wit. Een gesloten cykel gaat dus altijd door een even aantal vakjes. Als het bord een oneven aantal vak- jes heeft, dan is er dus minstens één open pad. En dit open pad heeft twee bijzondere driehoeken aan de uiteindes. □", "metadata": {"resource_path": "Dutch_TST/segmented/nl-2025-C2025_uitwerkingen.jsonl", "problem_match": "\nOpgave 2.", "solution_match": "\nOplossing."}}
+{"year": "2025", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "Dutch_TST", "problem": "Zij \\(\\triangle ABC\\) een scherphoekige driehoek zo dat \\(|AB| + |BC| = 4|AC|\\) en \\(|AB| < |BC|\\). Zij \\(D\\) het snijpunt van de bissectrice van \\(\\angle ABC\\) met de zijde \\(AC\\). Punten \\(P\\) en \\(Q\\) liggen op het lijnstuk \\(BD\\) zo dat \\(|BP| = 2|DQ|\\). Zij \\(\\ell\\) de lijn door \\(P\\) parallel aan \\(AC\\). De lijn door \\(Q\\) loodrecht op \\(BD\\) snijdt de lijnstukken \\(AB\\) en \\(BC\\) in respectievelijk \\(X\\) en \\(Y\\). Stel dat \\(X\\) en \\(Y\\) aan de andere kant liggen van \\(\\ell\\) dan \\(B\\). \n\nEen mier begint zijn reis in \\(X\\) en gaat vanaf daar naar een punt op \\(AC\\), dan een punt op \\(\\ell\\), dan weer terug naar een (eventueel ander) punt op \\(AC\\) en uiteindelijk naar \\(Y\\). Bewijs dat de lengte van de kortst mogelijke route van de mier gelijk is aan \\(4|XY|\\). \n\n", "solution": "Laat \\(R\\), \\(S\\) en \\(T\\) respectievelijk de punten zijn waar de mier de eerste keer op \\(AC\\) is, op \\(\\ell\\) is en de tweede keer op \\(AC\\) is. Laat \\(\\ell'\\) en \\(S'\\) de spiegelingen van \\(\\ell\\) en \\(S\\) in \\(AC\\) zijn, en zij \\(Y'\\) de spiegeling van \\(Y\\) in \\(\\ell'\\). Dan geldt wegens de driehoekongelijkheid voor de totale lengte van het pad van de mier dat \n\n\\[ \n\\begin{align*}\n|XR| + |RS| + |ST| + |TY| &= |XR| + |RS'| + |S'T| + |TY| \\\\\n&\\ge |XS'| + |S'Y| \\\\\n&= |XS'| + |S'Y'| \\\\\n&\\ge |XY'|,\n\\end{align*}\n\\]\n\nmet gelijkheid als \\(S'\\) op \\(XY'\\) ligt, \\(R\\) op \\(XS'\\) en \\(T\\) op \\(S'Y\\). Dus nu rest ons te bewijzen dat \\(|XY'| = 4|XY|\\). \n\nNu herdefiniëren we \\(S'\\) als het snijpunt van \\(XY'\\) en \\(\\ell'\\). Dan is \\(\\ell'\\) per definitie de buiten-bissectrice van \\(\\angle XS'Y\\). We definiëren ook \\(B'\\) als het snijpunt van \\(BQ\\) met \\(\\ell'\\). Dan ligt \\(B'\\) ook op de middelloodlijn van \\(XY\\), dus \\(B'\\) is het tweede BOM-punt van \\(\\triangle XYS'\\). (Merk op dat \\(BQ\\) niet parallel is aan \\(\\ell'\\) of er loodrecht op staat, want \\(\\ell' \\parallel AC\\) en \\(\\triangle BAC\\) is niet gelijkbenig. Dus dit BOM-punt is goed gedefinieerd en ongelijk aan \\(S'\\).) In het bijzonder is \\(XYB'S'\\) een koordenvierhoek. Wegens onze definitie van \\(\\ell'\\) is \\(B'\\) ook de spiegeling van\n\n\n\n\\(P\\) in \\(D\\). Dus samen met de voorwaarde dat \\(2|QD| = |BP|\\) rekenen we uit dat \n\n\\[ \n\\begin{align*} \n|QB'| &= |QD| + |DB'| \\\\ \n&= |QD| + |PD| \\\\ \n&= |QD| + |PQ| + |QD| \\\\ \n&= |BP| + |PQ| \\\\ \n&= |BQ|. \n\\end{align*} \n\\]\n\nOmdat \\(Q\\) ook het midden is van \\(XY\\) en de diagonalen loodrecht op elkaar staan, concludeer en we dat \\(BXB'Y\\) een ruit is. Nu gaan we hoekenjagen met deze koordenvierhoek, deze ruit en F-hoeken: \n\n\\[ \n\\begin{align*} \n\\angle S'XY &= \\angle S'B'Y \\\\ \n&= \\angle S'B'B - \\angle YB'B \\\\ \n&= \\angle CDB - \\angle YBB' \\\\ \n&= (180^\\circ - \\frac{1}{2}\\angle CBA - \\angle ACB) - \\frac{1}{2}\\angle CBA \\\\ \n&= \\angle BAC. \n\\end{align*} \n\\]\n\nAnalog geldt dat \\(\\angle S'YX = \\angle BCA\\). Dus \\(\\triangle XYS' \\sim \\triangle ACB\\), waardoor de voorwaarde \\(|AB| + |BC| = 4|AC|\\) ook in \\(\\triangle XYS'\\) geldt. We concluderen dat \\(|XY'| = |XS'| + |S'Y| = 4|XY'|\\), wat precies is wat we zochten. \\(\\square\\)", "metadata": {"resource_path": "Dutch_TST/segmented/nl-2025-C2025_uitwerkingen.jsonl", "problem_match": "\nOpgave 3.", "solution_match": "\nOplossing I."}}
+{"year": "2025", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "Dutch_TST", "problem": "Zij \\(\\triangle ABC\\) een scherphoekige driehoek zo dat \\(|AB| + |BC| = 4|AC|\\) en \\(|AB| < |BC|\\). Zij \\(D\\) het snijpunt van de bissectrice van \\(\\angle ABC\\) met de zijde \\(AC\\). Punten \\(P\\) en \\(Q\\) liggen op het lijnstuk \\(BD\\) zo dat \\(|BP| = 2|DQ|\\). Zij \\(\\ell\\) de lijn door \\(P\\) parallel aan \\(AC\\). De lijn door \\(Q\\) loodrecht op \\(BD\\) snijdt de lijnstukken \\(AB\\) en \\(BC\\) in respectievelijk \\(X\\) en \\(Y\\). Stel dat \\(X\\) en \\(Y\\) aan de andere kant liggen van \\(\\ell\\) dan \\(B\\). \n\nEen mier begint zijn reis in \\(X\\) en gaat vanaf daar naar een punt op \\(AC\\), dan een punt op \\(\\ell\\), dan weer terug naar een (eventueel ander) punt op \\(AC\\) en uiteindelijk naar \\(Y\\). Bewijs dat de lengte van de kortst mogelijke route van de mier gelijk is aan \\(4|XY|\\). \n\n", "solution": "We geven een alternatief voor het (subtiele gebruik van het) BOM-lemma. Definieer \\(B'\\) weer als het snijpunt van \\(BQ\\) met \\(\\ell'\\). Omdat \\(BXB'Y\\) een ruit is en \\(Y'\\) de spiegeling van \\(Y\\), geldt dat \\(|B'X| = |B'Y| = |B'Y'|\\). Dus \\(B'\\) is het middelpunt van de omgeschreven cirkel van \\(\\triangle XYY'\\). Daaruit volgt dat \\(\\angle XYY' = \\frac{1}{2}\\angle XB'Y\\). Dus met de buitenhoekstelling geldt nu dat \\(\\angle XS'Y = \\angle XYY' + \\angle Y'YS' = 2\\angle XYY' = \\angle XB'Y\\). Dus \\(XYB'S'\\) is een koordenvierhoek. Vanaf hier gaan we weer verder als Oplossing I. \\(\\square\\)", "metadata": {"resource_path": "Dutch_TST/segmented/nl-2025-C2025_uitwerkingen.jsonl", "problem_match": "\nOpgave 3.", "solution_match": "\nOplossing II."}}
{"year": "2025", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "Dutch_TST", "problem": "Bepaal alle gehele getallen \\(n \\ge 2\\) waarvoor \\(n\\) een deler is van \\(\\binom{2n-3}{n-1}\\).", "solution": "We gaan bewijzen dat het antwoord is: alle \\(n \\ge 2\\) die geen macht van 2 zijn. Er geldt \n\n\\[ \\binom{2n-3}{n-1} = \\frac{(2n-3)!}{(n-1)!(n-2)!} = \\frac{(2n-3)(2n-4)\\cdots(n+1)n}{(n-2)(n-3)\\cdots2\\cdot1}. \\qquad (1) \\]\n\nZij \\(p\\) een priemfactor van \\(n\\) en stel dat \\(p > 2\\). We weten dat de volgende uitdrukking geheel is: \n\n\\[ \\binom{2n-3}{n} = \\frac{(2n-3)!}{n!(n-3)!} = \\frac{(2n-3)(2n-4)\\cdots(n+1)}{(n-3)(n-4)\\cdots2\\cdot1}. \\]\n\nSchrijf \\(e_p(N)\\) voor het aantal factoren \\(p\\) in een geheel getal \\(N\\). Nu zien we dat \n\n\\[ e_p((2n-3)(2n-4)\\cdots(n+1)) \\ge e_p((n-3)(n-4)\\cdots2\\cdot1). \\]\n\nVerder geldt, aangezien \\(p \\mid n\\) en \\(p > 2\\), dat \\(p \\nmid n-2\\), dus vermenigvuldigen met \\(n-2\\) voegt geen factor \\(p\\) toe. We concluderen dat \n\n\\[ e_p((2n-3)(2n-4)\\cdots(n+1)n) - e_p((n-2)(n-3)\\cdots2\\cdot1) \\ge e_p(n). \\]\n\nDus \\(\\binom{2n-3}{n-1}\\) bevat minstens \\(e_p(n)\\) factoren \\(p\\). Nu is \\(n\\) een deler van \\(\\binom{2n-3}{n-1}\\) dan en slechts dan als dit resultaat ook geldt voor \\(p = 2\\). \n\nIn een product \\(a_1a_2\\cdots a_m\\) is het totaal aantal factoren 2 gelijk aan de som van het aantal \\(a_i\\) deelbaar door 2, het aantal \\(a_i\\) deelbaar door 4, het aantal \\(a_i\\) deelbaar door 8, ... Bekijk nu eerst \\(n = 2^k\\) met \\(k \\ge 1\\). Dan is in (1) de teller het product van \\(2^k, 2^k+1, 2^k+2, \\ldots, 2^{k+1}-3\\), terwijl de noemer het product is van \\(1, 2, 3, \\ldots, 2^k-2\\). Voor \\(1 \\le i \\le k-1\\) is het aantal getallen uit \\(1, 2, 3, \\ldots, 2^k-2\\) deelbaar door \\(2^i\\) precies gelijk aan het aantal getallen uit \\(2^k+1, 2^k+2, 2^k+3, \\ldots, 2^{k+1}-2\\) deelbaar door \\(2^i\\). Voor \\(i \\ge k\\) zijn beide aantallen 0. Dus \n\n\\[ e_2(1 \\cdot 2 \\cdot 3 \\cdots (2^k-2)) = e_2((2^k+1)(2^k+2)(2^k+3) \\cdots (2^{k+1}-2)). \\]\n\nAan de rechterkant delen we het product door \\(2^{k+1}-2\\) (één factor 2) en vermenigvuldigen met \\(2^k\\) (\\(k\\) factoren 2) om het product uit de teller te krijgen. We concluderen \n\n\\[ e_2(2^k(2^k+1)(2^k+2)(2^k+3)\\cdots(2^{k+1}-3)) - e_2(1 \\cdot 2 \\cdot 3 \\cdots (2^k-2)) = k-1, \\]\n\nen dus bevat \\(\\binom{2n-3}{n-1}\\) precies \\(k-1\\) factoren 2, wat niet genoeg is om deelbaar te zijn door \\(n\\). Dus \\(n = 2^k\\) voldoet niet.\n\n\n\nBekijk nu het geval dat \\(n\\) geen macht van 2 is. Zij \\(2^k\\) de grootste macht van 2 kleiner dan \\(n\\). We weten dat \\(\\binom{2n-3}{n-1}\\) geheel is, dus voor oneven \\(n\\) geldt dat \n\n\\[e_2((2n-3)(2n-4)\\cdots(n+1)n) - e_2((n-2)(n-3)\\cdots 2\\cdot 1) \\ge 0 = e_2(n).\\]\n\nAls \\(n\\) even is, dan is er een maximale \\(\\ell\\) zo dat \\(2^\\ell \\mid n\\). Aangezien \\(n\\) geen tweemacht is, merken we op dat \\(2^{\\ell+1} < 3 \\cdot 2^\\ell \\le n < 2^{k+1}\\). Dat betekent dat \\(\\ell < k\\). Verder is \\(2^k \\le n-2\\) en dus \\(n < 2^{k+1} \\le 2n-4\\), dus de teller van (1) bevat \\(2^{k+1}\\). Nu bekijken we voor elke \\(i\\) het aantal factoren in de teller en de noemer deelbaar door \\(2^i\\): \n\n- \\(i = 1\\): er zijn \\(n-2\\) gehele getallen in elk van de producten, en \\(n\\) is even, dus het aantal getallen in het product deelbaar door 2 is in teller en noemer gelijk. \n\n- \\(2 \\le i \\le \\ell\\): in de noemer zijn \\(\\lfloor \\frac{n-2}{2^i} \\rfloor\\) getallen deelbaar door \\(2^i\\); in de teller zijn er \\(\\lceil \\frac{n-2}{2^i} \\rceil\\) deelbaar door \\(2^i\\) aangezien het kleinste getal deelbaar is door \\(2^i\\); daarom is er in de teller één meer dan in de noemer. \n\n- \\(\\ell+1 \\le i \\le k\\): in de noemer zijn \\(\\lfloor \\frac{n-2}{2^i} \\rfloor\\) getallen deelbaar door \\(2^i\\); in de teller zijn het er minstens zoveel. \n\n- \\(i = k+1\\): in de noemer is geen getal deelbaar door \\(2^{k+1}\\); in de teller is het er precies één. \n\n- \\(i > k+1\\): zowel teller als noemer bevatten geen getallen deelbaar door \\(2^i\\). \n\nAlles bij elkaar concluderen we dat \n\n\\[e_2((2n-3)(2n-4)\\cdots(n+1)n) - e_2((n-2)(n-3)\\cdots 2\\cdot 1) \\ge \\ell - 1 + 1 = \\ell = e_2(n).\\]\n\nDus \\(n\\) deelt \\(\\binom{2n-3}{n-1}\\). Het antwoord op de vraag is dus: alle \\(n \\ge 2\\) die geen macht van 2 zijn. \\(\\square\\)", "metadata": {"resource_path": "Dutch_TST/segmented/nl-2025-C2025_uitwerkingen.jsonl", "problem_match": "\nOpgave 4.", "solution_match": "\nOplossing I."}}
{"year": "2025", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "Dutch_TST", "problem": "Bepaal alle gehele getallen \\(n \\ge 2\\) waarvoor \\(n\\) een deler is van \\(\\binom{2n-3}{n-1}\\).", "solution": "We merken als eerste op dat het aantal manieren om \\(n-2\\) identieke ballen te verdelen over \\(n\\) verschillende vakjes gelijk is aan \\(\\binom{(n-2)+(n-1)}{n-1} = \\binom{2n-3}{n-1}\\) wegens het paaseierenprincipe voor \\(n \\ge 2\\). We gaan kijken naar de symmetrieën in de verzameling van manieren. \n\nNoteer een verdeling als \\((x_1, x_2, \\ldots, x_n)\\) met \\(x_k\\) het aantal ballen in het \\(k\\)-de vakje. De kleinste \\(1 \\le p \\le n\\) zo dat \\((x_{1+p}, x_{2+p}, \\ldots, x_{n+p}) = (x_1, x_2, \\ldots, x_n)\\) noemen we de *periode* van deze verdeling, waarbij we de indices modulo \\(n\\) rekenen. Omdat je na \\(n\\) keer doordraaien ook weer bij de oorspronkelijke verdeling uitkomt en \\(p\\) minimaal is met deze eigenschap, geldt dat \\(p\\) een deler is van \\(n\\). We schrijven \\(d = n/p\\) en de verdeling bestaat nu uit \\(d\\) gelijke\n\n\n\ndelen \\((x_1, \\ldots, x_p) = (x_{p+1}, \\ldots, x_{2p}) = \\ldots = (x_{n-p+1}, \\ldots, x_n)\\). In het bijzonder heeft elk deel evenveel ballen, dus \\(d\\) is ook een deler van \\(n-2\\). Aangezien \\(\\text{gcd}(n, n-2) = \\text{gcd}(n, 2)\\) is \\(d\\) gelijk aan 1, of eventueel 2 als \\(n\\) even is. \n\nZij \\(A\\) de verzameling van manieren om \\(n-2\\) identieke ballen te verdelen over \\(n\\) verschillende vakjes, en zij \\(A_p\\) de deelverzameling van deze manieren met periode \\(p\\). Dan kunnen we het bovenstaande als volgt samenvatten: voor \\(n \\ge 2\\) geldt \\(|A| = \\binom{2n-3}{n-1}\\), als \\(n\\) oneven is geldt \\(A = A_n\\), en als \\(n\\) even is dan \\(A = A_n \\cup A_{n/2}\\). De crux van de oplossing is nu dat voor een verdeling \\((x_1, x_2, \\ldots, x_n)\\) met periode \\(p\\) de verdelingen \\((x_{1+i}, x_{2+i}, \\ldots, x_{n+i})\\) voor \\(0 \\le i \\le p-1\\) verschillende unieke verdelingen zijn (terwijl \\(i = p\\) dus juist dezelfde verdeling geeft als \\(i = 0\\)). Dus \\(p \\mid |A_p|\\). In het bijzonder geldt \\(n \\mid |A_n|\\), en voor alle oneven \\(n \\ge 3\\) dat \\(n\\) een deler is van \\(|A_n| = |A| = \\binom{2n-3}{n-1}\\). \n\nStel nu dat \\(n\\) even is, en schrijf \\(n = 2m\\). Zij \\(B\\) de verzameling van manieren om \\(m-1\\) identieke ballen te verdelen over \\(m\\) verschillende vakjes. Dan is de restrictie \\((x_1, x_2, \\ldots, x_{2m}) \\mapsto (x_1, x_2, \\ldots, x_m)\\) tot de eerste \\(m\\) vakjes een functie van \\(A_m\\) naar \\(B\\). De functie \\((x_1, x_2, \\ldots, x_m) \\mapsto (x_1, x_2, \\ldots, x_m, x_1, \\ldots, x_m)\\) die de eerste \\(m\\) vakjes herhaalt, is een tweezijdige inverse. Dus we hebben een bijectie tussen deze twee verzamelingen. Dit betekent dat \\(|A_m| = |B|\\), en \n\n\\[|A| = |A_n| + |A_m| \\equiv |A_m| = |B| \\mod n.\\]\n\nNu rekenen we met het paaseirenprincipe uit dat \n\n\\[|B| = \\binom{2m-2}{m-1} = \\binom{2m-3}{m-2} + \\binom{2m-3}{m-1} = 2\\binom{2m-3}{m-1}.\\]\n\nWe concluderen dat \\(n\\) een deler is van \\(|A| = \\binom{2n-3}{n-1}\\), dan en slechts dan als \\(n = 2m\\) een deler is van \\(|B| = 2\\binom{2m-3}{m-1}\\), dan en slechts dan als \\(m\\) een deler is van \\(\\binom{2m-3}{m-1}\\). \n\nWe hadden al bewezen dat \\(n\\) een deler is van \\(\\binom{2n-3}{n-1}\\) voor alle oneven \\(n \\ge 3\\). Door de conclusie van de vorige alinea herhaaldelijk toe te passen vinden we nu dus dat \\(n\\) een deler is van \\(\\binom{2n-3}{n-1}\\) voor alle \\(n\\) met een oneven deler groter of gelijk aan 3. Aan de andere kant rekenen we voor \\(n = 2\\) uit dat \\(\\binom{2n-3}{n-1} = \\binom{1}{1} = 1\\), waarvan 2 duidelijk geen deler is. Dus met herhaaldelijk toepassen van de conclusie van de vorige alinea is \\(n\\) géén deler van \\(\\binom{2n-3}{n-1}\\) wanneer \\(n\\) een tweemacht is. \\(\\square\\)", "metadata": {"resource_path": "Dutch_TST/segmented/nl-2025-C2025_uitwerkingen.jsonl", "problem_match": "\nOpgave 4.", "solution_match": "\nOplossing II."}}
{"year": "2025", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "Dutch_TST", "problem": "Bepaal alle gehele getallen \\(n \\ge 2\\) waarvoor \\(n\\) een deler is van \\(\\binom{2n-3}{n-1}\\).", "solution": "We gebruiken de notatie van oplossing II. We geven een alternatief voor de inductie voor even \\(n = 2m\\) vanaf de realisatie dat \\(|A| \\equiv |B| \\mod n\\). \n\nOmdat \\(\\text{gcd}(m, m-1) = 1\\), is er geen enkele verdeling in \\(B\\) met een periode kleiner dan \\(m\\). Dus \\(m \\mid |B|\\). Dat betekent dat \\(b = |B|/m\\) geheel is, en \\(n \\mid |A|\\) dan en slechts dan als \\(b\\)\n\n\n\neven is. Nu rekenen we uit dat \n\n\\[b = \\frac{1}{m} \\binom{2m-2}{m-1} = \\frac{1}{m} \\frac{(2m-2)!}{(m-1)!(m-1)!} = \\frac{1}{(2m-1)} \\frac{(2m-1)!}{m!(m-1)!} = \\frac{1}{(2m-1)} \\binom{2m-1}{m-1}.\\]\n\nAangezien \\(b\\) geheel is, volgt hieruit dat \\(2m-1\\) een deler is van \\(\\binom{2m-1}{m-1}\\). En omdat \\(2m-1\\) oneven is, is \\(b\\) even dan en slechts dan als \\(\\binom{2m-1}{m-1}\\) even is. Dit herschrijven we nog een keer als \\(\\binom{2m-1}{m-1} = \\frac{(2m-1)!}{m!(m-1)!} = \\frac{m}{2m} \\frac{(2m)!}{m!m!} = \\frac{1}{2} \\binom{2m}{m}\\). Dus \\(\\binom{2m-1}{m-1}\\) is even dan en slechts dan als 4 een deler is van \\(\\binom{2m}{m}\\). Zij \\(2^k\\) de grootste macht van 2 kleiner dan of gelijk aan \\(m\\). Met de notatie van oplossing I rekenen we dan uit \n\n\\[e_2(\\binom{2m}{m}) = e_2(2m!) - 2e_2(m!)\\] \\[= \\sum_{i=1}^{k+1} \\left\\lfloor \\frac{2m}{2^i} \\right\\rfloor - 2 \\sum_{i=1}^{k} \\left\\lfloor \\frac{m}{2^i} \\right\\rfloor\\] \\[= \\left( \\left\\lfloor \\frac{2m}{2} \\right\\rfloor + \\sum_{i=2}^{k+1} \\left\\lfloor \\frac{2m}{2^i} \\right\\rfloor \\right) - 2 \\sum_{i=1}^{k} \\left\\lfloor \\frac{m}{2^i} \\right\\rfloor\\] \\[= \\left( m + \\sum_{i=1}^{k} \\left\\lfloor \\frac{m}{2^i}\\right\\rfloor \\right) - 2 \\sum_{i=1}^{k} \\left\\lfloor \\frac{m}{2^i} \\right\\rfloor\\] \\[= m - \\sum_{i=1}^{k} \\left\\lfloor \\frac{m}{2^i}\\right\\rfloor\\] \\[\\geq m - \\sum_{i=1}^{k} \\frac{m}{2^i} = \\frac{m}{2^k} \\geq 1.\\]\n\nIn de twee ongelijkheden van de laatste regel geldt gelijkheid dan en slechts dan als \\(m = 2^k\\), en dus \\(n = 2^{k+1}\\). Aangezien 4 is een deler van \\(\\binom{2m}{m}\\) dan en slechts dan als \\(e_2(\\binom{2m}{m}) > 1\\), geldt dat dus dan en slechts dan als \\(n\\) géén tweemacht is. \\(\\square\\) \n\nOpmerking. De formule \\(m - \\sum_{i=1}^{k} \\left\\lfloor \\frac{m}{2^i} \\right\\rfloor\\) geeft het aantal enen in de binaire notatie van \\(m\\).", "metadata": {"resource_path": "Dutch_TST/segmented/nl-2025-C2025_uitwerkingen.jsonl", "problem_match": "\nOpgave 4.", "solution_match": "\nOplossing III."}}
diff --git a/Dutch_TST/segmented/nl-2025-D2025_uitwerkingen.jsonl b/Dutch_TST/segmented/nl-2025-D2025_uitwerkingen.jsonl
index c3e2159c974fa5b3548675a86b323b5881673f07..cd644f128ee17e8ee8c078b693213d801367d95f 100644
--- a/Dutch_TST/segmented/nl-2025-D2025_uitwerkingen.jsonl
+++ b/Dutch_TST/segmented/nl-2025-D2025_uitwerkingen.jsonl
@@ -1,7 +1,7 @@
{"year": "2025", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "Dutch_TST", "problem": "Kan een getal van de vorm 44...41, met een oneven aantal vieren gevolgd door een 1, een kwadraat zijn?", "solution": "Nee. We kunnen zo'n getal 44...41 schrijven als \\(4 \\cdot \\frac{10^{2m}-1}{9} - 3\\), met \\(m \\ge 1\\) geheel. Stel dat dit een kwadraat is. Schrijf \\(4 \\cdot \\frac{10^{2m}-1}{9} - 3 = n^2\\) met \\(n\\) positief geheel. Vermenigvuldigen met 9 geeft \\(4(10^{2m} - 1) - 27 = 9n^2\\), dus \\(4 \\cdot 10^{2m} = 9n^2 + 31\\), ofwel \\((2 \\cdot 10^m)^2 - (3n)^2 = 31\\). Schrijf nu \\(x = 2 \\cdot 10^m\\) en \\(y = 3n\\), dan krijgen we dus \\(x^2 - y^2 = 31\\). Dit kunnen we ontbinden als \\((x - y)(x + y) = 31\\). Dus \\(x + y \\mid 31\\), en omdat \\(x + y \\ge 2\\) krijgen we dus \\(x + y = 31\\) en \\(x - y = 1\\). Dit levert \\(x = 16\\), maar dat is niet van de vorm \\(2 \\cdot 10^m\\). We concluderen dat een getal van de vorm 44...41 met een oneven aantal vieren geen kwadraat kan zijn. \\(\\square\\)", "metadata": {"resource_path": "Dutch_TST/segmented/nl-2025-D2025_uitwerkingen.jsonl", "problem_match": "\nOpgave 1.", "solution_match": "\nOplossing I."}}
{"year": "2025", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "Dutch_TST", "problem": "Kan een getal van de vorm 44...41, met een oneven aantal vieren gevolgd door een 1, een kwadraat zijn?", "solution": "We merken op dat de restklassen van kwadraten modulo 11 gelijk zijn aan respectievelijk 0, 1, 4, 9, 5, 3, 3, 5, 9, 4, 1. Aan de andere kant kunnen we de restklasse van 44...41 uitrekenen door steeds twee vieren van het getal af te halen, waardoor de restklasse niet verandert wegens \\(44 \\cdot 10^\\ell \\equiv 0 \\mod 11\\). Omdat het gegeven getal een oneven aantal vieren heeft, is het dus gelijk aan \n\n\\[44...41 \\equiv 41 \\equiv 8 \\mod 11.\\]\n\nDit is niet een van de kwadratische restklassen, dus geen van deze getallen is een kwadraat. \\(\\square\\)", "metadata": {"resource_path": "Dutch_TST/segmented/nl-2025-D2025_uitwerkingen.jsonl", "problem_match": "\nOpgave 1.", "solution_match": "\nOplossing II."}}
-{"year": "2025", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "Dutch_TST", "problem": "Zij \\(\\triangle ABC\\) een scherphoekige driehoek met \\(|AB| > |AC|\\), zij \\(\\omega\\) de omgeschreven cirkel van \\(\\triangle ABC\\) met middelpunt \\(O\\). De hoogtelijn vanuit \\(A\\) snijdt \\(BC\\) in \\(D\\) en snijdt \\(\\omega\\) een tweede keer in \\(P\\). Definieer \\(H\\) als het hoogtepunt van \\(\\triangle ABC\\) en zij \\(K\\) het punt op het lijnstuk \\(BC\\) zodanig dat \\(|BD| = |KC|\\). De omgeschreven cirkel van \\(\\triangle PKH\\) snijdt \\(\\omega\\) een tweede keer in \\(Q\\) en snijdt de lijn \\(BC\\) een tweede keer in \\(N\\). Zij \\(T\\) het punt op de lijn \\(AD\\) zodanig dat \\(TN \\perp PQ\\). \n\nBewijs dat de lijn \\(KT\\) door \\(O\\) gaat. \n\n", "solution": "Zij \\(O\\) het middelpunt van \\(\\omega\\). Wegens \\(|BD| = |KC|\\) vallen de middelloodlijnen van \\(BC\\) en \\(KD\\) samen en in het bijzonder geldt dus dat \\(|OK| = |OD|\\). Zij \\(T'\\) de spiegeling van \\(K\\) in \\(O\\). Wegens Thales geldt dan dat \\(\\angle KDT' = 90^\\circ\\), dus \\(T'\\) ligt op \\(AD\\). \n\nHet is een standaardplaatje dat de spiegeling van \\(H\\) in \\(BC\\) op de omgeschreven cirkel ligt. (Eenvoudig te bewijzen door \\(\\angle BHC\\) uit te rekenen.) Hieruit volgt dat \\(\\triangle PKH\\) in zichzelf overgaat onder spiegeling in \\(BC\\). Dit betekent dat het middelpunt \\(M\\) van de omgeschreven cirkel van \\(\\triangle PKH\\) op \\(BC\\) ligt, en dat \\(M\\) het midden van \\(KN\\) is. Per definitie is \\(O\\) het midden van \\(KT'\\), dus \\(OM\\) is een middenparallel in \\(\\triangle KNT'\\); in het bijzonder geldt \\(OM \\parallel NT'\\). Omdat \\(P\\) en \\(Q\\) beide op zowel \\(\\omega\\) als de omgeschreven cirkel\n\n\n\nvan \\(\\triangle PKH\\) liggen, is \\(OM\\) de middelloodlijn van \\(PQ\\), dus \\(OM \\perp PQ\\). Nu volgt dat ook \\(NT' \\perp PQ\\), en kunnen we concluderen dat \\(T'\\) gelijk is aan \\(T\\). \\(\\square\\) \n\nOpmerking. Zij \\(\\ell\\) de lijn door \\(N\\) loodrecht op \\(PQ\\). We laten voor de volledigheid zien dat \\(\\ell\\) niet evenwijdig is met \\(AD\\) en in het bijzonder niet samenvalt (zodat \\(T\\) bestaat en uniek vastligt). Merk op dat \\(H\\) binnen \\(\\triangle ABC\\) ligt, omdat dit een scherphoekige driehoek is. Hieruit volgt dat \\(D \\neq N\\), en dus dat \\(\\ell\\) niet samenvalt met \\(AD\\). Omdat \\(D \\neq N\\) en \\(M\\) het midden is van \\(KN\\), is \\(M\\) niet het midden van \\(KD\\). Dus \\(M\\) is niet het midden van \\(BC\\). Hierdoor staat \\(OM\\) niet loodrecht op \\(BC\\). Omdat \\(OM \\parallel \\ell\\) en \\(AD \\perp BC\\) concluderen we dat \\(\\ell\\) niet evenwijdig is met \\(AD\\).", "metadata": {"resource_path": "Dutch_TST/segmented/nl-2025-D2025_uitwerkingen.jsonl", "problem_match": "\nOpgave 2.", "solution_match": "\nOplossing I."}}
-{"year": "2025", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "Dutch_TST", "problem": "Zij \\(\\triangle ABC\\) een scherphoekige driehoek met \\(|AB| > |AC|\\), zij \\(\\omega\\) de omgeschreven cirkel van \\(\\triangle ABC\\) met middelpunt \\(O\\). De hoogtelijn vanuit \\(A\\) snijdt \\(BC\\) in \\(D\\) en snijdt \\(\\omega\\) een tweede keer in \\(P\\). Definieer \\(H\\) als het hoogtepunt van \\(\\triangle ABC\\) en zij \\(K\\) het punt op het lijnstuk \\(BC\\) zodanig dat \\(|BD| = |KC|\\). De omgeschreven cirkel van \\(\\triangle PKH\\) snijdt \\(\\omega\\) een tweede keer in \\(Q\\) en snijdt de lijn \\(BC\\) een tweede keer in \\(N\\). Zij \\(T\\) het punt op de lijn \\(AD\\) zodanig dat \\(TN \\perp PQ\\). \n\nBewijs dat de lijn \\(KT\\) door \\(O\\) gaat. \n\n", "solution": "Voor een directere versie van de vorige oplossing, merken we wederom op dat \\(OM \\perp PQ \\perp TN\\). Dus \\(OM \\parallel TN\\), en omdat \\(M\\) het midden is van \\(KN\\) is de lijn \\(OM\\) dus de middenparallel van \\(\\triangle KNT\\) (maar \\(O\\) ligt a priori nog niet op \\(KT\\)). Dus \\(OM\\) gaat door het midden van \\(KT\\), zeg \\(O'\\). Verder is de middelloodlijn van \\(KD\\) een middenparallel in \\(\\triangle KDT\\), want hij gaat door het midden van \\(KD\\) en is evenwijdig aan \\(DT\\). Dus deze gaat ook door \\(O'\\). De middelloodlijn van \\(KD\\) is gelijk aan de middelloodlijn van \\(BC\\). Deze gaat ook door \\(O\\), omdat \\(BC\\) een koorde is van \\(\\omega\\). Omdat \\(MO\\) en de middelloodlijn van \\(BC\\) beide door zowel \\(O\\) als \\(O'\\) gaan, concluderen we dat \\(O' = O\\). Dus \\(KT\\) gaat door \\(O\\). \\(\\square\\)", "metadata": {"resource_path": "Dutch_TST/segmented/nl-2025-D2025_uitwerkingen.jsonl", "problem_match": "\nOpgave 2.", "solution_match": "\nOplossing II."}}
+{"year": "2025", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "Dutch_TST", "problem": "Zij \\(\\triangle ABC\\) een scherphoekige driehoek met \\(|AB| > |AC|\\), zij \\(\\omega\\) de omgeschreven cirkel van \\(\\triangle ABC\\) met middelpunt \\(O\\). De hoogtelijn vanuit \\(A\\) snijdt \\(BC\\) in \\(D\\) en snijdt \\(\\omega\\) een tweede keer in \\(P\\). Definieer \\(H\\) als het hoogtepunt van \\(\\triangle ABC\\) en zij \\(K\\) het punt op het lijnstuk \\(BC\\) zodanig dat \\(|BD| = |KC|\\). De omgeschreven cirkel van \\(\\triangle PKH\\) snijdt \\(\\omega\\) een tweede keer in \\(Q\\) en snijdt de lijn \\(BC\\) een tweede keer in \\(N\\). Zij \\(T\\) het punt op de lijn \\(AD\\) zodanig dat \\(TN \\perp PQ\\). \n\nBewijs dat de lijn \\(KT\\) door \\(O\\) gaat. \n\n", "solution": "Zij \\(O\\) het middelpunt van \\(\\omega\\). Wegens \\(|BD| = |KC|\\) vallen de middelloodlijnen van \\(BC\\) en \\(KD\\) samen en in het bijzonder geldt dus dat \\(|OK| = |OD|\\). Zij \\(T'\\) de spiegeling van \\(K\\) in \\(O\\). Wegens Thales geldt dan dat \\(\\angle KDT' = 90^\\circ\\), dus \\(T'\\) ligt op \\(AD\\). \n\nHet is een standaardplaatje dat de spiegeling van \\(H\\) in \\(BC\\) op de omgeschreven cirkel ligt. (Eenvoudig te bewijzen door \\(\\angle BHC\\) uit te rekenen.) Hieruit volgt dat \\(\\triangle PKH\\) in zichzelf overgaat onder spiegeling in \\(BC\\). Dit betekent dat het middelpunt \\(M\\) van de omgeschreven cirkel van \\(\\triangle PKH\\) op \\(BC\\) ligt, en dat \\(M\\) het midden van \\(KN\\) is. Per definitie is \\(O\\) het midden van \\(KT'\\), dus \\(OM\\) is een middenparallel in \\(\\triangle KNT'\\); in het bijzonder geldt \\(OM \\parallel NT'\\). Omdat \\(P\\) en \\(Q\\) beide op zowel \\(\\omega\\) als de omgeschreven cirkel\n\n\n\nvan \\(\\triangle PKH\\) liggen, is \\(OM\\) de middelloodlijn van \\(PQ\\), dus \\(OM \\perp PQ\\). Nu volgt dat ook \\(NT' \\perp PQ\\), en kunnen we concluderen dat \\(T'\\) gelijk is aan \\(T\\). \\(\\square\\) \n\nOpmerking. Zij \\(\\ell\\) de lijn door \\(N\\) loodrecht op \\(PQ\\). We laten voor de volledigheid zien dat \\(\\ell\\) niet evenwijdig is met \\(AD\\) en in het bijzonder niet samenvalt (zodat \\(T\\) bestaat en uniek vastligt). Merk op dat \\(H\\) binnen \\(\\triangle ABC\\) ligt, omdat dit een scherphoekige driehoek is. Hieruit volgt dat \\(D \\neq N\\), en dus dat \\(\\ell\\) niet samenvalt met \\(AD\\). Omdat \\(D \\neq N\\) en \\(M\\) het midden is van \\(KN\\), is \\(M\\) niet het midden van \\(KD\\). Dus \\(M\\) is niet het midden van \\(BC\\). Hierdoor staat \\(OM\\) niet loodrecht op \\(BC\\). Omdat \\(OM \\parallel \\ell\\) en \\(AD \\perp BC\\) concluderen we dat \\(\\ell\\) niet evenwijdig is met \\(AD\\).", "metadata": {"resource_path": "Dutch_TST/segmented/nl-2025-D2025_uitwerkingen.jsonl", "problem_match": "\nOpgave 2.", "solution_match": "\nOplossing I."}}
+{"year": "2025", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "Dutch_TST", "problem": "Zij \\(\\triangle ABC\\) een scherphoekige driehoek met \\(|AB| > |AC|\\), zij \\(\\omega\\) de omgeschreven cirkel van \\(\\triangle ABC\\) met middelpunt \\(O\\). De hoogtelijn vanuit \\(A\\) snijdt \\(BC\\) in \\(D\\) en snijdt \\(\\omega\\) een tweede keer in \\(P\\). Definieer \\(H\\) als het hoogtepunt van \\(\\triangle ABC\\) en zij \\(K\\) het punt op het lijnstuk \\(BC\\) zodanig dat \\(|BD| = |KC|\\). De omgeschreven cirkel van \\(\\triangle PKH\\) snijdt \\(\\omega\\) een tweede keer in \\(Q\\) en snijdt de lijn \\(BC\\) een tweede keer in \\(N\\). Zij \\(T\\) het punt op de lijn \\(AD\\) zodanig dat \\(TN \\perp PQ\\). \n\nBewijs dat de lijn \\(KT\\) door \\(O\\) gaat. \n\n", "solution": "Voor een directere versie van de vorige oplossing, merken we wederom op dat \\(OM \\perp PQ \\perp TN\\). Dus \\(OM \\parallel TN\\), en omdat \\(M\\) het midden is van \\(KN\\) is de lijn \\(OM\\) dus de middenparallel van \\(\\triangle KNT\\) (maar \\(O\\) ligt a priori nog niet op \\(KT\\)). Dus \\(OM\\) gaat door het midden van \\(KT\\), zeg \\(O'\\). Verder is de middelloodlijn van \\(KD\\) een middenparallel in \\(\\triangle KDT\\), want hij gaat door het midden van \\(KD\\) en is evenwijdig aan \\(DT\\). Dus deze gaat ook door \\(O'\\). De middelloodlijn van \\(KD\\) is gelijk aan de middelloodlijn van \\(BC\\). Deze gaat ook door \\(O\\), omdat \\(BC\\) een koorde is van \\(\\omega\\). Omdat \\(MO\\) en de middelloodlijn van \\(BC\\) beide door zowel \\(O\\) als \\(O'\\) gaan, concluderen we dat \\(O' = O\\). Dus \\(KT\\) gaat door \\(O\\). \\(\\square\\)", "metadata": {"resource_path": "Dutch_TST/segmented/nl-2025-D2025_uitwerkingen.jsonl", "problem_match": "\nOpgave 2.", "solution_match": "\nOplossing II."}}
{"year": "2025", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "Dutch_TST", "problem": "Johan en Quintijn spelen het volgende spel, waarbij ze om en om aan de beurt zijn en Johan begint. Aan het begin staan op een bord de getallen \\(1, 2, \\dots, 2024\\) geschreven. In elke beurt veegt de speler die aan de beurt is twee getallen \\(a\\) en \\(b\\) die op het bord staan uit, en schrijft het (mogelijk negatieve) verschil \\(a - b\\) op het bord. Het spel eindigt als er nog maar één getal op het bord staat. Als dit getal deelbaar is door 3, dan wint Johan, en anders wint Quintijn. \n\nBepaal welke van de twee spelers een winnende strategie heeft.", "solution": "We gaan bewijzen dat Quintijn een winnende strategie heeft. Merk allereerst op dat alleen het aantal getallen in elke restklasse modulo 3 van belang is. Als \\(a \\equiv i \\mod 3\\), \\(b \\equiv j \\mod 3\\) en \\(a - b \\equiv k \\mod 3\\), dan noteren we de zet met \\(a\\) en \\(b\\) als \\((i, j) \\to k\\). Zij \\(x\\) het aantal getallen op het bord die *niet* deelbaar zijn door 3. Merk op dat \\(x\\) in elke beurt met hoogstens 2 kan afnemen. \n\nQuintijn kan het spel willekeurig spelen totdat aan het begin van zijn beurt geldt dat \\(1 \\le x \\le 4\\). Vanaf dat moment houdt hij de volgende strategie aan. \n\n- Als \\(x = 4\\), dan doet Quintijn altijd een zet die er voor zorgt dat \\(x\\) gelijk blijft aan 4. Om aan te tonen dat dit altijd kan, merken we op dat Quintijn aan het begin van zijn beurt altijd een oneven aantal getallen over heeft. Als \\(x = 4\\), moet er dus ook nog minstens één getal op het bord staan dat deelbaar is door 3. Nu kan Quintijn dus een zet doen van de vorm \\((i, 0) \\to i\\) met \\(i \\in \\{1, 2\\}\\), en verandert de waarde van \\(x\\) dus niet. \n\nAls Quintijn deze strategie aanhoudt, moet Johan dus op een gegeven moment een zet doen waardoor \\(x \\in \\{2, 3\\}\\), en bevinden we ons in één van de volgende gevallen. \n\n- Als \\(x = 3\\), dan staan er zonder verlies van algemeenheid twee getallen op het bord die congruent zijn aan 1 modulo 3. Met de zet \\((1, 1) \\to 0\\) kan Quintijn er voor zorgen dat \\(x = 1\\), en dan blijft er altijd exact 1 getal dat niet deelbaar is door 3 op het bord staan. Dus is uiteindelijk ook het laatste getal niet deelbaar door 3, en wint Quintijn. \n\n- Als \\(x = 2\\), dan onderscheiden we twee deelgevallen. \n\n- De twee getallen zitten in verschillende restklassen modulo 3. Dan kan Quintijn de zet \\((2, 1) \\to 1\\) doen zodat \\(x = 1\\), en zoals we al zagen wint hij dan. \n\n- De twee getallen zitten in dezelfde restklass modulo 3, zeg dat ze beide congruent zijn aan 1 modulo 3. Omdat er een oneven aantal overgebleven is, staat er ook nog een getal op het bord dat deelbaar is door 3. Met de zet \\((0, 1) \\to 2\\) laat Quintijn een situatie over waarin opnieuw \\(x = 2\\), en de getallen niet congruent aan elkaar zijn modulo 3. Nu kan Johan er niet voor zorgen dat \\(x\\) gelijk wordt\n\n\n\naan 0, dus laat hij een situatie over waarin \\(x = 1\\) of \\(x = 2\\). Als Quintijn deze strategie aanhoudt, dan moet Johan uiteindelijk een zet doen waardoor we in het vorige deelgeval of in het geval \\(x = 1\\) terecht komen, en dus wint Quintijn. \n\n* Als \\(x = 1\\), dan blijft er zoals gezegd altijd een getal dat niet deelbaar is door 3 op het bord staan, dus wint Quintijn. \n\nWe concluderen dat Quintijn inderdaad een winnende strategie heeft.", "metadata": {"resource_path": "Dutch_TST/segmented/nl-2025-D2025_uitwerkingen.jsonl", "problem_match": "\nOpgave 3.", "solution_match": "\nOplossing I."}}
{"year": "2025", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "Dutch_TST", "problem": "Johan en Quintijn spelen het volgende spel, waarbij ze om en om aan de beurt zijn en Johan begint. Aan het begin staan op een bord de getallen \\(1, 2, \\dots, 2024\\) geschreven. In elke beurt veegt de speler die aan de beurt is twee getallen \\(a\\) en \\(b\\) die op het bord staan uit, en schrijft het (mogelijk negatieve) verschil \\(a - b\\) op het bord. Het spel eindigt als er nog maar één getal op het bord staat. Als dit getal deelbaar is door 3, dan wint Johan, en anders wint Quintijn. \n\nBepaal welke van de twee spelers een winnende strategie heeft.", "solution": "Voor \\(i = 0, 1, 2\\) noteren we het aantal getallen op het bord dat congruent is aan \\(i\\) modulo 3 met \\(x_i\\). Merk op dat Johan de laatst overgebleven getallen die niet deelbaar zijn door 3 alleen weg kan spelen als \\(x_1 = 2\\) en \\(x_2 = 0\\), of \\(x_1 = 0\\) en \\(x_2 = 2\\). We beweren nu dat, als \\(x_1\\) en \\(x_2\\) niet beide gelijk zijn aan 0, Quintijn altijd een zet kan doen die er voor zorgt dat minstens één van \\(x_1\\) en \\(x_2\\) oneven is. Zoals we net zagen, kan Johan dan niet alle getallen die niet deelbaar zijn door 3 wegspelen; en in zijn eerste beurt lukt dit natuurlijk ook niet. Door deze strategie te volgen, wint Quintijn dus. \n\nOm de bewering te bewijzen, stel eerst dat \\(x_1\\) en \\(x_2\\) beide even zijn. Dan moet minstens één van de twee minstens gelijk zijn aan 2, zeg \\(x_1 \\ge 2\\). Omdat Quintijn aan het begin van zijn beurt altijd een oneven aantal getallen over heeft, moet er ook nog een getal op het bord staan dat deelbaar is door 3. Met de zet \\((0, 1) \\to 2\\) zorgt Quintijn ervoor dat \\(x_1\\) met 1 afneemt en \\(x_2\\) met 1 toeneemt. Na deze beurt zijn \\(x_1\\) en \\(x_2\\) dus beide oneven. \n\nStel nu dat minstens één van \\(x_1\\) en \\(x_2\\) oneven is. Als er nog minstens 4 getallen over zijn, dan is er een \\(i\\) met \\(x_i \\ge 2\\). Door de zet \\((i, i) \\to 0\\) veranderen \\(x_1\\) en \\(x_2\\) niet modulo 2, dus volgt de bewering. Het laatste geval is dat er precies 3 getallen over zijn (want Quintijn heeft aan het begin van zijn beurt altijd een oneven aantal getallen over). Dan moet gelden dat \\(x_0 = x_1 = x_2 = 1\\), en kan Quintijn de zet \\((1, 0) \\to 1\\) doen. \n\nDit voltooit het bewijs van de bewering, dus Quintijn heeft inderdaad een winnende strategie.", "metadata": {"resource_path": "Dutch_TST/segmented/nl-2025-D2025_uitwerkingen.jsonl", "problem_match": "\nOpgave 3.", "solution_match": "\nOplossing II."}}
{"year": "2025", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "Dutch_TST", "problem": "Vind alle functies \\(f: \\mathbb{Z}_{>0} \\to \\mathbb{Z}_{>0}\\) zo dat voor alle positieve gehele getallen \\(m, n\\) geldt dat \n\n\\[(f(m))^2 + 2mf(n) + f(n^2)\\]\n\nhet kwadraat van een geheel getal is.", "solution": "Vul in \\(m = n = 1\\), dan moet \\(f(1)^2 + 3f(1)\\) een kwadraat zijn. Aangezien \n\n\\[(f(1) + 1)^2 \\le f(1)^2 + 3f(1) < (f(1) + 2)^2,\\]\n\nmoet er links gelijkheid gelden, en dus is \\(f(1) = 1\\). Laat nu \\(p = 2k+1\\) een oneven priemgetal en vul in \\(m = k = (p-1)/2\\) en \\(n = 1\\). Dan zien we dat \\(f(k)^2 + p\\) een kwadraat moet zijn, zeg \\(a^2\\) voor een positieve gehele \\(a\\). Dat geeft \\(p = a^2 - f(k)^2 = (a - f(k))(a + f(k))\\), en dit heeft als enige oplossing \\(a - f(k) = 1\\) en \\(a + f(k) = p = 2k + 1\\). Hiervan het verschil nemen geeft \\(f(k) = k\\). \n\nNu vullen we alleen \\(m = k\\) in (weer zodat \\(2k + 1\\) een oneven priemgetal is, dus zodat \\(f(k) = k\\)), dan zien we dat \\(k^2 + 2kf(n) + f(n^2) = (k + f(n))^2 + f(n^2) - f(n)^2\\) een kwadraat is. Als we \\(k\\) groot genoeg kiezen (voor gegeven \\(n\\)) zodat \\(1 - 2k - 2f(n) < f(n^2) - f(n)^2 < 1 + 2k + 2f(n)\\), dan zien we dat \n\n\\[(k + f(n) - 1)^2 < (k + f(n))^2 + f(n^2) - f(n)^2 < (k + f(n) + 1)^2.\\]\n\nDe middelste uitdrukking is een kwadraat en moet dus gelijk zijn aan \\((k + f(n))^2\\), dat betekent dat \\(f(n^2) = f(n)^2\\) voor alle \\(n\\). Tot slot vullen we \\(n = k\\) in, en zien we dat \\(f(m)^2 + 2mk + k^2 = (k + m)^2 + f(m)^2 - m^2\\) een kwadraat is. Als we nu \\(k\\) weer groot genoeg kiezen (voor gegeven \\(m\\)), zien we met dezelfde redenering dat \\(f(m)^2 = m^2\\), dus \\(f(m) = m\\) voor alle \\(m\\). \n\nDe functie \\(f(m) = m\\) voldoet inderdaad, want dan is het gevraagde gelijk aan \\((m+n)^2\\). □", "metadata": {"resource_path": "Dutch_TST/segmented/nl-2025-D2025_uitwerkingen.jsonl", "problem_match": "\nOpgave 4.", "solution_match": "\nOplossing."}}
diff --git a/Dutch_TST/segmented/nl-2025-E2025_uitwerkingen.jsonl b/Dutch_TST/segmented/nl-2025-E2025_uitwerkingen.jsonl
index 20e25eb672df549f03f6e725eae8eec223615acf..cda20740a11fac26026a6cce972c10b43ef66907 100644
--- a/Dutch_TST/segmented/nl-2025-E2025_uitwerkingen.jsonl
+++ b/Dutch_TST/segmented/nl-2025-E2025_uitwerkingen.jsonl
@@ -1,4 +1,4 @@
-{"year": "2025", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "Dutch_TST", "problem": "Zij \\(ABCD\\) een parallellogram en zij \\(M\\) het snijpunt van de diagonalen. De omgeschreven cirkel van \\(\\triangle ABM\\) snijdt het lijnstuk \\(AD\\) in \\(E \\neq A\\) en de omgeschreven cirkel van \\(\\triangle EMD\\) snijdt het lijnstuk \\(BE\\) in het punt \\(F \\neq E\\). \n\nBewijs dat \\(\\angle ACB = \\angle DCF\\). \n\n", "solution": "We laten eerst zien dat \\(CBFD\\) een koordenvierhoek is. Er geldt \n\n\\[ \n\\begin{align*} \n\\angle BCD &= \\angle BAD = \\angle BAE \\quad (\\text{parallellogram}) \\\\\n&= 180^\\circ - \\angle EMB \\quad (\\text{koordenvierhoekstelling in } EABM) \\\\\n&= \\angle EMD \\quad (\\text{gestrekte hoek}) \\\\\n&= \\angle EFD \\quad (\\text{omtrekshoekstelling in } EFMD) \\\\\n&= 180^\\circ - \\angle BFD \\quad (\\text{gestrekte hoek}). \n\\end{align*} \n\\]\n\nDus wegens de koordenvierhoekstelling is \\(CBFD\\) een koordenvierhoek. Daarmee vinden\n\n\n\nwe dat \n\n\\[\n\\begin{align*}\n\\angle ACD &= \\angle CAB = \\angle MAB \\tag{Z-hoeken} \\\\\n&= \\angle MEB = \\angle MEF \\tag{omtrekshoekstelling in EABM} \\\\\n&= \\angle MDF = \\angle BDF \\tag{omtrekshoekstelling in EFMD} \\\\\n&= \\angle BCF \\tag{omtrekshoekstelling in CBFD}.\n\\end{align*}\n\\]\n\nDus \\(\\angle ACF + \\angle FCD = \\angle ACD = \\angle BCF = \\angle BCA + \\angle ACF\\). Als we hier \\(\\angle ACF\\) van af\nhalen vinden we dat \\(\\angle ACB = \\angle DCF\\), wat we wilden bewijzen. \\(\\square\\)", "metadata": {"resource_path": "Dutch_TST/segmented/nl-2025-E2025_uitwerkingen.jsonl", "problem_match": "\nOpgave 1.", "solution_match": "\nOplossing."}}
+{"year": "2025", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "Dutch_TST", "problem": "Zij \\(ABCD\\) een parallellogram en zij \\(M\\) het snijpunt van de diagonalen. De omgeschreven cirkel van \\(\\triangle ABM\\) snijdt het lijnstuk \\(AD\\) in \\(E \\neq A\\) en de omgeschreven cirkel van \\(\\triangle EMD\\) snijdt het lijnstuk \\(BE\\) in het punt \\(F \\neq E\\). \n\nBewijs dat \\(\\angle ACB = \\angle DCF\\). \n\n", "solution": "We laten eerst zien dat \\(CBFD\\) een koordenvierhoek is. Er geldt \n\n\\[ \n\\begin{align*} \n\\angle BCD &= \\angle BAD = \\angle BAE \\quad (\\text{parallellogram}) \\\\\n&= 180^\\circ - \\angle EMB \\quad (\\text{koordenvierhoekstelling in } EABM) \\\\\n&= \\angle EMD \\quad (\\text{gestrekte hoek}) \\\\\n&= \\angle EFD \\quad (\\text{omtrekshoekstelling in } EFMD) \\\\\n&= 180^\\circ - \\angle BFD \\quad (\\text{gestrekte hoek}). \n\\end{align*} \n\\]\n\nDus wegens de koordenvierhoekstelling is \\(CBFD\\) een koordenvierhoek. Daarmee vinden\n\n\n\nwe dat \n\n\\[\n\\begin{align*}\n\\angle ACD &= \\angle CAB = \\angle MAB \\tag{Z-hoeken} \\\\\n&= \\angle MEB = \\angle MEF \\tag{omtrekshoekstelling in EABM} \\\\\n&= \\angle MDF = \\angle BDF \\tag{omtrekshoekstelling in EFMD} \\\\\n&= \\angle BCF \\tag{omtrekshoekstelling in CBFD}.\n\\end{align*}\n\\]\n\nDus \\(\\angle ACF + \\angle FCD = \\angle ACD = \\angle BCF = \\angle BCA + \\angle ACF\\). Als we hier \\(\\angle ACF\\) van af\nhalen vinden we dat \\(\\angle ACB = \\angle DCF\\), wat we wilden bewijzen. \\(\\square\\)", "metadata": {"resource_path": "Dutch_TST/segmented/nl-2025-E2025_uitwerkingen.jsonl", "problem_match": "\nOpgave 1.", "solution_match": "\nOplossing."}}
{"year": "2025", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "Dutch_TST", "problem": "We noemen een geheel getal \\(n \\ge 3\\) polypythagorees als er \\(n\\) verschillende positieve getallen zijn die je een cirkel achter elkaar kan zetten zo dat de som van de kwadraten van elk paar opvolgende getallen een kwadraat is. Zo is 3 een polypythagorees getal omdat je bijvoorbeeld met 44, 117 en 240 een drietal hebt waarvoor geldt dat \\(44^2 + 117^2 = 125^2\\), \\(117^2 + 240^2 = 267^2\\) en \\(240^2 + 44^2 = 244^2\\). \n\nVind alle polypythagorees getallen.", "solution": "We bewijzen met inductie dat alle gehele getallen groter of gelijk aan 2 polypythagorees zijn, waarbij we de definitie uitbreiden naar \\(n = 2\\) op de logische manier. Als inductiebasis nemen we (3, 4) voor \\(n = 2\\) en (44, 117, 240) uit het voorbeeld voor \\(n = 3\\). \n\nStel nu dat \\(n\\) polypythagorees is met als getuige de getallen \\((a_1, a_2, \\dots, a_n)\\). Kies nu een priemgetal \\(p\\) dat geen enkele van de \\(a_i\\) deelt. Dan definiëren we \\(x = p^2 - 1\\) en \\(y = 2p\\). Hiervoor geldt dat \\(x^2 + y^2 = (p^2 - 1)^2 + (2p)^2 = (p^2 + 1)^2\\). Door ons \\(n\\)-tal met \\(x\\) te vermenigvuldigen kunnen hiermee nu twee getallen invoegen: \n\n\\[(x a_1, x a_2, \\dots, x a_n, y a_n, y a_1).\\]\n\nWe controleren eenvoudig dat \n\n\\[ \n\\begin{align*} (x a_i)^2 + (x a_{i+1})^2 &= x^2 (a_i^2 + a_{i+1}^2), \\\\ (x a_n)^2 + (y a_n)^2 &= (x^2 + y^2) a_n, \\\\ (y a_n)^2 + (y a_1)^2 &= y^2 (a_n^2 + a_1^2), \\\\ (y a_1)^2 + (x a_1)^2 &= (y^2 + x^2) a_1^2, \\end{align*} \n\\]\n\ninderdaad allemaal kwadraten zijn wegens de inductiehypothese en de definitie van \\(x\\) en \\(y\\). Omdat de getallen \\(a_1, a_2, \\dots, a_n\\) allemaal verschillend zijn, zijn de getallen \\(x a_1, x a_2, \\dots, x a_n\\) ook allemaal verschillend. De getallen \\(y a_n\\) en \\(y a_1\\) zijn ook verschillend van elkaar. En omdat \\(y\\) deelbaar is door \\(p\\), maar \\(x\\) en \\(a_i\\) niet, kunnen \\(y a_n\\) of \\(y a_1\\) niet gelijk zijn aan een van de \\(x a_i\\). We concluderen dat \\(x a_1, x a_2, \\dots, x a_n, y a_n, y a_1\\) allemaal verschillend zijn, dus \\(n + 2\\) is polypythagorees. \n\nAangezien onze inductiebasis bestond uit \\(n = 2, 3\\) bewijst dit met stappen van twee dat alle getallen polypythagorees zijn. \\(\\square\\)", "metadata": {"resource_path": "Dutch_TST/segmented/nl-2025-E2025_uitwerkingen.jsonl", "problem_match": "\nOpgave 2.", "solution_match": "\nOplossing."}}
{"year": "2025", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "Dutch_TST", "problem": "Bepaal alle drietallen \\((x, y, p)\\) van positieve gehele getallen zo dat \\(p\\) een\npriemgetal is, \\(x^2 = p - 1\\) en \\(y^2 = 2p^2 - 1\\).", "solution": "Het enige drietal dat voldoet is \\((2, 7, 5)\\). \n\nWe rekenen eerst uit dat \n\n\\[ (y + x)(y - x) = y^2 - x^2 = (2p^2 - 1) - (p - 1) = 2p^2 - p = p(2p - 1). \\quad (1) \\]\n\nDat betekent in het bijzonder dat \\(p \\mid x + y\\) of \\(p \\mid x - y\\). \n\nStel dat \\(p \\mid y + x\\). Dan geldt dat \\(y = kp - x\\) voor een zekere \\(k \\in \\mathbb{Z}\\). Uit het gegeven volgt echter dat \\(2p > y > p\\) en \\(x < p\\). Dus \\(2p > y = kp - x > kp - p = (k - 1)p\\) en \\(kp > kp - x = y > p\\), waaruit we concluderen dat \\(k = 2\\). Dus \\(y = 2p - x\\). Als we dat invullen in (1) krijgen we \n\n\\[ 2p(2p - x - x) = p(2p - 1). \\]\n\nAls we \\(p\\) uitdelen, dan houden we over dat \\(4(p - x) = 2p - 1\\). Dit is onmogelijk aangezien de linkerkant even is en de rechterkant oneven. \n\nStel nu dat \\(p \\mid y - x\\). Dan hebben we \\(y = kp + x\\). Uit het gegeven volgt dan \\(2p > y = kp + x > kp\\) en \\((k + 1)p = kp + p > kp + x = y > p\\). Hieruit volgt dat \\(k = 1\\), dus \\(y - x = p\\), en wegens (1) ook dat \\(y + x = 2p - 1\\). Als we dat oplossen vinden we \\(2x = (y + x) - (y - x) = (2p - 1) - p = p - 1\\). We concluderen dat \\(4(p - 1) = 4x^2 = (2x)^2 = (p - 1)^2\\). Als kwadratische vergelijking in \\(p - 1\\) heeft dit de oplossingen \\(p - 1 = 0\\) en \\(p - 1 = 4\\). Alleen in het tweede geval is \\(p\\) een priemgetal, namelijk \\(p = 5\\). Dat geeft verder \\(x = (p-1)/2 = 2\\) en \\(y = p+x = 5+2 = 7\\). Deze voldoet. \\(\\square\\)", "metadata": {"resource_path": "Dutch_TST/segmented/nl-2025-E2025_uitwerkingen.jsonl", "problem_match": "\nOpgave 3.", "solution_match": "\nOplossing."}}
{"year": "2025", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "Dutch_TST", "problem": "We zeggen dat een rij \\(a_1, \\dots, a_n\\) van reële getallen afnemend stijgend is als voor alle \\(1 < i < n\\) geldt dat \\(0 < a_{i+1} - a_i < a_i - a_{i-1}\\). Vind voor elk positief geheel getal \\(m\\) het kleinste positieve gehele getal \\(k\\) waarvoor er een afnemend stijgende rij bestaat van lengte \\(k\\) zo dat 1 op zijn minst op \\(m\\) verschillende manieren geschreven kan worden als het verschil van twee elementen \\(a_i\\) en \\(a_j\\) uit de rij.", "solution": "We bewijzen eerst dat \\(k \\ge 2m\\). We definiëren \\(b_i = a_{i+1} - a_i\\). Dan is \\(b_1, b_2, \\dots, b_{k-1}\\) een dalende rij positieve reële getallen. En elk van de manieren om 1 te schrijven is in deze schrijfwijze een som van opvolgende elementen in deze rij \\(b_j + b_{j+1} + \\dots + b_{j+t-1} = a_{j+t} - a_j = 1\\) met \\(1 \\le j\\), \\(1 \\le t\\) en \\(j + t \\le k\\). Geen twee van deze manieren kunnen op dezelfde plek beginnen, want hun verschil is dan de som van de laatste paar \\(b_i\\) van de langere reeks, in tegenspraak met dat het verschil \\(1 - 1 = 0\\) moet zijn. Ook geen twee van deze manieren kunnen even lang zijn. Sterker nog als we de manieren ordenen op \\(1 \\le j_1 < j_2 < \\dots < j_m\\), dan geldt \\(1 \\le t_1 < t_2 < \\dots < t_m\\) want de termen in de reeks beginnend met \\(j_h\\) zijn stuk voor stuk groter dan de termen van de manier beginnend met \\(j_{h+1}\\). Dit betekent dat \n\n\\[k \\ge j_m + t_m \\ge m + m = 2m.\\]\n\nNu construeren we een voorbeeld voor \\(k = 2m\\). We definiëren de laatste \\(m\\) termen van de rij \\(b_1, \\dots, b_{2m-1}\\) als \\(b_{m+i} = \\frac{2(m-1)-i}{\\frac{3}{2}m(m-1)}\\) voor \\(0 \\le i \\le m-1\\). Dan is dit duidelijk een dalende rij en geldt er dat \n\n\\[b_m + b_{m+1} + \\dots + b_{2m-1} = \\frac{1}{\\frac{3}{2}m(m-1)} \\left((2m-2) + (2m-3) + \\dots + m + (m-1)\\right) = 1.\\]\n\nNu definiëren recursief \\(b_{m-1}, \\dots, b_1\\) als \\(b_i = b_{2i} + b_{2i+1}\\). Dan vinden we in het bijzonder dat\n\\[b_{m-1} = b_{m+(m-2)} + b_{m+(m-1)} = \\frac{m}{\\frac{3}{2}m(m-1)} + \\frac{m-1}{\\frac{3}{2}m(m-1)} = \\frac{2m-1}{\\frac{3}{2}m(m-1)} > \\frac{2m-2}{\\frac{3}{2}m(m-1)} = b_m.\\]\n\nEn voor \\(i < m-1\\) geldt recursief dat \\(b_i > b_{i+1}\\), omdat deze uitdrukking equivalent is met \\(b_{2i} + b_{2i+1} > b_{2i+2} + b_{2i+3}\\). We concluderen we dat \\(b_1, \\dots, b_{2m-1}\\) een dalende rij is. We bewijzen nu met inductie naar \\(j\\) (aflopend) dat \n\n\\[b_j + b_{j+1} + \\dots + b_{2j-1} = 1\\]\n\nvoor alle \\(1 \\le j \\le m\\). Voor de inductiebasis nemen we \\(b_m + b_{m+1} + \\dots + b_{2m-1} = 1\\). Voor de inductiestap merken we op dat \\(b_{j-1} + b_j + \\dots + b_{2j-3} = b_j + b_{j+1} + \\dots + b_{2j-1}\\). Dus als de formule geldt voor \\(j\\) dan geldt die ook voor \\(j-1\\). \\(\\square\\)", "metadata": {"resource_path": "Dutch_TST/segmented/nl-2025-E2025_uitwerkingen.jsonl", "problem_match": "\nOpgave 4.", "solution_match": "\nOplossing I."}}
diff --git a/EGMO/md/en-2025-solutions.md b/EGMO/md/en-2025-solutions.md
index a85870fdb26a1710e3f0f4105885f732bb437c8e..22f05f3589cee18e20288b56562c83c8465762bb 100644
--- a/EGMO/md/en-2025-solutions.md
+++ b/EGMO/md/en-2025-solutions.md
@@ -177,7 +177,7 @@ and so points \(C^{\prime}\) , \(B^{\prime}\) , \(Q\) , \(H\) are concyclic. Ana
which proves that \(P,Q,N,M\) are also concyclic.
-
+
Solution 1'. Introduce points \(B^{\prime}\) , \(C^{\prime}\) as above. Also define \(A^{\prime} = B^{\prime}E\cap C^{\prime}D\) , so that \(E A D A^{\prime}\) is a parallelogram and \(A D E\) is the medial triangle of \(A^{\prime}B^{\prime}C^{\prime}\) . It that follows that the orthocentre \(H\) of \(A D E\) is the circumcentre of \(A^{\prime}B^{\prime}C^{\prime}\) , and in particular
@@ -194,7 +194,7 @@ Lemma. Let \(A^{\prime}\) be the reflection of \(A\) around the orthocentre \(H\
Proof. Since \(OM \parallel AA'\) we have \(\frac{AT}{TM} = \frac{AA'}{OM} = \frac{2AH}{OM} = \frac{4OM}{OM} = 4\) . We used here that \(AH = 2OM\) . \(\square\)
-
+
Solution 2'. As in solution 2, we will prove that \(O'\) is both on line \(HT\) and the radical axis of the circles, hence \(T\) is on the radical axis, from which we conclude the required concyclicity. We present alternative proofs of both facts, discovered by contestants.
@@ -208,7 +208,7 @@ Now, let \(M_{2}\) , \(N_{2}\) be the second intersection points of \(O'M\) , \(
from which it follows that \(M_{2}H \parallel MN\) . Similarly, we have \(N_{2}H \parallel MN\) , and the two facts together imply that \(M_{2}, H, N_{2}\) are collinear and the line through them is parallel to \(MN\) , as claimed. \(\square\)
-
+
Solution 3. We compute using linear combinations with respect to \(A D E\) . We have \(B = 2D - E\) \(C = 2E - D\) \(M = \frac{A + D}{2}\) , and \(N = \frac{A + E}{2}\) , from which we immediately obtain that the intersection of \(T = B M\cap C N\) is \(T = \frac{3A + D + E}{5} = \frac{6M - B}{5} = \frac{6N - C}{5}\) . As in solution 2, we reduce to showing that \(T\) is on the radical axis of \(H D M\) and \(H E N\) , whence \(T M\cdot T P = T N\cdot T Q\) proves the concyclicity of \(P,Q,N,M\)
@@ -241,7 +241,7 @@ We also have the standard \(FD = \sin \gamma \cos \beta\) and \(EF = \sin \beta
\[f(L) = LD\cdot LD^{\prime} - LE\cdot LE^{\prime} = \frac{\sin\alpha}{2} (LD^{\prime} + LE^{\prime}) = \frac{\sin\alpha}{2} (FD^{\prime} + FE^{\prime} - FD - FE)\] \[\qquad = \frac{\sin\alpha}{4\cos\alpha} (\cos \gamma \sin (\beta -\alpha) - \cos \beta \sin (\gamma -\alpha) - 2\cos \alpha (\sin \gamma \cos \beta -\sin \beta \cos \gamma))\] \[\qquad = \frac{3\sin\alpha\sin(\beta - \gamma)}{4} = \frac{3}{4} (\sin (\beta +\gamma)\sin (\beta -\gamma)) = \frac{3}{8} (\cos (2\gamma) - \cos (2\beta))\] \[\qquad = \frac{3}{4} (\sin^{2}(\beta) - \sin^{2}(\gamma)) = -\frac{3}{2} f(A).\quad \square\]
-
+
Solution 4. Let \(T = BM\cap CN\) , let \(F\) be the foot of the altitude from \(A\) to \(BC\) , let \(O\) be the circumcentre of \((ADE)\) , let \(D^{\prime}\neq D\) and \(E^{\prime}\neq E\) be the second intersections of \((DHMP)\) and \((EHNQ)\) with line \(BC\) and let \(U\) and \(V\) be the antipodes of \(D\) and \(E\) on \((DHMP)\) and \((EHNQ)\) , respectively.
@@ -336,7 +336,7 @@ P4. Let \(A B C\) be an acute triangle with incentre \(I\) and \(A B \neq A C\)
Solution 1. We will prove that \(\triangle B I R\sim \triangle C I S\) , since the statement then follows from \(\angle T R I = \angle B R I = \angle C S I =\) \(\angle T S I\) .
-
+
Step 1. Let us prove \(\angle R B I = \angle S C I\) . We will use directed angles:
@@ -394,7 +394,7 @@ Remark. As shown above, \(E\) lies on \(Q P\) and \(A I\perp P Q\) . One can pro
Lemma. Let \(A B C D\) is a cyclic quadrilateral and \(E\) is the intersection of its diagonals. Then the midpoints of \(A B\) , \(B C\) , \(C D\) and the foot of the perpendicular from \(E\) to \(B C\) are concyclic.
-
+
Solution 5. Let \(O\) be the circumcenter of \((A B C)\) . Let \(M\) , \(N\) , and \(L\) be the midpoints of \(O D\) , \(P C\) , and \(Q B\) respectively.
@@ -422,27 +422,27 @@ This means that \(R T I S\) is cyclic with center \(K\) .
Remark. When \(K\) and \(T\) are on the same side of \(R S\) , it can be shown that \(\angle R K S = 2\angle R T S\) .
-
+
Solution 6. As shown above, we have that \(BTIC\) is cyclic. Let \(D\) and \(E\) be the second intersections of \(AC\) and \(AB\) with this circle, respectively. Since the center of this circle lies on \(AI\) (by symmetry about \(AI\) ), we have that \(AB = AD\) and \(AC = AE\) , therefore \(BE = CD\) . Note that since \(C - I - Q\) and \(A - B - E\) are collinear, by Reim's theorem we have that \(AQ \parallel EI\) and since \(AQ \parallel BT\) , we have that \(BT \parallel EI\) . Similarly, we get \(CT \parallel DI\) . Let \(F\) and \(G\) be the intersections of \(ID\) and \(IE\) with \(PS\) and \(QR\) , respectively. Clearly, \(RGEB\) and \(FSCD\) are parallelograms. Since \(RGEB\) is parallelogram and \(BEIT\) is isosceles trapezoid, we have that \(RGIT\) is isosceles trapezoid. Similarly, \(SFIT\) is isosceles trapezoid. Hence, both of them are cyclic. Note also that \(QR = AB = AD = PF\) and \(QG = AE = AC = PS\) . Since \(QR\) and \(PS\) are tangents to the circumcircle of \(\triangle ABC\) we have that \(R\) and \(F\) are symmetric (reflections) about the perpendicular bisector of \(PQ\) . Similarly, \(G\) and \(S\) are symmetric about the perpendicular bisector of \(PQ\) . This gives us that \(QP \parallel RF \parallel GS\) and that \(RFSG\) is an isosceles trapezoid, hence a cyclic quadrilateral with \(\angle RGS = 180^{\circ} - \angle GQP = 180^{\circ} - \angle QAP = 180^{\circ} - \angle RTS \Rightarrow R, G, S, T\) are concyclic. Combining all the facts about the cyclic quadrilaterals we proved above, we have that \(R, G, S, F, I, T\) are concyclic. Therefore \(R, T, I, S\) lie on a circle.
-
+
Solution 7. Let \(E\) be the \(A\) - excenter of \(\triangle ABC\) . Let the midpoints of \(AQ, QB, CP, PA\) be the points \(F, G, H, J\) , respectively. Both \(PD\) and \(EC\) are perpendicular to \(CI\) , hence \(PD \parallel CE\) .
Since \(PA = PC\) we have that \(AJHC\) is an isosceles trapezoid so it is cyclic. Let \(K\) be the second intersection of \((AJHC)\) and \(AI\) . Then \(\angle ADP = \angle ACP = \angle ACH = \angle AKH \Rightarrow DP \parallel KH\) . So \(KH\) is a line passing through the midpoint of the side \(CP\) of trapezoid \(DPCE\) and parallel to the bases, hence \(K\) is the midpoint of \(DE\) . Similarly, we show that the circle \((AFGB)\) passes through the midpoint of \(DE\) . Homothety centered at \(A\) with scale- factor 2 maps \((AJH)\) to \((APS)\) , \((AFG)\) to \((AQR)\) , and line \(AI\) to line \(AI\) . This means that the circles \((AQR)\) and \((APS)\) intersect on \(AI\) , call it point \(L\) .
-
+
Now, \(\angle I L R = 180^{\circ} - \angle A Q R = \angle Q A B = \angle Q C B = 180^{\circ} - \angle I T B = 180^{\circ} - \angle I T R\) , therefore \(R,L,I,T\) are concyclic. Similarly, we get that \(S,L,T,I\) are concyclic. Combining these, it means that \(R\) and \(S\) belong to the circle \((L I T)\) . The conclusion follows.
-
+
@@ -467,7 +467,7 @@ Let \((i,j)\) denote the cell in row \(i\) and column \(j\) . Consider the follo
\[\begin{array}{r l} & {(1,1)\rightarrow (1,2)\rightarrow (1,3)\rightarrow \ldots \rightarrow (1,n)}\\ & {\qquad \rightarrow (2,n)\rightarrow (2,n - 1)\rightarrow \ldots \rightarrow (2,2)}\\ & {\qquad \ldots}\\ & {\qquad \rightarrow (2i - 1,2)\rightarrow (2i - 1,3)\rightarrow \ldots \rightarrow (2i - 1,n)}\\ & {\qquad \rightarrow (2i,n)\rightarrow (2i,n - 1)\rightarrow \ldots \rightarrow (2i,2)}\\ & {\qquad \ldots}\\ & {\qquad \rightarrow (n,n)\rightarrow (n,n - 1)\rightarrow \ldots \rightarrow (n,2)}\\ & {\qquad \rightarrow (n,1)\rightarrow (n - 1,1)\rightarrow \ldots \rightarrow (2,1)\rightarrow (1,1).} \end{array} \quad (1,1)\]
-
+
Note that the cycle returns to the initial cell after visiting every cell exactly once. To prove that \((1,1)\) is good, we need to find a starting configuration such that Turbo traverses this cycle.
@@ -490,12 +490,12 @@ Consider the upper left corner of the board. With standard row and column number
Now suppose instead that \(b_{j + 1} = (2,1) = a_{i - 1}\) . Considering the \(a\) - route, the arrows in the upper left corner after \(i - 1\) steps must be like this:
-
+
Considering the \(b\) - route instead, the arrows after \(j - 1\) steps must be like this:
-
+
From the arrows in cell \((1,1)\) we see that \(i\equiv j + 1\) mod 4. However, for the cells \((2,1)\) and \((1,2)\) this gives a contradiction.
diff --git a/EGMO/segmented/en-2025-solutions.jsonl b/EGMO/segmented/en-2025-solutions.jsonl
index 3b65d8210fe1657f444c50c55d30c3d42ffd8a9b..dd39e7d05ca3cdbeb6065150857232cfc9d9b98c 100644
--- a/EGMO/segmented/en-2025-solutions.jsonl
+++ b/EGMO/segmented/en-2025-solutions.jsonl
@@ -4,22 +4,22 @@
{"year": "2025", "tier": "T2", "problem_label": "2", "problem_type": null, "exam": "EGMO", "problem": "An infinite increasing sequence \\(a_{1}< a_{2}< a_{3}< \\dots\\) of positive integers is called central if for every positive integer \\(n\\) , the arithmetic mean of the first \\(a_{n}\\) terms of the sequence is equal to \\(a_{n}\\) . \n\nShow that there exists an infinite sequence \\(b_{1},b_{2},b_{3},\\ldots\\) of positive integers such that for every central sequence \\(a_{1},a_{2},a_{3},\\ldots\\) , there are infinitely many positive integers \\(n\\) with \\(a_{n} = b_{n}\\) .", "solution": "We give an alternative proof that the sequence \\(b_{i} = 2i - 1\\) works. This proof is by contradiction, so we assume that there are only finitely many \\(a_{i}\\) such that \\(a_{i} = 2i - 1\\) . \n\nLet \\(S(n) = \\sum_{i = 1}^{n}a_{i}\\) . We have \\(S(a_{n}) = a_{n}^{2}\\) and \\(S(a_{n + 1}) = a_{n + 1}^{2}\\) . If \\(a_{n + 1} = a_{n} + 1\\) , then it follows that \n\n\\[S(a_{n + 1}) - S(a_{n}) = a_{n + 1}^{2} - a_{n}^{2} = a_{n + 1}^{2} - (a_{n + 1} - 1)^{2} = 2a_{n + 1} - 1.\\] \n\nOn the other hand, if \\(a_{n + 1} = a_{n} + 1\\) , then \\(S(a_{n + 1}) - S(a_{n})\\) is \\(a_{a_{n + 1}}\\) , so it follows that \\(a_{a_{n + 1}} = 2a_{n + 1} - 1\\) . By assumption, this can only happen finitely many times, so for all sufficiently large \\(n\\) we must have \\(a_{n + 1}\\geqslant a_{n} + 2\\) . \n\nFor large enough \\(n\\) , we now know that \\(a_{n} > 2n - 1\\) implies \\(a_{n + 1} > (2n - 1) + 2 = 2(n + 1) - 1\\) . This means that there are two cases possible: \n\n(A) For all sufficiently large \\(n\\) (say \\(n\\geqslant N_{A}\\) ) we have \\(a_{n} > 2n - 1\\) \n\n(B) For all sufficiently large \\(n\\) (say \\(n\\geqslant N_{B}\\) ) we have \\(a_{n}< 2n - 1\\) \n\nIn case (A), we know for \\(m > N_{A}\\) that \n\n\\[S(m) = S(N_{A}) + \\sum_{i = N_{A} + 1}^{m}a_{i}\\geqslant S(N_{A}) + \\sum_{i = N_{A} + 1}^{n}2i = S(N_{A}) + m(m + 1) - N_{A}(N_{A} + 1)\\] \\[\\qquad = m^{2} + m + S(N_{A}) - N_{A}(N_{A} + 1).\\] \n\nFor \\(m\\) large enough (e.g. \\(m > N_{A}(N_{A} + 1)\\) ), this expression is always larger than \\(m^{2}\\) , contradicting \\(S(a_{n}) = a_{n}^{2}\\) for all \\(n\\) . \n\nSimilarly, in case (B), we similarly know for \\(m > N_{B}\\) that \n\n\\[S(m) = S(N_{B}) + \\sum_{i = N_{B} + 1}^{m}a_{i}\\leqslant S(N_{B}) + \\sum_{i = N_{B} + 1}^{n}2(i - 1) = S(N_{B}) + m(m - 1) - N_{B}(N_{B} - 1)\\] \\[\\qquad = m^{2} - m + S(N_{B}) - N_{B}(N_{B} - 1).\\] \n\nFor \\(m\\) large enough (e.g. \\(m > S(N_{B})\\) ), this expression is always smaller than \\(m^{2}\\) , again contradicting \\(S(a_{n}) = a_{n}^{2}\\) for all \\(n\\) .", "metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P2.", "solution_match": "\nSolution 2. "}}
{"year": "2025", "tier": "T2", "problem_label": "2", "problem_type": null, "exam": "EGMO", "problem": "An infinite increasing sequence \\(a_{1}< a_{2}< a_{3}< \\dots\\) of positive integers is called central if for every positive integer \\(n\\) , the arithmetic mean of the first \\(a_{n}\\) terms of the sequence is equal to \\(a_{n}\\) . \n\nShow that there exists an infinite sequence \\(b_{1},b_{2},b_{3},\\ldots\\) of positive integers such that for every central sequence \\(a_{1},a_{2},a_{3},\\ldots\\) , there are infinitely many positive integers \\(n\\) with \\(a_{n} = b_{n}\\) .", "solution": "We claim that the sequence \\(b_{1}\\) , \\(b_{2}\\) , \\(b_{3}\\) , ... defined by \\(b_{i} = 2i - 1\\) has this property. \n\nLemma. If there are no terms \\(a_{j}\\) such that \\(a_{j} - a_{j - 1} = 1\\) , then \\(a_{j} = a_{j - 1} + 2\\) for all \\(j\\) . \n\nProof. Let \\(c\\) be such that \\(a_{d} = c\\) for some \\(d\\) . Now \n\n\\[a_{1} + a_{2} + \\dots +a_{c} = c^{2}.\\] \n\nEquality holds for \\(a_{i} = 2i - 1\\) for \\(1\\leq i\\leq c\\) , so if any difference between two consecutive terms is greater, the left- hand side of the equation is greater than \\(c^{2}\\) , a contradiction. \\(\\square\\) \n\nLemma. If both \\(d\\) and \\(d + 1\\) are terms of the sequence, i.e. \\(a_{c} = d\\) and \\(a_{c + 1} = d + 1\\) for some \\(c\\) , then \\(a_{d + 1} = 2d + 1 = b_{d + 1}\\) . \n\nProof. We have \\(a_{1} + a_{2} + \\dots +a_{d} = d^{2}\\) and \\(a_{1} + a_{2} + \\dots +a_{d + 1} = (d + 1)^{2}\\) . Hence \\(a_{d + 1} = (d + 1)^{2} - d^{2} = 2d + 1\\) . \\(\\square\\) \n\nFrom the observations above, we see that we are done if there are infinitely many gaps of size 1. The only remaining case is one with finitely many gaps of size 1. This will be the subject of the following lemma. \n\nLemma. If there are only finitely many indices \\(j\\) such that \\(a_{j + 1} - a_{j} = 1\\) , then there is an index \\(n_{0}\\) such that for all \\(k > n_{0}\\) , we have \\(a_{k} = 2k - 1\\) . \n\nProof. Let \\(r\\) and \\(s\\) be indices such that for all the \\(j\\) satisfying \\(a_{j + 1} - a_{j} = 1\\) , we have \\(j < r\\) , \\(s\\) . Furthermore, assume \\(s > r\\) and that there are \\(i_{1}\\) and \\(i_{2}\\) such that \\(a_{i_{1}} = r\\) and \\(a_{i_{2}} = s\\) . The first goal is to show that \\(a_{s} \\geq 2s - 1\\) . If \\(a_{r} \\geq 2r - 1\\) , this is clearly the case. Assume now \\(a_{r} < 2r - 1\\) . Now \\(a_{r} \\geq 2r - 1 - m\\) , where \\(m\\) is the number of indices \\(j\\) with \\(a_{j + 1} - a_{j} = 1\\) . Denote \\(a_{r + 1} = 2r + 1 - m + \\theta_{1}\\) , \\(a_{r + 2} = 2r + 3 - m + \\theta_{2}\\) , etc. Remember that \\(a_{r + j + 1} - a_{r + j} \\geq 2\\) always. Now \\(0 \\leq \\theta_{1} \\leq \\theta_{2} \\leq \\dots\\) . Furthermore, write \\(s = r + h\\) . Now \n\n\\[(r + h)^{2} - r^{2} = a_{r + 1} + a_{r + 2} + \\dots +a_{r + h} = \\sum_{j = 1}^{h}2r - 1 + 2j - m + \\theta_{j}.\\] \n\nFrom this we deduce \n\n\\[2r h + h^{2} = 2r h - h - m h + h(h + 1) + \\sum_{j = 1}^{h}\\theta_{j}.\\] \n\nSo we obtain \\(\\sum_{j = 1}^{h}\\theta_{j} = m h\\) . Since the sequence \\(\\theta_{j}\\) is increasing, we have \\(\\theta_{h} \\geq m\\) . Hence, \\(a_{s} = a_{r + h} \\geq 2r - 1 - m + 2h + m = 2r + 2h - 1 = 2s - 1\\) . \n\nNow \\(a_{s}\\) is exactly the desired shape. If for any \\(t > s\\) , we have \\(a_{t} - a_{t - 1} > 2\\) , then \n\n\\[a_{s} + a_{s + 1} + \\dots +a_{t} > t^{2} - s^{2},\\] \n\nagain a contradiction.", "metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P2.", "solution_match": "\nSolution 3. "}}
{"year": "2025", "tier": "T2", "problem_label": "2", "problem_type": null, "exam": "EGMO", "problem": "An infinite increasing sequence \\(a_{1}< a_{2}< a_{3}< \\dots\\) of positive integers is called central if for every positive integer \\(n\\) , the arithmetic mean of the first \\(a_{n}\\) terms of the sequence is equal to \\(a_{n}\\) . \n\nShow that there exists an infinite sequence \\(b_{1},b_{2},b_{3},\\ldots\\) of positive integers such that for every central sequence \\(a_{1},a_{2},a_{3},\\ldots\\) , there are infinitely many positive integers \\(n\\) with \\(a_{n} = b_{n}\\) .", "solution": "Note that \\(a_{1} = 1\\) because if it is not the case, then \\(a_{1}^{2} = a_{1} + \\dots +a_{a_{1}} > a_{1} + a_{1} + \\dots +a_{1} = a_{1}^{2}\\) . \n\nAssume by contradiction that there are only finitely many indices \\(k\\) such that \\(a_{k} = 2k - 1\\) . Set \\(i\\) to be the largest integer such that \\(a_{i} = 2i - 1\\) (which must exist as \\(a_{1} = 1\\) ). Assume that there exists \\(j \\geqslant i\\) such that \\(a_{j + 1} - a_{j} = 1\\) . Then \\(2a_{j + 1} - 1 = a_{j + 1}^{2} - a_{j}^{2} = a_{a_{j + 1}}\\) and since \\(a_{k} \\geqslant k\\) for all \\(k\\) , we have \\(a_{j + 1} \\geqslant j + 1 > i\\) , which contradicts the definition of \\(i\\) . Thus for all \\(j \\geqslant i\\) , we have \\(a_{j + 1} \\geqslant a_{j} + 2\\) , which implies by induction that \\(a_{j} \\geqslant 2j - 1\\) for \\(j \\geqslant i\\) , and even \\(a_{j} \\geq 2j\\) if \\(j > i\\) . \n\nThere are two ways to finish the solution from here. \n\n## First way to finish the solution \n\nFor all \\(n\\) such that \\(a_{n} \\geqslant i\\) , we have \n\n\\[a_{n + 1}^{2} - a_{n}^{2} = a_{a_{n + 1}} + a_{a_{n + 1} - 1} + \\dots +a_{a_{n + 1}}\\geqslant 2a_{n + 1} + 2(a_{n + 1} - 1) + \\dots +2(a_{n} + 1)\\] \\[\\qquad = (a_{n + 1} - a_{n})(a_{n + 1} + a_{n} + 1)\\] \\[\\qquad >a_{n + 1}^{2} - a_{n}^{2}.\\] \n\nThis gives a contradiction. \n\n## Second way to finish the solution \n\nFor all \\(n\\) such that \\(a_{n} \\geqslant i\\) , we introduce \\(x_{n} = a_{n + 1} - a_{n}\\) . We have \n\n\\[x_{n}^{2} + 2x_{n}a_{n} = a_{n + 1}^{2} - a_{n}^{2} = a_{a_{n + 1}} + a_{a_{n + 1} - 1} + \\dots +a_{a_{n} + 1}\\geqslant \\sum_{j = 1}^{x_{n}}(a_{a_{n}} + 2j)\\geq x_{n}a_{a_{n}} + x_{n}(x_{n} + 1).\\] \n\nBy simplifying, we get \\(a_{a_{n}} \\leq 2a_{n} - 1\\) , which gives a contradiction. \n\nComment. Proving that \\(a_{1} = 1\\) is not necessary for this solution. If there exists no \\(i\\) such that \\(a_{i} = 2i - 1\\) , then the same argument implies that there exists no \\(j\\) such that \\(a_{j + 1} - a_{j} = 1\\) , thus \\(a_{j} \\geqslant 2j - 1\\) for \\(j \\geqslant 1\\) .", "metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P2.", "solution_match": "\nSolution 4. "}}
-{"year": "2025", "tier": "T2", "problem_label": "3", "problem_type": null, "exam": "EGMO", "problem": "Let \\(A B C\\) be an acute triangle. Points \\(B\\) , \\(D\\) , \\(E\\) , and \\(C\\) lie on a line in this order and satisfy \\(B D =\\) \\(D E = E C\\) . Let \\(M\\) and \\(N\\) be the midpoints of \\(A D\\) and \\(A E\\) , respectively. Let \\(H\\) be the orthocentre of triangle \\(A D E\\) . Let \\(P\\) and \\(Q\\) be points on lines \\(B M\\) and \\(C N\\) , respectively, such that \\(D\\) , \\(H\\) , \\(M\\) , and \\(P\\) are concyclic and \\(E\\) , \\(H\\) , \\(N\\) , and \\(Q\\) are concyclic. Prove that \\(P\\) , \\(Q\\) , \\(N\\) , and \\(M\\) are concyclic. \n\nThe orthocentre of a triangle is the point of intersection of its altitudes.", "solution": "Denote by \\(B^{\\prime}\\) and \\(C^{\\prime}\\) the reflections of \\(B\\) and \\(C\\) in \\(M\\) and \\(N\\) , respectively. Points \\(C^{\\prime}\\) , \\(A\\) , \\(B^{\\prime}\\) are clearly collinear and \\(D E B^{\\prime}A\\) is a parallelogram. Since \\(E H\\perp A D\\) , we have \\(E H\\perp E B^{\\prime}\\) . Also \\(H A\\perp A B^{\\prime}\\) , so points \\(H,E,B^{\\prime},A\\) are concyclic. This gives \n\n\\[\\angle C^{\\prime}Q H = \\angle N Q H = \\angle N E H = \\angle A E H = \\angle A B^{\\prime}H = \\angle C^{\\prime}B^{\\prime}H,\\] \n\nand so points \\(C^{\\prime}\\) , \\(B^{\\prime}\\) , \\(Q\\) , \\(H\\) are concyclic. Analogously points \\(C^{\\prime}\\) , \\(B^{\\prime}\\) , \\(P\\) , \\(H\\) are concyclic, and so all points \\(B^{\\prime}\\) , \\(C^{\\prime}\\) , \\(P\\) , \\(Q\\) , \\(H\\) are. Now we have \n\n\\[\\angle N M B^{\\prime} = \\angle A B^{\\prime}M = \\angle C^{\\prime}B^{\\prime}P = \\angle C^{\\prime}Q P = \\angle N Q P,\\] \n\nwhich proves that \\(P,Q,N,M\\) are also concyclic. \n\n", "metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P3.", "solution_match": "\nSolution 1. "}}
+{"year": "2025", "tier": "T2", "problem_label": "3", "problem_type": null, "exam": "EGMO", "problem": "Let \\(A B C\\) be an acute triangle. Points \\(B\\) , \\(D\\) , \\(E\\) , and \\(C\\) lie on a line in this order and satisfy \\(B D =\\) \\(D E = E C\\) . Let \\(M\\) and \\(N\\) be the midpoints of \\(A D\\) and \\(A E\\) , respectively. Let \\(H\\) be the orthocentre of triangle \\(A D E\\) . Let \\(P\\) and \\(Q\\) be points on lines \\(B M\\) and \\(C N\\) , respectively, such that \\(D\\) , \\(H\\) , \\(M\\) , and \\(P\\) are concyclic and \\(E\\) , \\(H\\) , \\(N\\) , and \\(Q\\) are concyclic. Prove that \\(P\\) , \\(Q\\) , \\(N\\) , and \\(M\\) are concyclic. \n\nThe orthocentre of a triangle is the point of intersection of its altitudes.", "solution": "Denote by \\(B^{\\prime}\\) and \\(C^{\\prime}\\) the reflections of \\(B\\) and \\(C\\) in \\(M\\) and \\(N\\) , respectively. Points \\(C^{\\prime}\\) , \\(A\\) , \\(B^{\\prime}\\) are clearly collinear and \\(D E B^{\\prime}A\\) is a parallelogram. Since \\(E H\\perp A D\\) , we have \\(E H\\perp E B^{\\prime}\\) . Also \\(H A\\perp A B^{\\prime}\\) , so points \\(H,E,B^{\\prime},A\\) are concyclic. This gives \n\n\\[\\angle C^{\\prime}Q H = \\angle N Q H = \\angle N E H = \\angle A E H = \\angle A B^{\\prime}H = \\angle C^{\\prime}B^{\\prime}H,\\] \n\nand so points \\(C^{\\prime}\\) , \\(B^{\\prime}\\) , \\(Q\\) , \\(H\\) are concyclic. Analogously points \\(C^{\\prime}\\) , \\(B^{\\prime}\\) , \\(P\\) , \\(H\\) are concyclic, and so all points \\(B^{\\prime}\\) , \\(C^{\\prime}\\) , \\(P\\) , \\(Q\\) , \\(H\\) are. Now we have \n\n\\[\\angle N M B^{\\prime} = \\angle A B^{\\prime}M = \\angle C^{\\prime}B^{\\prime}P = \\angle C^{\\prime}Q P = \\angle N Q P,\\] \n\nwhich proves that \\(P,Q,N,M\\) are also concyclic. \n\n", "metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P3.", "solution_match": "\nSolution 1. "}}
{"year": "2025", "tier": "T2", "problem_label": "3", "problem_type": null, "exam": "EGMO", "problem": "Let \\(A B C\\) be an acute triangle. Points \\(B\\) , \\(D\\) , \\(E\\) , and \\(C\\) lie on a line in this order and satisfy \\(B D =\\) \\(D E = E C\\) . Let \\(M\\) and \\(N\\) be the midpoints of \\(A D\\) and \\(A E\\) , respectively. Let \\(H\\) be the orthocentre of triangle \\(A D E\\) . Let \\(P\\) and \\(Q\\) be points on lines \\(B M\\) and \\(C N\\) , respectively, such that \\(D\\) , \\(H\\) , \\(M\\) , and \\(P\\) are concyclic and \\(E\\) , \\(H\\) , \\(N\\) , and \\(Q\\) are concyclic. Prove that \\(P\\) , \\(Q\\) , \\(N\\) , and \\(M\\) are concyclic. \n\nThe orthocentre of a triangle is the point of intersection of its altitudes.", "solution": "Introduce points \\(B^{\\prime}\\) , \\(C^{\\prime}\\) as above. Also define \\(A^{\\prime} = B^{\\prime}E\\cap C^{\\prime}D\\) , so that \\(E A D A^{\\prime}\\) is a parallelogram and \\(A D E\\) is the medial triangle of \\(A^{\\prime}B^{\\prime}C^{\\prime}\\) . It that follows that the orthocentre \\(H\\) of \\(A D E\\) is the circumcentre of \\(A^{\\prime}B^{\\prime}C^{\\prime}\\) , and in particular \n\n\\[\\angle C^{\\prime}B^{\\prime}H = 90^{\\circ} - \\angle B^{\\prime}A^{\\prime}C^{\\prime} = 90^{\\circ} - \\angle D A E = \\angle A E H = \\angle N E H = \\angle N Q H = \\angle C^{\\prime}Q H.\\] \n\nSo again we have that \\(C^{\\prime}\\) , \\(B^{\\prime}\\) , \\(Q\\) , \\(H\\) are concyclic and conclude as in Solution 1.", "metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P3.", "solution_match": "\nSolution 1'."}}
-{"year": "2025", "tier": "T2", "problem_label": "3", "problem_type": null, "exam": "EGMO", "problem": "Let \\(A B C\\) be an acute triangle. Points \\(B\\) , \\(D\\) , \\(E\\) , and \\(C\\) lie on a line in this order and satisfy \\(B D =\\) \\(D E = E C\\) . Let \\(M\\) and \\(N\\) be the midpoints of \\(A D\\) and \\(A E\\) , respectively. Let \\(H\\) be the orthocentre of triangle \\(A D E\\) . Let \\(P\\) and \\(Q\\) be points on lines \\(B M\\) and \\(C N\\) , respectively, such that \\(D\\) , \\(H\\) , \\(M\\) , and \\(P\\) are concyclic and \\(E\\) , \\(H\\) , \\(N\\) , and \\(Q\\) are concyclic. Prove that \\(P\\) , \\(Q\\) , \\(N\\) , and \\(M\\) are concyclic. \n\nThe orthocentre of a triangle is the point of intersection of its altitudes.", "solution": "Let \\(X\\) be the second intersection of \\((D H M)\\) and \\((E H N)\\) and let \\(O^{\\prime}\\) be the circumcentre of \\((A M N)\\) Note that \\(\\angle M X N = \\angle M D H + \\angle N E H = 180^{\\circ} - 2\\angle D A E\\) and since \\(\\angle M O^{\\prime}N = 2\\angle D A E\\) we have that \\(X,M,O^{\\prime},N\\) is cyclic and since \\(\\angle M X H = \\angle N X H\\) it means that \\(H X\\) is the angle bisector of \\(\\angle M X N\\) but since \\(O^{\\prime}M = O^{\\prime}N\\) it means that \\(H,X,O^{\\prime}\\) are collinear. Let \\(B M\\) and \\(C N\\) intersect at \\(T\\) and let \\(K\\) and \\(L\\) be the midpoints of \\(M N\\) and \\(B C\\) . Note that \\(L\\) is also the midpoint of \\(D E\\) . Since \\(M N\\) is parallel to \\(B C\\) it means that \\(T\\) , \\(K\\) , and \\(L\\) are collinear, but since \\(A\\) , \\(K\\) , and \\(L\\) are collinear we get that \\(A\\) , \\(T\\) , \\(K\\) , and \\(L\\) are collinear. Now, \\(\\frac{T L}{T K} = \\frac{B C}{M N} = 6\\) . Since \\(K L = K A\\) it means that \\(\\frac{A T}{T K} = \\frac{T L - 2T K}{T K} = 4\\) so by the lemma below, \\(T\\) lies on \\(H O^{\\prime}\\) . Since \\(H O^{\\prime}\\) is the radical axis of \\((D H M)\\) and \\((E H N)\\) we finish the problem using the Radical Axes Theorem \\((T M\\cdot T P = T N\\cdot T Q)\\) .\n\n\n\nLemma. Let \\(A^{\\prime}\\) be the reflection of \\(A\\) around the orthocentre \\(H\\) of \\(\\triangle ABC\\) and \\(O\\) and \\(M\\) be the circumcentre of \\(\\triangle ABC\\) and the midpoint of \\(BC\\) , respectively. Let \\(T\\) be the intersection of \\(A^{\\prime}O\\) and \\(AM\\) . Then \\(\\frac{AT}{TM} = 4\\) . \n\nProof. Since \\(OM \\parallel AA'\\) we have \\(\\frac{AT}{TM} = \\frac{AA'}{OM} = \\frac{2AH}{OM} = \\frac{4OM}{OM} = 4\\) . We used here that \\(AH = 2OM\\) . \\(\\square\\) \n\n", "metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P3.", "solution_match": "\nSolution 2. "}}
-{"year": "2025", "tier": "T2", "problem_label": "3", "problem_type": null, "exam": "EGMO", "problem": "Let \\(A B C\\) be an acute triangle. Points \\(B\\) , \\(D\\) , \\(E\\) , and \\(C\\) lie on a line in this order and satisfy \\(B D =\\) \\(D E = E C\\) . Let \\(M\\) and \\(N\\) be the midpoints of \\(A D\\) and \\(A E\\) , respectively. Let \\(H\\) be the orthocentre of triangle \\(A D E\\) . Let \\(P\\) and \\(Q\\) be points on lines \\(B M\\) and \\(C N\\) , respectively, such that \\(D\\) , \\(H\\) , \\(M\\) , and \\(P\\) are concyclic and \\(E\\) , \\(H\\) , \\(N\\) , and \\(Q\\) are concyclic. Prove that \\(P\\) , \\(Q\\) , \\(N\\) , and \\(M\\) are concyclic. \n\nThe orthocentre of a triangle is the point of intersection of its altitudes.", "solution": "As in solution 2, we will prove that \\(O'\\) is both on line \\(HT\\) and the radical axis of the circles, hence \\(T\\) is on the radical axis, from which we conclude the required concyclicity. We present alternative proofs of both facts, discovered by contestants. \n\nLet \\(M_{1}\\) , \\(N_{1}\\) be the midpoints of \\(AM\\) , \\(AN\\) , respectively, so that \\(AM_{1}:M_{1}D = AN_{1}:N_{1}E = 1:3\\) . It is easy to verify (e.g. by applying the theorems of Ceva or Menelaus, or by computing in barycentric coordinates as in Solution 3) that \\(T\\) lies on \\(EM_{1}\\) and \\(DN_{1}\\) . Note that \\(M_{1}N_{1} \\parallel DE\\) , and also \\(M_{1}O' \\parallel HE\\) as both are perpendicular to \\(AMD\\) , and similarly \\(N_{1}O' \\parallel HD\\) . It follows that \\(DEH\\) and \\(N_{1}M_{1}O'\\) are homothetically similar triangles, and the center of their (negative) homothety is \\(T = DN_{1} \\cap EM_{1}\\) . Therefore \\(T\\) also lies on \\(HO'\\) , as claimed. (We also have that the homothety is by factor \\(\\frac{M_{1}N_{1}}{ED} = -\\frac{1}{4}\\) .) \n\nNow, let \\(M_{2}\\) , \\(N_{2}\\) be the second intersection points of \\(O'M\\) , \\(O'N\\) with the circumcircles of \\(HMD\\) , \\(HNE\\) , respectively. To prove that \\(O'\\) is on the radical axis, it suffices to show that \\(O'M \\cdot O'M_{2} = O'N \\cdot O'N_{2}\\) . But by definition of \\(O'\\) we have \\(O'M = O'N\\) , so we must show \\(O'M_{2} = O'N_{2}\\) , which is equivalent to \\(M_{2}N_{2} \\parallel MN\\) . Angle chasing in circle \\(MM_{2}DH\\) gives \n\n\\[\\angle OM_{2}H = \\angle MM_{2}H = \\angle MDH = \\angle ADH = 90^{\\circ} - \\angle EAD = 90^{\\circ} - \\angle NAM = \\angle O'M N,\\] \n\nfrom which it follows that \\(M_{2}H \\parallel MN\\) . Similarly, we have \\(N_{2}H \\parallel MN\\) , and the two facts together imply that \\(M_{2}, H, N_{2}\\) are collinear and the line through them is parallel to \\(MN\\) , as claimed. \\(\\square\\)\n\n\n", "metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P3.", "solution_match": "\nSolution 2'."}}
-{"year": "2025", "tier": "T2", "problem_label": "3", "problem_type": null, "exam": "EGMO", "problem": "Let \\(A B C\\) be an acute triangle. Points \\(B\\) , \\(D\\) , \\(E\\) , and \\(C\\) lie on a line in this order and satisfy \\(B D =\\) \\(D E = E C\\) . Let \\(M\\) and \\(N\\) be the midpoints of \\(A D\\) and \\(A E\\) , respectively. Let \\(H\\) be the orthocentre of triangle \\(A D E\\) . Let \\(P\\) and \\(Q\\) be points on lines \\(B M\\) and \\(C N\\) , respectively, such that \\(D\\) , \\(H\\) , \\(M\\) , and \\(P\\) are concyclic and \\(E\\) , \\(H\\) , \\(N\\) , and \\(Q\\) are concyclic. Prove that \\(P\\) , \\(Q\\) , \\(N\\) , and \\(M\\) are concyclic. \n\nThe orthocentre of a triangle is the point of intersection of its altitudes.", "solution": "We compute using linear combinations with respect to \\(A D E\\) . We have \\(B = 2D - E\\) \\(C = 2E - D\\) \\(M = \\frac{A + D}{2}\\) , and \\(N = \\frac{A + E}{2}\\) , from which we immediately obtain that the intersection of \\(T = B M\\cap C N\\) is \\(T = \\frac{3A + D + E}{5} = \\frac{6M - B}{5} = \\frac{6N - C}{5}\\) . As in solution 2, we reduce to showing that \\(T\\) is on the radical axis of \\(H D M\\) and \\(H E N\\) , whence \\(T M\\cdot T P = T N\\cdot T Q\\) proves the concyclicity of \\(P,Q,N,M\\) \n\nSynthetic finish, similar to Solution 2. Let \\(O\\) be the circumcentre of \\(A D E\\) and \\(O^{\\prime} = \\frac{A + O}{2}\\) be the circumcentre of \\(A M N\\) . As in Solution 2, we have that \\(O^{\\prime}\\) is on the desired radical axis, so it is enough to show \\(T\\in O^{\\prime}H\\) . Let \\(G = \\frac{A + D + E}{3}\\) be the barycentre of \\(A D E\\) . By properties of the Euler line, we also have \\(G = \\frac{H + 2O}{3}\\) . Now using our known identities we find \n\n\\[T = \\frac{3A + D + E}{5} = \\frac{2A + 3G}{5} = \\frac{2A + H + 2O}{5} = \\frac{H + 4O^{\\prime}}{5}\\] \n\nand in particular \\(T\\in H O^{\\prime}\\) , as we wanted to show. \n\nComputational finish. Let \\(f(T)\\) be the power difference at \\(T\\) w.r.t. \\(D H M\\) and \\(E H N\\) . We compute \\(f(A)\\) and \\(f(L)\\) where \\(L = \\frac{D + E}{2}\\) . Since \\(T = \\frac{3A + 2L}{5}\\) , it is enough to show that \\(3P(A) + 2P(L) = 0\\) . In the following all lengths are directed. We compute trigonometrically: Let \\(\\alpha ,\\beta ,\\gamma\\) be the angles of \\(A D E\\) and assume the diameter of its circumcircle is 1. We have \n\n\\[f(A) = \\frac{A D^{2} - A E^{2}}{2} = \\frac{\\sin^{2}(\\gamma) - \\sin^{2}(\\beta)}{2}.\\] \n\nTo compute \\(P(L)\\) , let \\(D^{\\prime},E^{\\prime}\\) be the second intersection points of \\(H M D,H N E\\) with \\(D E\\) , and let \\(M^{\\prime},N^{\\prime},F\\) be the feet of the perpendiculars from \\(H\\) to \\(A D,A E,D E\\) , respectively. Note that \\(D M^{\\prime} =\\) \\(\\sin \\alpha \\cos \\beta\\) , thus \n\n\\[M^{\\prime}M = D M - D M^{\\prime} = \\frac{\\sin(\\alpha + \\beta)}{2} - \\sin \\alpha \\cos \\beta = \\frac{\\sin(\\beta - \\alpha)}{2}.\\] \n\nWe also have \\(H M^{\\prime} = \\cos \\alpha \\cos \\beta\\) , \\(H F = \\cos \\beta \\cos \\gamma\\) , and \\(H M^{\\prime}M\\sim H F D^{\\prime}\\) , therefore \n\n\\[F D^{\\prime} = \\frac{H F}{H M^{\\prime}} M^{\\prime}M = \\frac{\\cos\\gamma\\sin(\\beta - \\alpha)}{\\cos\\alpha}\\] \n\nand similarly \n\n\\[E^{\\prime}F = \\frac{\\cos\\beta\\sin(\\gamma - \\alpha)}{\\cos\\alpha}.\\]\n\n\n\nWe also have the standard \\(FD = \\sin \\gamma \\cos \\beta\\) and \\(EF = \\sin \\beta \\cos \\gamma\\) . We can finally compute \n\n\\[f(L) = LD\\cdot LD^{\\prime} - LE\\cdot LE^{\\prime} = \\frac{\\sin\\alpha}{2} (LD^{\\prime} + LE^{\\prime}) = \\frac{\\sin\\alpha}{2} (FD^{\\prime} + FE^{\\prime} - FD - FE)\\] \\[\\qquad = \\frac{\\sin\\alpha}{4\\cos\\alpha} (\\cos \\gamma \\sin (\\beta -\\alpha) - \\cos \\beta \\sin (\\gamma -\\alpha) - 2\\cos \\alpha (\\sin \\gamma \\cos \\beta -\\sin \\beta \\cos \\gamma))\\] \\[\\qquad = \\frac{3\\sin\\alpha\\sin(\\beta - \\gamma)}{4} = \\frac{3}{4} (\\sin (\\beta +\\gamma)\\sin (\\beta -\\gamma)) = \\frac{3}{8} (\\cos (2\\gamma) - \\cos (2\\beta))\\] \\[\\qquad = \\frac{3}{4} (\\sin^{2}(\\beta) - \\sin^{2}(\\gamma)) = -\\frac{3}{2} f(A).\\quad \\square\\] \n\n", "metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P3.", "solution_match": "\nSolution 3. "}}
+{"year": "2025", "tier": "T2", "problem_label": "3", "problem_type": null, "exam": "EGMO", "problem": "Let \\(A B C\\) be an acute triangle. Points \\(B\\) , \\(D\\) , \\(E\\) , and \\(C\\) lie on a line in this order and satisfy \\(B D =\\) \\(D E = E C\\) . Let \\(M\\) and \\(N\\) be the midpoints of \\(A D\\) and \\(A E\\) , respectively. Let \\(H\\) be the orthocentre of triangle \\(A D E\\) . Let \\(P\\) and \\(Q\\) be points on lines \\(B M\\) and \\(C N\\) , respectively, such that \\(D\\) , \\(H\\) , \\(M\\) , and \\(P\\) are concyclic and \\(E\\) , \\(H\\) , \\(N\\) , and \\(Q\\) are concyclic. Prove that \\(P\\) , \\(Q\\) , \\(N\\) , and \\(M\\) are concyclic. \n\nThe orthocentre of a triangle is the point of intersection of its altitudes.", "solution": "Let \\(X\\) be the second intersection of \\((D H M)\\) and \\((E H N)\\) and let \\(O^{\\prime}\\) be the circumcentre of \\((A M N)\\) Note that \\(\\angle M X N = \\angle M D H + \\angle N E H = 180^{\\circ} - 2\\angle D A E\\) and since \\(\\angle M O^{\\prime}N = 2\\angle D A E\\) we have that \\(X,M,O^{\\prime},N\\) is cyclic and since \\(\\angle M X H = \\angle N X H\\) it means that \\(H X\\) is the angle bisector of \\(\\angle M X N\\) but since \\(O^{\\prime}M = O^{\\prime}N\\) it means that \\(H,X,O^{\\prime}\\) are collinear. Let \\(B M\\) and \\(C N\\) intersect at \\(T\\) and let \\(K\\) and \\(L\\) be the midpoints of \\(M N\\) and \\(B C\\) . Note that \\(L\\) is also the midpoint of \\(D E\\) . Since \\(M N\\) is parallel to \\(B C\\) it means that \\(T\\) , \\(K\\) , and \\(L\\) are collinear, but since \\(A\\) , \\(K\\) , and \\(L\\) are collinear we get that \\(A\\) , \\(T\\) , \\(K\\) , and \\(L\\) are collinear. Now, \\(\\frac{T L}{T K} = \\frac{B C}{M N} = 6\\) . Since \\(K L = K A\\) it means that \\(\\frac{A T}{T K} = \\frac{T L - 2T K}{T K} = 4\\) so by the lemma below, \\(T\\) lies on \\(H O^{\\prime}\\) . Since \\(H O^{\\prime}\\) is the radical axis of \\((D H M)\\) and \\((E H N)\\) we finish the problem using the Radical Axes Theorem \\((T M\\cdot T P = T N\\cdot T Q)\\) .\n\n\n\nLemma. Let \\(A^{\\prime}\\) be the reflection of \\(A\\) around the orthocentre \\(H\\) of \\(\\triangle ABC\\) and \\(O\\) and \\(M\\) be the circumcentre of \\(\\triangle ABC\\) and the midpoint of \\(BC\\) , respectively. Let \\(T\\) be the intersection of \\(A^{\\prime}O\\) and \\(AM\\) . Then \\(\\frac{AT}{TM} = 4\\) . \n\nProof. Since \\(OM \\parallel AA'\\) we have \\(\\frac{AT}{TM} = \\frac{AA'}{OM} = \\frac{2AH}{OM} = \\frac{4OM}{OM} = 4\\) . We used here that \\(AH = 2OM\\) . \\(\\square\\) \n\n", "metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P3.", "solution_match": "\nSolution 2. "}}
+{"year": "2025", "tier": "T2", "problem_label": "3", "problem_type": null, "exam": "EGMO", "problem": "Let \\(A B C\\) be an acute triangle. Points \\(B\\) , \\(D\\) , \\(E\\) , and \\(C\\) lie on a line in this order and satisfy \\(B D =\\) \\(D E = E C\\) . Let \\(M\\) and \\(N\\) be the midpoints of \\(A D\\) and \\(A E\\) , respectively. Let \\(H\\) be the orthocentre of triangle \\(A D E\\) . Let \\(P\\) and \\(Q\\) be points on lines \\(B M\\) and \\(C N\\) , respectively, such that \\(D\\) , \\(H\\) , \\(M\\) , and \\(P\\) are concyclic and \\(E\\) , \\(H\\) , \\(N\\) , and \\(Q\\) are concyclic. Prove that \\(P\\) , \\(Q\\) , \\(N\\) , and \\(M\\) are concyclic. \n\nThe orthocentre of a triangle is the point of intersection of its altitudes.", "solution": "As in solution 2, we will prove that \\(O'\\) is both on line \\(HT\\) and the radical axis of the circles, hence \\(T\\) is on the radical axis, from which we conclude the required concyclicity. We present alternative proofs of both facts, discovered by contestants. \n\nLet \\(M_{1}\\) , \\(N_{1}\\) be the midpoints of \\(AM\\) , \\(AN\\) , respectively, so that \\(AM_{1}:M_{1}D = AN_{1}:N_{1}E = 1:3\\) . It is easy to verify (e.g. by applying the theorems of Ceva or Menelaus, or by computing in barycentric coordinates as in Solution 3) that \\(T\\) lies on \\(EM_{1}\\) and \\(DN_{1}\\) . Note that \\(M_{1}N_{1} \\parallel DE\\) , and also \\(M_{1}O' \\parallel HE\\) as both are perpendicular to \\(AMD\\) , and similarly \\(N_{1}O' \\parallel HD\\) . It follows that \\(DEH\\) and \\(N_{1}M_{1}O'\\) are homothetically similar triangles, and the center of their (negative) homothety is \\(T = DN_{1} \\cap EM_{1}\\) . Therefore \\(T\\) also lies on \\(HO'\\) , as claimed. (We also have that the homothety is by factor \\(\\frac{M_{1}N_{1}}{ED} = -\\frac{1}{4}\\) .) \n\nNow, let \\(M_{2}\\) , \\(N_{2}\\) be the second intersection points of \\(O'M\\) , \\(O'N\\) with the circumcircles of \\(HMD\\) , \\(HNE\\) , respectively. To prove that \\(O'\\) is on the radical axis, it suffices to show that \\(O'M \\cdot O'M_{2} = O'N \\cdot O'N_{2}\\) . But by definition of \\(O'\\) we have \\(O'M = O'N\\) , so we must show \\(O'M_{2} = O'N_{2}\\) , which is equivalent to \\(M_{2}N_{2} \\parallel MN\\) . Angle chasing in circle \\(MM_{2}DH\\) gives \n\n\\[\\angle OM_{2}H = \\angle MM_{2}H = \\angle MDH = \\angle ADH = 90^{\\circ} - \\angle EAD = 90^{\\circ} - \\angle NAM = \\angle O'M N,\\] \n\nfrom which it follows that \\(M_{2}H \\parallel MN\\) . Similarly, we have \\(N_{2}H \\parallel MN\\) , and the two facts together imply that \\(M_{2}, H, N_{2}\\) are collinear and the line through them is parallel to \\(MN\\) , as claimed. \\(\\square\\)\n\n\n", "metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P3.", "solution_match": "\nSolution 2'."}}
+{"year": "2025", "tier": "T2", "problem_label": "3", "problem_type": null, "exam": "EGMO", "problem": "Let \\(A B C\\) be an acute triangle. Points \\(B\\) , \\(D\\) , \\(E\\) , and \\(C\\) lie on a line in this order and satisfy \\(B D =\\) \\(D E = E C\\) . Let \\(M\\) and \\(N\\) be the midpoints of \\(A D\\) and \\(A E\\) , respectively. Let \\(H\\) be the orthocentre of triangle \\(A D E\\) . Let \\(P\\) and \\(Q\\) be points on lines \\(B M\\) and \\(C N\\) , respectively, such that \\(D\\) , \\(H\\) , \\(M\\) , and \\(P\\) are concyclic and \\(E\\) , \\(H\\) , \\(N\\) , and \\(Q\\) are concyclic. Prove that \\(P\\) , \\(Q\\) , \\(N\\) , and \\(M\\) are concyclic. \n\nThe orthocentre of a triangle is the point of intersection of its altitudes.", "solution": "We compute using linear combinations with respect to \\(A D E\\) . We have \\(B = 2D - E\\) \\(C = 2E - D\\) \\(M = \\frac{A + D}{2}\\) , and \\(N = \\frac{A + E}{2}\\) , from which we immediately obtain that the intersection of \\(T = B M\\cap C N\\) is \\(T = \\frac{3A + D + E}{5} = \\frac{6M - B}{5} = \\frac{6N - C}{5}\\) . As in solution 2, we reduce to showing that \\(T\\) is on the radical axis of \\(H D M\\) and \\(H E N\\) , whence \\(T M\\cdot T P = T N\\cdot T Q\\) proves the concyclicity of \\(P,Q,N,M\\) \n\nSynthetic finish, similar to Solution 2. Let \\(O\\) be the circumcentre of \\(A D E\\) and \\(O^{\\prime} = \\frac{A + O}{2}\\) be the circumcentre of \\(A M N\\) . As in Solution 2, we have that \\(O^{\\prime}\\) is on the desired radical axis, so it is enough to show \\(T\\in O^{\\prime}H\\) . Let \\(G = \\frac{A + D + E}{3}\\) be the barycentre of \\(A D E\\) . By properties of the Euler line, we also have \\(G = \\frac{H + 2O}{3}\\) . Now using our known identities we find \n\n\\[T = \\frac{3A + D + E}{5} = \\frac{2A + 3G}{5} = \\frac{2A + H + 2O}{5} = \\frac{H + 4O^{\\prime}}{5}\\] \n\nand in particular \\(T\\in H O^{\\prime}\\) , as we wanted to show. \n\nComputational finish. Let \\(f(T)\\) be the power difference at \\(T\\) w.r.t. \\(D H M\\) and \\(E H N\\) . We compute \\(f(A)\\) and \\(f(L)\\) where \\(L = \\frac{D + E}{2}\\) . Since \\(T = \\frac{3A + 2L}{5}\\) , it is enough to show that \\(3P(A) + 2P(L) = 0\\) . In the following all lengths are directed. We compute trigonometrically: Let \\(\\alpha ,\\beta ,\\gamma\\) be the angles of \\(A D E\\) and assume the diameter of its circumcircle is 1. We have \n\n\\[f(A) = \\frac{A D^{2} - A E^{2}}{2} = \\frac{\\sin^{2}(\\gamma) - \\sin^{2}(\\beta)}{2}.\\] \n\nTo compute \\(P(L)\\) , let \\(D^{\\prime},E^{\\prime}\\) be the second intersection points of \\(H M D,H N E\\) with \\(D E\\) , and let \\(M^{\\prime},N^{\\prime},F\\) be the feet of the perpendiculars from \\(H\\) to \\(A D,A E,D E\\) , respectively. Note that \\(D M^{\\prime} =\\) \\(\\sin \\alpha \\cos \\beta\\) , thus \n\n\\[M^{\\prime}M = D M - D M^{\\prime} = \\frac{\\sin(\\alpha + \\beta)}{2} - \\sin \\alpha \\cos \\beta = \\frac{\\sin(\\beta - \\alpha)}{2}.\\] \n\nWe also have \\(H M^{\\prime} = \\cos \\alpha \\cos \\beta\\) , \\(H F = \\cos \\beta \\cos \\gamma\\) , and \\(H M^{\\prime}M\\sim H F D^{\\prime}\\) , therefore \n\n\\[F D^{\\prime} = \\frac{H F}{H M^{\\prime}} M^{\\prime}M = \\frac{\\cos\\gamma\\sin(\\beta - \\alpha)}{\\cos\\alpha}\\] \n\nand similarly \n\n\\[E^{\\prime}F = \\frac{\\cos\\beta\\sin(\\gamma - \\alpha)}{\\cos\\alpha}.\\]\n\n\n\nWe also have the standard \\(FD = \\sin \\gamma \\cos \\beta\\) and \\(EF = \\sin \\beta \\cos \\gamma\\) . We can finally compute \n\n\\[f(L) = LD\\cdot LD^{\\prime} - LE\\cdot LE^{\\prime} = \\frac{\\sin\\alpha}{2} (LD^{\\prime} + LE^{\\prime}) = \\frac{\\sin\\alpha}{2} (FD^{\\prime} + FE^{\\prime} - FD - FE)\\] \\[\\qquad = \\frac{\\sin\\alpha}{4\\cos\\alpha} (\\cos \\gamma \\sin (\\beta -\\alpha) - \\cos \\beta \\sin (\\gamma -\\alpha) - 2\\cos \\alpha (\\sin \\gamma \\cos \\beta -\\sin \\beta \\cos \\gamma))\\] \\[\\qquad = \\frac{3\\sin\\alpha\\sin(\\beta - \\gamma)}{4} = \\frac{3}{4} (\\sin (\\beta +\\gamma)\\sin (\\beta -\\gamma)) = \\frac{3}{8} (\\cos (2\\gamma) - \\cos (2\\beta))\\] \\[\\qquad = \\frac{3}{4} (\\sin^{2}(\\beta) - \\sin^{2}(\\gamma)) = -\\frac{3}{2} f(A).\\quad \\square\\] \n\n", "metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P3.", "solution_match": "\nSolution 3. "}}
{"year": "2025", "tier": "T2", "problem_label": "3", "problem_type": null, "exam": "EGMO", "problem": "Let \\(A B C\\) be an acute triangle. Points \\(B\\) , \\(D\\) , \\(E\\) , and \\(C\\) lie on a line in this order and satisfy \\(B D =\\) \\(D E = E C\\) . Let \\(M\\) and \\(N\\) be the midpoints of \\(A D\\) and \\(A E\\) , respectively. Let \\(H\\) be the orthocentre of triangle \\(A D E\\) . Let \\(P\\) and \\(Q\\) be points on lines \\(B M\\) and \\(C N\\) , respectively, such that \\(D\\) , \\(H\\) , \\(M\\) , and \\(P\\) are concyclic and \\(E\\) , \\(H\\) , \\(N\\) , and \\(Q\\) are concyclic. Prove that \\(P\\) , \\(Q\\) , \\(N\\) , and \\(M\\) are concyclic. \n\nThe orthocentre of a triangle is the point of intersection of its altitudes.", "solution": "Let \\(T = BM\\cap CN\\) , let \\(F\\) be the foot of the altitude from \\(A\\) to \\(BC\\) , let \\(O\\) be the circumcentre of \\((ADE)\\) , let \\(D^{\\prime}\\neq D\\) and \\(E^{\\prime}\\neq E\\) be the second intersections of \\((DHMP)\\) and \\((EHNQ)\\) with line \\(BC\\) and let \\(U\\) and \\(V\\) be the antipodes of \\(D\\) and \\(E\\) on \\((DHMP)\\) and \\((EHNQ)\\) , respectively. \n\nWe begin with a bit of barycentric coordinates. Set barycentric coordinates in \\(\\triangle ABC\\) , set so that \\(A = (1,0,0)\\) , \\(B = (0,1,0)\\) , and \\(C = (0,0,1)\\) . The definitions of \\(D\\) and \\(E\\) give \\(D = (0,2 / 3,1 / 3)\\) and \\(E = (0,1 / 3,2 / 3)\\) , whence \\(M = (1 / 2,1 / 3,1 / 6)\\) and \\(N = (1 / 2,1 / 6,1 / 3)\\) . This means that line \\(BM\\) is given by \\((1 / 2:y:1 / 6)\\) while line \\(CN\\) is given by \\((1 / 2:1 / 6:z)\\) . So their intersection \\(T\\) is \\((1 / 2:1 / 6:1 / 6) = (3:1:1)\\) , giving \\(T = \\frac{3A + B + C}{5} = \\frac{3A + D + E}{5}\\) . \n\nOur next tool is the linearity of the power of a point. Let \\(f:\\mathbb{R}^{2}\\to \\mathbb{R}\\) be defined by \n\n\\[f(Z) = \\mathrm{Pow}_{(DHMP)}(Z) - \\mathrm{Pow}_{(EHNQ)}(Z).\\] \n\nIt suffices to show that \\(f(T) = 0\\) ; from there, the required concyclicity will follow from \\(TM\\cdot TP =\\) \\(TN\\cdot TQ\\) . The key observation is that \\(f\\) is a linear function. In particular, \\(f(T) = \\frac{3f(A) + f(D) + f(E)}{5}\\) So, we need only show that \\(3f(A) + f(D) + f(E) = 0\\) . Pick an arbitrary one- dimensional coordinate system on the line \\(BC\\) and let \\(\\tau\\) be the map from a point on \\(BC\\) to its coordinate. We compute \n\n\\[f(A) = AM\\cdot AD - AN\\cdot AE = \\frac{AD^2 - AE^2}{2} = \\frac{FD^2 - FE^2}{2}\\] \\[\\qquad = (\\tau (E) - \\tau (D))\\left(\\tau (F) - \\tau \\left(\\frac{D + E}{2}\\right)\\right),\\] \\[f(D) = -DE\\cdot DE^{\\prime} = (\\tau (E) - \\tau (D))(\\tau (D) - \\tau (E^{\\prime})),\\] \\[f(E) = ED\\cdot ED^{\\prime} = (\\tau (E) - \\tau (D))(\\tau (E) - \\tau (D^{\\prime})).\\] \n\nRearranging, it suffices to show that \\(3\\tau (F) = \\tau (D^{\\prime}) + \\tau (E^{\\prime}) + \\tau ((D + E) / 2)\\) . This can be rewritten as \\(3F = D^{\\prime} + E^{\\prime} + (D + E) / 2\\) . By projecting down to line \\(BC\\) , it suffices to show that the displacement vector \\(H + 2A - (O + U + V)\\) is perpendicular to line \\(BC\\) .\n\n\n\nWe do this using complex numbers. Let \\((A D E)\\) be the unit circle with \\(A = a\\) , \\(D = d\\) , and \\(E = e\\) , so that \\(O = 0\\) and \\(H = h = a + d + e\\) . Note that \\(U\\) satisfies \\(U M\\perp A D\\) and \\(U H\\perp D H\\perp A E\\) . Translating these conditions into equations, we have \\(u = a d\\overline{{u}}\\) and \\(u + a e\\overline{{u}} = h + a e\\overline{{h}}\\) . Rearranging gives \n\n\\[(d + e)u = d u + e(a d\\overline{{u}}) = d(h + a e\\overline{{h}}) \\Rightarrow u = \\frac{d h + a d e\\overline{{h}}}{d + e}.\\] \n\nComputing \\(v\\) similarly, we get that \n\n\\[v:= H + 2A - (O + U + V) = h + 2a - \\frac{(d + e)h + 2a d e\\overline{{h}}}{d + e} = 2\\left(a - \\frac{a d e\\overline{{h}}}{d + e}\\right) = -\\frac{2d e}{d + e}.\\] \n\nThis displacement vector \\(v\\) satisfies \\(v = d e\\overline{{v}}\\) and so it is orthogonal to line \\(D E\\) , as desired.", "metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P3.", "solution_match": "\nSolution 4. "}}
{"year": "2025", "tier": "T2", "problem_label": "3", "problem_type": null, "exam": "EGMO", "problem": "Let \\(A B C\\) be an acute triangle. Points \\(B\\) , \\(D\\) , \\(E\\) , and \\(C\\) lie on a line in this order and satisfy \\(B D =\\) \\(D E = E C\\) . Let \\(M\\) and \\(N\\) be the midpoints of \\(A D\\) and \\(A E\\) , respectively. Let \\(H\\) be the orthocentre of triangle \\(A D E\\) . Let \\(P\\) and \\(Q\\) be points on lines \\(B M\\) and \\(C N\\) , respectively, such that \\(D\\) , \\(H\\) , \\(M\\) , and \\(P\\) are concyclic and \\(E\\) , \\(H\\) , \\(N\\) , and \\(Q\\) are concyclic. Prove that \\(P\\) , \\(Q\\) , \\(N\\) , and \\(M\\) are concyclic. \n\nThe orthocentre of a triangle is the point of intersection of its altitudes.", "solution": "This solution uses almost exclusively complex numbers. As in other solutions, we reduce to showing that \\(H\\) , \\(S:= (D H M)\\cap (E H N)\\) , and \\(T:= B M\\cap C N\\) are collinear; this is all of the synthetic information we shall use. (If one computes \\(T = \\frac{3A + D + E}{5}\\) using means other than complex numbers, the solution becomes much shorter.) \n\nWe use complex numbers with \\(A = a\\) , \\(D = d\\) , and \\(E = e\\) on the unit circle. Note that \\(H = a + d + e\\) , \\(M = \\frac{a + d}{2}\\) , and \\(B = 2d - e\\) . We will make great use of the \"arbitrary line intersection formula,\" which says that the intersection between lines \\(W X\\) and \\(Y Z\\) can be written explicitly as \n\n\\[\\frac{(\\overline{{u}} x - w\\overline{{x}})(y - z) - (w - x)(\\overline{{y}} z - y\\overline{{z}})}{(\\overline{{w}} - \\overline{{x}})(y - z) - (w - x)(\\overline{{y}} - \\overline{{z}})}.\\] \n\nWe first use this to find \\(T = t\\) . We compute \n\n\\[b - m = (2d - e) - \\frac{a + d}{2} = \\frac{3d - a - 2e}{2}\\] \\[\\overline{{b}} - m = \\frac{3a e - d e - 2a d}{2a d e}\\] \\[\\overline{{b}} m - b\\overline{{m}} = \\frac{2e - d}{d e}\\cdot \\frac{a + d}{2} -(2d - e)\\cdot \\frac{a + d}{2a d}\\] \\[\\qquad = \\frac{a(a + d)(2e - d) - (2d - e)(a + d)e}{2a d e} = \\frac{(a + d)(2a e + e^{2} - a d - 2d e)}{2a d e}.\\] \n\nIf \\(\\mathcal{E}\\) is some expression, we use the notation \\(\\mathcal{E} - \\{\\sim \\}\\) to denote \\(\\mathcal{E}\\) minus the expression formed by swapping \\(d\\) and \\(e\\) in \\(\\mathcal{E}\\) . We now have \n\n\\[t = \\frac{(\\overline{{b}}m - b\\overline{{m}})(c - n) - \\{\\sim\\}}{(\\overline{{b}} - \\overline{{m}})(c - n) - \\{\\sim\\}}\\] \\[\\quad = \\frac{(a + d)(2a e + e^{2} - a d - 2d e)(3e - a - 2d) - \\{\\sim\\}}{(3a e - d e - 2a d)(3e - a - 2d) - \\{\\sim\\}}\\] \\[\\quad = \\frac{[a^{3}(d - 2e) + a^{2}(3d^{2} - 7d e + 5e^{2}) + a(2d^{3} - d^{2}e - 3d e^{2} + 3e^{3}) + d e(4d^{2} - 8d e + 3e^{2})] - \\{\\sim\\}}{[a^{2}(2d - 3e) + a(4d^{2} - 11d e + 9e^{2}) + d e(2d - 3e)] - \\{\\sim\\}}\\] \\[\\quad = \\frac{a^{3}(3(d - e)) - a^{2}(2(d^{2} - e^{2})) + a(-(d^{3} - e^{3}) + 2d e(d - e)) + d e(d^{2} - e^{2})}{a^{2}(5(d - e)) - a(5(d^{2} - e^{2})) + d e(5(d - e))}\\] \\[\\quad = \\frac{3a^{3} - 2a^{2}(d + e) - a(d^{2} - d e + e^{2}) + d e(d + e)}{5(a^{2} - a(d + e) + d e)}\\] \\[\\quad = \\frac{(a - d)(a - e)(3a + d + e)}{5(a - d)(a - e)} = \\frac{3a + d + e}{5}.\\] \n\n(The factorization in the last line can be motivated by noting that the expression, while cubic in \\(a\\) , is only quadratic in \\(d\\) . When written out as a polynomial in \\(d\\) , each coefficient is divisible by \\(a - e\\) ; by symmetry, the numerator is divisible by \\(a - d\\) as well, and the factorization follows.)\n\n\n\nComputing \\(S = s\\) is slightly harder, as it is the intersection of two circles rather than of two lines. We get around this by noting that \\(\\{h,d,m,s\\}\\) are concyclic if and only if \\(\\{0,h - d,h - m,h - s\\}\\) are concyclic, which happens if and only if \\(\\{\\frac{1}{h - d},\\frac{1}{h - m},\\frac{1}{h - s}\\}\\) are collinear. (One can see this by inversion, or just by writing out the cross- ratio in the special case when one of the points is zero.) Thus \\(\\frac{1}{h - s}\\) is the intersection of the line through \\(w:= \\frac{1}{h - d}\\) and \\(x:= \\frac{1}{h - m}\\) and the line through \\(y:= \\frac{1}{h - e}\\) and \\(z:= \\frac{1}{h - n}\\) . We compute \n\n\\[w - x = \\frac{1}{a + e} -\\frac{1}{\\frac{a + d}{2} + e} = -\\frac{a - d}{(a + e)(a + d + 2e)}\\] \\[\\overline{w - x} = \\frac{ae^{2}(a - d)}{(a + e)(2ad + ae + de)}\\] \\[\\overline{w} x - w\\overline{x} = \\frac{ae}{a + e}\\cdot \\frac{2}{a + d + 2e} -\\frac{1}{a + e}\\cdot \\frac{2ade}{2ad + ae + de}\\] \\[\\qquad = 2ae\\frac{(2ad + ae + de) - d(a + d + 2e)}{(a + e)(a + d + 2e)(2ad + ae + de)}\\] \\[\\qquad = \\frac{2ae(a - d)(d + e)}{(a + e)(a + d + 2e)(2ad + ae + de)}.\\] \n\nUsing the line intersection formula, we have \n\n\\[\\frac{1}{h - s} = \\frac{(\\overline{w} x - w\\overline{x})(y - z) - \\{\\sim\\}}{(\\overline{w} - \\overline{x})(y - z) - \\{\\sim\\}}\\] \\[= \\frac{[\\frac{2ae(a - d)(d + e)}{(a + e)(a + d + 2e)(2ad + ae + de)}] - \\{\\sim\\}}{[\\frac{ae^{2}(a - d)}{(a + e)(2ad + ae + de)}] - \\{\\sim\\}}\\] \\[= 2(d + e)\\frac{[e(ad + 2ae + de)] - \\{\\sim\\}}{[e^{2}(a + d + 2e)(ad + 2ae + de)] - \\{\\sim\\}}\\] \\[= 2(d + e)\\frac{[a(de + 2e^{2}) + de^{2}] - \\{\\sim\\}}{[a^{2}(de^{2} + 2e^{3}) + a(d^{2}e^{2} + 5de^{3} + 4e^{4}) + de(de^{2} + 2e^{3})] - \\{\\sim\\}}\\] \\[= \\frac{2(d + e)(a(2(e^{2} - d^{2})) + de(e - d))}{(a^{2} + de)(2(e^{3} - d^{3}) + de(e - d)) + a(4(e^{4} - d^{4}) + 5de(e^{2} - e^{2}))}\\] \\[= \\frac{2(d + e)(2a(d + e) + de)}{(a^{2} + de)(2d^{2} + 3de + 2e^{2}) + a(d + e)(4d^{2} + 5de + 4e^{2})}.\\] \n\nSince \\(h - t = \\frac{2a + 4d + 4e}{5}\\) , this gives us \n\n\\[\\frac{h - s}{h - t} = \\frac{5}{4}\\cdot \\frac{(a^{2} + de)(2d^{2} + 3de + 2e^{2}) + a(d + e)(4d^{2} + 5de + 4e^{2})}{(d + e)(2ad + 2ae + de)(a + 2d + 2e)}.\\] \n\nIt is easy to see that this is real by using the symmetry of the expressions (both the numerator and denominator satisfy \\(\\mathcal{E} = a^{2}d^{3}e^{3}\\overline{\\mathcal{E}}\\) . We conclude that \\(H\\) , \\(S\\) , and \\(T\\) are collinear, as desired.", "metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P3.", "solution_match": "\nSolution 5. "}}
{"year": "2025", "tier": "T2", "problem_label": "3", "problem_type": null, "exam": "EGMO", "problem": "Let \\(A B C\\) be an acute triangle. Points \\(B\\) , \\(D\\) , \\(E\\) , and \\(C\\) lie on a line in this order and satisfy \\(B D =\\) \\(D E = E C\\) . Let \\(M\\) and \\(N\\) be the midpoints of \\(A D\\) and \\(A E\\) , respectively. Let \\(H\\) be the orthocentre of triangle \\(A D E\\) . Let \\(P\\) and \\(Q\\) be points on lines \\(B M\\) and \\(C N\\) , respectively, such that \\(D\\) , \\(H\\) , \\(M\\) , and \\(P\\) are concyclic and \\(E\\) , \\(H\\) , \\(N\\) , and \\(Q\\) are concyclic. Prove that \\(P\\) , \\(Q\\) , \\(N\\) , and \\(M\\) are concyclic. \n\nThe orthocentre of a triangle is the point of intersection of its altitudes.", "solution": "As usual, we reduce to proving \\(T = (3A + D + E) / 5\\) is on the radical axis and compute; this time in Cartesian coordinates. \n\nLet \\(A(0,h)\\) , \\(D(b,0)\\) , \\(E(c,0)\\) be coordinates for \\(ADE\\) . Then \\(H(0, - \\frac{b c}{h})\\) , \\(M(\\frac{b}{2},\\frac{h}{2})\\) , \\(N(\\frac{c}{2},\\frac{h}{2})\\) and \\(T(\\frac{b + c}{5},\\frac{3h}{5})\\) . We compute \\(O_{D}(x_{D},y_{D})\\) the circumcentre of \\(DMH\\) and obtain \\(O_{E}\\) by symmetry. We then have to verify that \\(O_{D}O_{E} \\perp HT\\) , which can be done by comparing slopes. \n\nThe centre \\(O_{D}\\) can be given by the intersection of perpendicular bisectors of \\(DM\\) and \\(DH\\) . This gives the following system of equations on \\(x_{D},y_{D}\\) :\n\n\n\n\\[h\\cdot x_{D} + c\\cdot y_{D} = \\frac{b h}{2} -\\frac{b c^{2}}{2 h}\\] \\[-b\\cdot x_{D} + h\\cdot y_{D} = -\\frac{3 b^{2}}{4} +\\frac{h^{2}}{4}\\] \n\nSolving the system gives \n\n\\[(h^{2} + b c)x_{D} = \\frac{3 b^{2}c}{4} -\\frac{b c^{2}}{2} +\\frac{h^{2}b}{2} -\\frac{h^{2}c}{4}\\] \\[(h^{2} + b c)y_{D} = -\\frac{h b^{2}}{4} +\\frac{h^{3}}{4} -\\frac{b^{2}c^{2}}{2 h}\\] \n\nThe formulas for \\(x_{E}, y_{E}\\) will be the same, swapping \\(b \\leftrightarrow c\\) by symmetry; thus \\(y_{D} - y_{E}\\) and \\(x_{D} - x_{E}\\) will be antisymmetric in \\(b, c\\) and divisible by \\(b - c\\) , and explicitly: \n\n\\[(h^{2} + b c)(x_{D} - x_{E}) = \\frac{b - c}{4} (5b c + 3h^{2})\\] \\[(h^{2} + b c)(y_{D} - y_{E}) = -\\frac{b - c}{4} h(b + c)\\] \n\nSo \\(\\frac{y_{D} - y_{E}}{x_{D} - x_{E}} = -\\frac{h(b + c)}{5b c + 3h^{2}}\\) . \n\nThe other slope is more immediate: \n\n\\[\\frac{y_{T} - y_{H}}{x_{T} - x_{H}} = \\frac{3h / 5 + b c / h}{(b + c) / 5} = \\frac{5b c + 3h^{2}}{h(b + c)} = -\\frac{x_{D} - x_{E}}{y_{D} - y_{E}}\\] \n\nso indeed the two slopes correspond to perpendicular lines.\n\n\n\n## Day 2", "metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P3.", "solution_match": "\nSolution 6. "}}
-{"year": "2025", "tier": "T2", "problem_label": "4", "problem_type": null, "exam": "EGMO", "problem": "Let \\(A B C\\) be an acute triangle with incentre \\(I\\) and \\(A B \\neq A C\\) . Let lines \\(B I\\) and \\(C I\\) intersect the circumcircle of \\(A B C\\) at \\(P \\neq B\\) and \\(Q \\neq C\\) , respectively. Consider points \\(R\\) and \\(S\\) such that \\(A Q R B\\) and \\(A C S P\\) are parallelograms (with \\(A Q \\parallel R B\\) , \\(A B \\parallel Q R\\) , \\(A C \\parallel S P\\) , and \\(A P \\parallel C S\\) ). Let \\(T\\) be the point of intersection of lines \\(R B\\) and \\(S C\\) . Prove that points \\(R\\) , \\(S\\) , \\(T\\) , and \\(I\\) are concyclic.", "solution": "We will prove that \\(\\triangle B I R\\sim \\triangle C I S\\) , since the statement then follows from \\(\\angle T R I = \\angle B R I = \\angle C S I =\\) \\(\\angle T S I\\) . \n\n\n \n\nStep 1. Let us prove \\(\\angle R B I = \\angle S C I\\) . We will use directed angles: \n\n\\[(B R,B I) = (B R,A B) + (A B,B I) = (A Q,A B) + (B I,B C)\\] \\[\\qquad = (C Q,B C) + (B I,B C) = (C I,C B) + (B I,B C),\\] \n\nwhich is symmetric in \\(B,C\\) . Therefore, analogously we would obtain the same expression for \\((C S,C I)\\) \n\nStep 2. Let us prove \\(B R / B I = C S / C I\\) . Clearly \\(B R = A Q\\) and \\(C S = A P\\) . Angle chasing gives \\(\\angle I C B = \\angle Q C B = \\angle A P Q\\) , and similarly \\(\\angle P Q A = \\angle C B I\\) , and so \\(\\triangle I B C\\sim \\triangle A Q P\\) , from which the desired \\(A Q / B I = A P / C I\\) follows. This finishes the solution. \n\nRemark. In the alternative solutions below, the notation \\((A B C)\\) refers to the circle passing through points \\(A\\) , \\(B\\) , and \\(C\\) . \\(D\\) will refer to the midpoint of arc \\(B C\\) not containing point \\(A\\) , unless stated otherwise. \n\nLet \\(I_{A}\\) be the \\(A\\) - excenter of \\(\\triangle A B C\\) . It is well- known that points \\(B\\) , \\(C\\) , \\(I\\) , and \\(I_{A}\\) lie on a circle with diameter \\(I I_{A}\\) . The center of this circle is point \\(D\\) . It is clear that \\(A\\) , \\(I\\) and \\(D\\) are collinear. \n\nIn some of the solutions, some of the facts below may be used:\n\n\n\n\\(\\cdot B,T,I,C\\) are concyclic. Indeed, \\(\\angle B T C = \\angle Q A P = A + \\frac{B + C}{2} = 90^{\\circ} + \\frac{A}{2} = \\angle B I C\\) , so \\(B,T,I,C\\) are concyclic. \n\n\\(\\angle Q A P = \\angle B T C = \\angle R T S\\) . Indeed, since \\(Q A\\parallel B T\\) and \\(P A\\parallel C T\\) , then \\(\\angle Q A P = \\angle B T C =\\) \\(\\angle R T S\\) \n\n\\(\\cdot R Q\\) and \\(P S\\) are tangents to \\((A B C)\\) . Indeed, \\(\\angle R Q B = \\angle Q B A = \\angle Q A B\\) and similarly for \\(P S\\) \n\n\\(\\cdot \\triangle A Q P\\sim I B C\\) (see Solution 1 for the proof). \n\n\\(\\cdot Q P\\) is the perpendicular bisector of \\(A I\\) . Indeed, \\(Q A = Q I\\) and \\(P A = P I\\) so \\(P Q\\perp A I\\) and \\(P Q\\) bisects \\(A I\\) .", "metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P4.", "solution_match": "\nSolution 1. "}}
+{"year": "2025", "tier": "T2", "problem_label": "4", "problem_type": null, "exam": "EGMO", "problem": "Let \\(A B C\\) be an acute triangle with incentre \\(I\\) and \\(A B \\neq A C\\) . Let lines \\(B I\\) and \\(C I\\) intersect the circumcircle of \\(A B C\\) at \\(P \\neq B\\) and \\(Q \\neq C\\) , respectively. Consider points \\(R\\) and \\(S\\) such that \\(A Q R B\\) and \\(A C S P\\) are parallelograms (with \\(A Q \\parallel R B\\) , \\(A B \\parallel Q R\\) , \\(A C \\parallel S P\\) , and \\(A P \\parallel C S\\) ). Let \\(T\\) be the point of intersection of lines \\(R B\\) and \\(S C\\) . Prove that points \\(R\\) , \\(S\\) , \\(T\\) , and \\(I\\) are concyclic.", "solution": "We will prove that \\(\\triangle B I R\\sim \\triangle C I S\\) , since the statement then follows from \\(\\angle T R I = \\angle B R I = \\angle C S I =\\) \\(\\angle T S I\\) . \n\n\n \n\nStep 1. Let us prove \\(\\angle R B I = \\angle S C I\\) . We will use directed angles: \n\n\\[(B R,B I) = (B R,A B) + (A B,B I) = (A Q,A B) + (B I,B C)\\] \\[\\qquad = (C Q,B C) + (B I,B C) = (C I,C B) + (B I,B C),\\] \n\nwhich is symmetric in \\(B,C\\) . Therefore, analogously we would obtain the same expression for \\((C S,C I)\\) \n\nStep 2. Let us prove \\(B R / B I = C S / C I\\) . Clearly \\(B R = A Q\\) and \\(C S = A P\\) . Angle chasing gives \\(\\angle I C B = \\angle Q C B = \\angle A P Q\\) , and similarly \\(\\angle P Q A = \\angle C B I\\) , and so \\(\\triangle I B C\\sim \\triangle A Q P\\) , from which the desired \\(A Q / B I = A P / C I\\) follows. This finishes the solution. \n\nRemark. In the alternative solutions below, the notation \\((A B C)\\) refers to the circle passing through points \\(A\\) , \\(B\\) , and \\(C\\) . \\(D\\) will refer to the midpoint of arc \\(B C\\) not containing point \\(A\\) , unless stated otherwise. \n\nLet \\(I_{A}\\) be the \\(A\\) - excenter of \\(\\triangle A B C\\) . It is well- known that points \\(B\\) , \\(C\\) , \\(I\\) , and \\(I_{A}\\) lie on a circle with diameter \\(I I_{A}\\) . The center of this circle is point \\(D\\) . It is clear that \\(A\\) , \\(I\\) and \\(D\\) are collinear. \n\nIn some of the solutions, some of the facts below may be used:\n\n\n\n\\(\\cdot B,T,I,C\\) are concyclic. Indeed, \\(\\angle B T C = \\angle Q A P = A + \\frac{B + C}{2} = 90^{\\circ} + \\frac{A}{2} = \\angle B I C\\) , so \\(B,T,I,C\\) are concyclic. \n\n\\(\\angle Q A P = \\angle B T C = \\angle R T S\\) . Indeed, since \\(Q A\\parallel B T\\) and \\(P A\\parallel C T\\) , then \\(\\angle Q A P = \\angle B T C =\\) \\(\\angle R T S\\) \n\n\\(\\cdot R Q\\) and \\(P S\\) are tangents to \\((A B C)\\) . Indeed, \\(\\angle R Q B = \\angle Q B A = \\angle Q A B\\) and similarly for \\(P S\\) \n\n\\(\\cdot \\triangle A Q P\\sim I B C\\) (see Solution 1 for the proof). \n\n\\(\\cdot Q P\\) is the perpendicular bisector of \\(A I\\) . Indeed, \\(Q A = Q I\\) and \\(P A = P I\\) so \\(P Q\\perp A I\\) and \\(P Q\\) bisects \\(A I\\) .", "metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P4.", "solution_match": "\nSolution 1. "}}
{"year": "2025", "tier": "T2", "problem_label": "4", "problem_type": null, "exam": "EGMO", "problem": "Let \\(A B C\\) be an acute triangle with incentre \\(I\\) and \\(A B \\neq A C\\) . Let lines \\(B I\\) and \\(C I\\) intersect the circumcircle of \\(A B C\\) at \\(P \\neq B\\) and \\(Q \\neq C\\) , respectively. Consider points \\(R\\) and \\(S\\) such that \\(A Q R B\\) and \\(A C S P\\) are parallelograms (with \\(A Q \\parallel R B\\) , \\(A B \\parallel Q R\\) , \\(A C \\parallel S P\\) , and \\(A P \\parallel C S\\) ). Let \\(T\\) be the point of intersection of lines \\(R B\\) and \\(S C\\) . Prove that points \\(R\\) , \\(S\\) , \\(T\\) , and \\(I\\) are concyclic.", "solution": "We use complex numbers, with \\((A B C)\\) as the unit circle. Set \\(D = d\\) , \\(P = p\\) , \\(Q = q\\) , so that \\(A = a = - \\frac{p q}{d}\\) , \\(B = b = - \\frac{d q}{p}\\) , and \\(C = c = - \\frac{d q}{q}\\) . Write \\(z\\sim w\\) if \\(z / w\\) is a nonzero real number. We observe that \n\n\\[\\frac{R - T}{S - T}\\sim \\frac{R - B}{S - C} = \\frac{Q - A}{P - A} = \\frac{q + \\frac{p q}{d}}{p + \\frac{p q}{d}} = \\frac{(d + p)q}{(d + q)p}.\\] \n\nSo, it suffices to show that \\(\\frac{I - R}{I - S}\\sim \\frac{(d + p)q}{(d + q)p}\\) . Indeed, \n\n\\[I - R = (d + p + q) - (Q + B - A) = d + p + \\frac{d q}{p} - \\frac{p q}{d} = (d + p)\\left(1 + \\frac{(d - p)q}{d p}\\right) = \\frac{(d + p)(d p + d q - p q)}{d p},\\] \n\nso \n\n\\[\\frac{I - R}{I - S} = \\frac{\\frac{d + p}{d p}}{\\frac{d + q}{d q}} = \\frac{(d + p)q}{(d + q)p}.\\]", "metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P4.", "solution_match": "\nSolution 2. "}}
{"year": "2025", "tier": "T2", "problem_label": "4", "problem_type": null, "exam": "EGMO", "problem": "Let \\(A B C\\) be an acute triangle with incentre \\(I\\) and \\(A B \\neq A C\\) . Let lines \\(B I\\) and \\(C I\\) intersect the circumcircle of \\(A B C\\) at \\(P \\neq B\\) and \\(Q \\neq C\\) , respectively. Consider points \\(R\\) and \\(S\\) such that \\(A Q R B\\) and \\(A C S P\\) are parallelograms (with \\(A Q \\parallel R B\\) , \\(A B \\parallel Q R\\) , \\(A C \\parallel S P\\) , and \\(A P \\parallel C S\\) ). Let \\(T\\) be the point of intersection of lines \\(R B\\) and \\(S C\\) . Prove that points \\(R\\) , \\(S\\) , \\(T\\) , and \\(I\\) are concyclic.", "solution": "In the following, all segment notations denote vectors. \n\nAs mentioned above, we find \\(\\triangle A Q P\\sim \\triangle I B C\\) , and by definitions of the parallelograms we have \\(B R = A Q\\) and \\(C S = A P\\) as well as \\(\\angle R T S = \\angle Q A P\\) , so it suffices to show \\(\\angle R I S = \\angle Q A P\\) . From the similarity \\(\\triangle A Q P\\sim \\triangle I B C\\) , we have a spiral map \\(\\lambda\\) such that \\(I B = \\lambda A Q\\) and \\(I C = \\lambda A P\\) . It follows that \\(I R = I B + B R = (\\lambda +1)A Q\\) and \\(I S = I C + C S = (\\lambda +1)A P\\) . Because \\(\\lambda +1\\) is also a spiral map, we have \\(\\triangle I R S\\sim \\triangle A Q P\\) and in particular \\(\\angle R I S = \\angle Q A P\\) , as we wanted to show. \n\nRemark. This solution is deeply related to the complex numbers solution; indeed, the vectors can be interpreted as complex numbers and the spiral map as a complex scalar multiplication. But it only relies on the additive structure of the complex numbers as a real plane and the linear map acting on them (rather than, e.g., multiplying two points together), making vectors a slightly more natural language for the claims. \n\nRemark. The number 1 in the solution above represents the identity map.", "metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P4.", "solution_match": "\nSolution 3. "}}
-{"year": "2025", "tier": "T2", "problem_label": "4", "problem_type": null, "exam": "EGMO", "problem": "Let \\(A B C\\) be an acute triangle with incentre \\(I\\) and \\(A B \\neq A C\\) . Let lines \\(B I\\) and \\(C I\\) intersect the circumcircle of \\(A B C\\) at \\(P \\neq B\\) and \\(Q \\neq C\\) , respectively. Consider points \\(R\\) and \\(S\\) such that \\(A Q R B\\) and \\(A C S P\\) are parallelograms (with \\(A Q \\parallel R B\\) , \\(A B \\parallel Q R\\) , \\(A C \\parallel S P\\) , and \\(A P \\parallel C S\\) ). Let \\(T\\) be the point of intersection of lines \\(R B\\) and \\(S C\\) . Prove that points \\(R\\) , \\(S\\) , \\(T\\) , and \\(I\\) are concyclic.", "solution": "Let \\(E\\) , \\(F\\) , \\(G\\) be the midpoints of \\(A I\\) , \\(B Q\\) , \\(C P\\) . As in Solution 1, angle chase shows that \\(\\triangle A Q P\\sim\\) \\(\\triangle I B C\\) . \n\nNote that by the Mean Geometry Theorem we have that \\(\\frac{1}{2} A Q P + \\frac{1}{2} I B C = E F G\\) is similar to \\(\\triangle I B C\\) . Homothety with center \\(A\\) and scale- factor 2 maps \\(E F G\\) to \\(I R S\\) . Hence \\(\\angle R I S = \\angle F E G = \\angle Q A P =\\) \\(\\angle B T C = \\angle R T S\\) , so \\(R,T,I,S\\) are concyclic. \n\nRemark. As shown above, \\(E\\) lies on \\(Q P\\) and \\(A I\\perp P Q\\) . One can prove that \\(\\angle F E G = \\angle B I C\\) in another way. Let \\(J\\) be the midpoint of \\(P Q\\) . Then \\(\\angle B I C = \\angle F J G\\) by midlines and \\(\\angle F J G = \\angle F E G\\) by the lemma below applied in \\(B C P Q\\) . \n\nLemma. Let \\(A B C D\\) is a cyclic quadrilateral and \\(E\\) is the intersection of its diagonals. Then the midpoints of \\(A B\\) , \\(B C\\) , \\(C D\\) and the foot of the perpendicular from \\(E\\) to \\(B C\\) are concyclic.\n\n\n", "metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P4.", "solution_match": "\nSolution 4. "}}
-{"year": "2025", "tier": "T2", "problem_label": "4", "problem_type": null, "exam": "EGMO", "problem": "Let \\(A B C\\) be an acute triangle with incentre \\(I\\) and \\(A B \\neq A C\\) . Let lines \\(B I\\) and \\(C I\\) intersect the circumcircle of \\(A B C\\) at \\(P \\neq B\\) and \\(Q \\neq C\\) , respectively. Consider points \\(R\\) and \\(S\\) such that \\(A Q R B\\) and \\(A C S P\\) are parallelograms (with \\(A Q \\parallel R B\\) , \\(A B \\parallel Q R\\) , \\(A C \\parallel S P\\) , and \\(A P \\parallel C S\\) ). Let \\(T\\) be the point of intersection of lines \\(R B\\) and \\(S C\\) . Prove that points \\(R\\) , \\(S\\) , \\(T\\) , and \\(I\\) are concyclic.", "solution": "Let \\(O\\) be the circumcenter of \\((A B C)\\) . Let \\(M\\) , \\(N\\) , and \\(L\\) be the midpoints of \\(O D\\) , \\(P C\\) , and \\(Q B\\) respectively. \n\nClaim 1. \\(\\triangle O P Q\\) and \\(\\triangle D C B\\) are directly similar. \n\nProof. Clearly \\(D B = D C\\) and \\(O Q = O P\\) . Also note that \\(\\angle Q O P = 2\\angle Q D P = 2\\angle Q D A + 2\\angle P D A =\\) \\(\\angle B D A + \\angle C D A = \\angle B D C\\) . So the two triangles are directly similar by SAS. \n\nClaim 2. \\(M L = M N\\) and \\(\\angle L M N = 180^{\\circ} - \\angle B A C\\) \n\nProof. Note that since \\(\\triangle O Q P\\sim \\triangle D B C\\) by the Mean Geometry Theorem, we have that the average of the two triangles is also similar to them, therefore \\(\\triangle M L N\\sim \\triangle D B C\\Rightarrow M L = M N\\) and \\(\\angle L M N =\\) \\(\\angle B D C = 180^{\\circ} - \\angle B A C\\) \n\nLet \\(K\\) be the reflection of \\(A\\) over \\(M\\) \n\nClaim 3. \\(K\\) is the circumcenter of \\(\\triangle R T S\\) \n\nProof. Note that since \\(A Q R B\\) and \\(A P S C\\) are parallelograms we have that \\(A - L - R\\) are collinear and that \\(A - N - S\\) are collinear. The homothety centered at \\(A\\) with scale- factor 2 maps \\(\\triangle L M N\\) to \\(\\triangle R K S\\) , therefore \\(K R = K S\\) and \\(\\angle R K S = \\angle L M N = \\angle B D C = 2(180^{\\circ} - \\angle R T S)\\) (and \\(K\\) and \\(T\\) are in opposite sides of \\(R S\\) ), implying that \\(K\\) is the circumcenter of \\(\\triangle R T S\\) \n\nClaim 4. \\(K T = K I\\) \n\nProof. Note that \\(A O K D\\) is a parallelogram. Let \\(B T\\) intersect the \\((A B C)\\) again at point \\(G\\) . Since \\(\\angle A B G = \\angle A B T = \\angle Q A B = \\angle Q C A\\Rightarrow A Q = A G\\) and also \\(O Q = O G\\) hence \\(A O\\perp Q G\\) . Then by Reim's theorem we have that \\(Q G\\parallel T I\\) and also that \\(A O\\parallel D K\\) , so \\(D K\\perp T I\\) . Since \\(D I = D T\\) , it means that \\(K D\\) is the perpendicular bisector of \\(T I\\) , therefore \\(K T = K I\\) . \n\nThis means that \\(R T I S\\) is cyclic with center \\(K\\) . \n\nRemark. When \\(K\\) and \\(T\\) are on the same side of \\(R S\\) , it can be shown that \\(\\angle R K S = 2\\angle R T S\\) .\n\n\n", "metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P4.", "solution_match": "\nSolution 5. "}}
-{"year": "2025", "tier": "T2", "problem_label": "4", "problem_type": null, "exam": "EGMO", "problem": "Let \\(A B C\\) be an acute triangle with incentre \\(I\\) and \\(A B \\neq A C\\) . Let lines \\(B I\\) and \\(C I\\) intersect the circumcircle of \\(A B C\\) at \\(P \\neq B\\) and \\(Q \\neq C\\) , respectively. Consider points \\(R\\) and \\(S\\) such that \\(A Q R B\\) and \\(A C S P\\) are parallelograms (with \\(A Q \\parallel R B\\) , \\(A B \\parallel Q R\\) , \\(A C \\parallel S P\\) , and \\(A P \\parallel C S\\) ). Let \\(T\\) be the point of intersection of lines \\(R B\\) and \\(S C\\) . Prove that points \\(R\\) , \\(S\\) , \\(T\\) , and \\(I\\) are concyclic.", "solution": "As shown above, we have that \\(BTIC\\) is cyclic. Let \\(D\\) and \\(E\\) be the second intersections of \\(AC\\) and \\(AB\\) with this circle, respectively. Since the center of this circle lies on \\(AI\\) (by symmetry about \\(AI\\) ), we have that \\(AB = AD\\) and \\(AC = AE\\) , therefore \\(BE = CD\\) . Note that since \\(C - I - Q\\) and \\(A - B - E\\) are collinear, by Reim's theorem we have that \\(AQ \\parallel EI\\) and since \\(AQ \\parallel BT\\) , we have that \\(BT \\parallel EI\\) . Similarly, we get \\(CT \\parallel DI\\) . Let \\(F\\) and \\(G\\) be the intersections of \\(ID\\) and \\(IE\\) with \\(PS\\) and \\(QR\\) , respectively. Clearly, \\(RGEB\\) and \\(FSCD\\) are parallelograms. Since \\(RGEB\\) is parallelogram and \\(BEIT\\) is isosceles trapezoid, we have that \\(RGIT\\) is isosceles trapezoid. Similarly, \\(SFIT\\) is isosceles trapezoid. Hence, both of them are cyclic. Note also that \\(QR = AB = AD = PF\\) and \\(QG = AE = AC = PS\\) . Since \\(QR\\) and \\(PS\\) are tangents to the circumcircle of \\(\\triangle ABC\\) we have that \\(R\\) and \\(F\\) are symmetric (reflections) about the perpendicular bisector of \\(PQ\\) . Similarly, \\(G\\) and \\(S\\) are symmetric about the perpendicular bisector of \\(PQ\\) . This gives us that \\(QP \\parallel RF \\parallel GS\\) and that \\(RFSG\\) is an isosceles trapezoid, hence a cyclic quadrilateral with \\(\\angle RGS = 180^{\\circ} - \\angle GQP = 180^{\\circ} - \\angle QAP = 180^{\\circ} - \\angle RTS \\Rightarrow R, G, S, T\\) are concyclic. Combining all the facts about the cyclic quadrilaterals we proved above, we have that \\(R, G, S, F, I, T\\) are concyclic. Therefore \\(R, T, I, S\\) lie on a circle.\n\n\n", "metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P4.", "solution_match": "\nSolution 6. "}}
-{"year": "2025", "tier": "T2", "problem_label": "4", "problem_type": null, "exam": "EGMO", "problem": "Let \\(A B C\\) be an acute triangle with incentre \\(I\\) and \\(A B \\neq A C\\) . Let lines \\(B I\\) and \\(C I\\) intersect the circumcircle of \\(A B C\\) at \\(P \\neq B\\) and \\(Q \\neq C\\) , respectively. Consider points \\(R\\) and \\(S\\) such that \\(A Q R B\\) and \\(A C S P\\) are parallelograms (with \\(A Q \\parallel R B\\) , \\(A B \\parallel Q R\\) , \\(A C \\parallel S P\\) , and \\(A P \\parallel C S\\) ). Let \\(T\\) be the point of intersection of lines \\(R B\\) and \\(S C\\) . Prove that points \\(R\\) , \\(S\\) , \\(T\\) , and \\(I\\) are concyclic.", "solution": "Let \\(E\\) be the \\(A\\) - excenter of \\(\\triangle ABC\\) . Let the midpoints of \\(AQ, QB, CP, PA\\) be the points \\(F, G, H, J\\) , respectively. Both \\(PD\\) and \\(EC\\) are perpendicular to \\(CI\\) , hence \\(PD \\parallel CE\\) . \n\nSince \\(PA = PC\\) we have that \\(AJHC\\) is an isosceles trapezoid so it is cyclic. Let \\(K\\) be the second intersection of \\((AJHC)\\) and \\(AI\\) . Then \\(\\angle ADP = \\angle ACP = \\angle ACH = \\angle AKH \\Rightarrow DP \\parallel KH\\) . So \\(KH\\) is a line passing through the midpoint of the side \\(CP\\) of trapezoid \\(DPCE\\) and parallel to the bases, hence \\(K\\) is the midpoint of \\(DE\\) . Similarly, we show that the circle \\((AFGB)\\) passes through the midpoint of \\(DE\\) . Homothety centered at \\(A\\) with scale- factor 2 maps \\((AJH)\\) to \\((APS)\\) , \\((AFG)\\) to \\((AQR)\\) , and line \\(AI\\) to line \\(AI\\) . This means that the circles \\((AQR)\\) and \\((APS)\\) intersect on \\(AI\\) , call it point \\(L\\) . \n\n\n\n\n\n\nNow, \\(\\angle I L R = 180^{\\circ} - \\angle A Q R = \\angle Q A B = \\angle Q C B = 180^{\\circ} - \\angle I T B = 180^{\\circ} - \\angle I T R\\) , therefore \\(R,L,I,T\\) are concyclic. Similarly, we get that \\(S,L,T,I\\) are concyclic. Combining these, it means that \\(R\\) and \\(S\\) belong to the circle \\((L I T)\\) . The conclusion follows. \n\n", "metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P4.", "solution_match": "\nSolution 7. "}}
-{"year": "2025", "tier": "T2", "problem_label": "5", "problem_type": null, "exam": "EGMO", "problem": "Let \\(n > 1\\) be an integer. In a configuration of an \\(n \\times n\\) board, each of the \\(n^{2}\\) cells contains an arrow, either pointing up, down, left, or right. Given a starting configuration, Turbo the snail starts in one of the cells of the board and travels from cell to cell. In each move, Turbo moves one square unit in the direction indicated by the arrow in her cell (possibly leaving the board). After each move, the arrows in all of the cells rotate \\(90^{\\circ}\\) counterclockwise. We call a cell good if, starting from that cell, Turbo visits each cell of the board exactly once, without leaving the board, and returns to her initial cell at the end. Determine, in terms of \\(n\\) , the maximum number of good cells over all possible starting configurations.", "solution": "We will show that the maximum number of good cells over all possible starting configurations is \n\n\\[\\frac{n^{2}}{4} \\quad \\text{if} n \\text{ is even and}\\] \\[0 \\quad \\text{if} n \\text{ is odd.}\\] \n\n## Odd \\(n\\) \n\nFirst, we will prove that there are no good cells if \\(n\\) is an odd number. \n\nFor Turbo to reach her goal, she must return to her initial cell after visiting every cell exactly once. Consider the chessboard coloring of the board. Without loss of generality, we assume that Turbo starts in a black cell. Since, at every step, Turbo moves to a cell of a different color; she will be in a white cell after \\(n^{2} \\equiv 1 \\mod 2\\) moves. Thus, it is impossible for Turbo to come back to her initial black cell on her \\(n^{2}\\) - th move, which is a contradiction. Thus there are no good cells. \n\n## Lower bound for even \\(n\\) \n\nWe will now construct a starting configuration with \\(\\frac{n^{2}}{4}\\) good cells for even \\(n\\) . \n\nLet \\((i,j)\\) denote the cell in row \\(i\\) and column \\(j\\) . Consider the following cycle \n\n\\[\\begin{array}{r l} & {(1,1)\\rightarrow (1,2)\\rightarrow (1,3)\\rightarrow \\ldots \\rightarrow (1,n)}\\\\ & {\\qquad \\rightarrow (2,n)\\rightarrow (2,n - 1)\\rightarrow \\ldots \\rightarrow (2,2)}\\\\ & {\\qquad \\ldots}\\\\ & {\\qquad \\rightarrow (2i - 1,2)\\rightarrow (2i - 1,3)\\rightarrow \\ldots \\rightarrow (2i - 1,n)}\\\\ & {\\qquad \\rightarrow (2i,n)\\rightarrow (2i,n - 1)\\rightarrow \\ldots \\rightarrow (2i,2)}\\\\ & {\\qquad \\ldots}\\\\ & {\\qquad \\rightarrow (n,n)\\rightarrow (n,n - 1)\\rightarrow \\ldots \\rightarrow (n,2)}\\\\ & {\\qquad \\rightarrow (n,1)\\rightarrow (n - 1,1)\\rightarrow \\ldots \\rightarrow (2,1)\\rightarrow (1,1).} \\end{array} \\quad (1,1)\\] \n\n\n \n\nNote that the cycle returns to the initial cell after visiting every cell exactly once. To prove that \\((1,1)\\) is good, we need to find a starting configuration such that Turbo traverses this cycle. \n\nLet \\(c_{i}\\) be the \\((i - 1)\\) - th cell on the cycle: so we have \\(c_{0} = (1,1)\\) , \\(c_{2} = (1,2)\\) , ..., \\(c_{n^{2} - 1} = (2,1)\\) . For every \\(i\\) , we draw an arrow in cell \\(c_{i}\\) pointing towards cell \\(c_{i + 1}\\) (or pointing towards \\(c_{0}\\) if \\(i = n^{2} - 1\\) ) and then rotate this arrow \\(i\\) times \\(90^{\\circ}\\) in the clockwise direction. After \\(i\\) moves, the arrow in \\(c_{i}\\) will have rotated \\(i\\) times \\(90^{\\circ}\\) counterclockwise and be in the same direction as on the path defined above. Thus, Turbo will traverse the cycle \\(c_{0}, c_{1}, c_{2}, \\ldots , c_{n^{2} - 1}, c_{0}\\) and \\((1,1)\\) is good. \n\nEvery four moves, all arrows point in the same direction as in the beginning. Moreover, the board will return to its initial configuration after traversing the full cycle, since \\(n^{2}\\) , the length of the cycle, is divisible by 4. Therefore Turbo can also start at any \\(c_{i}\\) with 4 \\(| i\\) and follow the same route. Hence the cells \\(c_{0}, c_{4}, c_{8}, \\ldots , c_{n^{2} - 4}\\) are good and there are \\(\\frac{n^{2}}{4}\\) of such cells.\n\n\n\n## Upper bound for even \\(n\\) \n\nWe will prove that for even \\(n\\) and any start configuration there are at most \\(\\frac{n^{2}}{4}\\) good cells. \n\nLet \\(a_{0}\\) be a good cell. Let \\(a_{0},a_{1},a_{2},\\ldots ,a_{n^{2} - 1},a_{n^{2}} = a_{0}\\) be the sequence of cells that Turbo visits when she starts at \\(a_{0}\\) . Now suppose there is another good cell \\(b_{0}\\) and let \\(b_{0},b_{1},b_{2},\\ldots ,b_{n^{2} - 1},b_{n^{2}} = b_{0}\\) be the sequence of cells that Turbo visits when she starts at \\(b_{0}\\) . \n\nNote that, since \\(4\\mid n^{2}\\) , the arrows are back to their initial configuration after \\(n^{2}\\) steps. Thus, if Turbo keeps walking after returning to her initial cell, she would just traverse the same cycle over and over again. \n\nConsider the upper left corner of the board. With standard row and column numbering, the corner cell is \\((1,1)\\) . This cell has only two neighbours, so both the \\(a\\) - route and the \\(b\\) - route must have cells \\((2,1),(1,1),(1,2)\\) in that order or \\((1,2),(1,1),(2,1)\\) in that order. Without loss of generality, \\(a_{i - 1} =\\) \\((2,1)\\) , \\(a_{i} = (1,1)\\) and \\(a_{i + 1} = (1,2)\\) for some \\(i\\) . Let \\(j\\) be such that \\(b_{j} = (1,1)\\) . If \\(b_{j - 1} = (2,1) = a_{i - 1}\\) then the arrow in cell \\((2,1)\\) must be pointed in the same direction after \\(i - 1\\) steps and after \\(j - 1\\) steps, so \\(i\\equiv j\\) mod 4. But then the arrow in cell \\(b_{j} = (1,1) = a_{i}\\) must also be pointed in the same direction after \\(i\\) and after \\(j\\) steps, so Turbo moves to \\(b_{j + 1} = a_{i + 1}\\) in both cases, and again finds the arrow pointed in the same direction in both cases. Continuing, we find that the \\(b\\) - route is actually identical to \\(a_{4t},a_{4t + 1},\\ldots ,a_{n^{2}} = a_{0},a_{1},\\ldots ,a_{4t - 1},a_{4t}\\) for some \\(t\\) , as any other starting point would have the arrows in the wrong direction initially. \n\nNow suppose instead that \\(b_{j + 1} = (2,1) = a_{i - 1}\\) . Considering the \\(a\\) - route, the arrows in the upper left corner after \\(i - 1\\) steps must be like this: \n\n\n \n\nConsidering the \\(b\\) - route instead, the arrows after \\(j - 1\\) steps must be like this: \n\n\n \n\nFrom the arrows in cell \\((1,1)\\) we see that \\(i\\equiv j + 1\\) mod 4. However, for the cells \\((2,1)\\) and \\((1,2)\\) this gives a contradiction. \n\nWe conclude that the only possible good cells are \\(a_{4t}\\) for \\(t = 0,1,\\ldots ,\\frac{n^{2}}{4} - 1\\) , which gives at most \\(\\frac{n^{2}}{4}\\) good cells.", "metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P5.", "solution_match": "\nSolution."}}
+{"year": "2025", "tier": "T2", "problem_label": "4", "problem_type": null, "exam": "EGMO", "problem": "Let \\(A B C\\) be an acute triangle with incentre \\(I\\) and \\(A B \\neq A C\\) . Let lines \\(B I\\) and \\(C I\\) intersect the circumcircle of \\(A B C\\) at \\(P \\neq B\\) and \\(Q \\neq C\\) , respectively. Consider points \\(R\\) and \\(S\\) such that \\(A Q R B\\) and \\(A C S P\\) are parallelograms (with \\(A Q \\parallel R B\\) , \\(A B \\parallel Q R\\) , \\(A C \\parallel S P\\) , and \\(A P \\parallel C S\\) ). Let \\(T\\) be the point of intersection of lines \\(R B\\) and \\(S C\\) . Prove that points \\(R\\) , \\(S\\) , \\(T\\) , and \\(I\\) are concyclic.", "solution": "Let \\(E\\) , \\(F\\) , \\(G\\) be the midpoints of \\(A I\\) , \\(B Q\\) , \\(C P\\) . As in Solution 1, angle chase shows that \\(\\triangle A Q P\\sim\\) \\(\\triangle I B C\\) . \n\nNote that by the Mean Geometry Theorem we have that \\(\\frac{1}{2} A Q P + \\frac{1}{2} I B C = E F G\\) is similar to \\(\\triangle I B C\\) . Homothety with center \\(A\\) and scale- factor 2 maps \\(E F G\\) to \\(I R S\\) . Hence \\(\\angle R I S = \\angle F E G = \\angle Q A P =\\) \\(\\angle B T C = \\angle R T S\\) , so \\(R,T,I,S\\) are concyclic. \n\nRemark. As shown above, \\(E\\) lies on \\(Q P\\) and \\(A I\\perp P Q\\) . One can prove that \\(\\angle F E G = \\angle B I C\\) in another way. Let \\(J\\) be the midpoint of \\(P Q\\) . Then \\(\\angle B I C = \\angle F J G\\) by midlines and \\(\\angle F J G = \\angle F E G\\) by the lemma below applied in \\(B C P Q\\) . \n\nLemma. Let \\(A B C D\\) is a cyclic quadrilateral and \\(E\\) is the intersection of its diagonals. Then the midpoints of \\(A B\\) , \\(B C\\) , \\(C D\\) and the foot of the perpendicular from \\(E\\) to \\(B C\\) are concyclic.\n\n\n", "metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P4.", "solution_match": "\nSolution 4. "}}
+{"year": "2025", "tier": "T2", "problem_label": "4", "problem_type": null, "exam": "EGMO", "problem": "Let \\(A B C\\) be an acute triangle with incentre \\(I\\) and \\(A B \\neq A C\\) . Let lines \\(B I\\) and \\(C I\\) intersect the circumcircle of \\(A B C\\) at \\(P \\neq B\\) and \\(Q \\neq C\\) , respectively. Consider points \\(R\\) and \\(S\\) such that \\(A Q R B\\) and \\(A C S P\\) are parallelograms (with \\(A Q \\parallel R B\\) , \\(A B \\parallel Q R\\) , \\(A C \\parallel S P\\) , and \\(A P \\parallel C S\\) ). Let \\(T\\) be the point of intersection of lines \\(R B\\) and \\(S C\\) . Prove that points \\(R\\) , \\(S\\) , \\(T\\) , and \\(I\\) are concyclic.", "solution": "Let \\(O\\) be the circumcenter of \\((A B C)\\) . Let \\(M\\) , \\(N\\) , and \\(L\\) be the midpoints of \\(O D\\) , \\(P C\\) , and \\(Q B\\) respectively. \n\nClaim 1. \\(\\triangle O P Q\\) and \\(\\triangle D C B\\) are directly similar. \n\nProof. Clearly \\(D B = D C\\) and \\(O Q = O P\\) . Also note that \\(\\angle Q O P = 2\\angle Q D P = 2\\angle Q D A + 2\\angle P D A =\\) \\(\\angle B D A + \\angle C D A = \\angle B D C\\) . So the two triangles are directly similar by SAS. \n\nClaim 2. \\(M L = M N\\) and \\(\\angle L M N = 180^{\\circ} - \\angle B A C\\) \n\nProof. Note that since \\(\\triangle O Q P\\sim \\triangle D B C\\) by the Mean Geometry Theorem, we have that the average of the two triangles is also similar to them, therefore \\(\\triangle M L N\\sim \\triangle D B C\\Rightarrow M L = M N\\) and \\(\\angle L M N =\\) \\(\\angle B D C = 180^{\\circ} - \\angle B A C\\) \n\nLet \\(K\\) be the reflection of \\(A\\) over \\(M\\) \n\nClaim 3. \\(K\\) is the circumcenter of \\(\\triangle R T S\\) \n\nProof. Note that since \\(A Q R B\\) and \\(A P S C\\) are parallelograms we have that \\(A - L - R\\) are collinear and that \\(A - N - S\\) are collinear. The homothety centered at \\(A\\) with scale- factor 2 maps \\(\\triangle L M N\\) to \\(\\triangle R K S\\) , therefore \\(K R = K S\\) and \\(\\angle R K S = \\angle L M N = \\angle B D C = 2(180^{\\circ} - \\angle R T S)\\) (and \\(K\\) and \\(T\\) are in opposite sides of \\(R S\\) ), implying that \\(K\\) is the circumcenter of \\(\\triangle R T S\\) \n\nClaim 4. \\(K T = K I\\) \n\nProof. Note that \\(A O K D\\) is a parallelogram. Let \\(B T\\) intersect the \\((A B C)\\) again at point \\(G\\) . Since \\(\\angle A B G = \\angle A B T = \\angle Q A B = \\angle Q C A\\Rightarrow A Q = A G\\) and also \\(O Q = O G\\) hence \\(A O\\perp Q G\\) . Then by Reim's theorem we have that \\(Q G\\parallel T I\\) and also that \\(A O\\parallel D K\\) , so \\(D K\\perp T I\\) . Since \\(D I = D T\\) , it means that \\(K D\\) is the perpendicular bisector of \\(T I\\) , therefore \\(K T = K I\\) . \n\nThis means that \\(R T I S\\) is cyclic with center \\(K\\) . \n\nRemark. When \\(K\\) and \\(T\\) are on the same side of \\(R S\\) , it can be shown that \\(\\angle R K S = 2\\angle R T S\\) .\n\n\n", "metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P4.", "solution_match": "\nSolution 5. "}}
+{"year": "2025", "tier": "T2", "problem_label": "4", "problem_type": null, "exam": "EGMO", "problem": "Let \\(A B C\\) be an acute triangle with incentre \\(I\\) and \\(A B \\neq A C\\) . Let lines \\(B I\\) and \\(C I\\) intersect the circumcircle of \\(A B C\\) at \\(P \\neq B\\) and \\(Q \\neq C\\) , respectively. Consider points \\(R\\) and \\(S\\) such that \\(A Q R B\\) and \\(A C S P\\) are parallelograms (with \\(A Q \\parallel R B\\) , \\(A B \\parallel Q R\\) , \\(A C \\parallel S P\\) , and \\(A P \\parallel C S\\) ). Let \\(T\\) be the point of intersection of lines \\(R B\\) and \\(S C\\) . Prove that points \\(R\\) , \\(S\\) , \\(T\\) , and \\(I\\) are concyclic.", "solution": "As shown above, we have that \\(BTIC\\) is cyclic. Let \\(D\\) and \\(E\\) be the second intersections of \\(AC\\) and \\(AB\\) with this circle, respectively. Since the center of this circle lies on \\(AI\\) (by symmetry about \\(AI\\) ), we have that \\(AB = AD\\) and \\(AC = AE\\) , therefore \\(BE = CD\\) . Note that since \\(C - I - Q\\) and \\(A - B - E\\) are collinear, by Reim's theorem we have that \\(AQ \\parallel EI\\) and since \\(AQ \\parallel BT\\) , we have that \\(BT \\parallel EI\\) . Similarly, we get \\(CT \\parallel DI\\) . Let \\(F\\) and \\(G\\) be the intersections of \\(ID\\) and \\(IE\\) with \\(PS\\) and \\(QR\\) , respectively. Clearly, \\(RGEB\\) and \\(FSCD\\) are parallelograms. Since \\(RGEB\\) is parallelogram and \\(BEIT\\) is isosceles trapezoid, we have that \\(RGIT\\) is isosceles trapezoid. Similarly, \\(SFIT\\) is isosceles trapezoid. Hence, both of them are cyclic. Note also that \\(QR = AB = AD = PF\\) and \\(QG = AE = AC = PS\\) . Since \\(QR\\) and \\(PS\\) are tangents to the circumcircle of \\(\\triangle ABC\\) we have that \\(R\\) and \\(F\\) are symmetric (reflections) about the perpendicular bisector of \\(PQ\\) . Similarly, \\(G\\) and \\(S\\) are symmetric about the perpendicular bisector of \\(PQ\\) . This gives us that \\(QP \\parallel RF \\parallel GS\\) and that \\(RFSG\\) is an isosceles trapezoid, hence a cyclic quadrilateral with \\(\\angle RGS = 180^{\\circ} - \\angle GQP = 180^{\\circ} - \\angle QAP = 180^{\\circ} - \\angle RTS \\Rightarrow R, G, S, T\\) are concyclic. Combining all the facts about the cyclic quadrilaterals we proved above, we have that \\(R, G, S, F, I, T\\) are concyclic. Therefore \\(R, T, I, S\\) lie on a circle.\n\n\n", "metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P4.", "solution_match": "\nSolution 6. "}}
+{"year": "2025", "tier": "T2", "problem_label": "4", "problem_type": null, "exam": "EGMO", "problem": "Let \\(A B C\\) be an acute triangle with incentre \\(I\\) and \\(A B \\neq A C\\) . Let lines \\(B I\\) and \\(C I\\) intersect the circumcircle of \\(A B C\\) at \\(P \\neq B\\) and \\(Q \\neq C\\) , respectively. Consider points \\(R\\) and \\(S\\) such that \\(A Q R B\\) and \\(A C S P\\) are parallelograms (with \\(A Q \\parallel R B\\) , \\(A B \\parallel Q R\\) , \\(A C \\parallel S P\\) , and \\(A P \\parallel C S\\) ). Let \\(T\\) be the point of intersection of lines \\(R B\\) and \\(S C\\) . Prove that points \\(R\\) , \\(S\\) , \\(T\\) , and \\(I\\) are concyclic.", "solution": "Let \\(E\\) be the \\(A\\) - excenter of \\(\\triangle ABC\\) . Let the midpoints of \\(AQ, QB, CP, PA\\) be the points \\(F, G, H, J\\) , respectively. Both \\(PD\\) and \\(EC\\) are perpendicular to \\(CI\\) , hence \\(PD \\parallel CE\\) . \n\nSince \\(PA = PC\\) we have that \\(AJHC\\) is an isosceles trapezoid so it is cyclic. Let \\(K\\) be the second intersection of \\((AJHC)\\) and \\(AI\\) . Then \\(\\angle ADP = \\angle ACP = \\angle ACH = \\angle AKH \\Rightarrow DP \\parallel KH\\) . So \\(KH\\) is a line passing through the midpoint of the side \\(CP\\) of trapezoid \\(DPCE\\) and parallel to the bases, hence \\(K\\) is the midpoint of \\(DE\\) . Similarly, we show that the circle \\((AFGB)\\) passes through the midpoint of \\(DE\\) . Homothety centered at \\(A\\) with scale- factor 2 maps \\((AJH)\\) to \\((APS)\\) , \\((AFG)\\) to \\((AQR)\\) , and line \\(AI\\) to line \\(AI\\) . This means that the circles \\((AQR)\\) and \\((APS)\\) intersect on \\(AI\\) , call it point \\(L\\) . \n\n\n\n\n\n\nNow, \\(\\angle I L R = 180^{\\circ} - \\angle A Q R = \\angle Q A B = \\angle Q C B = 180^{\\circ} - \\angle I T B = 180^{\\circ} - \\angle I T R\\) , therefore \\(R,L,I,T\\) are concyclic. Similarly, we get that \\(S,L,T,I\\) are concyclic. Combining these, it means that \\(R\\) and \\(S\\) belong to the circle \\((L I T)\\) . The conclusion follows. \n\n", "metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P4.", "solution_match": "\nSolution 7. "}}
+{"year": "2025", "tier": "T2", "problem_label": "5", "problem_type": null, "exam": "EGMO", "problem": "Let \\(n > 1\\) be an integer. In a configuration of an \\(n \\times n\\) board, each of the \\(n^{2}\\) cells contains an arrow, either pointing up, down, left, or right. Given a starting configuration, Turbo the snail starts in one of the cells of the board and travels from cell to cell. In each move, Turbo moves one square unit in the direction indicated by the arrow in her cell (possibly leaving the board). After each move, the arrows in all of the cells rotate \\(90^{\\circ}\\) counterclockwise. We call a cell good if, starting from that cell, Turbo visits each cell of the board exactly once, without leaving the board, and returns to her initial cell at the end. Determine, in terms of \\(n\\) , the maximum number of good cells over all possible starting configurations.", "solution": "We will show that the maximum number of good cells over all possible starting configurations is \n\n\\[\\frac{n^{2}}{4} \\quad \\text{if} n \\text{ is even and}\\] \\[0 \\quad \\text{if} n \\text{ is odd.}\\] \n\n## Odd \\(n\\) \n\nFirst, we will prove that there are no good cells if \\(n\\) is an odd number. \n\nFor Turbo to reach her goal, she must return to her initial cell after visiting every cell exactly once. Consider the chessboard coloring of the board. Without loss of generality, we assume that Turbo starts in a black cell. Since, at every step, Turbo moves to a cell of a different color; she will be in a white cell after \\(n^{2} \\equiv 1 \\mod 2\\) moves. Thus, it is impossible for Turbo to come back to her initial black cell on her \\(n^{2}\\) - th move, which is a contradiction. Thus there are no good cells. \n\n## Lower bound for even \\(n\\) \n\nWe will now construct a starting configuration with \\(\\frac{n^{2}}{4}\\) good cells for even \\(n\\) . \n\nLet \\((i,j)\\) denote the cell in row \\(i\\) and column \\(j\\) . Consider the following cycle \n\n\\[\\begin{array}{r l} & {(1,1)\\rightarrow (1,2)\\rightarrow (1,3)\\rightarrow \\ldots \\rightarrow (1,n)}\\\\ & {\\qquad \\rightarrow (2,n)\\rightarrow (2,n - 1)\\rightarrow \\ldots \\rightarrow (2,2)}\\\\ & {\\qquad \\ldots}\\\\ & {\\qquad \\rightarrow (2i - 1,2)\\rightarrow (2i - 1,3)\\rightarrow \\ldots \\rightarrow (2i - 1,n)}\\\\ & {\\qquad \\rightarrow (2i,n)\\rightarrow (2i,n - 1)\\rightarrow \\ldots \\rightarrow (2i,2)}\\\\ & {\\qquad \\ldots}\\\\ & {\\qquad \\rightarrow (n,n)\\rightarrow (n,n - 1)\\rightarrow \\ldots \\rightarrow (n,2)}\\\\ & {\\qquad \\rightarrow (n,1)\\rightarrow (n - 1,1)\\rightarrow \\ldots \\rightarrow (2,1)\\rightarrow (1,1).} \\end{array} \\quad (1,1)\\] \n\n\n \n\nNote that the cycle returns to the initial cell after visiting every cell exactly once. To prove that \\((1,1)\\) is good, we need to find a starting configuration such that Turbo traverses this cycle. \n\nLet \\(c_{i}\\) be the \\((i - 1)\\) - th cell on the cycle: so we have \\(c_{0} = (1,1)\\) , \\(c_{2} = (1,2)\\) , ..., \\(c_{n^{2} - 1} = (2,1)\\) . For every \\(i\\) , we draw an arrow in cell \\(c_{i}\\) pointing towards cell \\(c_{i + 1}\\) (or pointing towards \\(c_{0}\\) if \\(i = n^{2} - 1\\) ) and then rotate this arrow \\(i\\) times \\(90^{\\circ}\\) in the clockwise direction. After \\(i\\) moves, the arrow in \\(c_{i}\\) will have rotated \\(i\\) times \\(90^{\\circ}\\) counterclockwise and be in the same direction as on the path defined above. Thus, Turbo will traverse the cycle \\(c_{0}, c_{1}, c_{2}, \\ldots , c_{n^{2} - 1}, c_{0}\\) and \\((1,1)\\) is good. \n\nEvery four moves, all arrows point in the same direction as in the beginning. Moreover, the board will return to its initial configuration after traversing the full cycle, since \\(n^{2}\\) , the length of the cycle, is divisible by 4. Therefore Turbo can also start at any \\(c_{i}\\) with 4 \\(| i\\) and follow the same route. Hence the cells \\(c_{0}, c_{4}, c_{8}, \\ldots , c_{n^{2} - 4}\\) are good and there are \\(\\frac{n^{2}}{4}\\) of such cells.\n\n\n\n## Upper bound for even \\(n\\) \n\nWe will prove that for even \\(n\\) and any start configuration there are at most \\(\\frac{n^{2}}{4}\\) good cells. \n\nLet \\(a_{0}\\) be a good cell. Let \\(a_{0},a_{1},a_{2},\\ldots ,a_{n^{2} - 1},a_{n^{2}} = a_{0}\\) be the sequence of cells that Turbo visits when she starts at \\(a_{0}\\) . Now suppose there is another good cell \\(b_{0}\\) and let \\(b_{0},b_{1},b_{2},\\ldots ,b_{n^{2} - 1},b_{n^{2}} = b_{0}\\) be the sequence of cells that Turbo visits when she starts at \\(b_{0}\\) . \n\nNote that, since \\(4\\mid n^{2}\\) , the arrows are back to their initial configuration after \\(n^{2}\\) steps. Thus, if Turbo keeps walking after returning to her initial cell, she would just traverse the same cycle over and over again. \n\nConsider the upper left corner of the board. With standard row and column numbering, the corner cell is \\((1,1)\\) . This cell has only two neighbours, so both the \\(a\\) - route and the \\(b\\) - route must have cells \\((2,1),(1,1),(1,2)\\) in that order or \\((1,2),(1,1),(2,1)\\) in that order. Without loss of generality, \\(a_{i - 1} =\\) \\((2,1)\\) , \\(a_{i} = (1,1)\\) and \\(a_{i + 1} = (1,2)\\) for some \\(i\\) . Let \\(j\\) be such that \\(b_{j} = (1,1)\\) . If \\(b_{j - 1} = (2,1) = a_{i - 1}\\) then the arrow in cell \\((2,1)\\) must be pointed in the same direction after \\(i - 1\\) steps and after \\(j - 1\\) steps, so \\(i\\equiv j\\) mod 4. But then the arrow in cell \\(b_{j} = (1,1) = a_{i}\\) must also be pointed in the same direction after \\(i\\) and after \\(j\\) steps, so Turbo moves to \\(b_{j + 1} = a_{i + 1}\\) in both cases, and again finds the arrow pointed in the same direction in both cases. Continuing, we find that the \\(b\\) - route is actually identical to \\(a_{4t},a_{4t + 1},\\ldots ,a_{n^{2}} = a_{0},a_{1},\\ldots ,a_{4t - 1},a_{4t}\\) for some \\(t\\) , as any other starting point would have the arrows in the wrong direction initially. \n\nNow suppose instead that \\(b_{j + 1} = (2,1) = a_{i - 1}\\) . Considering the \\(a\\) - route, the arrows in the upper left corner after \\(i - 1\\) steps must be like this: \n\n\n \n\nConsidering the \\(b\\) - route instead, the arrows after \\(j - 1\\) steps must be like this: \n\n\n \n\nFrom the arrows in cell \\((1,1)\\) we see that \\(i\\equiv j + 1\\) mod 4. However, for the cells \\((2,1)\\) and \\((1,2)\\) this gives a contradiction. \n\nWe conclude that the only possible good cells are \\(a_{4t}\\) for \\(t = 0,1,\\ldots ,\\frac{n^{2}}{4} - 1\\) , which gives at most \\(\\frac{n^{2}}{4}\\) good cells.", "metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P5.", "solution_match": "\nSolution."}}
{"year": "2025", "tier": "T2", "problem_label": "6", "problem_type": null, "exam": "EGMO", "problem": "In each cell of a \\(2025 \\times 2025\\) board, a nonnegative real number is written in such a way that the sum of the numbers in each row is equal to 1, and the sum of the numbers in each column is equal to 1. Define \\(r_{i}\\) to be the largest value in row \\(i\\) , and let \\(R = r_{1} + r_{2} + \\dots + r_{2025}\\) . Similarly, define \\(c_{i}\\) to be the largest value in column \\(i\\) , and let \\(C = c_{1} + c_{2} + \\dots + c_{2025}\\) .What is the largest possible value of \\(\\frac{R}{C}\\) ? \n\nWhat is the largest possible value of \\(\\frac{R}{C}\\) ?", "solution": "Answer: \\(\\frac{2025}{89}\\) . \n\nIn general, if the table is \\(m^{2} \\times m^{2}\\) , the answer is \\(\\frac{m^{2}}{2m - 1}\\) . \n\nThe example is as follows: label rows and columns from 1 to \\(m^{2}\\) , from top to bottom and left to right. For the first \\(m\\) columns, write \\(\\frac{1}{m}\\) in all squares whose coordinates have the same residue modulo \\(m\\) and place 0 everywhere else. For the remaining \\(m^{2} - m\\) columns, place \\(\\frac{1}{m^{2}}\\) everywhere. Then \\(R = m^{2} \\cdot \\frac{1}{m} = m\\) , and \\(C = m \\cdot \\frac{1}{m} + (m^{2} - m) \\cdot \\frac{1}{m^{2}} = 2 - \\frac{1}{m}\\) . So the ratio is as claimed. \n\n| 1/2 | 0 | 1/4 | 1/4 |
| 0 | 1/2 | 1/4 | 1/4 |
| 1/2 | 0 | 1/4 | 1/4 |
| 0 | 1/2 | 1/4 | 1/4 |
\n\nIn particular, when \\(n := m^{2} = 2025\\) , we get \\(\\frac{2025}{89}\\) . Now we need to show that \\(\\frac{R}{C} \\leq \\frac{n}{2\\sqrt{n} - 1}\\) . \n\nFor each row, select one cell having the largest value appearing in said row and colour it red. Then, without loss of generality, we may rearrange the columns such that the red cells appear in the first \\(k\\) columns from the left, and each column contains at least one red cell, for some \\(k \\leq n\\) . \n\nFor the \\(j^{\\mathrm{th}}\\) column, for all \\(1 \\leq j \\leq k\\) , let \\(p_{j}\\) and \\(n_{j}\\) denote the sum and number of red cells in it, respectively. We observe that \\(c_{j}\\) , the biggest number in the \\(j^{\\mathrm{th}}\\) column, is at least \\(\\frac{p_{j}}{n_{j}}\\) , for all \\(1 \\leq j \\leq k\\) . For all other columns, the largest value they contain is at least \\(\\frac{1}{n}\\) , as their sum is 1. Thus, \\(C \\geq \\frac{p_{1}}{n_{1}} + \\frac{p_{2}}{n_{2}} + \\dots + \\frac{p_{k}}{n_{k}} + \\frac{n - k}{n}\\) . \n\nWe can also observe that \\(R = p_{1} + p_{2} + \\dots + p_{k}\\) . \n\nTherefore we have to show that: \n\n\\[p_{1} + p_{2} + \\dots +p_{k} \\leq \\frac{n}{2\\sqrt{n} - 1} \\cdot \\left(\\frac{p_{1}}{n_{1}} + \\frac{p_{2}}{n_{2}} + \\dots + \\frac{p_{k}}{n_{k}} + \\frac{n - k}{n}\\right). \\quad (*)\\] \n\nBy construction, \\(n_{1} + n_{2} + \\dots + n_{k} = n\\) , and, as the numbers in every column are nonnegative, we see that \\(p_{j} \\leq 1\\) for every \\(j\\) . Also, since each number in a red cell is at least \\(\\frac{1}{n}\\) , we also have \\(p_{j} \\geq \\frac{n_{j}}{n}\\) . \n\nSince our inequality is linear in each \\(p_{j}\\) , it suffices to prove it when each variable equals one of its two critical values. By relabeling, we may assume that \\(p_{j} = \\frac{n_{j}}{n}\\) for \\(1 \\leq j \\leq t\\) , and \\(p_{j} = 1\\) for \\(t + 1 \\leq j \\leq k\\) , for an integer \\(0 \\leq t \\leq k\\) . \n\nFirst, if \\(t = k\\) , we observe that \\(p_{1} + p_{2} + \\dots +p_{k} = \\frac{n_{1}}{n} + \\frac{n_{2}}{n} + \\dots + \\frac{n_{k}}{n} = 1\\) , and that \\(\\frac{p_{1}}{n_{1}} + \\frac{p_{2}}{n_{2}} + \\dots + \\frac{p_{k}}{n_{k}} = \\frac{k}{n}\\) , so the inequality becomes \\(1 \\leq \\frac{n}{2\\sqrt{n} - 1}\\) , which is true. \n\nFrom now on we may assume that \\(t < k\\) . We need to show that: \n\n\\[\\frac{n_{1} + \\dots + n_{t}}{n} +k - t \\leq \\frac{n}{2\\sqrt{n} - 1} \\cdot \\left(\\frac{t}{n} + \\frac{1}{n_{t + 1}} + \\dots + \\frac{1}{n_{k}} + \\frac{n - k}{n}\\right).\\] \n\nBy Cauchy- Schwarz inequality we have that: \n\n\\[\\frac{1}{n_{t + 1}} + \\dots + \\frac{1}{n_{k}} \\geq \\frac{(k - t)^{2}}{n_{t + 1} + \\dots + n_{k}} = \\frac{(k - t)^{2}}{n - (n_{1} + \\dots + n_{t})}. \\quad (CS)\\]\n\n\n\nLet \\(n_{1} + \\dots +n_{t} = n\\cdot q\\) , where \\(0\\leq q< 1\\) . Thus, it is now enough to show that: \n\n\\[q + k - t\\leq \\frac{n}{2\\sqrt{n - 1}}\\left(\\frac{t}{n} +\\frac{(k - t)^{2}}{n - nq} +\\frac{n - k}{n}\\right).\\] \n\nLet \\(k - t = \\ell \\geq 1\\) . The inequality becomes: \n\n\\[q + \\ell \\leq \\frac{1}{2\\sqrt{n} - 1}\\cdot \\left(n - \\ell +\\frac{\\ell^{2}}{1 - q}\\right).\\] \n\nRearranging this we get: \n\n\\[n + q + \\frac{\\ell^{2}}{1 - q}\\geq 2(q + \\ell)\\sqrt{n}.\\] \n\nIf \\(q = 0\\) the inequality is trivially true by \\(A M - G M\\) . Suppose now that \\(0< q< 1\\) . Then, by Cauchy- Schwarz, we have: \n\n\\[q + \\frac{\\ell^{2}}{1 - q} = \\frac{q^{2}}{q} +\\frac{\\ell^{2}}{1 - q}\\geq (q + \\ell)^{2}.\\] \n\nIt is therefore enough to show that \\(n + (q + \\ell)^{2}\\geq 2(q + \\ell)\\sqrt{n}\\) , which is true by \\(A M - G M\\) , completing the proof. \n\nRemark. Another way to show \\(n + q + \\frac{\\ell^{2}}{1 - q}\\geq 2(q + \\ell)\\sqrt{n}\\) is to split it into two \\(A M - G M\\) 's, as: \n\n\\[\\left(n(1 - q) + \\frac{\\ell^{2}}{1 - q}\\right) + (nq + q)\\geq 2\\ell \\sqrt{n} +2q\\sqrt{n}.\\]", "metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P6.", "solution_match": "\nSolution 1. "}}
{"year": "2025", "tier": "T2", "problem_label": "6", "problem_type": null, "exam": "EGMO", "problem": "In each cell of a \\(2025 \\times 2025\\) board, a nonnegative real number is written in such a way that the sum of the numbers in each row is equal to 1, and the sum of the numbers in each column is equal to 1. Define \\(r_{i}\\) to be the largest value in row \\(i\\) , and let \\(R = r_{1} + r_{2} + \\dots + r_{2025}\\) . Similarly, define \\(c_{i}\\) to be the largest value in column \\(i\\) , and let \\(C = c_{1} + c_{2} + \\dots + c_{2025}\\) .What is the largest possible value of \\(\\frac{R}{C}\\) ? \n\nWhat is the largest possible value of \\(\\frac{R}{C}\\) ?", "solution": "We prove the main inequality \\((\\ast)\\) in a slightly different manner. Instead of the strong lower bound \\(p_{j}\\geq \\frac{n_{j}}{n}\\) , we use the weaker, simpler and more immediate lower bound \\(p_{j}\\geq 0\\) (thus proving the inequality in a larger regime). \n\nAs in Solution 1, suppose \\(p_{j} = 0\\) for \\(1\\leq j\\leq t\\) and \\(p_{j} = 1\\) for \\(t + 1\\leq j\\leq k\\) , with \\(\\ell = k - t\\) . We also denote by \\(m = n_{t + 1} + \\dots +n_{k}\\) and note that \\(m\\leq n - t\\) , since \\(n_{j}\\geq 1\\) for each \\(i\\leq t\\) . We need to prove that: \n\n\\[\\ell \\leq \\frac{n}{2\\sqrt{n} - 1}\\cdot \\left(\\frac{1}{n_{t + 1}} +\\dots +\\frac{1}{n_{k}} +\\frac{n - k}{n}\\right).\\] \n\nRearranging and using the same Cauchy- Schwarz (CS) as in Solution 1, we see it suffices to show that: \n\n\\[(2\\sqrt{n} -1)\\ell \\leq n - k + \\frac{n\\ell^{2}}{m},\\] \n\nor equivalently, that: \n\n\\[2\\sqrt{n}\\ell \\leq n - t + \\frac{n\\ell^{2}}{m}.\\] \n\nBut since \\(n - t\\geq m\\) this immediately follows from \\(2\\sqrt{n}\\ell \\leq m + \\frac{n\\ell^{2}}{m}\\) , which is a simple application of \\(A M - G M\\) . \n\n(Note that in Solution 1 the case \\(t = k\\) , which is equivalent to \\(m = \\ell = 0\\) , was dealt with separately, to avoid the appearance of \\(\\frac{0}{0}\\) terms such as \\(\\frac{n\\ell^{2}}{m}\\) . It is easy to verify that the corresponding term should in fact be 0 and the final \\(A M - G M\\) replaced with \\(0\\leq 0\\) , and all transitions are valid. Alternatively, the case can be argued directly by simply noting that \\((\\ast)\\) evaluates to \\(0\\leq \\frac{n - k}{2\\sqrt{n} - 1}\\) , which is obvious.)", "metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P6.", "solution_match": "\nSolution 1'."}}
{"year": "2025", "tier": "T2", "problem_label": "6", "problem_type": null, "exam": "EGMO", "problem": "In each cell of a \\(2025 \\times 2025\\) board, a nonnegative real number is written in such a way that the sum of the numbers in each row is equal to 1, and the sum of the numbers in each column is equal to 1. Define \\(r_{i}\\) to be the largest value in row \\(i\\) , and let \\(R = r_{1} + r_{2} + \\dots + r_{2025}\\) . Similarly, define \\(c_{i}\\) to be the largest value in column \\(i\\) , and let \\(C = c_{1} + c_{2} + \\dots + c_{2025}\\) .What is the largest possible value of \\(\\frac{R}{C}\\) ? \n\nWhat is the largest possible value of \\(\\frac{R}{C}\\) ?", "solution": "1\". This is an alternative way of getting the upper bound on \\(\\frac{R}{C}\\) from \n\n\\[\\frac{R}{C}\\leq \\frac{p_{1} + p_{2} + \\ldots + p_{k}}{\\frac{p_{1}}{n_{1}} + \\frac{p_{2}}{n_{2}} + \\ldots + \\frac{p_{k}}{n_{k}} + \\frac{n - k}{n}}.\\] \n\nUsing the fact that \\(\\sum_{j = 1}^{k}n_{j} = n\\) , we can rewrite the above right hand side as follows:\n\n\n\n\\[\\frac{\\sum_{j = 1}^{k}p_{j}}{\\sum_{j = 1}^{k}\\left(\\frac{p_{j}}{n_{j}} +\\frac{n_{j} - 1}{n}\\right)}.\\] \n\nWe notice that this is a quotient of affine functions in the \\(p_{j}\\) 's, for which the denominator does not vanish over the set defined by \\(0 \\leq p_{j} \\leq 1\\) . Therefore the maximum of this function is attained when a certain number of \\(p_{j}\\) 's are 1 and the others are 0. Without loss of generality we may assume that the first \\(t\\) are equal to 1 and the other \\(k - t\\) are 0 for some \\(0 \\leq t \\leq k\\) . Then one has that the previous expression is at most \n\n\\[\\frac{t}{\\sum_{1 \\leq j \\leq t} \\left(\\frac{1}{n_{j}} + \\frac{n_{j} - 1}{n}\\right) + \\sum_{t < j \\leq k} \\frac{n_{j} - 1}{n}}.\\] \n\nWe now lower bound the denominator by observing that the second sum is non negative, while each term of the first sum can be bounded by \\(AM - GM\\) as follows: \n\n\\[\\frac{1}{n_{j}} + \\frac{n_{j} - 1}{n} \\geq \\frac{2}{\\sqrt{n}} - \\frac{1}{n}.\\] \n\nWe therefore have \n\n\\[\\frac{R}{C} \\leq \\max_{1 \\leq t \\leq k} \\frac{t}{\\sum_{1 \\leq j \\leq t} \\left(\\frac{2}{\\sqrt{n}} - \\frac{1}{n}\\right)} = \\frac{n}{2\\sqrt{n} - 1},\\] \n\nwhich finishes the proof.", "metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P6.", "solution_match": "\nSolution "}}
diff --git a/HarvardMIT/md/en-282-2025-feb-comb-solutions.md b/HarvardMIT/md/en-282-2025-feb-comb-solutions.md
index 6a0ce7b178c03ad21eea2bb9772c1ccc93c6a3c7..5cf2d2cf47d944fe7c71728ee80308334162294b 100644
--- a/HarvardMIT/md/en-282-2025-feb-comb-solutions.md
+++ b/HarvardMIT/md/en-282-2025-feb-comb-solutions.md
@@ -45,7 +45,7 @@ Answer: 2304
Solution: Let a lateral move refer to one which is either up or right. Then the lateral moves are the only ones which increase Kelvin's sum of coordinates by 1, while all other moves do not change the sum, so Kelvin must make 6 of them, one to increase this sum from \(i\) to \(i + 1\) for each \(i \in [0, 5]\) .
-
+
We claim there exists a unique path corresponding to each set of 6 lateral moves chosen in this way. Indeed, the diagonal moves allow Kelvin to get from any point on \(x + y = i\) to any second point on \(x + y = i\) in exactly one way.
@@ -60,7 +60,7 @@ Answer: 200
Solution:
-
+
This is simply asking for the longest circuit in the adjacency graph of this grid. Note that this grid has \(4 \cdot 9 = 36\) cells of odd degree, 9 along each side. If we color the cells with checkerboard colors so that the corners are black, then 20 of these 36 cells are white. An Eulerian circuit uses an even number of doors from each cell, so at least one door from each of these cells goes unused. No door connects two white cells, so at least 20 doors are unused, leaving at most \(2 \cdot 10 \cdot 11 - 20 = \lfloor 200 \rfloor\) doors crossed.
@@ -165,7 +165,7 @@ Answer: \(\frac{4}{9}\)
Solution:
-
+
First, we compute the probability that the line through two random points in a triangle \(A B C\) passes through segments \(\overline{{A B}}\) and \(\overline{{A C}}\) . We can take an affine transform of the two random points and the triangle such that \(A B C\) becomes equilateral. Since the distribution of the two points is still uniform and independent, the probability of the line intersecting any two given sides is \(\frac{1}{3}\) by symmetry.
@@ -173,12 +173,12 @@ First, we compute the probability that the line through two random points in a t
Next, we compute the probability that the line through two random points in a rectangle \(A B C D\) passes through opposite edges \(\overline{{A B}}\) and \(\overline{{C D}}\) .
-
+
If the line passes through \(\overline{{A B}}\) and \(\overline{{B C}}\) , the points must both lie in triangle \(A B C\) , whose area is half that of \(A B C D\) . Given this, the probability the line passes through those two sides is \(\frac{1}{3}\) , as computed before. Thus the probability the line passes through \(\overline{{A B}}\) and \(\overline{{B C}}\) is \(\left(\frac{1}{2}\right)^{2}\cdot \frac{1}{3} = \frac{1}{12}\) . The same goes for the other pairs of adjacent edges. By symmetry, the line is equally likely to pass through either pair of opposite edges, each with probability \(\frac{1}{2}\left(1 - 4\cdot \frac{1}{12}\right) = \frac{1}{3}\) .
-
+
We now return to the original problem. If the line passes through a pair of opposite edges, then both points must be in the rectangle formed by these edges, which has area \(\frac{2}{3}\) that of the hexagon. Given this, the probability the line passes through those two edges is \(\frac{1}{3}\) as computed before. Thus, the probability that the line passes through the given pair of opposite edges is \(\left(\frac{2}{3}\right)^{2}\cdot \frac{1}{3} = \frac{4}{27}\) . Hence, the probability the line passes through any of the three pairs of opposite edges is \(3\cdot \frac{4}{27} = \left[\frac{4}{9}\right]\) .
diff --git a/HarvardMIT/md/en-282-2025-feb-geo-solutions.md b/HarvardMIT/md/en-282-2025-feb-geo-solutions.md
index 0e339281c403d8901a3505788f7ffa3c952450ed..c31fe5e00927e2e8b712e3577d55eb3eecb47186 100644
--- a/HarvardMIT/md/en-282-2025-feb-geo-solutions.md
+++ b/HarvardMIT/md/en-282-2025-feb-geo-solutions.md
@@ -13,7 +13,7 @@ Answer: 26
Solution:
-
+
Extend \(DE\) to meet \(AC\) at \(X\) . Observe that \(ABEX\) and \(DFCX\) are isosceles trapezoids (both with base angles of \(60^{\circ}\) ), so we have
@@ -33,7 +33,7 @@ Answer: 63
Solution: Note that the difference in heights between the two stalactites does not equal the difference in heights between the stalagmites. This tells us that the two stalactites form a pair of opposite vertices of the square, and likewise for the stalagmites. As the midpoint of the pairs of structures must then coincide, we know that it is \(\frac{16 + 36}{2} = 26\) from the ceiling due to the stalactites, and \(\frac{25 + 49}{2} = 37\) from the ground due to the stalagmites. Therefore the height of the cave is just the sum of these two values, i.e. 63.
-
+
3. Point \(P\) lies inside square \(ABCD\) such that the areas of \(\triangle PAB\) , \(\triangle PBC\) , \(\triangle PCD\) , and \(\triangle PDA\) are 1, 2, 3, and 4, in some order. Compute \(PA \cdot PB \cdot PC \cdot PD\) .
@@ -44,7 +44,7 @@ Answer: \(\boxed{8\sqrt{10}}\)
Solution:
-
+
Let \(h_{1}\) , \(h_{2}\) , \(h_{3}\) , and \(h_{4}\) be the lengths of the altitudes from \(P\) to sides \(AB\) , \(BC\) , \(CD\) , and \(DA\) , respectively. Then, the problem statement implies that \(\{h_{1}, h_{2}, h_{3}, h_{4}\} = \{x, 2x, 3x, 4x\}\) for some \(x\) . Furthermore, the area of the square is \(1 + 2 + 3 + 4 = 10\) , so we have
@@ -61,7 +61,7 @@ Hence either \((\{h_{1}, h_{3}\} = \{x, 4x\}\) and \(\{h_{2}, h_{4}\} = \{2x, 3x
Semicircle \(S_{1}\) is inscribed in a semicircle \(S_{2}\) , which is inscribed in another semicircle \(S_{3}\) . The radii of \(S_{1}\) and \(S_{3}\) are 1 and 10, respectively, and the diameters of \(S_{1}\) and \(S_{3}\) are parallel. The endpoints of the diameter of \(S_{3}\) are \(A\) and \(B\) , and \(S_{2}\) 's arc is tangent to \(AB\) at \(C\) . Compute \(AC \cdot CB\) .
-
+
Proposed by: Karthik Venkata Vedula
@@ -70,7 +70,7 @@ Answer: 20
Solution:
-
+
Let \(P\) , \(Q\) , and \(R\) be the midpoints of the diameters (i.e., the center of the circular arcs) of \(S_{3}\) , \(S_{2}\) , and \(S_{1}\) , respectively. Observe that if one fixes \(S_{3}\) , the location of \(S_{2}\) is uniquely determined by the angle between the diameters of \(S_{2}\) and \(S_{3}\) . The same holds for \(S_{2}\) and \(S_{1}\) . Thus, the figures \(S_{3} \cup S_{2}\) and \(S_{2} \cup S_{1}\) are similar. This gives us that the radius of \(S_{2}\) is \(\sqrt{10}\) .
@@ -86,7 +86,7 @@ Answer: \(\sqrt{23} - 2\sqrt{3}\)
Solution:
-
+
Let \(A^{\prime}\) be the antipode of \(A\) . As \(\angle A X A^{\prime} = 90^{\circ}\) , we have \(X\) , \(P\) , and \(A^{\prime}\) are collinear. As \(\angle B P C = 120^{\circ}\) and \(\angle B A^{\prime}C = 180^{\circ} - \angle B A C = 120^{\circ}\) , it follows that \(P\) lies on the circle with center \(A^{\prime}\) passing through \(B\) and \(C\) , so \(A^{\prime}P = \frac{B C}{\sqrt{3}} = 2\sqrt{3}\) and \(A A^{\prime} = 2A^{\prime}B = 4\sqrt{3}\) . By the Pythagorean theorem, \(X A^{\prime} = \sqrt{(4\sqrt{3})^{2} - 5^{2}} = \sqrt{23}\) , so the answer is \(X A^{\prime} - P A^{\prime} = \left[\sqrt{23} - 2\sqrt{3}\right]\) .
@@ -99,7 +99,7 @@ Answer: \(\boxed{9\sqrt{15}}\)
Solution:
-
+
Let \(O_{A}\) , \(O_{B}\) , \(O_{C}\) , and \(O_{D}\) be the circumcenters of \(\triangle B C D\) , \(\triangle C D A\) , \(\triangle D A B\) , and \(\triangle A B C\) , respectively. Note that \(O_{B}O_{C}\) is the perpendicular bisector of \(\overline{{A D}}\) . Similarly, \(O_{B}O_{D}\perp A C\) and \(O_{C}O_{D}\perp A B\) . As \(A B\parallel C D\) , we have \(O_{C}O_{D}\perp C D\) . Then, \(\triangle O_{B}O_{C}O_{D}\stackrel {\star}{\sim}\triangle A D C\) , as their corresponding sides are perpendicular. Likewise, \(\triangle O_{D}O_{A}O_{B}\stackrel {\star}{\sim}\triangle C B A\) , so \(O_{A}O_{B}O_{C}O_{D}\stackrel {\star}{\sim}B A D C\) .
@@ -126,7 +126,7 @@ Answer: \(\frac{7}{18}\)
Solution 1:
-
+
Let the internal and external bisectors of \(\angle B A C\) meet \(B C\) at \(D\) and \(E\) . Similarly, let the internal and external bisectors of \(\angle B P C\) meet \(B C\) at \(D^{\prime}\) and \(E^{\prime}\) . The angle condition implies that \(A D\parallel P D^{\prime}\) and \(A E\parallel P E^{\prime}\) . Thus, triangles \(A D E\) and \(P D^{\prime}E^{\prime}\) are homothetic. Hence, the requested ratio is \(\frac{D E}{D^{\prime}E^{\prime}}\) . Repeated applications of the angle bisector theorem yield
@@ -138,7 +138,7 @@ so \(D E = 24\) and \(D^{\prime}E^{\prime} = 28 / 3\) . Hence, the answer is \(\
Solution 2:
-
+
Let \(D = A P\cap B C\) \(E = B P\cap A C\) and \(F = C P\cap A B\) . Then, \(B C E F\) is cyclic, so by Power of a Point, \(\frac{A E}{A F} = \frac{A B}{A C} = \frac{3}{4}\) . Let \(A E = 3x\) and \(A F = 4x\) . Then,
@@ -175,7 +175,7 @@ Answer: \(\sqrt{6}\)
Solution:
-
+
Let \(D^{\prime}\) be the reflection of \(D\) across \(M\) . Then, \(A D E D^{\prime}\) is a parallelogram. Hence, \(D^{\prime}A = 2\) , so \(D^{\prime}B = 6\) . Thus, if \(D^{\prime}M = M D = x\) , then Power of a Point at \(D^{\prime}\) gives \(x\cdot (2x) = 2\cdot 6\) , so \(x = \sqrt{6}\) .
@@ -197,12 +197,12 @@ Let \(P_{1}\) and \(P_{2}\) be the projections of \(O_{1}\) and \(O_{2}\) onto s
Solving this, we get \(x = 7 + 2\sqrt{37}\) , which implies that \(A B = 2x = \boxed{14 + 4\sqrt{37}}\) . (The condition that \(\triangle A X D\) and \(\triangle B X C\) are acute rules out \(14 - 4\sqrt{37}\) .)
-
+
Solution 2:
-
+
Let \(P\) be the antipode of \(A\) in \(\odot (AXD)\) and \(Q\) be the antipode of \(B\) in \(\odot (BXC)\) . From \(\angle PDA = \angle QCB = 90^{\circ}\) , we get that \(P\) and \(Q\) lie on \(CD\) . Moreover, from \(\angle PXA = 90^{\circ}\) , we get that \(P \in BX\) , and similarly \(Q \in AX\) .
@@ -226,7 +226,7 @@ from which it follows that \(t_{a} = 5\) , \(t_{b} = 6\) , and \(t_{c} = 7\) .
Now, a translation of a plane can be written in one of three equivalent forms: it can be viewed as a translation in the \(x\) direction by a distance \(d_{x}\) , a translation in the \(y\) direction by a distance \(d_{y}\) , or a translation in the \(z\) direction by a distance \(d_{z}\) (with coordinate axes chosen as shown below).
-
+
As shown above, we can express \(t_{a}\) , \(t_{b}\) , and \(t_{c}\) in terms of \(d_{x}\) , \(d_{y}\) , and \(d_{z}\) using the Pythagorean theorem, which yields \(t_{a} = \sqrt{d_{y}^{2} + d_{z}^{2}}\) , \(t_{b} = \sqrt{d_{z}^{2} + d_{x}^{2}}\) , and \(t_{c} = \sqrt{d_{x}^{2} + d_{y}^{2}}\) . Hence,
@@ -239,7 +239,7 @@ We can draw a right pyramid with legs \(d_{x}\) , \(d_{y}\) , and \(d_{z}\) whic
Solution 2: Let the vertices of the hexagon be \(ABCDEF\) , where \(AB = 45\) , \(BC = 66\) , etc. Note that \(AB \parallel DE\) , \(BC \parallel EF\) , and \(CD \parallel FA\) . Let \(O\) be the center of the prism, and let \(M\) , \(N\) , and \(P\) be the midpoints of \(AD\) , \(BE\) , and \(CF\) , respectively.
-
+
diff --git a/HarvardMIT/md/en-282-2025-feb-guts-solutions.md b/HarvardMIT/md/en-282-2025-feb-guts-solutions.md
index f9b465f04f189d95476620d433e282d69a819728..5ca6aa72e3c436f15104d75d21ee89d8896f283b 100644
--- a/HarvardMIT/md/en-282-2025-feb-guts-solutions.md
+++ b/HarvardMIT/md/en-282-2025-feb-guts-solutions.md
@@ -52,7 +52,7 @@ Answer: \(\sqrt{3}\)
Solution:
-
+
The area condition implies \([ABC] = 2[PBC]\) . Hence, \(P\) lies on the \(A\) - midline of \(\triangle ABC\) . Therefore, the minimum possible value of \(PA\) is the distance from \(A\) to this midline. This is achieved by taking \(P\) to be the foot of the perpendicular from \(A\) to the \(A\) - midline. This distance is half the altitude of \(ABC\) , which has side length 4, so the answer is \(\frac{1}{2} (2\sqrt{3}) = \boxed{\sqrt{3}}\) .
@@ -65,7 +65,7 @@ Answer: \(\boxed {2\sqrt{2} - 2 = 2(\sqrt{2} - 1)}\)
Solution:
-
+
After drawing the graph, it's clear that the circle should pass through \((- 1,0)\) and be tangent to \(y = x - 1\) and \(y = - x + 1\) . Letting the radius of this circle be \(r\) , we have \(r\sqrt{2} +r = 2\) , so \(\boxed {r = 2\sqrt{2} - 2}\) .
@@ -80,7 +80,7 @@ Answer: \(\boxed {2\sqrt {13}}\)
Solution:
-
+
Let \(C^{\prime}\) be the reflection of \(C\) over \(D\) . Then, \(\overline{{E C}}\parallel \overline{{C^{\prime}F}}\) since \(E C F C^{\prime}\) is a parallelogram. Thus, \(B F C^{\prime}\) is an equilateral triangle, so \(B F = B C^{\prime} = 3\) and \(\angle F B D = 120^{\circ}\) . By Law of Cosines, we get \(D F = \sqrt{3^{2} + 3\cdot1 + 1^{2}} = \sqrt{13}\) and \(E F = \boxed {2\sqrt {13}}\) .
@@ -121,7 +121,7 @@ Answer: \(\boxed{5}\)
Solution:
-
+
Let \(O\) be the center of the hexagon. Without loss of generality, assume \(P\) is in \(\triangle ABO\) . Then, segment \(PQ\) is entirely contained in one of the given quadrilaterals if and only if \(Q\) is not in \(\triangle DEO\) . The probability that \(Q\) is in \(\triangle DEO\) is \(\frac{|DEO|}{|ABCDEF|} = \frac{1}{6}\) , so the answer is \(\boxed{5}\) .
@@ -134,7 +134,7 @@ Answer: \(\boxed{2 + 3\sqrt{2}}\)
Solution:
-
+
Let \(P_{1}\) and \(P_{2}\) be the two congruent pentagons. Let \(p(P)\) denote the perimeter of polygon \(P\) .
@@ -209,7 +209,7 @@ Answer: \(\boxed{5 + \sqrt{5}}\)
Solution:
-
+
In regular pentagon \(ABCDE\) (labeled clockwise), reflect \(ABDE\) across \(AB\) to obtain \(ABD'E'\) . Then, \(CDE'D'\) is one such parallelogram \(P\) . The length of \(CD'\) is
@@ -228,14 +228,14 @@ Answer: \(\sqrt{13}\)
Solution 1:
-
+
From \(\angle E = 60^{\circ}\) , we get that \(\angle AEF = 120^{\circ} - \angle CED = \angle CDE\) . Therefore, \(\triangle AEF \sim \triangle CDE\) . Since \(EF:DE = 2:1\) , the ratio of similarity must be \(2:1\) , so \(AE = 2CD = 8\) . Recall \(ABC\) has side length \(7 + 4 = 11\) , so \(EC = 11 - 8 = 3\) . Law of Cosines on \(\triangle CDE\) gives \(DE^2 = \sqrt{3^2 + 4^2 - 3\cdot 4} = \sqrt{13}\) .
Solution 2:
-
+
Let \(\odot (DEF)\) meet \(AC\) again at point \(X\) . Then, \(\angle FXA = 180^{\circ} - \angle FXE = \angle FDE = 90^{\circ}\) and \(\angle XDC = 180^{\circ} - \angle DCX - \angle DXC = 120^{\circ} - \angle DXE = 120^{\circ} - \angle DFE = 90^{\circ}\) . It follows that \(CX = 2CD = 8\) , so \(AX = 11 - CX = 3\) , and \(AF = 2AX = 6\) . Thus, Law of Cosines on \(\triangle AEF\) gives \(EF = \sqrt{8^2 + 6^2 - 8\cdot 6} = 2\sqrt{13}\) , implying that \(DE = \sqrt{13}\) .
@@ -370,7 +370,7 @@ Answer: \(\frac{3\sqrt{3} - 3}{2} = \frac{3}{2} (\sqrt{3} - 1)\)
Solution 1:
-
+
Let \(O\) be the center of \(\omega\) , and let the two tangent lines intersect at \(P\) . Note that \(O\) lies on the external angle bisector of \(\angle CPE\) because the tangents are symmetric about line \(PO\) . Additionally, \(O\) lies on the perpendicular bisector of \(CE\) by symmetry. By Fact 5, \(COPE\) is cyclic and \(\angle COE = 90^{\circ}\) . To finish, observe that \(\angle COD = 45^{\circ}\) . Dropping the altitude \(CH\) down to \(AD\) gives \(OH = CH = \sqrt{3}\) . So, \(AO = AH - OH = 3 - \sqrt{3}\) . The desired answer is then \(\frac{\sqrt{3}}{2} \cdot AO = \left[\frac{3\sqrt{3} - 3}{2}\right]\) .
@@ -417,7 +417,7 @@ Answer: 4
Solution:
-
+
Construct point \(P\) on \(A B\) such that \(X A = X P\) and point \(Q\) on \(C D\) such that \(X D = X Q\) . The angle condition gives \(Q C = X Q = X D\) and \(P B = X P = X A\) . Moreover, \(A D Q P\) is an isosceles trapezoid.
@@ -541,12 +541,12 @@ Let \(M\) be the midpoint of \(AB\) . Note that because \(\triangle OXO^{\prime}
by the Pythagorean theorem, which means that \(XY = AX^{\prime} = \left\lfloor \frac{11}{5} \right\rfloor\) .
-
+
Solution 2:
-
+
Let \(M\) be the midpoint of \(AB\) . We will find \(MO\) first.
@@ -603,7 +603,7 @@ Answer: \(\frac{\sqrt{5}}{2}\)
Solution:
-
+
We can square both sides of the second curve to get \(x^{2} = \frac{y^{3}}{1 - y}\) , which further rearranges to
diff --git a/HarvardMIT/md/en-282-2025-feb-team-solutions.md b/HarvardMIT/md/en-282-2025-feb-team-solutions.md
index 342bc28d28977bd0394e85a59f618003692054e3..ed4403a9cffdcc05000fd13d0368d6c672072e92 100644
--- a/HarvardMIT/md/en-282-2025-feb-team-solutions.md
+++ b/HarvardMIT/md/en-282-2025-feb-team-solutions.md
@@ -40,7 +40,7 @@ Proposed by: Pitchayut Saengrungkonkka
Solution 1:
-
+
Let \(O_{1}\) , \(O_{2}\) , and \(O\) be the centers of \(\omega_{1}\) , \(\omega_{2}\) , and the circumcircle of \(\triangle AXY\) , respectively.
@@ -53,7 +53,7 @@ Remark. One may also consider the antipodes of \(A\) on \(\omega_{1}\) and \(\om
Solution 2:
-
+
Let \(A^{\prime}\) be the \(A\) - antipode in circle \((A X Y)\) . It suffices to show that \(A^{\prime}\) lies on a fixed line. We will show that this line is one that is parallel to \(A B\) .
@@ -115,7 +115,7 @@ Proposed by: Pitchayut Saengrungkonka
Solution 1:
-
+
We use \(\angle\) to denote directed angles. Let \(Y\) and \(Z\) be the feet of the altitudes from \(B\) and \(C\) to \(AC\) and \(AB\) , respectively. Then \(\angle HZF = \angle HXF = 90^{\circ}\) , so \(HZFX\) is cyclic. Similarly, \(HYEX\) is cyclic. Therefore,
@@ -133,7 +133,7 @@ Proof. Because \(HX \perp EF\) and \(HE \perp HF\) , the quadrilateral \(HE'X'F'
From the claim, \(X'\) lies on the circle with diameter \(BC\) (which \(B'\) and \(C'\) also lie on). Since this circle is invariant under the inversion, \(X\) lies on the circle with diameter \(BC\) as well, and \(\angle BXC = 90^{\circ}\) .
-
+
Solution 3: We begin by proving the following lemma.
@@ -146,7 +146,7 @@ Proof. Let \(P_{A},P_{B},P_{C},P_{D}\) the feet of the altitudes from \(P\) to \
Therefore, \(P_{A}P_{B}P_{C}P_{D}\) is cyclic as desired.
-
+
@@ -237,7 +237,7 @@ Claim 1. Line \(IY\) is tangent to \(\omega\) .
Proof. Let the line through \(I\) parallel to \(BC\) meet \(AB\) and \(AC\) at \(B^{\prime}\) and \(C^{\prime}\) , respectively. Notice that \(B^{\prime}L\) is the perpendicular bisector of \(BI\) , so \(B^{\prime}L\) externally bisects \(\angle AB^{\prime}C^{\prime}\) . Similarly, \(C^{\prime}L\) externally bisects \(\angle AC^{\prime}B^{\prime}\) . Hence, \(L\) is the excenter of \(\triangle AB^{\prime}C^{\prime}\) , which means that \(B^{\prime}C^{\prime}\) is tangent to \(\omega\) . \(\square\)
-
+
Now, we note that \(LY \perp BC\) , so \(L\) , \(Y\) , and \(M\) are collinear (on the perpendicular bisector of \(BC\) ). Since \(\angle YPX = 90^{\circ}\) and \(\angle YMD = 90^{\circ}\) , \(PDMY\) is cyclic. However, \(IYMD\) is a rectangle, so \(IPDMY\) is a cyclic pentagon. Hence, \(\angle IPM = \angle IDM = 90^{\circ}\) .
@@ -250,7 +250,7 @@ Claim 2. \(P M \parallel D^{\prime}X\) .
Proof. Consider the homothety at \(A\) that sends \(\omega\) to the incircle. It sends \(L\) to \(I\) and \(P\) to \(D^{\prime}\) . Furthermore, it's well- known that \(X\) lies on \(A M\) . Because \(I X \parallel L M\) , we also have that the homothety sends \(M\) to \(X\) . These facts imply that \(D^{\prime}X \parallel P M\) . \(\square\)
-
+
Let \(T\) be the antipode of \(D\) on the incircle. Let \(A T\) intersect the incircle again at \(T^{\prime}\) . Since \(X\) lies on the polar of \(A\) with respect to the incircle, by Brocard's theorem, we have \(D^{\prime}\) , \(X\) , and \(T^{\prime}\) are collinear. It is well- known that \(A T \parallel I M\) . Therefore, \(\angle D P M = \angle D D^{\prime}X = \angle D D^{\prime}T^{\prime} = \angle D T T^{\prime} = \angle D I M\) . Consequently, \(I M D P\) is cyclic, and \(\angle I P M = \angle I D M = 90^{\circ}\) .
@@ -260,7 +260,7 @@ Solution 3: Let \(\omega\) be tangent to \(A B\) and \(A C\) at \(E\) and \(F\)
Lastly, we want \(P\) to be on the circle with diameter \(I M\) . This circle intersects \(E F\) again at the foot from \(M\) to \(A I\) , which is the midpoint of \(E F\) . Let this point be \(M^{\prime}\) . Consider the homothety sending the incircle to \(\omega\) . This clearly sends \(D\) to the second intersection of \(A D\) and \(\omega\) , which is \(P^{\prime}\) , and it sends \(I\) to \(L\) . Note that \(A P\cdot A P^{\prime} = A E^{2} = A M\cdot A L\) , as the circle with diameter \(L E\) is tangent to \(A E\) . Thus, \(P P^{\prime}M^{\prime}L\) is cyclic. Since \(I D \parallel L P^{\prime}\) , \(I\) lies on \(M^{\prime}L\) , and \(D\) lies on \(P P^{\prime}\) . By Reim's, we also have \(P D M^{\prime}I\) is cyclic. As \(I M\) is a diameter of \((D M^{\prime}I)\) , we have \(\angle I P M = 90^{\circ}\) .
-
+
9. [60] Let \(\mathbb{Z}\) be the set of integers. Determine, with proof, all primes \(p\) for which there exists a function \(f\colon \mathbb{Z}\to \mathbb{Z}\) such that for any integer \(x\) ,
diff --git a/HarvardMIT/md/en-284-tournaments-2025-hmic-solutions.md b/HarvardMIT/md/en-284-tournaments-2025-hmic-solutions.md
index 34e8e8ee01bf3d78558cf7ba5d636f78098cf97c..7611897daa196c5465a81b7a56273bf677590763 100644
--- a/HarvardMIT/md/en-284-tournaments-2025-hmic-solutions.md
+++ b/HarvardMIT/md/en-284-tournaments-2025-hmic-solutions.md
@@ -7,7 +7,7 @@
Proposed by: Albert Wang
-
+
Solution 1: Let \(d(P, X Y)\) be the distance from \(P\) to line \(X Y\) . We will first prove \(\mathcal{P}_{A}\) and \(\mathcal{P}_{B}\) intersect at most once.
@@ -75,7 +75,7 @@ The right- hand side is bounded in \(t\) , so the left- hand side must also be b
Proposed by: Derek Liu
-
+
Solution: In what follows, all angles are directed.
diff --git a/HarvardMIT/segmented/en-282-2025-feb-comb-solutions.jsonl b/HarvardMIT/segmented/en-282-2025-feb-comb-solutions.jsonl
index 7cfffa2da6d8f9f9df1c3939a8b16845eb9f43b1..39296670defd322fed2a658d57cd779e76aa6182 100644
--- a/HarvardMIT/segmented/en-282-2025-feb-comb-solutions.jsonl
+++ b/HarvardMIT/segmented/en-282-2025-feb-comb-solutions.jsonl
@@ -5,9 +5,9 @@
{"year": "2025", "tier": "T4", "problem_label": "3", "problem_type": "Combinatorics", "exam": "HMMT", "problem": "Ben has 16 balls labeled 1, 2, 3, ..., 16, as well as 4 indistinguishable boxes. Two balls are neighbors if their labels differ by 1. Compute the number of ways for him to put 4 balls in each box such that each ball is in the same box as at least one of its neighbors. (The order in which the balls are placed does not matter.)", "solution": "Answer: 105", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-comb-solutions.jsonl", "problem_match": "\n3. ", "solution_match": "\nProposed by: Benjamin Shimabukuro \n\n"}}
{"year": "2025", "tier": "T4", "problem_label": "3", "problem_type": "Combinatorics", "exam": "HMMT", "problem": "Ben has 16 balls labeled 1, 2, 3, ..., 16, as well as 4 indistinguishable boxes. Two balls are neighbors if their labels differ by 1. Compute the number of ways for him to put 4 balls in each box such that each ball is in the same box as at least one of its neighbors. (The order in which the balls are placed does not matter.)", "solution": "Each box must either contain a single group of four consecutive balls (e.g. 5, 6, 7, 8) or two groups of two consecutive balls (e.g. 5, 6, 9, 10). Since all groups have even lengths, this means that 1 and 2 are in the same group, 3 and 4 are in the same group, and so on. We can think of each of these 8 pairs of balls as an individual unit, so the answer is equal to the number of ways to put 8 objects in 4 indistinguishable boxes, where each box has 2 objects without any additional restrictions. The number of ways to do this is \\(\\frac{8!}{2^{4}\\cdot 4!} = \\boxed{105}\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-comb-solutions.jsonl", "problem_match": "\n3. ", "solution_match": "\nSolution: "}}
{"year": "2025", "tier": "T4", "problem_label": "4", "problem_type": "Combinatorics", "exam": "HMMT", "problem": "Sophie is at \\((0,0)\\) on a coordinate grid and would like to get to \\((3,3)\\) . If Sophie is at \\((x,y)\\) , in a single step she can move to one of \\((x + 1,y)\\) , \\((x,y + 1)\\) , \\((x - 1,y + 1)\\) , or \\((x + 1,y - 1)\\) . She cannot revisit any points along her path, and neither her \\(x\\) - coordinate nor her \\(y\\) - coordinate can ever be less than 0 or greater than 3. Compute the number of ways for Sophie to reach \\((3,3)\\) .", "solution": "Answer: 2304", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-comb-solutions.jsonl", "problem_match": "\n4. ", "solution_match": "\nProposed by: Derek Liu \n\n"}}
-{"year": "2025", "tier": "T4", "problem_label": "4", "problem_type": "Combinatorics", "exam": "HMMT", "problem": "Sophie is at \\((0,0)\\) on a coordinate grid and would like to get to \\((3,3)\\) . If Sophie is at \\((x,y)\\) , in a single step she can move to one of \\((x + 1,y)\\) , \\((x,y + 1)\\) , \\((x - 1,y + 1)\\) , or \\((x + 1,y - 1)\\) . She cannot revisit any points along her path, and neither her \\(x\\) - coordinate nor her \\(y\\) - coordinate can ever be less than 0 or greater than 3. Compute the number of ways for Sophie to reach \\((3,3)\\) .", "solution": "Let a lateral move refer to one which is either up or right. Then the lateral moves are the only ones which increase Kelvin's sum of coordinates by 1, while all other moves do not change the sum, so Kelvin must make 6 of them, one to increase this sum from \\(i\\) to \\(i + 1\\) for each \\(i \\in [0, 5]\\) . \n\n\n \n\nWe claim there exists a unique path corresponding to each set of 6 lateral moves chosen in this way. Indeed, the diagonal moves allow Kelvin to get from any point on \\(x + y = i\\) to any second point on \\(x + y = i\\) in exactly one way. \n\nObserve that when \\(i \\leq 2\\) , the number of lateral moves that increase the sum of coordinates from \\(i\\) to \\(i + 1\\) is \\(2i\\) , as is the number that increase it from \\(5 - i\\) to \\(6 - i\\) . Thus the answer is \\(2 \\cdot 4 \\cdot 6 \\cdot 6 \\cdot 4 \\cdot 2 = \\lfloor 2304 \\rfloor\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-comb-solutions.jsonl", "problem_match": "\n4. ", "solution_match": "\nSolution: "}}
+{"year": "2025", "tier": "T4", "problem_label": "4", "problem_type": "Combinatorics", "exam": "HMMT", "problem": "Sophie is at \\((0,0)\\) on a coordinate grid and would like to get to \\((3,3)\\) . If Sophie is at \\((x,y)\\) , in a single step she can move to one of \\((x + 1,y)\\) , \\((x,y + 1)\\) , \\((x - 1,y + 1)\\) , or \\((x + 1,y - 1)\\) . She cannot revisit any points along her path, and neither her \\(x\\) - coordinate nor her \\(y\\) - coordinate can ever be less than 0 or greater than 3. Compute the number of ways for Sophie to reach \\((3,3)\\) .", "solution": "Let a lateral move refer to one which is either up or right. Then the lateral moves are the only ones which increase Kelvin's sum of coordinates by 1, while all other moves do not change the sum, so Kelvin must make 6 of them, one to increase this sum from \\(i\\) to \\(i + 1\\) for each \\(i \\in [0, 5]\\) . \n\n\n \n\nWe claim there exists a unique path corresponding to each set of 6 lateral moves chosen in this way. Indeed, the diagonal moves allow Kelvin to get from any point on \\(x + y = i\\) to any second point on \\(x + y = i\\) in exactly one way. \n\nObserve that when \\(i \\leq 2\\) , the number of lateral moves that increase the sum of coordinates from \\(i\\) to \\(i + 1\\) is \\(2i\\) , as is the number that increase it from \\(5 - i\\) to \\(6 - i\\) . Thus the answer is \\(2 \\cdot 4 \\cdot 6 \\cdot 6 \\cdot 4 \\cdot 2 = \\lfloor 2304 \\rfloor\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-comb-solutions.jsonl", "problem_match": "\n4. ", "solution_match": "\nSolution: "}}
{"year": "2025", "tier": "T4", "problem_label": "5", "problem_type": "Combinatorics", "exam": "HMMT", "problem": "In an \\(11 \\times 11\\) grid of cells, each pair of edge-adjacent cells is connected by a door. Karthik wants to walk a path in this grid. He can start in any cell, but he must end in the same cell he started in, and he cannot go through any door more than once (not even in opposite directions). Compute the maximum number of doors he can go through in such a path.", "solution": "Answer: 200", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-comb-solutions.jsonl", "problem_match": "\n5. ", "solution_match": "\nProposed by: Derek Liu \n\n"}}
-{"year": "2025", "tier": "T4", "problem_label": "5", "problem_type": "Combinatorics", "exam": "HMMT", "problem": "In an \\(11 \\times 11\\) grid of cells, each pair of edge-adjacent cells is connected by a door. Karthik wants to walk a path in this grid. He can start in any cell, but he must end in the same cell he started in, and he cannot go through any door more than once (not even in opposite directions). Compute the maximum number of doors he can go through in such a path.", "solution": "\n \n\nThis is simply asking for the longest circuit in the adjacency graph of this grid. Note that this grid has \\(4 \\cdot 9 = 36\\) cells of odd degree, 9 along each side. If we color the cells with checkerboard colors so that the corners are black, then 20 of these 36 cells are white. An Eulerian circuit uses an even number of doors from each cell, so at least one door from each of these cells goes unused. No door connects two white cells, so at least 20 doors are unused, leaving at most \\(2 \\cdot 10 \\cdot 11 - 20 = \\lfloor 200 \\rfloor\\) doors crossed.\n\n\n\nTo see this is achievable, we first delete the bottom- right cell and its 2 doors, as well as the top- left cell and its 2 doors. This leaves 8 odd- degree cells along each side of the grid; we can delete 4 doors along each to cover all remaining odd- degree cells, for 20 total doors deleted. The resulting graph is connected and has no odd- degree cells, so it must have an Eulerian circuit. This circuit is our desired path.", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-comb-solutions.jsonl", "problem_match": "\n5. ", "solution_match": "\nSolution: \n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "5", "problem_type": "Combinatorics", "exam": "HMMT", "problem": "In an \\(11 \\times 11\\) grid of cells, each pair of edge-adjacent cells is connected by a door. Karthik wants to walk a path in this grid. He can start in any cell, but he must end in the same cell he started in, and he cannot go through any door more than once (not even in opposite directions). Compute the maximum number of doors he can go through in such a path.", "solution": "\n \n\nThis is simply asking for the longest circuit in the adjacency graph of this grid. Note that this grid has \\(4 \\cdot 9 = 36\\) cells of odd degree, 9 along each side. If we color the cells with checkerboard colors so that the corners are black, then 20 of these 36 cells are white. An Eulerian circuit uses an even number of doors from each cell, so at least one door from each of these cells goes unused. No door connects two white cells, so at least 20 doors are unused, leaving at most \\(2 \\cdot 10 \\cdot 11 - 20 = \\lfloor 200 \\rfloor\\) doors crossed.\n\n\n\nTo see this is achievable, we first delete the bottom- right cell and its 2 doors, as well as the top- left cell and its 2 doors. This leaves 8 odd- degree cells along each side of the grid; we can delete 4 doors along each to cover all remaining odd- degree cells, for 20 total doors deleted. The resulting graph is connected and has no odd- degree cells, so it must have an Eulerian circuit. This circuit is our desired path.", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-comb-solutions.jsonl", "problem_match": "\n5. ", "solution_match": "\nSolution: \n\n"}}
{"year": "2025", "tier": "T4", "problem_label": "6", "problem_type": "Combinatorics", "exam": "HMMT", "problem": "Compute the number of ways to pick two rectangles in a \\(5 \\times 5\\) grid of squares such that the edges of the rectangles lie on the lines of the grid and the rectangles do not overlap at their interiors, edges, or vertices. The order in which the rectangles are chosen does not matter.", "solution": "Answer: 6300", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-comb-solutions.jsonl", "problem_match": "\n6. ", "solution_match": "\nProposed by: Jacob Paltrowitz \n\n"}}
{"year": "2025", "tier": "T4", "problem_label": "6", "problem_type": "Combinatorics", "exam": "HMMT", "problem": "Compute the number of ways to pick two rectangles in a \\(5 \\times 5\\) grid of squares such that the edges of the rectangles lie on the lines of the grid and the rectangles do not overlap at their interiors, edges, or vertices. The order in which the rectangles are chosen does not matter.", "solution": "A rectangle can be specified by two intervals, one specifying its horizontal extent ( \\(x\\) - coordinates of left and right sides) and one specifying its vertical extent ( \\(y\\) - coordinates of bottom and top sides). For the rectangles to not overlap, we need either the horizontal intervals or the vertical intervals to be disjoint (possibly both). \n\nFirst, we will count the number of ways for the horizontal intervals to be disjoint. Let these intervals be \\([a,b]\\) and \\([c,d]\\) . As the order of the rectangles does not matter, we can assume without loss of generality that \\(a< c\\) , so \\(a< b< c< d\\) . Then there are \\(\\binom{6}{4}\\) choices for \\(a\\) , \\(b\\) , \\(c\\) , and \\(d\\) . There are no restrictions on the vertical intervals, so the number of ways to choose them is \\(\\binom{6}{2}^{2}\\) . Thus, the total number of pairs of rectangles for which the horizontal intervals are disjoint is \\(\\binom{6}{4}\\binom{6}{2}^{2}\\) . \n\nBy symmetry, the total number of pairs of rectangles for which the vertical intervals are disjoint is the same. It remains to count the number of ways for both the horizontal and vertical intervals to be disjoint. Again, let the horizontal intervals be \\([a,b]\\) and \\([c,d]\\) , and let the vertical intervals be \\([e,f]\\) and \\([g,h]\\) . We can assume \\(a< b< c< d\\) without loss of generality, so there are \\(\\binom{6}{4}\\) ways to choose the horizontal intervals. However, the cases \\(e< f< g< h\\) and \\(g< h< e< f\\) are now distinct, so there are \\(2\\binom{6}{4}\\) ways to choose the vertical intervals. Therefore, there are \\(2\\binom{6}{4}^{2}\\) pairs of rectangles for which both the horizontal and vertical intervals are disjoint. \n\nThrough inclusion- exclusion, we get the final answer of \n\n\\[2\\cdot \\binom{6}{4}\\binom{6}{2}^{2} - 2\\cdot \\binom{6}{4}^{2} = \\boxed{6300}.\\]", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-comb-solutions.jsonl", "problem_match": "\n6. ", "solution_match": "\nSolution: "}}
{"year": "2025", "tier": "T4", "problem_label": "7", "problem_type": "Combinatorics", "exam": "HMMT", "problem": "Compute the number of ways to arrange 3 copies of each of the 26 lowercase letters of the English alphabet such that for any two distinct letters \\(x_{1}\\) and \\(x_{2}\\) , the number of \\(x_{2}\\) 's between the first and second occurrences of \\(x_{1}\\) equals the number of \\(x_{2}\\) 's between the second and third occurrences of \\(x_{1}\\) .", "solution": "Answer: 22526!", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-comb-solutions.jsonl", "problem_match": "\n7. ", "solution_match": "\nProposed by: Derek Liu \n\n"}}
@@ -15,6 +15,6 @@
{"year": "2025", "tier": "T4", "problem_label": "8", "problem_type": "Combinatorics", "exam": "HMMT", "problem": "Albert writes 2025 numbers \\(a_{1}\\) , ..., \\(a_{2025}\\) in a circle on a blackboard. Initially, each of the numbers is uniformly and independently sampled at random from the interval \\([0,1]\\) . Then, each second, he simultaneously replaces \\(a_{i}\\) with \\(\\max (a_{i - 1}, a_{i}, a_{i + 1})\\) for all \\(i = 1, 2, \\ldots , 2025\\) (where \\(a_{0} = a_{2025}\\) and \\(a_{2026} = a_{1}\\) ). Compute the expected value of the number of distinct values remaining after 100 seconds.", "solution": "Answer: \\(\\boxed{2025}\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-comb-solutions.jsonl", "problem_match": "\n8. ", "solution_match": "\nProposed by: Isaac Zhu \n\n"}}
{"year": "2025", "tier": "T4", "problem_label": "8", "problem_type": "Combinatorics", "exam": "HMMT", "problem": "Albert writes 2025 numbers \\(a_{1}\\) , ..., \\(a_{2025}\\) in a circle on a blackboard. Initially, each of the numbers is uniformly and independently sampled at random from the interval \\([0,1]\\) . Then, each second, he simultaneously replaces \\(a_{i}\\) with \\(\\max (a_{i - 1}, a_{i}, a_{i + 1})\\) for all \\(i = 1, 2, \\ldots , 2025\\) (where \\(a_{0} = a_{2025}\\) and \\(a_{2026} = a_{1}\\) ). Compute the expected value of the number of distinct values remaining after 100 seconds.", "solution": "We can assume that the initial numbers are all distinct, since this occurs with probability 1. For clarity, we denote the value of \\(a_{i}\\) after \\(t\\) seconds as \\(a_{i,t}\\) . The index \\(i\\) is taken mod 2025. \n\nIn general, after \\(k< 1012\\) seconds, we claim the expected number of distinct values remaining is \\(\\frac{2025}{k + 1}\\) . To show this, we first prove that for any remaining value, its appearances are consecutive. Indeed, note that for all \\(i\\) and \\(k\\) , \n\n\\[a_{i,k} = \\max (a_{i - 1,k - 1}, a_{i,k - 1}, a_{i + 1,k - 1}) = \\max (a_{i - 2,k - 2}, \\ldots , a_{i + 2,k - 2}) = \\dots = \\max (a_{i - k,0}, \\ldots , a_{i + k,0}).\\] \n\nGiven an initial number \\(a_{c,0}\\) , let \\(j_{1}\\) and \\(j_{2}\\) be the smallest positive integers such that \\(a_{c - j_{1},0} > a_{c,0}\\) and \\(a_{c + j_{2},0} > a_{c,0}\\) . As the initial numbers are distinct, we conclude \\(a_{i,k} = a_{c,0}\\) if and only if \\(\\{i - k, i - k + 1, \\ldots , i + k\\}\\) contains \\(c\\) but neither \\(c - j_{1}\\) nor \\(c + j_{2}\\) (mod 2025). The indices \\(i\\) that satisfy this are clearly consecutive.\n\n\n\nNow consider the indicator variables \n\n\\[\\mathbf{1}_{i} = \\left\\{ \\begin{array}{ll}1 & \\mathrm{if~}a_{i,k}\\neq a_{i + 1,k}\\\\ 0 & \\mathrm{otherwise} \\end{array} \\right.\\] \n\nfor \\(1\\leq i\\leq 2025\\) . Note that the number of distinct values on the board after \\(k\\) seconds is simply \\(\\textstyle \\sum_{i = 1}^{2025}\\mathbf{1}_{i}\\) . By linearity of expectation, it suffices to compute \\(\\mathbb{E}[\\mathbf{1}_{i}]\\) for each \\(i\\) . Recall \n\n\\[a_{i,k} = \\max (a_{i - k,0},\\ldots ,a_{i + k,0})\\] \n\nfor each \\(1\\leq i\\leq 2025\\) . Hence, the condition \\(a_{i,k}\\neq a_{i + 1,k}\\) can be written as \n\n\\[\\max (a_{i - k,0},\\ldots ,a_{i + k,0})\\neq \\max (a_{i + 1 - k,0},\\ldots ,a_{i + 1 + k,0}).\\] \n\nThis is the case if and only if \\(\\max (a_{i - k,0},\\ldots ,a_{i + k + 1,0})\\) is either \\(a_{i - k,0}\\) or \\(a_{i + k + 1,0}\\) (as \\(a_{i - k,0}\\neq a_{i + k + 1,0}\\) and \\(2k + 2< 2025\\) ). Since the \\(2k + 2\\) values \\(a_{i - k,0}\\) , ..., \\(a_{i + k + 1,0}\\) were sampled independently from the same distribution, each of \\(a_{i - k,0}\\) and \\(a_{i + k + 1,0}\\) has a \\(\\frac{1}{2k + 2}\\) probability of being their maximum. Hence, \n\n\\[\\mathbb{E}[\\mathbf{1}_{i}] = \\operatorname *{Pr}[a_{i,k}\\neq a_{i + 1,k}] = \\frac{2}{2k + 2} = \\frac{1}{k + 1}.\\] \n\nThus, the expected number of distinct values is \n\n\\[\\sum_{i = 1}^{2025}\\mathbb{E}[\\mathbf{1}_{i}] = \\frac{2025}{k + 1}.\\] \n\nSubstituting \\(k = 100\\) yields the answer \\(\\left[\\frac{2025}{101}\\right]\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-comb-solutions.jsonl", "problem_match": "\n8. ", "solution_match": "\nSolution: "}}
{"year": "2025", "tier": "T4", "problem_label": "9", "problem_type": "Combinatorics", "exam": "HMMT", "problem": "Two points are selected independently and uniformly at random inside a regular hexagon. Compute the probability that a line passing through both of the points intersects a pair of opposite edges of the hexagon.", "solution": "Answer: \\(\\frac{4}{9}\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-comb-solutions.jsonl", "problem_match": "\n9. ", "solution_match": "\nProposed by: Albert Wang \n\n"}}
-{"year": "2025", "tier": "T4", "problem_label": "9", "problem_type": "Combinatorics", "exam": "HMMT", "problem": "Two points are selected independently and uniformly at random inside a regular hexagon. Compute the probability that a line passing through both of the points intersects a pair of opposite edges of the hexagon.", "solution": "\n \n\nFirst, we compute the probability that the line through two random points in a triangle \\(A B C\\) passes through segments \\(\\overline{{A B}}\\) and \\(\\overline{{A C}}\\) . We can take an affine transform of the two random points and the triangle such that \\(A B C\\) becomes equilateral. Since the distribution of the two points is still uniform and independent, the probability of the line intersecting any two given sides is \\(\\frac{1}{3}\\) by symmetry. \n\nNext, we compute the probability that the line through two random points in a rectangle \\(A B C D\\) passes through opposite edges \\(\\overline{{A B}}\\) and \\(\\overline{{C D}}\\) .\n\n\n\n \n\nIf the line passes through \\(\\overline{{A B}}\\) and \\(\\overline{{B C}}\\) , the points must both lie in triangle \\(A B C\\) , whose area is half that of \\(A B C D\\) . Given this, the probability the line passes through those two sides is \\(\\frac{1}{3}\\) , as computed before. Thus the probability the line passes through \\(\\overline{{A B}}\\) and \\(\\overline{{B C}}\\) is \\(\\left(\\frac{1}{2}\\right)^{2}\\cdot \\frac{1}{3} = \\frac{1}{12}\\) . The same goes for the other pairs of adjacent edges. By symmetry, the line is equally likely to pass through either pair of opposite edges, each with probability \\(\\frac{1}{2}\\left(1 - 4\\cdot \\frac{1}{12}\\right) = \\frac{1}{3}\\) . \n\n\n \n\nWe now return to the original problem. If the line passes through a pair of opposite edges, then both points must be in the rectangle formed by these edges, which has area \\(\\frac{2}{3}\\) that of the hexagon. Given this, the probability the line passes through those two edges is \\(\\frac{1}{3}\\) as computed before. Thus, the probability that the line passes through the given pair of opposite edges is \\(\\left(\\frac{2}{3}\\right)^{2}\\cdot \\frac{1}{3} = \\frac{4}{27}\\) . Hence, the probability the line passes through any of the three pairs of opposite edges is \\(3\\cdot \\frac{4}{27} = \\left[\\frac{4}{9}\\right]\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-comb-solutions.jsonl", "problem_match": "\n9. ", "solution_match": "\nSolution: \n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "9", "problem_type": "Combinatorics", "exam": "HMMT", "problem": "Two points are selected independently and uniformly at random inside a regular hexagon. Compute the probability that a line passing through both of the points intersects a pair of opposite edges of the hexagon.", "solution": "\n \n\nFirst, we compute the probability that the line through two random points in a triangle \\(A B C\\) passes through segments \\(\\overline{{A B}}\\) and \\(\\overline{{A C}}\\) . We can take an affine transform of the two random points and the triangle such that \\(A B C\\) becomes equilateral. Since the distribution of the two points is still uniform and independent, the probability of the line intersecting any two given sides is \\(\\frac{1}{3}\\) by symmetry. \n\nNext, we compute the probability that the line through two random points in a rectangle \\(A B C D\\) passes through opposite edges \\(\\overline{{A B}}\\) and \\(\\overline{{C D}}\\) .\n\n\n\n \n\nIf the line passes through \\(\\overline{{A B}}\\) and \\(\\overline{{B C}}\\) , the points must both lie in triangle \\(A B C\\) , whose area is half that of \\(A B C D\\) . Given this, the probability the line passes through those two sides is \\(\\frac{1}{3}\\) , as computed before. Thus the probability the line passes through \\(\\overline{{A B}}\\) and \\(\\overline{{B C}}\\) is \\(\\left(\\frac{1}{2}\\right)^{2}\\cdot \\frac{1}{3} = \\frac{1}{12}\\) . The same goes for the other pairs of adjacent edges. By symmetry, the line is equally likely to pass through either pair of opposite edges, each with probability \\(\\frac{1}{2}\\left(1 - 4\\cdot \\frac{1}{12}\\right) = \\frac{1}{3}\\) . \n\n\n \n\nWe now return to the original problem. If the line passes through a pair of opposite edges, then both points must be in the rectangle formed by these edges, which has area \\(\\frac{2}{3}\\) that of the hexagon. Given this, the probability the line passes through those two edges is \\(\\frac{1}{3}\\) as computed before. Thus, the probability that the line passes through the given pair of opposite edges is \\(\\left(\\frac{2}{3}\\right)^{2}\\cdot \\frac{1}{3} = \\frac{4}{27}\\) . Hence, the probability the line passes through any of the three pairs of opposite edges is \\(3\\cdot \\frac{4}{27} = \\left[\\frac{4}{9}\\right]\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-comb-solutions.jsonl", "problem_match": "\n9. ", "solution_match": "\nSolution: \n\n"}}
{"year": "2025", "tier": "T4", "problem_label": "10", "problem_type": "Combinatorics", "exam": "HMMT", "problem": "The circumference of a circle is divided into 45 arcs, each of length 1. Initially, there are 15 snakes, each of length 1, occupying every third arc. Every second, each snake independently moves either one arc left or one arc right, each with probability \\(\\frac{1}{2}\\) . If two snakes ever touch, they merge to form a single snake occupying the arcs of both of the previous snakes, and the merged snake moves as one snake. Compute the expected number of seconds until there is only one snake left.", "solution": "Answer: \\(\\frac{448}{3}\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-comb-solutions.jsonl", "problem_match": "\n10. ", "solution_match": "\nProposed by: Srinivas Arun \n\n"}}
{"year": "2025", "tier": "T4", "problem_label": "10", "problem_type": "Combinatorics", "exam": "HMMT", "problem": "The circumference of a circle is divided into 45 arcs, each of length 1. Initially, there are 15 snakes, each of length 1, occupying every third arc. Every second, each snake independently moves either one arc left or one arc right, each with probability \\(\\frac{1}{2}\\) . If two snakes ever touch, they merge to form a single snake occupying the arcs of both of the previous snakes, and the merged snake moves as one snake. Compute the expected number of seconds until there is only one snake left.", "solution": "We solve the problem generally for \\(n\\) snakes and \\(3n\\) arcs. Without loss of generality, fix the two snakes \\(A\\) and \\(B\\) that will eventually form the ends of the last snake. Note that \\(A\\) and \\(B\\) must be consecutive in the initial configuration; assume \\(B\\) lies immediately clockwise from \\(A\\) . \n\nLet \\(d\\) be the arclength between \\(A\\) and \\(B\\) (measuring clockwise from \\(A\\) to \\(B\\) ). As long as \\(d \\neq 0\\) and \\(d \\neq 2n\\) , we know that \\(A\\) and \\(B\\) lie in different snakes and thus move independently. Therefore, we can consider \\(d\\) to be on a random walk starting at 2, where", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-comb-solutions.jsonl", "problem_match": "\n10. ", "solution_match": "\nSolution: "}}
diff --git a/HarvardMIT/segmented/en-282-2025-feb-geo-solutions.jsonl b/HarvardMIT/segmented/en-282-2025-feb-geo-solutions.jsonl
index 8700152df31aaf298f3df7f117fe56e42e204f6c..3679bd22da7a2ee084416b93d7fdc2816a2ef5af 100644
--- a/HarvardMIT/segmented/en-282-2025-feb-geo-solutions.jsonl
+++ b/HarvardMIT/segmented/en-282-2025-feb-geo-solutions.jsonl
@@ -1,24 +1,24 @@
{"year": "2025", "tier": "T4", "problem_label": "1", "problem_type": "Geometry", "exam": "HMMT", "problem": "Equilateral triangles \\(\\triangle ABC\\) and \\(\\triangle DEF\\) are drawn such that points \\(B\\) , \\(E\\) , \\(F\\) , and \\(C\\) lie on a line in this order, and point \\(D\\) lies inside triangle \\(\\triangle ABC\\) . If \\(BE = 14\\) , \\(EF = 15\\) , and \\(FC = 16\\) , compute \\(AD\\) .", "solution": "Answer: 26", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-geo-solutions.jsonl", "problem_match": "\n1. ", "solution_match": "\nProposed by: Jackson Dryg, Karthik Venkata Vedula \n\n"}}
-{"year": "2025", "tier": "T4", "problem_label": "1", "problem_type": "Geometry", "exam": "HMMT", "problem": "Equilateral triangles \\(\\triangle ABC\\) and \\(\\triangle DEF\\) are drawn such that points \\(B\\) , \\(E\\) , \\(F\\) , and \\(C\\) lie on a line in this order, and point \\(D\\) lies inside triangle \\(\\triangle ABC\\) . If \\(BE = 14\\) , \\(EF = 15\\) , and \\(FC = 16\\) , compute \\(AD\\) .", "solution": "\n \n\nExtend \\(DE\\) to meet \\(AC\\) at \\(X\\) . Observe that \\(ABEX\\) and \\(DFCX\\) are isosceles trapezoids (both with base angles of \\(60^{\\circ}\\) ), so we have \n\n\\(AX = BE = 14\\) \\(DX = FC = 16\\) , and \\(\\angle AXD = 120^{\\circ}\\) \n\nBy Law of Cosines on \\(\\triangle ADX\\) , the answer is \n\n\\[{A D=\\sqrt{A X^{2}+D X^{2}-2\\cos(120^{\\circ})A X}\\cdot D X}\\] \\[{=\\sqrt{14^{2}+16^{2}+14\\cdot16}=\\boxed{26}.}\\]", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-geo-solutions.jsonl", "problem_match": "\n1. ", "solution_match": "\nSolution: \n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "1", "problem_type": "Geometry", "exam": "HMMT", "problem": "Equilateral triangles \\(\\triangle ABC\\) and \\(\\triangle DEF\\) are drawn such that points \\(B\\) , \\(E\\) , \\(F\\) , and \\(C\\) lie on a line in this order, and point \\(D\\) lies inside triangle \\(\\triangle ABC\\) . If \\(BE = 14\\) , \\(EF = 15\\) , and \\(FC = 16\\) , compute \\(AD\\) .", "solution": "\n \n\nExtend \\(DE\\) to meet \\(AC\\) at \\(X\\) . Observe that \\(ABEX\\) and \\(DFCX\\) are isosceles trapezoids (both with base angles of \\(60^{\\circ}\\) ), so we have \n\n\\(AX = BE = 14\\) \\(DX = FC = 16\\) , and \\(\\angle AXD = 120^{\\circ}\\) \n\nBy Law of Cosines on \\(\\triangle ADX\\) , the answer is \n\n\\[{A D=\\sqrt{A X^{2}+D X^{2}-2\\cos(120^{\\circ})A X}\\cdot D X}\\] \\[{=\\sqrt{14^{2}+16^{2}+14\\cdot16}=\\boxed{26}.}\\]", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-geo-solutions.jsonl", "problem_match": "\n1. ", "solution_match": "\nSolution: \n\n"}}
{"year": "2025", "tier": "T4", "problem_label": "2", "problem_type": "Geometry", "exam": "HMMT", "problem": "In a two-dimensional cave with a parallel floor and ceiling, two stalactites of lengths 16 and 36 hang perpendicularly from the ceiling, while two stalagmites of heights 25 and 49 grow perpendicularly from the ground. If the tips of these four structures form the vertices of a square in some order, compute the height of the cave.", "solution": "Answer: 63", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-geo-solutions.jsonl", "problem_match": "\n2. ", "solution_match": "\nProposed by: Albert Wang \n\n"}}
-{"year": "2025", "tier": "T4", "problem_label": "2", "problem_type": "Geometry", "exam": "HMMT", "problem": "In a two-dimensional cave with a parallel floor and ceiling, two stalactites of lengths 16 and 36 hang perpendicularly from the ceiling, while two stalagmites of heights 25 and 49 grow perpendicularly from the ground. If the tips of these four structures form the vertices of a square in some order, compute the height of the cave.", "solution": "Note that the difference in heights between the two stalactites does not equal the difference in heights between the stalagmites. This tells us that the two stalactites form a pair of opposite vertices of the square, and likewise for the stalagmites. As the midpoint of the pairs of structures must then coincide, we know that it is \\(\\frac{16 + 36}{2} = 26\\) from the ceiling due to the stalactites, and \\(\\frac{25 + 49}{2} = 37\\) from the ground due to the stalagmites. Therefore the height of the cave is just the sum of these two values, i.e. 63.\n\n\n", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-geo-solutions.jsonl", "problem_match": "\n2. ", "solution_match": "\nSolution: "}}
+{"year": "2025", "tier": "T4", "problem_label": "2", "problem_type": "Geometry", "exam": "HMMT", "problem": "In a two-dimensional cave with a parallel floor and ceiling, two stalactites of lengths 16 and 36 hang perpendicularly from the ceiling, while two stalagmites of heights 25 and 49 grow perpendicularly from the ground. If the tips of these four structures form the vertices of a square in some order, compute the height of the cave.", "solution": "Note that the difference in heights between the two stalactites does not equal the difference in heights between the stalagmites. This tells us that the two stalactites form a pair of opposite vertices of the square, and likewise for the stalagmites. As the midpoint of the pairs of structures must then coincide, we know that it is \\(\\frac{16 + 36}{2} = 26\\) from the ceiling due to the stalactites, and \\(\\frac{25 + 49}{2} = 37\\) from the ground due to the stalagmites. Therefore the height of the cave is just the sum of these two values, i.e. 63.\n\n\n", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-geo-solutions.jsonl", "problem_match": "\n2. ", "solution_match": "\nSolution: "}}
{"year": "2025", "tier": "T4", "problem_label": "3", "problem_type": "Geometry", "exam": "HMMT", "problem": "Point \\(P\\) lies inside square \\(ABCD\\) such that the areas of \\(\\triangle PAB\\) , \\(\\triangle PBC\\) , \\(\\triangle PCD\\) , and \\(\\triangle PDA\\) are 1, 2, 3, and 4, in some order. Compute \\(PA \\cdot PB \\cdot PC \\cdot PD\\) .", "solution": "Answer: \\(\\boxed{8\\sqrt{10}}\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-geo-solutions.jsonl", "problem_match": "\n3. ", "solution_match": "\nProposed by: Karthik Venkata Vedula \n\n"}}
-{"year": "2025", "tier": "T4", "problem_label": "3", "problem_type": "Geometry", "exam": "HMMT", "problem": "Point \\(P\\) lies inside square \\(ABCD\\) such that the areas of \\(\\triangle PAB\\) , \\(\\triangle PBC\\) , \\(\\triangle PCD\\) , and \\(\\triangle PDA\\) are 1, 2, 3, and 4, in some order. Compute \\(PA \\cdot PB \\cdot PC \\cdot PD\\) .", "solution": "\n \n\nLet \\(h_{1}\\) , \\(h_{2}\\) , \\(h_{3}\\) , and \\(h_{4}\\) be the lengths of the altitudes from \\(P\\) to sides \\(AB\\) , \\(BC\\) , \\(CD\\) , and \\(DA\\) , respectively. Then, the problem statement implies that \\(\\{h_{1}, h_{2}, h_{3}, h_{4}\\} = \\{x, 2x, 3x, 4x\\}\\) for some \\(x\\) . Furthermore, the area of the square is \\(1 + 2 + 3 + 4 = 10\\) , so we have \n\n\\[h_{1} + h_{3} = \\sqrt{10} = h_{2} + h_{4}.\\] \n\nHence either \\((\\{h_{1}, h_{3}\\} = \\{x, 4x\\}\\) and \\(\\{h_{2}, h_{4}\\} = \\{2x, 3x\\}\\) , or \\((\\{h_{2}, h_{4}\\} = \\{x, 4x\\}\\) and \\(\\{h_{1}, h_{3}\\} = \\{2x, 3x\\}\\) . In any case, we get \\(5x = \\sqrt{10}\\) from above, and finish via the Pythagorean theorem: \n\n\\[P A\\cdot P B\\cdot P C\\cdot P D = \\sqrt{x^{2} + (2x)^{2}}\\cdot \\sqrt{(2x)^{2} + (4x)^{2}}\\cdot \\sqrt{(4x)^{2} + (3x)^{2}}\\cdot \\sqrt{(3x)^{2} + x^{2}}\\] \\[\\qquad = 50\\sqrt{10}x^{4}\\] \\[\\qquad = (50\\sqrt{10})\\cdot \\left(\\frac{2}{5}\\right)^{2}\\] \\[\\qquad = \\boxed{8\\sqrt{10}}.\\]", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-geo-solutions.jsonl", "problem_match": "\n3. ", "solution_match": "\nSolution: \n\n"}}
-{"year": "2025", "tier": "T4", "problem_label": "4", "problem_type": "Geometry", "exam": "HMMT", "problem": "A semicircle is inscribed in another semicircle if the smaller semicircle's diameter is a chord of the larger semicircle, and the smaller semicircle's arc is tangent to the diameter of the larger semicircle. \n\nSemicircle \\(S_{1}\\) is inscribed in a semicircle \\(S_{2}\\) , which is inscribed in another semicircle \\(S_{3}\\) . The radii of \\(S_{1}\\) and \\(S_{3}\\) are 1 and 10, respectively, and the diameters of \\(S_{1}\\) and \\(S_{3}\\) are parallel. The endpoints of the diameter of \\(S_{3}\\) are \\(A\\) and \\(B\\) , and \\(S_{2}\\) 's arc is tangent to \\(AB\\) at \\(C\\) . Compute \\(AC \\cdot CB\\) . \n\n", "solution": "Answer: 20", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-geo-solutions.jsonl", "problem_match": "\n4. ", "solution_match": "\nProposed by: Karthik Venkata Vedula \n\n"}}
-{"year": "2025", "tier": "T4", "problem_label": "4", "problem_type": "Geometry", "exam": "HMMT", "problem": "A semicircle is inscribed in another semicircle if the smaller semicircle's diameter is a chord of the larger semicircle, and the smaller semicircle's arc is tangent to the diameter of the larger semicircle. \n\nSemicircle \\(S_{1}\\) is inscribed in a semicircle \\(S_{2}\\) , which is inscribed in another semicircle \\(S_{3}\\) . The radii of \\(S_{1}\\) and \\(S_{3}\\) are 1 and 10, respectively, and the diameters of \\(S_{1}\\) and \\(S_{3}\\) are parallel. The endpoints of the diameter of \\(S_{3}\\) are \\(A\\) and \\(B\\) , and \\(S_{2}\\) 's arc is tangent to \\(AB\\) at \\(C\\) . Compute \\(AC \\cdot CB\\) . \n\n", "solution": "\n \n\nLet \\(P\\) , \\(Q\\) , and \\(R\\) be the midpoints of the diameters (i.e., the center of the circular arcs) of \\(S_{3}\\) , \\(S_{2}\\) , and \\(S_{1}\\) , respectively. Observe that if one fixes \\(S_{3}\\) , the location of \\(S_{2}\\) is uniquely determined by the angle between the diameters of \\(S_{2}\\) and \\(S_{3}\\) . The same holds for \\(S_{2}\\) and \\(S_{1}\\) . Thus, the figures \\(S_{3} \\cup S_{2}\\) and \\(S_{2} \\cup S_{1}\\) are similar. This gives us that the radius of \\(S_{2}\\) is \\(\\sqrt{10}\\) . \n\nTo compute the answer, we define \\(V\\) to be either intersection of the arcs of \\(S_{2}\\) and \\(S_{3}\\) . By the Pythagorean theorem, \\(PQ = \\sqrt{PV^{2} - VQ^{2}} = \\sqrt{100 - 10} = \\sqrt{90}\\) . By the Pythagorean theorem again, \\(PC = \\sqrt{PQ^{2} - QC^{2}} = \\sqrt{90 - 10} = \\sqrt{80}\\) . Thus \\(AC \\cdot CB = (10 + PC)(10 - PC) = 100 - PC^{2} = \\left[20\\right]\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-geo-solutions.jsonl", "problem_match": "\n4. ", "solution_match": "\nSolution: \n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "3", "problem_type": "Geometry", "exam": "HMMT", "problem": "Point \\(P\\) lies inside square \\(ABCD\\) such that the areas of \\(\\triangle PAB\\) , \\(\\triangle PBC\\) , \\(\\triangle PCD\\) , and \\(\\triangle PDA\\) are 1, 2, 3, and 4, in some order. Compute \\(PA \\cdot PB \\cdot PC \\cdot PD\\) .", "solution": "\n \n\nLet \\(h_{1}\\) , \\(h_{2}\\) , \\(h_{3}\\) , and \\(h_{4}\\) be the lengths of the altitudes from \\(P\\) to sides \\(AB\\) , \\(BC\\) , \\(CD\\) , and \\(DA\\) , respectively. Then, the problem statement implies that \\(\\{h_{1}, h_{2}, h_{3}, h_{4}\\} = \\{x, 2x, 3x, 4x\\}\\) for some \\(x\\) . Furthermore, the area of the square is \\(1 + 2 + 3 + 4 = 10\\) , so we have \n\n\\[h_{1} + h_{3} = \\sqrt{10} = h_{2} + h_{4}.\\] \n\nHence either \\((\\{h_{1}, h_{3}\\} = \\{x, 4x\\}\\) and \\(\\{h_{2}, h_{4}\\} = \\{2x, 3x\\}\\) , or \\((\\{h_{2}, h_{4}\\} = \\{x, 4x\\}\\) and \\(\\{h_{1}, h_{3}\\} = \\{2x, 3x\\}\\) . In any case, we get \\(5x = \\sqrt{10}\\) from above, and finish via the Pythagorean theorem: \n\n\\[P A\\cdot P B\\cdot P C\\cdot P D = \\sqrt{x^{2} + (2x)^{2}}\\cdot \\sqrt{(2x)^{2} + (4x)^{2}}\\cdot \\sqrt{(4x)^{2} + (3x)^{2}}\\cdot \\sqrt{(3x)^{2} + x^{2}}\\] \\[\\qquad = 50\\sqrt{10}x^{4}\\] \\[\\qquad = (50\\sqrt{10})\\cdot \\left(\\frac{2}{5}\\right)^{2}\\] \\[\\qquad = \\boxed{8\\sqrt{10}}.\\]", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-geo-solutions.jsonl", "problem_match": "\n3. ", "solution_match": "\nSolution: \n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "4", "problem_type": "Geometry", "exam": "HMMT", "problem": "A semicircle is inscribed in another semicircle if the smaller semicircle's diameter is a chord of the larger semicircle, and the smaller semicircle's arc is tangent to the diameter of the larger semicircle. \n\nSemicircle \\(S_{1}\\) is inscribed in a semicircle \\(S_{2}\\) , which is inscribed in another semicircle \\(S_{3}\\) . The radii of \\(S_{1}\\) and \\(S_{3}\\) are 1 and 10, respectively, and the diameters of \\(S_{1}\\) and \\(S_{3}\\) are parallel. The endpoints of the diameter of \\(S_{3}\\) are \\(A\\) and \\(B\\) , and \\(S_{2}\\) 's arc is tangent to \\(AB\\) at \\(C\\) . Compute \\(AC \\cdot CB\\) . \n\n", "solution": "Answer: 20", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-geo-solutions.jsonl", "problem_match": "\n4. ", "solution_match": "\nProposed by: Karthik Venkata Vedula \n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "4", "problem_type": "Geometry", "exam": "HMMT", "problem": "A semicircle is inscribed in another semicircle if the smaller semicircle's diameter is a chord of the larger semicircle, and the smaller semicircle's arc is tangent to the diameter of the larger semicircle. \n\nSemicircle \\(S_{1}\\) is inscribed in a semicircle \\(S_{2}\\) , which is inscribed in another semicircle \\(S_{3}\\) . The radii of \\(S_{1}\\) and \\(S_{3}\\) are 1 and 10, respectively, and the diameters of \\(S_{1}\\) and \\(S_{3}\\) are parallel. The endpoints of the diameter of \\(S_{3}\\) are \\(A\\) and \\(B\\) , and \\(S_{2}\\) 's arc is tangent to \\(AB\\) at \\(C\\) . Compute \\(AC \\cdot CB\\) . \n\n", "solution": "\n \n\nLet \\(P\\) , \\(Q\\) , and \\(R\\) be the midpoints of the diameters (i.e., the center of the circular arcs) of \\(S_{3}\\) , \\(S_{2}\\) , and \\(S_{1}\\) , respectively. Observe that if one fixes \\(S_{3}\\) , the location of \\(S_{2}\\) is uniquely determined by the angle between the diameters of \\(S_{2}\\) and \\(S_{3}\\) . The same holds for \\(S_{2}\\) and \\(S_{1}\\) . Thus, the figures \\(S_{3} \\cup S_{2}\\) and \\(S_{2} \\cup S_{1}\\) are similar. This gives us that the radius of \\(S_{2}\\) is \\(\\sqrt{10}\\) . \n\nTo compute the answer, we define \\(V\\) to be either intersection of the arcs of \\(S_{2}\\) and \\(S_{3}\\) . By the Pythagorean theorem, \\(PQ = \\sqrt{PV^{2} - VQ^{2}} = \\sqrt{100 - 10} = \\sqrt{90}\\) . By the Pythagorean theorem again, \\(PC = \\sqrt{PQ^{2} - QC^{2}} = \\sqrt{90 - 10} = \\sqrt{80}\\) . Thus \\(AC \\cdot CB = (10 + PC)(10 - PC) = 100 - PC^{2} = \\left[20\\right]\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-geo-solutions.jsonl", "problem_match": "\n4. ", "solution_match": "\nSolution: \n\n"}}
{"year": "2025", "tier": "T4", "problem_label": "5", "problem_type": "Geometry", "exam": "HMMT", "problem": "Let \\(\\triangle ABC\\) be an equilateral triangle with side length 6. Let \\(P\\) be a point inside triangle \\(\\triangle ABC\\) such that \\(\\angle BPC = 120^{\\circ}\\) . The circle with diameter \\(\\overline{AP}\\) meets the circumcircle of \\(\\triangle ABC\\) again at \\(X \\neq A\\) . Given that \\(AX = 5\\) , compute \\(XP\\) .", "solution": "Answer: \\(\\sqrt{23} - 2\\sqrt{3}\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-geo-solutions.jsonl", "problem_match": "\n5. ", "solution_match": "\nProposed by: Pitchayut Saengrungkonkha \n\n"}}
-{"year": "2025", "tier": "T4", "problem_label": "5", "problem_type": "Geometry", "exam": "HMMT", "problem": "Let \\(\\triangle ABC\\) be an equilateral triangle with side length 6. Let \\(P\\) be a point inside triangle \\(\\triangle ABC\\) such that \\(\\angle BPC = 120^{\\circ}\\) . The circle with diameter \\(\\overline{AP}\\) meets the circumcircle of \\(\\triangle ABC\\) again at \\(X \\neq A\\) . Given that \\(AX = 5\\) , compute \\(XP\\) .", "solution": "\n \n\nLet \\(A^{\\prime}\\) be the antipode of \\(A\\) . As \\(\\angle A X A^{\\prime} = 90^{\\circ}\\) , we have \\(X\\) , \\(P\\) , and \\(A^{\\prime}\\) are collinear. As \\(\\angle B P C = 120^{\\circ}\\) and \\(\\angle B A^{\\prime}C = 180^{\\circ} - \\angle B A C = 120^{\\circ}\\) , it follows that \\(P\\) lies on the circle with center \\(A^{\\prime}\\) passing through \\(B\\) and \\(C\\) , so \\(A^{\\prime}P = \\frac{B C}{\\sqrt{3}} = 2\\sqrt{3}\\) and \\(A A^{\\prime} = 2A^{\\prime}B = 4\\sqrt{3}\\) . By the Pythagorean theorem, \\(X A^{\\prime} = \\sqrt{(4\\sqrt{3})^{2} - 5^{2}} = \\sqrt{23}\\) , so the answer is \\(X A^{\\prime} - P A^{\\prime} = \\left[\\sqrt{23} - 2\\sqrt{3}\\right]\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-geo-solutions.jsonl", "problem_match": "\n5. ", "solution_match": "\nSolution:\n\n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "5", "problem_type": "Geometry", "exam": "HMMT", "problem": "Let \\(\\triangle ABC\\) be an equilateral triangle with side length 6. Let \\(P\\) be a point inside triangle \\(\\triangle ABC\\) such that \\(\\angle BPC = 120^{\\circ}\\) . The circle with diameter \\(\\overline{AP}\\) meets the circumcircle of \\(\\triangle ABC\\) again at \\(X \\neq A\\) . Given that \\(AX = 5\\) , compute \\(XP\\) .", "solution": "\n \n\nLet \\(A^{\\prime}\\) be the antipode of \\(A\\) . As \\(\\angle A X A^{\\prime} = 90^{\\circ}\\) , we have \\(X\\) , \\(P\\) , and \\(A^{\\prime}\\) are collinear. As \\(\\angle B P C = 120^{\\circ}\\) and \\(\\angle B A^{\\prime}C = 180^{\\circ} - \\angle B A C = 120^{\\circ}\\) , it follows that \\(P\\) lies on the circle with center \\(A^{\\prime}\\) passing through \\(B\\) and \\(C\\) , so \\(A^{\\prime}P = \\frac{B C}{\\sqrt{3}} = 2\\sqrt{3}\\) and \\(A A^{\\prime} = 2A^{\\prime}B = 4\\sqrt{3}\\) . By the Pythagorean theorem, \\(X A^{\\prime} = \\sqrt{(4\\sqrt{3})^{2} - 5^{2}} = \\sqrt{23}\\) , so the answer is \\(X A^{\\prime} - P A^{\\prime} = \\left[\\sqrt{23} - 2\\sqrt{3}\\right]\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-geo-solutions.jsonl", "problem_match": "\n5. ", "solution_match": "\nSolution:\n\n\n"}}
{"year": "2025", "tier": "T4", "problem_label": "6", "problem_type": "Geometry", "exam": "HMMT", "problem": "Trapezoid \\(A B C D\\) , with \\(A B\\parallel C D\\) , has side lengths \\(A B = 11\\) , \\(B C = 8\\) , \\(C D = 19\\) , and \\(D A = 4\\) . Compute the area of the convex quadrilateral whose vertices are the circumcenters of \\(\\triangle A B C\\) , \\(\\triangle B C D\\) , \\(\\triangle C D A\\) , and \\(\\triangle D A B\\) .", "solution": "Answer: \\(\\boxed{9\\sqrt{15}}\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-geo-solutions.jsonl", "problem_match": "\n6. ", "solution_match": "\nProposed by: Karthik Venkata Vedula, Pitchayut Saengrungkonkka \n\n"}}
-{"year": "2025", "tier": "T4", "problem_label": "6", "problem_type": "Geometry", "exam": "HMMT", "problem": "Trapezoid \\(A B C D\\) , with \\(A B\\parallel C D\\) , has side lengths \\(A B = 11\\) , \\(B C = 8\\) , \\(C D = 19\\) , and \\(D A = 4\\) . Compute the area of the convex quadrilateral whose vertices are the circumcenters of \\(\\triangle A B C\\) , \\(\\triangle B C D\\) , \\(\\triangle C D A\\) , and \\(\\triangle D A B\\) .", "solution": "\n \n\nLet \\(O_{A}\\) , \\(O_{B}\\) , \\(O_{C}\\) , and \\(O_{D}\\) be the circumcenters of \\(\\triangle B C D\\) , \\(\\triangle C D A\\) , \\(\\triangle D A B\\) , and \\(\\triangle A B C\\) , respectively. Note that \\(O_{B}O_{C}\\) is the perpendicular bisector of \\(\\overline{{A D}}\\) . Similarly, \\(O_{B}O_{D}\\perp A C\\) and \\(O_{C}O_{D}\\perp A B\\) . As \\(A B\\parallel C D\\) , we have \\(O_{C}O_{D}\\perp C D\\) . Then, \\(\\triangle O_{B}O_{C}O_{D}\\stackrel {\\star}{\\sim}\\triangle A D C\\) , as their corresponding sides are perpendicular. Likewise, \\(\\triangle O_{D}O_{A}O_{B}\\stackrel {\\star}{\\sim}\\triangle C B A\\) , so \\(O_{A}O_{B}O_{C}O_{D}\\stackrel {\\star}{\\sim}B A D C\\) . \n\nTherefore, we only need to compute the area of \\(A B C D\\) and the ratio of similarity between the two trapezoids. Draw a line parallel to \\(\\overline{{A D}}\\) passing through \\(B\\) . Let this line intersect \\(\\overline{{C D}}\\) at \\(X\\) . Then, \\(B X = A D = 4\\) , \\(C B = 8\\) , and \\(C X = C D - A B = 8\\) . The height \\(h\\) of \\(A B C D\\) is given by \n\n\\[h = d(B,\\overline{{X C}}) = \\frac{2[\\triangle B X C]}{X C} = \\frac{B X\\cdot d(C,\\overline{{B X}})}{X C} = \\frac{4\\sqrt{8^{2} - 2^{2}}}{8} = \\sqrt{15}.\\]\n\n\n\nOn the other hand, the height of \\(O_{A}O_{B}O_{C}O_{D}\\) is the distance between the perpendicular bisectors of \\(\\overline{{A B}}\\) and \\(\\overline{{C D}}\\) , which pass through the midpoints \\(M\\) of \\(\\overline{{A B}}\\) and \\(N\\) of \\(\\overline{{C D}}\\) . Let \\(A^{\\prime}\\) and \\(M^{\\prime}\\) be the projections of \\(A\\) and \\(M\\) onto \\(\\overline{{C D}}\\) , respectively. Then, the height of \\(O_{A}O_{B}O_{C}O_{D}\\) is given by \n\n\\[M^{\\prime}N = D N - D M^{\\prime} = \\frac{C D}{2} -\\frac{A B}{2} -D A^{\\prime} = \\frac{19 - 11}{2} -\\sqrt{4^{2} - h^{2}} = 3.\\] \n\nTherefore, the similarity ratio between the two trapezoids is \\(\\frac{3}{\\sqrt{15}}\\) . We know that the area of \\(A B C D\\) is \\(\\frac{1}{2} (11 + 19)\\sqrt{15} = 15\\sqrt{15}\\) , so the area of \\(O_{A}O_{B}O_{C}O_{D}\\) is \\(\\left(\\frac{3}{\\sqrt{15}}\\right)^{2}\\cdot 15\\sqrt{15} = \\boxed{9\\sqrt{15}}\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-geo-solutions.jsonl", "problem_match": "\n6. ", "solution_match": "\nSolution: \n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "6", "problem_type": "Geometry", "exam": "HMMT", "problem": "Trapezoid \\(A B C D\\) , with \\(A B\\parallel C D\\) , has side lengths \\(A B = 11\\) , \\(B C = 8\\) , \\(C D = 19\\) , and \\(D A = 4\\) . Compute the area of the convex quadrilateral whose vertices are the circumcenters of \\(\\triangle A B C\\) , \\(\\triangle B C D\\) , \\(\\triangle C D A\\) , and \\(\\triangle D A B\\) .", "solution": "\n \n\nLet \\(O_{A}\\) , \\(O_{B}\\) , \\(O_{C}\\) , and \\(O_{D}\\) be the circumcenters of \\(\\triangle B C D\\) , \\(\\triangle C D A\\) , \\(\\triangle D A B\\) , and \\(\\triangle A B C\\) , respectively. Note that \\(O_{B}O_{C}\\) is the perpendicular bisector of \\(\\overline{{A D}}\\) . Similarly, \\(O_{B}O_{D}\\perp A C\\) and \\(O_{C}O_{D}\\perp A B\\) . As \\(A B\\parallel C D\\) , we have \\(O_{C}O_{D}\\perp C D\\) . Then, \\(\\triangle O_{B}O_{C}O_{D}\\stackrel {\\star}{\\sim}\\triangle A D C\\) , as their corresponding sides are perpendicular. Likewise, \\(\\triangle O_{D}O_{A}O_{B}\\stackrel {\\star}{\\sim}\\triangle C B A\\) , so \\(O_{A}O_{B}O_{C}O_{D}\\stackrel {\\star}{\\sim}B A D C\\) . \n\nTherefore, we only need to compute the area of \\(A B C D\\) and the ratio of similarity between the two trapezoids. Draw a line parallel to \\(\\overline{{A D}}\\) passing through \\(B\\) . Let this line intersect \\(\\overline{{C D}}\\) at \\(X\\) . Then, \\(B X = A D = 4\\) , \\(C B = 8\\) , and \\(C X = C D - A B = 8\\) . The height \\(h\\) of \\(A B C D\\) is given by \n\n\\[h = d(B,\\overline{{X C}}) = \\frac{2[\\triangle B X C]}{X C} = \\frac{B X\\cdot d(C,\\overline{{B X}})}{X C} = \\frac{4\\sqrt{8^{2} - 2^{2}}}{8} = \\sqrt{15}.\\]\n\n\n\nOn the other hand, the height of \\(O_{A}O_{B}O_{C}O_{D}\\) is the distance between the perpendicular bisectors of \\(\\overline{{A B}}\\) and \\(\\overline{{C D}}\\) , which pass through the midpoints \\(M\\) of \\(\\overline{{A B}}\\) and \\(N\\) of \\(\\overline{{C D}}\\) . Let \\(A^{\\prime}\\) and \\(M^{\\prime}\\) be the projections of \\(A\\) and \\(M\\) onto \\(\\overline{{C D}}\\) , respectively. Then, the height of \\(O_{A}O_{B}O_{C}O_{D}\\) is given by \n\n\\[M^{\\prime}N = D N - D M^{\\prime} = \\frac{C D}{2} -\\frac{A B}{2} -D A^{\\prime} = \\frac{19 - 11}{2} -\\sqrt{4^{2} - h^{2}} = 3.\\] \n\nTherefore, the similarity ratio between the two trapezoids is \\(\\frac{3}{\\sqrt{15}}\\) . We know that the area of \\(A B C D\\) is \\(\\frac{1}{2} (11 + 19)\\sqrt{15} = 15\\sqrt{15}\\) , so the area of \\(O_{A}O_{B}O_{C}O_{D}\\) is \\(\\left(\\frac{3}{\\sqrt{15}}\\right)^{2}\\cdot 15\\sqrt{15} = \\boxed{9\\sqrt{15}}\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-geo-solutions.jsonl", "problem_match": "\n6. ", "solution_match": "\nSolution: \n\n"}}
{"year": "2025", "tier": "T4", "problem_label": "7", "problem_type": "Geometry", "exam": "HMMT", "problem": "Point \\(P\\) is inside triangle \\(\\triangle A B C\\) such that \\(\\angle A B P = \\angle A C P\\) . Given that \\(A B = 6\\) , \\(A C = 8\\) , \\(B C = 7\\) , and \\(\\frac{B P}{P C} = \\frac{1}{2}\\) , compute \\(\\frac{|B P C|}{|A B C|}\\) . \n\n(Here, \\([X Y Z]\\) denotes the area of \\(\\triangle X Y Z\\) ).", "solution": "Answer: \\(\\frac{7}{18}\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-geo-solutions.jsonl", "problem_match": "\n7. ", "solution_match": "\nProposed by: Pitchayut Saengrungkonkha \n\n"}}
-{"year": "2025", "tier": "T4", "problem_label": "7", "problem_type": "Geometry", "exam": "HMMT", "problem": "Point \\(P\\) is inside triangle \\(\\triangle A B C\\) such that \\(\\angle A B P = \\angle A C P\\) . Given that \\(A B = 6\\) , \\(A C = 8\\) , \\(B C = 7\\) , and \\(\\frac{B P}{P C} = \\frac{1}{2}\\) , compute \\(\\frac{|B P C|}{|A B C|}\\) . \n\n(Here, \\([X Y Z]\\) denotes the area of \\(\\triangle X Y Z\\) ).", "solution": "\n \n\nLet the internal and external bisectors of \\(\\angle B A C\\) meet \\(B C\\) at \\(D\\) and \\(E\\) . Similarly, let the internal and external bisectors of \\(\\angle B P C\\) meet \\(B C\\) at \\(D^{\\prime}\\) and \\(E^{\\prime}\\) . The angle condition implies that \\(A D\\parallel P D^{\\prime}\\) and \\(A E\\parallel P E^{\\prime}\\) . Thus, triangles \\(A D E\\) and \\(P D^{\\prime}E^{\\prime}\\) are homothetic. Hence, the requested ratio is \\(\\frac{D E}{D^{\\prime}E^{\\prime}}\\) . Repeated applications of the angle bisector theorem yield \n\n\\[B D = 7\\cdot {\\frac{6}{6 + 8}} = 3,\\] \\[B E = 7\\cdot {\\frac{6}{8 - 6}} = 21,\\] \\[B D^{\\prime} = 7\\cdot {\\frac{1}{1 + 2}} = {\\frac{7}{3}},\\] \\[B E^{\\prime} = 7\\cdot {\\frac{1}{2 - 1}} = 7,\\] \n\nso \\(D E = 24\\) and \\(D^{\\prime}E^{\\prime} = 28 / 3\\) . Hence, the answer is \\(\\frac{28 / 3}{24} = \\left[\\frac{7}{18}\\right]\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-geo-solutions.jsonl", "problem_match": "\n7. ", "solution_match": "\nSolution 1: \n\n"}}
-{"year": "2025", "tier": "T4", "problem_label": "7", "problem_type": "Geometry", "exam": "HMMT", "problem": "Point \\(P\\) is inside triangle \\(\\triangle A B C\\) such that \\(\\angle A B P = \\angle A C P\\) . Given that \\(A B = 6\\) , \\(A C = 8\\) , \\(B C = 7\\) , and \\(\\frac{B P}{P C} = \\frac{1}{2}\\) , compute \\(\\frac{|B P C|}{|A B C|}\\) . \n\n(Here, \\([X Y Z]\\) denotes the area of \\(\\triangle X Y Z\\) ).", "solution": "\n \n\nLet \\(D = A P\\cap B C\\) \\(E = B P\\cap A C\\) and \\(F = C P\\cap A B\\) . Then, \\(B C E F\\) is cyclic, so by Power of a Point, \\(\\frac{A E}{A F} = \\frac{A B}{A C} = \\frac{3}{4}\\) . Let \\(A E = 3x\\) and \\(A F = 4x\\) . Then, \n\n\\[\\triangle P F B\\sim \\triangle P E C\\Longrightarrow \\frac{B F}{C E} = \\frac{B P}{C P}\\Longrightarrow \\frac{6 - 4x}{8 - 3x} = \\frac{1}{2}.\\] \n\nSolving for \\(x\\) gives \\(x = \\frac{4}{5}\\) , so we get that \\(A F = \\frac{16}{5}\\) , \\(F B = \\frac{14}{5}\\) , \\(A E = \\frac{12}{5}\\) , and \\(E C = \\frac{40}{5}\\) . By Ceva's theorem on \\(\\triangle A B C\\) and point \\(P\\) , we have \n\n\\[\\frac{B D}{D C} = \\frac{A E}{E C}\\cdot \\frac{B F}{F A} = \\frac{A E}{A F}\\cdot \\frac{B F}{C E} = \\frac{3}{4}\\cdot \\frac{1}{2} = \\frac{3}{8}.\\] \n\nFinally, by Menelaus's theorem on \\(\\triangle A D C\\) and line \\(B E\\) , we get that \n\n\\[\\frac{A P}{P D} = \\frac{C B}{B D}\\cdot \\frac{A E}{E C} = \\frac{11}{3}\\cdot \\frac{12}{28} = \\frac{11}{7},\\] \n\nwhich implies that \\(\\frac{|B P C|}{|A B C|} = \\left[\\frac{7}{18}\\right]\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-geo-solutions.jsonl", "problem_match": "\n7. ", "solution_match": "\nSolution 2:\n\n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "7", "problem_type": "Geometry", "exam": "HMMT", "problem": "Point \\(P\\) is inside triangle \\(\\triangle A B C\\) such that \\(\\angle A B P = \\angle A C P\\) . Given that \\(A B = 6\\) , \\(A C = 8\\) , \\(B C = 7\\) , and \\(\\frac{B P}{P C} = \\frac{1}{2}\\) , compute \\(\\frac{|B P C|}{|A B C|}\\) . \n\n(Here, \\([X Y Z]\\) denotes the area of \\(\\triangle X Y Z\\) ).", "solution": "\n \n\nLet the internal and external bisectors of \\(\\angle B A C\\) meet \\(B C\\) at \\(D\\) and \\(E\\) . Similarly, let the internal and external bisectors of \\(\\angle B P C\\) meet \\(B C\\) at \\(D^{\\prime}\\) and \\(E^{\\prime}\\) . The angle condition implies that \\(A D\\parallel P D^{\\prime}\\) and \\(A E\\parallel P E^{\\prime}\\) . Thus, triangles \\(A D E\\) and \\(P D^{\\prime}E^{\\prime}\\) are homothetic. Hence, the requested ratio is \\(\\frac{D E}{D^{\\prime}E^{\\prime}}\\) . Repeated applications of the angle bisector theorem yield \n\n\\[B D = 7\\cdot {\\frac{6}{6 + 8}} = 3,\\] \\[B E = 7\\cdot {\\frac{6}{8 - 6}} = 21,\\] \\[B D^{\\prime} = 7\\cdot {\\frac{1}{1 + 2}} = {\\frac{7}{3}},\\] \\[B E^{\\prime} = 7\\cdot {\\frac{1}{2 - 1}} = 7,\\] \n\nso \\(D E = 24\\) and \\(D^{\\prime}E^{\\prime} = 28 / 3\\) . Hence, the answer is \\(\\frac{28 / 3}{24} = \\left[\\frac{7}{18}\\right]\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-geo-solutions.jsonl", "problem_match": "\n7. ", "solution_match": "\nSolution 1: \n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "7", "problem_type": "Geometry", "exam": "HMMT", "problem": "Point \\(P\\) is inside triangle \\(\\triangle A B C\\) such that \\(\\angle A B P = \\angle A C P\\) . Given that \\(A B = 6\\) , \\(A C = 8\\) , \\(B C = 7\\) , and \\(\\frac{B P}{P C} = \\frac{1}{2}\\) , compute \\(\\frac{|B P C|}{|A B C|}\\) . \n\n(Here, \\([X Y Z]\\) denotes the area of \\(\\triangle X Y Z\\) ).", "solution": "\n \n\nLet \\(D = A P\\cap B C\\) \\(E = B P\\cap A C\\) and \\(F = C P\\cap A B\\) . Then, \\(B C E F\\) is cyclic, so by Power of a Point, \\(\\frac{A E}{A F} = \\frac{A B}{A C} = \\frac{3}{4}\\) . Let \\(A E = 3x\\) and \\(A F = 4x\\) . Then, \n\n\\[\\triangle P F B\\sim \\triangle P E C\\Longrightarrow \\frac{B F}{C E} = \\frac{B P}{C P}\\Longrightarrow \\frac{6 - 4x}{8 - 3x} = \\frac{1}{2}.\\] \n\nSolving for \\(x\\) gives \\(x = \\frac{4}{5}\\) , so we get that \\(A F = \\frac{16}{5}\\) , \\(F B = \\frac{14}{5}\\) , \\(A E = \\frac{12}{5}\\) , and \\(E C = \\frac{40}{5}\\) . By Ceva's theorem on \\(\\triangle A B C\\) and point \\(P\\) , we have \n\n\\[\\frac{B D}{D C} = \\frac{A E}{E C}\\cdot \\frac{B F}{F A} = \\frac{A E}{A F}\\cdot \\frac{B F}{C E} = \\frac{3}{4}\\cdot \\frac{1}{2} = \\frac{3}{8}.\\] \n\nFinally, by Menelaus's theorem on \\(\\triangle A D C\\) and line \\(B E\\) , we get that \n\n\\[\\frac{A P}{P D} = \\frac{C B}{B D}\\cdot \\frac{A E}{E C} = \\frac{11}{3}\\cdot \\frac{12}{28} = \\frac{11}{7},\\] \n\nwhich implies that \\(\\frac{|B P C|}{|A B C|} = \\left[\\frac{7}{18}\\right]\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-geo-solutions.jsonl", "problem_match": "\n7. ", "solution_match": "\nSolution 2:\n\n\n"}}
{"year": "2025", "tier": "T4", "problem_label": "7", "problem_type": "Geometry", "exam": "HMMT", "problem": "Point \\(P\\) is inside triangle \\(\\triangle A B C\\) such that \\(\\angle A B P = \\angle A C P\\) . Given that \\(A B = 6\\) , \\(A C = 8\\) , \\(B C = 7\\) , and \\(\\frac{B P}{P C} = \\frac{1}{2}\\) , compute \\(\\frac{|B P C|}{|A B C|}\\) . \n\n(Here, \\([X Y Z]\\) denotes the area of \\(\\triangle X Y Z\\) ).", "solution": "We use barycentric coordinates with respect to \\(\\triangle A B C\\) . From \\(\\angle A B P = \\angle A C P\\) , we get \n\n\\[\\frac{|A B P|}{|A C P|} = \\frac{A B\\cdot B P}{A C\\cdot C P} = \\frac{3}{8},\\] \n\nso \\(P\\) has coordinate \\((- :8:3)\\) . Let \\(Q\\) be the isogonal conjugate of \\(P\\) , so \\(Q\\) lies on perpendicular bisector of \\(B C\\) . By Steiner ratio theorem, \\(Q\\) has coordinate \\(\\left(- :8^{2} / 8:6^{2} / 3\\right) = \\left(- :8:12\\right) = \\left(- :2:3\\right)\\) . Let the coordinate be \\((t:2:3)\\) for some real number \\(t\\) . Then, recall the equation for the perpendicular bisector (from Corollary 6 of https://web.evanchem.cc/handouts/bary/bary- full.pdf): \n\n\\[a^{2}(z - y) + x(c^{2} - b^{2}) = 0\\] \\[\\Rightarrow 7^{2}(3 - 2) + t(6^{2} - 8^{2}) = 0,\\] \n\nso \\(t = \\frac{7^{2}(3 - 2)}{8^{2} - 6^{2}} = \\frac{7}{4}\\) . Hence, point \\(Q\\) has coordinates \\((7:8:12)\\) , so point \\(P\\) has coordinates \\((7^{2} / 7:8^{2} / 8:6^{2} / 12) = (7:8:3)\\) . Therefore, \\(|B P C| / |A B C| = \\left[\\frac{7}{18}\\right]\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-geo-solutions.jsonl", "problem_match": "\n7. ", "solution_match": "\nSolution 3: "}}
{"year": "2025", "tier": "T4", "problem_label": "8", "problem_type": "Geometry", "exam": "HMMT", "problem": "Let \\(A B C D\\) be an isosceles trapezoid such that \\(C D > A B = 4\\) . Let \\(E\\) be a point on line \\(C D\\) such that \\(D E = 2\\) and \\(D\\) lies between \\(E\\) and \\(C\\) . Let \\(M\\) be the midpoint of \\(\\overline{{A E}}\\) . Given that points \\(A\\) , \\(B\\) , \\(C\\) , \\(D\\) , and \\(M\\) lie on a circle with radius 5, compute \\(M D\\) .", "solution": "Answer: \\(\\sqrt{6}\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-geo-solutions.jsonl", "problem_match": "\n8. ", "solution_match": "\nProposed by: Sarunyu Thongjarast \n\n"}}
-{"year": "2025", "tier": "T4", "problem_label": "8", "problem_type": "Geometry", "exam": "HMMT", "problem": "Let \\(A B C D\\) be an isosceles trapezoid such that \\(C D > A B = 4\\) . Let \\(E\\) be a point on line \\(C D\\) such that \\(D E = 2\\) and \\(D\\) lies between \\(E\\) and \\(C\\) . Let \\(M\\) be the midpoint of \\(\\overline{{A E}}\\) . Given that points \\(A\\) , \\(B\\) , \\(C\\) , \\(D\\) , and \\(M\\) lie on a circle with radius 5, compute \\(M D\\) .", "solution": "\n \n\nLet \\(D^{\\prime}\\) be the reflection of \\(D\\) across \\(M\\) . Then, \\(A D E D^{\\prime}\\) is a parallelogram. Hence, \\(D^{\\prime}A = 2\\) , so \\(D^{\\prime}B = 6\\) . Thus, if \\(D^{\\prime}M = M D = x\\) , then Power of a Point at \\(D^{\\prime}\\) gives \\(x\\cdot (2x) = 2\\cdot 6\\) , so \\(x = \\sqrt{6}\\) . \n\nRemark. The radius of the circle is unnecessary.", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-geo-solutions.jsonl", "problem_match": "\n8. ", "solution_match": "\nSolution: \n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "8", "problem_type": "Geometry", "exam": "HMMT", "problem": "Let \\(A B C D\\) be an isosceles trapezoid such that \\(C D > A B = 4\\) . Let \\(E\\) be a point on line \\(C D\\) such that \\(D E = 2\\) and \\(D\\) lies between \\(E\\) and \\(C\\) . Let \\(M\\) be the midpoint of \\(\\overline{{A E}}\\) . Given that points \\(A\\) , \\(B\\) , \\(C\\) , \\(D\\) , and \\(M\\) lie on a circle with radius 5, compute \\(M D\\) .", "solution": "\n \n\nLet \\(D^{\\prime}\\) be the reflection of \\(D\\) across \\(M\\) . Then, \\(A D E D^{\\prime}\\) is a parallelogram. Hence, \\(D^{\\prime}A = 2\\) , so \\(D^{\\prime}B = 6\\) . Thus, if \\(D^{\\prime}M = M D = x\\) , then Power of a Point at \\(D^{\\prime}\\) gives \\(x\\cdot (2x) = 2\\cdot 6\\) , so \\(x = \\sqrt{6}\\) . \n\nRemark. The radius of the circle is unnecessary.", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-geo-solutions.jsonl", "problem_match": "\n8. ", "solution_match": "\nSolution: \n\n"}}
{"year": "2025", "tier": "T4", "problem_label": "9", "problem_type": "Geometry", "exam": "HMMT", "problem": "Let \\(A B C D\\) be a rectangle with \\(B C = 24\\) . Point \\(X\\) lies inside the rectangle such that \\(\\angle A X B = 90^{\\circ}\\) . Given that triangles \\(\\triangle A X D\\) and \\(\\triangle B X C\\) are both acute and have circumradii 13 and 15, respectively, compute \\(A B\\) .", "solution": "Answer: \\(\\boxed{14 + 4\\sqrt{37}}\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-geo-solutions.jsonl", "problem_match": "\n9. ", "solution_match": "\nProposed by: Pitchayut Saengrungkongka \n\n"}}
-{"year": "2025", "tier": "T4", "problem_label": "9", "problem_type": "Geometry", "exam": "HMMT", "problem": "Let \\(A B C D\\) be a rectangle with \\(B C = 24\\) . Point \\(X\\) lies inside the rectangle such that \\(\\angle A X B = 90^{\\circ}\\) . Given that triangles \\(\\triangle A X D\\) and \\(\\triangle B X C\\) are both acute and have circumradii 13 and 15, respectively, compute \\(A B\\) .", "solution": "Let \\(M\\) be the midpoint of \\(A B\\) . Let \\(O_{1}\\) and \\(O_{2}\\) be the circumcenters of \\(\\triangle A X D\\) and \\(\\triangle B X C\\) , respectively. Since \\(O_{1}M\\) is the perpendicular bisector of \\(A X\\) and \\(O_{2}M\\) is the perpendicular bisector of \\(B X\\) , we get that \\(\\angle O_{1}M O_{2} = 90^{\\circ}\\) . \n\nLet \\(P_{1}\\) and \\(P_{2}\\) be the projections of \\(O_{1}\\) and \\(O_{2}\\) onto segment \\(A B\\) , respectively, and let \\(A B = 2x\\) . By the Pythagorean theorem, \\(P_{1}A = \\sqrt{O_{1}A^{2} - O_{1}P_{1}^{2}} = \\sqrt{15^{2} - 12^{2}} = 5\\) , so \\(M P_{1} = M A - P_{1}A = x - 5\\) . Likewise, \\(M P_{2} = M B - \\sqrt{15^{2} - 12^{2}} = x - 9\\) . Since \\(\\triangle M P_{1}O_{1}\\sim \\triangle O_{2}P_{2}M\\) , we know \n\n\\[(x - 5)(x - 9) = M P_{1}\\cdot M P_{2} = O_{2}P_{2}\\cdot P_{1}O_{1} = 12^{2}.\\] \n\nSolving this, we get \\(x = 7 + 2\\sqrt{37}\\) , which implies that \\(A B = 2x = \\boxed{14 + 4\\sqrt{37}}\\) . (The condition that \\(\\triangle A X D\\) and \\(\\triangle B X C\\) are acute rules out \\(14 - 4\\sqrt{37}\\) .)\n\n\n", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-geo-solutions.jsonl", "problem_match": "\n9. ", "solution_match": "\nSolution 1: "}}
-{"year": "2025", "tier": "T4", "problem_label": "9", "problem_type": "Geometry", "exam": "HMMT", "problem": "Let \\(A B C D\\) be a rectangle with \\(B C = 24\\) . Point \\(X\\) lies inside the rectangle such that \\(\\angle A X B = 90^{\\circ}\\) . Given that triangles \\(\\triangle A X D\\) and \\(\\triangle B X C\\) are both acute and have circumradii 13 and 15, respectively, compute \\(A B\\) .", "solution": "\n \n\nLet \\(P\\) be the antipode of \\(A\\) in \\(\\odot (AXD)\\) and \\(Q\\) be the antipode of \\(B\\) in \\(\\odot (BXC)\\) . From \\(\\angle PDA = \\angle QCB = 90^{\\circ}\\) , we get that \\(P\\) and \\(Q\\) lie on \\(CD\\) . Moreover, from \\(\\angle PXA = 90^{\\circ}\\) , we get that \\(P \\in BX\\) , and similarly \\(Q \\in AX\\) . \n\nBeing a diameter, \\(AP = 2\\cdot 13 = 26\\) , so by the Pythagorean theorem, \\(DP = \\sqrt{26^{2} - 24^{2}} = 10\\) . Similarly, \\(BQ = 30\\) and \\(CQ = \\sqrt{30^{2} - 24^{2}} = 18\\) . Letting \\(AB = x\\) , we get \\(PQ = x - 28\\) . Quadrilateral \\(ABQP\\) has perpendicular diagonals, so \\(AB^{2} + PQ^{2} = AP^{2} + BQ^{2}\\) , which means that \\(x^{2} + (x - 28)^{2} = 26^{2} + 30^{2}\\) . \n\nSolving this quadratic gives \\(x = \\boxed {14 + 4\\sqrt{37}}\\) . (The condition that \\(\\triangle AXD\\) and \\(\\triangle BXC\\) are acute rules out \\(14 - 4\\sqrt{37}\\) .)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-geo-solutions.jsonl", "problem_match": "\n9. ", "solution_match": "\nSolution 2: \n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "9", "problem_type": "Geometry", "exam": "HMMT", "problem": "Let \\(A B C D\\) be a rectangle with \\(B C = 24\\) . Point \\(X\\) lies inside the rectangle such that \\(\\angle A X B = 90^{\\circ}\\) . Given that triangles \\(\\triangle A X D\\) and \\(\\triangle B X C\\) are both acute and have circumradii 13 and 15, respectively, compute \\(A B\\) .", "solution": "Let \\(M\\) be the midpoint of \\(A B\\) . Let \\(O_{1}\\) and \\(O_{2}\\) be the circumcenters of \\(\\triangle A X D\\) and \\(\\triangle B X C\\) , respectively. Since \\(O_{1}M\\) is the perpendicular bisector of \\(A X\\) and \\(O_{2}M\\) is the perpendicular bisector of \\(B X\\) , we get that \\(\\angle O_{1}M O_{2} = 90^{\\circ}\\) . \n\nLet \\(P_{1}\\) and \\(P_{2}\\) be the projections of \\(O_{1}\\) and \\(O_{2}\\) onto segment \\(A B\\) , respectively, and let \\(A B = 2x\\) . By the Pythagorean theorem, \\(P_{1}A = \\sqrt{O_{1}A^{2} - O_{1}P_{1}^{2}} = \\sqrt{15^{2} - 12^{2}} = 5\\) , so \\(M P_{1} = M A - P_{1}A = x - 5\\) . Likewise, \\(M P_{2} = M B - \\sqrt{15^{2} - 12^{2}} = x - 9\\) . Since \\(\\triangle M P_{1}O_{1}\\sim \\triangle O_{2}P_{2}M\\) , we know \n\n\\[(x - 5)(x - 9) = M P_{1}\\cdot M P_{2} = O_{2}P_{2}\\cdot P_{1}O_{1} = 12^{2}.\\] \n\nSolving this, we get \\(x = 7 + 2\\sqrt{37}\\) , which implies that \\(A B = 2x = \\boxed{14 + 4\\sqrt{37}}\\) . (The condition that \\(\\triangle A X D\\) and \\(\\triangle B X C\\) are acute rules out \\(14 - 4\\sqrt{37}\\) .)\n\n\n", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-geo-solutions.jsonl", "problem_match": "\n9. ", "solution_match": "\nSolution 1: "}}
+{"year": "2025", "tier": "T4", "problem_label": "9", "problem_type": "Geometry", "exam": "HMMT", "problem": "Let \\(A B C D\\) be a rectangle with \\(B C = 24\\) . Point \\(X\\) lies inside the rectangle such that \\(\\angle A X B = 90^{\\circ}\\) . Given that triangles \\(\\triangle A X D\\) and \\(\\triangle B X C\\) are both acute and have circumradii 13 and 15, respectively, compute \\(A B\\) .", "solution": "\n \n\nLet \\(P\\) be the antipode of \\(A\\) in \\(\\odot (AXD)\\) and \\(Q\\) be the antipode of \\(B\\) in \\(\\odot (BXC)\\) . From \\(\\angle PDA = \\angle QCB = 90^{\\circ}\\) , we get that \\(P\\) and \\(Q\\) lie on \\(CD\\) . Moreover, from \\(\\angle PXA = 90^{\\circ}\\) , we get that \\(P \\in BX\\) , and similarly \\(Q \\in AX\\) . \n\nBeing a diameter, \\(AP = 2\\cdot 13 = 26\\) , so by the Pythagorean theorem, \\(DP = \\sqrt{26^{2} - 24^{2}} = 10\\) . Similarly, \\(BQ = 30\\) and \\(CQ = \\sqrt{30^{2} - 24^{2}} = 18\\) . Letting \\(AB = x\\) , we get \\(PQ = x - 28\\) . Quadrilateral \\(ABQP\\) has perpendicular diagonals, so \\(AB^{2} + PQ^{2} = AP^{2} + BQ^{2}\\) , which means that \\(x^{2} + (x - 28)^{2} = 26^{2} + 30^{2}\\) . \n\nSolving this quadratic gives \\(x = \\boxed {14 + 4\\sqrt{37}}\\) . (The condition that \\(\\triangle AXD\\) and \\(\\triangle BXC\\) are acute rules out \\(14 - 4\\sqrt{37}\\) .)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-geo-solutions.jsonl", "problem_match": "\n9. ", "solution_match": "\nSolution 2: \n\n"}}
{"year": "2025", "tier": "T4", "problem_label": "10", "problem_type": "Geometry", "exam": "HMMT", "problem": "A plane \\(\\mathcal{P}\\) intersects a rectangular prism at a hexagon which has side lengths 45, 66, 63, 55, 54, and 77, in that order. Compute the distance from the center of the rectangular prism to \\(\\mathcal{P}\\) .", "solution": "Answer: \\(\\boxed{\\sqrt{\\frac{95}{24}}}\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-geo-solutions.jsonl", "problem_match": "\n10. ", "solution_match": "\nProposed by: Albert Wang, Karthik Venkata Vedula \n\n"}}
-{"year": "2025", "tier": "T4", "problem_label": "10", "problem_type": "Geometry", "exam": "HMMT", "problem": "A plane \\(\\mathcal{P}\\) intersects a rectangular prism at a hexagon which has side lengths 45, 66, 63, 55, 54, and 77, in that order. Compute the distance from the center of the rectangular prism to \\(\\mathcal{P}\\) .", "solution": "Translate \\(\\mathcal{P}\\) so that it contains the center. The intersection of the translated plane with the rectangular prism is a centrally symmetric hexagon. Let its side lengths be \\(a\\) , \\(b\\) , \\(c\\) , \\(a\\) , \\(b\\) , and \\(c\\) , in that order. Then, for some \\(t_{a}\\) , \\(t_{b}\\) , and \\(t_{c}\\) , the side lengths of the hexagon before the translation were \n\n\\[(a - t_{a},b + t_{b},c - t_{c},a + t_{a},b - t_{b},c + t_{c}) = (45,66,63,55,54,77),\\] \n\nfrom which it follows that \\(t_{a} = 5\\) , \\(t_{b} = 6\\) , and \\(t_{c} = 7\\) . \n\nNow, a translation of a plane can be written in one of three equivalent forms: it can be viewed as a translation in the \\(x\\) direction by a distance \\(d_{x}\\) , a translation in the \\(y\\) direction by a distance \\(d_{y}\\) , or a translation in the \\(z\\) direction by a distance \\(d_{z}\\) (with coordinate axes chosen as shown below).\n\n\n\n \n\nAs shown above, we can express \\(t_{a}\\) , \\(t_{b}\\) , and \\(t_{c}\\) in terms of \\(d_{x}\\) , \\(d_{y}\\) , and \\(d_{z}\\) using the Pythagorean theorem, which yields \\(t_{a} = \\sqrt{d_{y}^{2} + d_{z}^{2}}\\) , \\(t_{b} = \\sqrt{d_{z}^{2} + d_{x}^{2}}\\) , and \\(t_{c} = \\sqrt{d_{x}^{2} + d_{y}^{2}}\\) . Hence, \n\n\\[(d_{x}^{2}, d_{y}^{2}, d_{z}^{2}) = \\left(\\frac{5^{2} + 6^{2} - 7^{2}}{2}, \\frac{6^{2} + 7^{2} - 5^{2}}{2}, \\frac{7^{2} + 5^{2} - 6^{2}}{2}\\right) = (6, 30, 19).\\] \n\nWe can draw a right pyramid with legs \\(d_{x}\\) , \\(d_{y}\\) , and \\(d_{z}\\) which has the center of the prism as a vertex with opposite face on \\(\\mathcal{P}\\) . Then, the height of this pyramid, or the distance from the center to \\(\\mathcal{P}\\) , is \n\n\\[\\sqrt{\\frac{1}{d_{x}^{-2} + d_{y}^{-2} + d_{z}^{-2}}} = \\sqrt{\\frac{1}{6^{-1} + 30^{-1} + 19^{-1}}} = \\sqrt{\\frac{95}{24}};\\]", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-geo-solutions.jsonl", "problem_match": "\n10. ", "solution_match": "\nSolution 1: "}}
-{"year": "2025", "tier": "T4", "problem_label": "10", "problem_type": "Geometry", "exam": "HMMT", "problem": "A plane \\(\\mathcal{P}\\) intersects a rectangular prism at a hexagon which has side lengths 45, 66, 63, 55, 54, and 77, in that order. Compute the distance from the center of the rectangular prism to \\(\\mathcal{P}\\) .", "solution": "Let the vertices of the hexagon be \\(ABCDEF\\) , where \\(AB = 45\\) , \\(BC = 66\\) , etc. Note that \\(AB \\parallel DE\\) , \\(BC \\parallel EF\\) , and \\(CD \\parallel FA\\) . Let \\(O\\) be the center of the prism, and let \\(M\\) , \\(N\\) , and \\(P\\) be the midpoints of \\(AD\\) , \\(BE\\) , and \\(CF\\) , respectively. \n\n\n\n\n\n\nThe key observation is that \\(M N\\) is the midline between \\(A B\\) and \\(D E\\) . Hence, plane \\(O M N\\) is the midplane between the faces of the prism containing sides \\(A B\\) and \\(D E\\) . Similarly, planes \\(O M P\\) and \\(O N P\\) are the other two midplanes of the prism. Thus, \\(O M\\) , \\(O N\\) , and \\(O P\\) are mutually orthogonal. \n\nObserve \n\n\\[M N = \\frac{|A B - D E|}{2} = 5, \\quad N P = \\frac{|B C - E F|}{2} = 6, \\quad \\text{and} \\quad P M = \\frac{|C D - F A|}{2} = 7,\\] \n\nso by Heron's formula, we can compute the area of \\(M N P\\) to be \\(\\sqrt{9(9 - 5)(9 - 6)(9 - 7)} = 6\\sqrt{6}\\) . \n\nMoreover, if \\(x = O M\\) , \\(y = O N\\) , and \\(z = O P\\) , then, \n\n\\[x^{2} + y^{2} = 5^{2}, \\quad y^{2} + z^{2} = 6^{2}, \\quad \\text{and} \\quad z^{2} + x^{2} = 7^{2}.\\] \n\nSolving this system of equations gives \\(x = \\sqrt{19}\\) , \\(y = \\sqrt{6}\\) , and \\(z = \\sqrt{30}\\) . Therefore, if \\(d\\) is the distance from \\(O\\) to plane \\(M N P\\) (i.e., the answer), the volume of tetrahedron \\(O M N P\\) can be written as \n\n\\[\\frac{1}{6} \\cdot \\sqrt{19} \\cdot \\sqrt{6} \\cdot \\sqrt{30} = \\frac{1}{3} \\cdot (6\\sqrt{6}) \\cdot d,\\] \n\nSO \n\n\\[d = \\frac{\\sqrt{19 \\cdot 6 \\cdot 30}}{2 \\cdot 6 \\sqrt{6}} = \\sqrt{\\frac{95}{24}}.\\]", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-geo-solutions.jsonl", "problem_match": "\n10. ", "solution_match": "\nSolution 2: "}}
+{"year": "2025", "tier": "T4", "problem_label": "10", "problem_type": "Geometry", "exam": "HMMT", "problem": "A plane \\(\\mathcal{P}\\) intersects a rectangular prism at a hexagon which has side lengths 45, 66, 63, 55, 54, and 77, in that order. Compute the distance from the center of the rectangular prism to \\(\\mathcal{P}\\) .", "solution": "Translate \\(\\mathcal{P}\\) so that it contains the center. The intersection of the translated plane with the rectangular prism is a centrally symmetric hexagon. Let its side lengths be \\(a\\) , \\(b\\) , \\(c\\) , \\(a\\) , \\(b\\) , and \\(c\\) , in that order. Then, for some \\(t_{a}\\) , \\(t_{b}\\) , and \\(t_{c}\\) , the side lengths of the hexagon before the translation were \n\n\\[(a - t_{a},b + t_{b},c - t_{c},a + t_{a},b - t_{b},c + t_{c}) = (45,66,63,55,54,77),\\] \n\nfrom which it follows that \\(t_{a} = 5\\) , \\(t_{b} = 6\\) , and \\(t_{c} = 7\\) . \n\nNow, a translation of a plane can be written in one of three equivalent forms: it can be viewed as a translation in the \\(x\\) direction by a distance \\(d_{x}\\) , a translation in the \\(y\\) direction by a distance \\(d_{y}\\) , or a translation in the \\(z\\) direction by a distance \\(d_{z}\\) (with coordinate axes chosen as shown below).\n\n\n\n \n\nAs shown above, we can express \\(t_{a}\\) , \\(t_{b}\\) , and \\(t_{c}\\) in terms of \\(d_{x}\\) , \\(d_{y}\\) , and \\(d_{z}\\) using the Pythagorean theorem, which yields \\(t_{a} = \\sqrt{d_{y}^{2} + d_{z}^{2}}\\) , \\(t_{b} = \\sqrt{d_{z}^{2} + d_{x}^{2}}\\) , and \\(t_{c} = \\sqrt{d_{x}^{2} + d_{y}^{2}}\\) . Hence, \n\n\\[(d_{x}^{2}, d_{y}^{2}, d_{z}^{2}) = \\left(\\frac{5^{2} + 6^{2} - 7^{2}}{2}, \\frac{6^{2} + 7^{2} - 5^{2}}{2}, \\frac{7^{2} + 5^{2} - 6^{2}}{2}\\right) = (6, 30, 19).\\] \n\nWe can draw a right pyramid with legs \\(d_{x}\\) , \\(d_{y}\\) , and \\(d_{z}\\) which has the center of the prism as a vertex with opposite face on \\(\\mathcal{P}\\) . Then, the height of this pyramid, or the distance from the center to \\(\\mathcal{P}\\) , is \n\n\\[\\sqrt{\\frac{1}{d_{x}^{-2} + d_{y}^{-2} + d_{z}^{-2}}} = \\sqrt{\\frac{1}{6^{-1} + 30^{-1} + 19^{-1}}} = \\sqrt{\\frac{95}{24}};\\]", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-geo-solutions.jsonl", "problem_match": "\n10. ", "solution_match": "\nSolution 1: "}}
+{"year": "2025", "tier": "T4", "problem_label": "10", "problem_type": "Geometry", "exam": "HMMT", "problem": "A plane \\(\\mathcal{P}\\) intersects a rectangular prism at a hexagon which has side lengths 45, 66, 63, 55, 54, and 77, in that order. Compute the distance from the center of the rectangular prism to \\(\\mathcal{P}\\) .", "solution": "Let the vertices of the hexagon be \\(ABCDEF\\) , where \\(AB = 45\\) , \\(BC = 66\\) , etc. Note that \\(AB \\parallel DE\\) , \\(BC \\parallel EF\\) , and \\(CD \\parallel FA\\) . Let \\(O\\) be the center of the prism, and let \\(M\\) , \\(N\\) , and \\(P\\) be the midpoints of \\(AD\\) , \\(BE\\) , and \\(CF\\) , respectively. \n\n\n\n\n\n\nThe key observation is that \\(M N\\) is the midline between \\(A B\\) and \\(D E\\) . Hence, plane \\(O M N\\) is the midplane between the faces of the prism containing sides \\(A B\\) and \\(D E\\) . Similarly, planes \\(O M P\\) and \\(O N P\\) are the other two midplanes of the prism. Thus, \\(O M\\) , \\(O N\\) , and \\(O P\\) are mutually orthogonal. \n\nObserve \n\n\\[M N = \\frac{|A B - D E|}{2} = 5, \\quad N P = \\frac{|B C - E F|}{2} = 6, \\quad \\text{and} \\quad P M = \\frac{|C D - F A|}{2} = 7,\\] \n\nso by Heron's formula, we can compute the area of \\(M N P\\) to be \\(\\sqrt{9(9 - 5)(9 - 6)(9 - 7)} = 6\\sqrt{6}\\) . \n\nMoreover, if \\(x = O M\\) , \\(y = O N\\) , and \\(z = O P\\) , then, \n\n\\[x^{2} + y^{2} = 5^{2}, \\quad y^{2} + z^{2} = 6^{2}, \\quad \\text{and} \\quad z^{2} + x^{2} = 7^{2}.\\] \n\nSolving this system of equations gives \\(x = \\sqrt{19}\\) , \\(y = \\sqrt{6}\\) , and \\(z = \\sqrt{30}\\) . Therefore, if \\(d\\) is the distance from \\(O\\) to plane \\(M N P\\) (i.e., the answer), the volume of tetrahedron \\(O M N P\\) can be written as \n\n\\[\\frac{1}{6} \\cdot \\sqrt{19} \\cdot \\sqrt{6} \\cdot \\sqrt{30} = \\frac{1}{3} \\cdot (6\\sqrt{6}) \\cdot d,\\] \n\nSO \n\n\\[d = \\frac{\\sqrt{19 \\cdot 6 \\cdot 30}}{2 \\cdot 6 \\sqrt{6}} = \\sqrt{\\frac{95}{24}}.\\]", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-geo-solutions.jsonl", "problem_match": "\n10. ", "solution_match": "\nSolution 2: "}}
diff --git a/HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl b/HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl
index 0d62ab7cadb55888b3bafc8061cb4db5e2e7792b..3b0a3a97e9ec2d9493466b85da0d874bf2b8eeb5 100644
--- a/HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl
+++ b/HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl
@@ -5,19 +5,19 @@
{"year": "2025", "tier": "T4", "problem_label": "3", "problem_type": null, "exam": "HMMT", "problem": "Jacob rolls two fair six-sided dice. If the outcomes of these dice rolls are the same, he rolls a third fair six-sided die. Compute the probability that the sum of outcomes of all the dice he rolls is even.", "solution": "Answer: \\(\\frac{5}{12}\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n3. [5]", "solution_match": "\nProposed by: Rishabh Das \n\n"}}
{"year": "2025", "tier": "T4", "problem_label": "3", "problem_type": null, "exam": "HMMT", "problem": "Jacob rolls two fair six-sided dice. If the outcomes of these dice rolls are the same, he rolls a third fair six-sided die. Compute the probability that the sum of outcomes of all the dice he rolls is even.", "solution": "There's a \\(\\frac{1}{2} - \\frac{1}{6} = \\frac{1}{3}\\) probability that he rolls an even number without getting doubles: whatever the first roll is, there is a \\(\\frac{1}{2}\\) chance that the second roll is of opposite parity, and we subtract the \\(\\frac{1}{6}\\) chance that the second roll is the same. \n\nThere's a \\(\\frac{1}{6} \\cdot \\frac{1}{2} = \\frac{1}{12}\\) probability that he gets doubles and then rolls an even number. \n\nSumming \\(\\frac{1}{3}\\) and \\(\\frac{1}{12}\\) gets us \\(\\left[\\frac{5}{12}\\right]\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n3. [5]", "solution_match": "\nSolution: "}}
{"year": "2025", "tier": "T4", "problem_label": "4", "problem_type": null, "exam": "HMMT", "problem": "Let \\(\\triangle ABC\\) be an equilateral triangle with side length 4. Across all points \\(P\\) inside triangle \\(\\triangle ABC\\) satisfying \\([PAB] + [PAC] = [PBC]\\) , compute the minimum possible length of \\(PA\\) . \n\n(Here, \\([XYZ]\\) denotes the area of triangle \\(\\triangle XYZ\\) .)", "solution": "Answer: \\(\\sqrt{3}\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n4. [5]", "solution_match": "\nProposed by: Isabella Zhu \n\n"}}
-{"year": "2025", "tier": "T4", "problem_label": "4", "problem_type": null, "exam": "HMMT", "problem": "Let \\(\\triangle ABC\\) be an equilateral triangle with side length 4. Across all points \\(P\\) inside triangle \\(\\triangle ABC\\) satisfying \\([PAB] + [PAC] = [PBC]\\) , compute the minimum possible length of \\(PA\\) . \n\n(Here, \\([XYZ]\\) denotes the area of triangle \\(\\triangle XYZ\\) .)", "solution": "\n \n\nThe area condition implies \\([ABC] = 2[PBC]\\) . Hence, \\(P\\) lies on the \\(A\\) - midline of \\(\\triangle ABC\\) . Therefore, the minimum possible value of \\(PA\\) is the distance from \\(A\\) to this midline. This is achieved by taking \\(P\\) to be the foot of the perpendicular from \\(A\\) to the \\(A\\) - midline. This distance is half the altitude of \\(ABC\\) , which has side length 4, so the answer is \\(\\frac{1}{2} (2\\sqrt{3}) = \\boxed{\\sqrt{3}}\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n4. [5]", "solution_match": "\nSolution:\n\n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "4", "problem_type": null, "exam": "HMMT", "problem": "Let \\(\\triangle ABC\\) be an equilateral triangle with side length 4. Across all points \\(P\\) inside triangle \\(\\triangle ABC\\) satisfying \\([PAB] + [PAC] = [PBC]\\) , compute the minimum possible length of \\(PA\\) . \n\n(Here, \\([XYZ]\\) denotes the area of triangle \\(\\triangle XYZ\\) .)", "solution": "\n \n\nThe area condition implies \\([ABC] = 2[PBC]\\) . Hence, \\(P\\) lies on the \\(A\\) - midline of \\(\\triangle ABC\\) . Therefore, the minimum possible value of \\(PA\\) is the distance from \\(A\\) to this midline. This is achieved by taking \\(P\\) to be the foot of the perpendicular from \\(A\\) to the \\(A\\) - midline. This distance is half the altitude of \\(ABC\\) , which has side length 4, so the answer is \\(\\frac{1}{2} (2\\sqrt{3}) = \\boxed{\\sqrt{3}}\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n4. [5]", "solution_match": "\nSolution:\n\n\n"}}
{"year": "2025", "tier": "T4", "problem_label": "5", "problem_type": null, "exam": "HMMT", "problem": "Compute the largest possible radius of a circle contained in the region defined by \\(|x + |y||\\leq 1\\) in the coordinate plane.", "solution": "Answer: \\(\\boxed {2\\sqrt{2} - 2 = 2(\\sqrt{2} - 1)}\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n5. [6]", "solution_match": "\nProposed by: Rishabh Das \n\n"}}
-{"year": "2025", "tier": "T4", "problem_label": "5", "problem_type": null, "exam": "HMMT", "problem": "Compute the largest possible radius of a circle contained in the region defined by \\(|x + |y||\\leq 1\\) in the coordinate plane.", "solution": "\n \n\nAfter drawing the graph, it's clear that the circle should pass through \\((- 1,0)\\) and be tangent to \\(y = x - 1\\) and \\(y = - x + 1\\) . Letting the radius of this circle be \\(r\\) , we have \\(r\\sqrt{2} +r = 2\\) , so \\(\\boxed {r = 2\\sqrt{2} - 2}\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n5. [6]", "solution_match": "\nSolution: \n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "5", "problem_type": null, "exam": "HMMT", "problem": "Compute the largest possible radius of a circle contained in the region defined by \\(|x + |y||\\leq 1\\) in the coordinate plane.", "solution": "\n \n\nAfter drawing the graph, it's clear that the circle should pass through \\((- 1,0)\\) and be tangent to \\(y = x - 1\\) and \\(y = - x + 1\\) . Letting the radius of this circle be \\(r\\) , we have \\(r\\sqrt{2} +r = 2\\) , so \\(\\boxed {r = 2\\sqrt{2} - 2}\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n5. [6]", "solution_match": "\nSolution: \n\n"}}
{"year": "2025", "tier": "T4", "problem_label": "6", "problem_type": null, "exam": "HMMT", "problem": "Let \\(\\triangle ABC\\) be an equilateral triangle. Point \\(D\\) lies on segment \\(\\overline{BC}\\) such that \\(BD = 1\\) and \\(DC = 4\\) . Points \\(E\\) and \\(F\\) lie on rays \\(\\overrightarrow{AC}\\) and \\(\\overrightarrow{AB}\\) , respectively, such that \\(D\\) is the midpoint of \\(\\overline{EF}\\) . Compute \\(EF\\) .", "solution": "Answer: \\(\\boxed {2\\sqrt {13}}\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n6. [6]", "solution_match": "\nProposed by: Pitchayut Saengrungkonkka\n\n\n\n"}}
-{"year": "2025", "tier": "T4", "problem_label": "6", "problem_type": null, "exam": "HMMT", "problem": "Let \\(\\triangle ABC\\) be an equilateral triangle. Point \\(D\\) lies on segment \\(\\overline{BC}\\) such that \\(BD = 1\\) and \\(DC = 4\\) . Points \\(E\\) and \\(F\\) lie on rays \\(\\overrightarrow{AC}\\) and \\(\\overrightarrow{AB}\\) , respectively, such that \\(D\\) is the midpoint of \\(\\overline{EF}\\) . Compute \\(EF\\) .", "solution": "\n \n\nLet \\(C^{\\prime}\\) be the reflection of \\(C\\) over \\(D\\) . Then, \\(\\overline{{E C}}\\parallel \\overline{{C^{\\prime}F}}\\) since \\(E C F C^{\\prime}\\) is a parallelogram. Thus, \\(B F C^{\\prime}\\) is an equilateral triangle, so \\(B F = B C^{\\prime} = 3\\) and \\(\\angle F B D = 120^{\\circ}\\) . By Law of Cosines, we get \\(D F = \\sqrt{3^{2} + 3\\cdot1 + 1^{2}} = \\sqrt{13}\\) and \\(E F = \\boxed {2\\sqrt {13}}\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n6. [6]", "solution_match": "\nSolution: \n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "6", "problem_type": null, "exam": "HMMT", "problem": "Let \\(\\triangle ABC\\) be an equilateral triangle. Point \\(D\\) lies on segment \\(\\overline{BC}\\) such that \\(BD = 1\\) and \\(DC = 4\\) . Points \\(E\\) and \\(F\\) lie on rays \\(\\overrightarrow{AC}\\) and \\(\\overrightarrow{AB}\\) , respectively, such that \\(D\\) is the midpoint of \\(\\overline{EF}\\) . Compute \\(EF\\) .", "solution": "\n \n\nLet \\(C^{\\prime}\\) be the reflection of \\(C\\) over \\(D\\) . Then, \\(\\overline{{E C}}\\parallel \\overline{{C^{\\prime}F}}\\) since \\(E C F C^{\\prime}\\) is a parallelogram. Thus, \\(B F C^{\\prime}\\) is an equilateral triangle, so \\(B F = B C^{\\prime} = 3\\) and \\(\\angle F B D = 120^{\\circ}\\) . By Law of Cosines, we get \\(D F = \\sqrt{3^{2} + 3\\cdot1 + 1^{2}} = \\sqrt{13}\\) and \\(E F = \\boxed {2\\sqrt {13}}\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n6. [6]", "solution_match": "\nSolution: \n\n"}}
{"year": "2025", "tier": "T4", "problem_label": "7", "problem_type": null, "exam": "HMMT", "problem": "The number \n\n\\[\\frac{9^{9} - 8^{8}}{1001}\\] \n\nis an integer. Compute the sum of its prime factors.", "solution": "Answer: 231", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n7. [6]", "solution_match": "\nProposed by: Derek Liu \n\n"}}
{"year": "2025", "tier": "T4", "problem_label": "7", "problem_type": null, "exam": "HMMT", "problem": "The number \n\n\\[\\frac{9^{9} - 8^{8}}{1001}\\] \n\nis an integer. Compute the sum of its prime factors.", "solution": "Observe \n\n\\[{9^{9}-8^{8}=27^{6}-16^{6}}\\] \\[{=(27^{2}-16^{2})(27^{4}+27^{2}\\cdot16^{2}+16^{4})}\\] \\[{=(27-16)(27+16)(27^{2}-27\\cdot16+16^{2})(27^{2}+27\\cdot16+16^{2})}\\] \\[{=11\\cdot43\\cdot553\\cdot1417.}\\] \n\nThe remaining factorizations are motivated by the fact that \\(1001 = 7\\cdot 11\\cdot 13\\) . We see that \\(553 = 7\\cdot 79\\) and \\(1417 = 13\\cdot 109\\) , so the answer is \\(43 + 79 + 109 = \\boxed {231}\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n7. [6]", "solution_match": "\nSolution: "}}
{"year": "2025", "tier": "T4", "problem_label": "8", "problem_type": null, "exam": "HMMT", "problem": "A checkerboard is a rectangular grid of cells colored black and white such that the top-left corner is black and no two cells of the same color share an edge. Two checkerboards are distinct if and only if they have a different number of rows or columns. For example, a \\(20 \\times 25\\) checkerboard and a \\(25 \\times 20\\) checkerboard are considered distinct. \n\nCompute the number of distinct checkerboards that have exactly 41 black cells.", "solution": "Answer: 9", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n8. [6]", "solution_match": "\nProposed by: Albert Wang \n\n"}}
{"year": "2025", "tier": "T4", "problem_label": "8", "problem_type": null, "exam": "HMMT", "problem": "A checkerboard is a rectangular grid of cells colored black and white such that the top-left corner is black and no two cells of the same color share an edge. Two checkerboards are distinct if and only if they have a different number of rows or columns. For example, a \\(20 \\times 25\\) checkerboard and a \\(25 \\times 20\\) checkerboard are considered distinct. \n\nCompute the number of distinct checkerboards that have exactly 41 black cells.", "solution": "Since there is a black corner on the checkerboard, the number of white squares is at most the number of black squares. So, the board either has 40 or 41 white squares. Therefore, we want to compute the number of ordered pairs \\((r,c)\\) with a product of 81 or 82. Since \\(81 = 3^{4}\\) has 5 divisors and \\(82 = 41\\cdot 2\\) has 4 divisors, there are \\(\\boxed{9}\\) checkerboards with exactly 41 black cells.", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n8. [6]", "solution_match": "\nSolution: "}}
{"year": "2025", "tier": "T4", "problem_label": "9", "problem_type": null, "exam": "HMMT", "problem": "Let \\(P\\) and \\(Q\\) be points selected uniformly and independently at random inside a regular hexagon \\(ABCDEF\\) . Compute the probability that segment \\(\\overline{PQ}\\) is entirely contained in at least one of the quadrilaterals \\(ABCD\\) , \\(BCDE\\) , \\(CDEF\\) , \\(DEFA\\) , \\(EFAB\\) , or \\(FABC\\) .", "solution": "Answer: \\(\\boxed{5}\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n9. [7]", "solution_match": "\nProposed by: Isabella Zhu \n\n"}}
-{"year": "2025", "tier": "T4", "problem_label": "9", "problem_type": null, "exam": "HMMT", "problem": "Let \\(P\\) and \\(Q\\) be points selected uniformly and independently at random inside a regular hexagon \\(ABCDEF\\) . Compute the probability that segment \\(\\overline{PQ}\\) is entirely contained in at least one of the quadrilaterals \\(ABCD\\) , \\(BCDE\\) , \\(CDEF\\) , \\(DEFA\\) , \\(EFAB\\) , or \\(FABC\\) .", "solution": "\n \n\nLet \\(O\\) be the center of the hexagon. Without loss of generality, assume \\(P\\) is in \\(\\triangle ABO\\) . Then, segment \\(PQ\\) is entirely contained in one of the given quadrilaterals if and only if \\(Q\\) is not in \\(\\triangle DEO\\) . The probability that \\(Q\\) is in \\(\\triangle DEO\\) is \\(\\frac{|DEO|}{|ABCDEF|} = \\frac{1}{6}\\) , so the answer is \\(\\boxed{5}\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n9. [7]", "solution_match": "\nSolution: \n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "9", "problem_type": null, "exam": "HMMT", "problem": "Let \\(P\\) and \\(Q\\) be points selected uniformly and independently at random inside a regular hexagon \\(ABCDEF\\) . Compute the probability that segment \\(\\overline{PQ}\\) is entirely contained in at least one of the quadrilaterals \\(ABCD\\) , \\(BCDE\\) , \\(CDEF\\) , \\(DEFA\\) , \\(EFAB\\) , or \\(FABC\\) .", "solution": "\n \n\nLet \\(O\\) be the center of the hexagon. Without loss of generality, assume \\(P\\) is in \\(\\triangle ABO\\) . Then, segment \\(PQ\\) is entirely contained in one of the given quadrilaterals if and only if \\(Q\\) is not in \\(\\triangle DEO\\) . The probability that \\(Q\\) is in \\(\\triangle DEO\\) is \\(\\frac{|DEO|}{|ABCDEF|} = \\frac{1}{6}\\) , so the answer is \\(\\boxed{5}\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n9. [7]", "solution_match": "\nSolution: \n\n"}}
{"year": "2025", "tier": "T4", "problem_label": "10", "problem_type": null, "exam": "HMMT", "problem": "A square of side length 1 is dissected into two congruent pentagons. Compute the least upper bound of the perimeter of one of these pentagons.", "solution": "Answer: \\(\\boxed{2 + 3\\sqrt{2}}\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n10. [7]", "solution_match": "\nProposed by: Isabella Zhu \n\n"}}
-{"year": "2025", "tier": "T4", "problem_label": "10", "problem_type": null, "exam": "HMMT", "problem": "A square of side length 1 is dissected into two congruent pentagons. Compute the least upper bound of the perimeter of one of these pentagons.", "solution": "\n \n\nLet \\(P_{1}\\) and \\(P_{2}\\) be the two congruent pentagons. Let \\(p(P)\\) denote the perimeter of polygon \\(P\\) .\n\n\n\nWe give an upper bound for \\(p(P_{1}) + p(P_{2})\\) . Note that since a square has four sides, at least four sides of \\(P_{1}\\) and \\(P_{2}\\) combined lie on the sides of the square. These sides have total length at most 4, the perimeter of \\(ABCD\\) . \n\nEach of the remaining sides has length at most \\(\\sqrt{2}\\) , since the longest possible length of a segment inside \\(ABCD\\) is \\(\\sqrt{2}\\) . There are at most 6 remaining sides, so \n\n\\[p(P_{1}) + p(P_{2})\\leq 4 + 6\\sqrt{2}.\\] \n\nSince \\(P_{1}\\) and \\(P_{2}\\) are congruent, this implies \n\n\\[p(P_{1}) = p(P_{2})\\leq \\boxed {2 + 3\\sqrt{2}}.\\] \n\nThis least upper bound can be achieved by placing \\(X\\) close to \\(C\\) and \\(Y\\) close to \\(A\\) , as seen in the diagram.", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n10. [7]", "solution_match": "\nSolution: \n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "10", "problem_type": null, "exam": "HMMT", "problem": "A square of side length 1 is dissected into two congruent pentagons. Compute the least upper bound of the perimeter of one of these pentagons.", "solution": "\n \n\nLet \\(P_{1}\\) and \\(P_{2}\\) be the two congruent pentagons. Let \\(p(P)\\) denote the perimeter of polygon \\(P\\) .\n\n\n\nWe give an upper bound for \\(p(P_{1}) + p(P_{2})\\) . Note that since a square has four sides, at least four sides of \\(P_{1}\\) and \\(P_{2}\\) combined lie on the sides of the square. These sides have total length at most 4, the perimeter of \\(ABCD\\) . \n\nEach of the remaining sides has length at most \\(\\sqrt{2}\\) , since the longest possible length of a segment inside \\(ABCD\\) is \\(\\sqrt{2}\\) . There are at most 6 remaining sides, so \n\n\\[p(P_{1}) + p(P_{2})\\leq 4 + 6\\sqrt{2}.\\] \n\nSince \\(P_{1}\\) and \\(P_{2}\\) are congruent, this implies \n\n\\[p(P_{1}) = p(P_{2})\\leq \\boxed {2 + 3\\sqrt{2}}.\\] \n\nThis least upper bound can be achieved by placing \\(X\\) close to \\(C\\) and \\(Y\\) close to \\(A\\) , as seen in the diagram.", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n10. [7]", "solution_match": "\nSolution: \n\n"}}
{"year": "2025", "tier": "T4", "problem_label": "11", "problem_type": null, "exam": "HMMT", "problem": "Let \\(f(n) = n^{2} + 100\\) . Compute the remainder when \\(f(f(\\dots f(f(1))\\dots))\\) is divided by \\(10^{4}\\) .", "solution": "Answer: 3101", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n11. [7]", "solution_match": "\nProposed by: Pitchayut Saengrungkonkha \n\n"}}
{"year": "2025", "tier": "T4", "problem_label": "11", "problem_type": null, "exam": "HMMT", "problem": "Let \\(f(n) = n^{2} + 100\\) . Compute the remainder when \\(f(f(\\dots f(f(1))\\dots))\\) is divided by \\(10^{4}\\) .", "solution": "We claim that \\(f^{k}(n)\\equiv 1 + 100(2^{n} - 1)\\bmod 10^{4}\\) . We can see this by induction, as \n\n\\[f(1 + 100(2^{n} - 1)) = (1 + 100(2^{n} - 1))^{2} + 100\\] \\[\\qquad \\equiv 1 + 200(2^{n} - 1) + 100\\pmod {10^{4}}\\] \\[\\qquad \\equiv 1 + 100(2^{n + 1} - 1)\\pmod {10^{4}}.\\] \n\nThus, it suffices to compute \\(2^{2025}\\bmod 100\\) . We note that \\(2^{2025}\\equiv 0\\) (mod 4) and by Euler's Totient theorem, \\(2^{2025}\\equiv 2^{5}\\) (mod 25), so \\(2^{2025}\\equiv 32\\) (mod 100). Hence, we can compute \n\n\\[f^{2025}(1)\\equiv 1 + 100(31)\\equiv \\boxed {3101}\\bmod 10^{4}.\\]", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n11. [7]", "solution_match": "\nSolution: "}}
{"year": "2025", "tier": "T4", "problem_label": "12", "problem_type": null, "exam": "HMMT", "problem": "Holden has a collection of polygons. He writes down a list containing the measure of each interior angle of each of his polygons. He writes down the list \\(30^{\\circ}\\) , \\(50^{\\circ}\\) , \\(60^{\\circ}\\) , \\(70^{\\circ}\\) , \\(90^{\\circ}\\) , \\(100^{\\circ}\\) , \\(120^{\\circ}\\) , \\(160^{\\circ}\\) , and \\(x^{\\circ}\\) , in some order. Compute \\(x\\) .", "solution": "Answer: 220", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n12. [7]", "solution_match": "\nProposed by: Rishabh Das \n\n"}}
@@ -25,10 +25,10 @@
{"year": "2025", "tier": "T4", "problem_label": "13", "problem_type": null, "exam": "HMMT", "problem": "A number is upwards if its digits in base 10 are nondecreasing when read from left to right. Compute the number of positive integers less than \\(10^{6}\\) that are both upwards and multiples of 11.", "solution": "Answer: 219", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n13. [9]", "solution_match": "\nProposed by: Srinivas Arun \n\n"}}
{"year": "2025", "tier": "T4", "problem_label": "13", "problem_type": null, "exam": "HMMT", "problem": "A number is upwards if its digits in base 10 are nondecreasing when read from left to right. Compute the number of positive integers less than \\(10^{6}\\) that are both upwards and multiples of 11.", "solution": "For a number \\(d_{5}d_{4}d_{3}d_{2}d_{1}d_{0}\\) (allowing leading 0s) to be upwards and a multiple of 11, we must have \n\n\\[d_{5}\\leq d_{4}\\leq d_{3}\\leq d_{2}\\leq d_{1}\\leq d_{0},\\] \n\n\\[d_{0} - d_{1} + d_{2} - d_{3} + d_{4} - d_{5}\\equiv 0\\pmod {11}.\\] \n\nNote that \\(d_{0} - d_{1}\\) , \\(d_{2} - d_{3}\\) , and \\(d_{4} - d_{5}\\) are all nonnegative. Thus, \n\n\\[0\\leq (d_{0} - d_{1}) + (d_{2} - d_{3}) + (d_{4} - d_{5})\\] \\[\\leq (d_{0} - d_{1}) + (d_{1} - d_{2}) + (d_{2} - d_{3}) + (d_{3} - d_{4}) + (d_{4} - d_{5})\\] \\[= d_{0} - d_{5}\\] \\[\\leq 9.\\] \n\nTherefore, \n\n\\[(d_{0} - d_{1}) + (d_{2} - d_{3}) + (d_{4} - d_{5}) = 0,\\] \n\nwhich can only occur when \\(d_{0} = d_{1}\\) , \\(d_{2} = d_{3}\\) , and \\(d_{4} = d_{5}\\) , i.e. the number is of the form \\(aabccc\\) . We can easily verify that all numbers of the form \\(aabccc\\) for digits \\(a \\leq b \\leq c\\) satisfy our conditions, so we simply have to count them. \n\nThere are \\(\\binom{12}{3} = 220\\) such triples of digits \\((a,b,c)\\) . However, one of these triples is \\((0,0,0)\\) , which corresponds to the number 0. Thus our answer is \\(220 - 1 = \\left\\lfloor \\frac{219}{2}\\right\\rfloor\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n13. [9]", "solution_match": "\nSolution: "}}
{"year": "2025", "tier": "T4", "problem_label": "14", "problem_type": null, "exam": "HMMT", "problem": "A parallelogram \\(P\\) can be folded over a straight line so that the resulting shape is a regular pentagon with side length 1. Compute the perimeter of \\(P\\) .", "solution": "Answer: \\(\\boxed{5 + \\sqrt{5}}\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n14. [9]", "solution_match": "\nProposed by: Arul Kolla \n\n"}}
-{"year": "2025", "tier": "T4", "problem_label": "14", "problem_type": null, "exam": "HMMT", "problem": "A parallelogram \\(P\\) can be folded over a straight line so that the resulting shape is a regular pentagon with side length 1. Compute the perimeter of \\(P\\) .", "solution": "\n \n\nIn regular pentagon \\(ABCDE\\) (labeled clockwise), reflect \\(ABDE\\) across \\(AB\\) to obtain \\(ABD'E'\\) . Then, \\(CDE'D'\\) is one such parallelogram \\(P\\) . The length of \\(CD'\\) is \n\n\\[CB + BD = 1 + 2\\cos \\angle CBD = 1 + 2\\cos (\\pi /5) = 1 + \\frac{\\sqrt{5} + 1}{2} = \\frac{\\sqrt{5} + 3}{2}.\\] \n\nHence, the perimeter of the desired parallelogram is \\(2\\left(1 + \\frac{\\sqrt{5} + 3}{2}\\right) = \\left\\lfloor \\frac{5 + \\sqrt{5}}{2}\\right\\rfloor\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n14. [9]", "solution_match": "\nSolution: \n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "14", "problem_type": null, "exam": "HMMT", "problem": "A parallelogram \\(P\\) can be folded over a straight line so that the resulting shape is a regular pentagon with side length 1. Compute the perimeter of \\(P\\) .", "solution": "\n \n\nIn regular pentagon \\(ABCDE\\) (labeled clockwise), reflect \\(ABDE\\) across \\(AB\\) to obtain \\(ABD'E'\\) . Then, \\(CDE'D'\\) is one such parallelogram \\(P\\) . The length of \\(CD'\\) is \n\n\\[CB + BD = 1 + 2\\cos \\angle CBD = 1 + 2\\cos (\\pi /5) = 1 + \\frac{\\sqrt{5} + 1}{2} = \\frac{\\sqrt{5} + 3}{2}.\\] \n\nHence, the perimeter of the desired parallelogram is \\(2\\left(1 + \\frac{\\sqrt{5} + 3}{2}\\right) = \\left\\lfloor \\frac{5 + \\sqrt{5}}{2}\\right\\rfloor\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n14. [9]", "solution_match": "\nSolution: \n\n"}}
{"year": "2025", "tier": "T4", "problem_label": "15", "problem_type": null, "exam": "HMMT", "problem": "Right triangle \\(\\triangle DEF\\) with \\(\\angle D = 90^{\\circ}\\) and \\(\\angle F = 30^{\\circ}\\) is inscribed in equilateral triangle \\(\\triangle ABC\\) such that \\(D\\) , \\(E\\) , and \\(F\\) lie on segments \\(\\overline{BC}\\) , \\(\\overline{CA}\\) , and \\(\\overline{AB}\\) , respectively. Given that \\(BD = 7\\) and \\(DC = 4\\) , compute \\(DE\\) .", "solution": "Answer: \\(\\sqrt{13}\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n15. [9]", "solution_match": "\nProposed by: Pitchayut Saengrungkonka \n\n"}}
-{"year": "2025", "tier": "T4", "problem_label": "15", "problem_type": null, "exam": "HMMT", "problem": "Right triangle \\(\\triangle DEF\\) with \\(\\angle D = 90^{\\circ}\\) and \\(\\angle F = 30^{\\circ}\\) is inscribed in equilateral triangle \\(\\triangle ABC\\) such that \\(D\\) , \\(E\\) , and \\(F\\) lie on segments \\(\\overline{BC}\\) , \\(\\overline{CA}\\) , and \\(\\overline{AB}\\) , respectively. Given that \\(BD = 7\\) and \\(DC = 4\\) , compute \\(DE\\) .", "solution": "\n \n\nFrom \\(\\angle E = 60^{\\circ}\\) , we get that \\(\\angle AEF = 120^{\\circ} - \\angle CED = \\angle CDE\\) . Therefore, \\(\\triangle AEF \\sim \\triangle CDE\\) . Since \\(EF:DE = 2:1\\) , the ratio of similarity must be \\(2:1\\) , so \\(AE = 2CD = 8\\) . Recall \\(ABC\\) has side length \\(7 + 4 = 11\\) , so \\(EC = 11 - 8 = 3\\) . Law of Cosines on \\(\\triangle CDE\\) gives \\(DE^2 = \\sqrt{3^2 + 4^2 - 3\\cdot 4} = \\sqrt{13}\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n15. [9]", "solution_match": "\nSolution 1: \n\n"}}
-{"year": "2025", "tier": "T4", "problem_label": "15", "problem_type": null, "exam": "HMMT", "problem": "Right triangle \\(\\triangle DEF\\) with \\(\\angle D = 90^{\\circ}\\) and \\(\\angle F = 30^{\\circ}\\) is inscribed in equilateral triangle \\(\\triangle ABC\\) such that \\(D\\) , \\(E\\) , and \\(F\\) lie on segments \\(\\overline{BC}\\) , \\(\\overline{CA}\\) , and \\(\\overline{AB}\\) , respectively. Given that \\(BD = 7\\) and \\(DC = 4\\) , compute \\(DE\\) .", "solution": "\n \n\nLet \\(\\odot (DEF)\\) meet \\(AC\\) again at point \\(X\\) . Then, \\(\\angle FXA = 180^{\\circ} - \\angle FXE = \\angle FDE = 90^{\\circ}\\) and \\(\\angle XDC = 180^{\\circ} - \\angle DCX - \\angle DXC = 120^{\\circ} - \\angle DXE = 120^{\\circ} - \\angle DFE = 90^{\\circ}\\) . It follows that \\(CX = 2CD = 8\\) , so \\(AX = 11 - CX = 3\\) , and \\(AF = 2AX = 6\\) . Thus, Law of Cosines on \\(\\triangle AEF\\) gives \\(EF = \\sqrt{8^2 + 6^2 - 8\\cdot 6} = 2\\sqrt{13}\\) , implying that \\(DE = \\sqrt{13}\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n15. [9]", "solution_match": "\nSolution 2: \n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "15", "problem_type": null, "exam": "HMMT", "problem": "Right triangle \\(\\triangle DEF\\) with \\(\\angle D = 90^{\\circ}\\) and \\(\\angle F = 30^{\\circ}\\) is inscribed in equilateral triangle \\(\\triangle ABC\\) such that \\(D\\) , \\(E\\) , and \\(F\\) lie on segments \\(\\overline{BC}\\) , \\(\\overline{CA}\\) , and \\(\\overline{AB}\\) , respectively. Given that \\(BD = 7\\) and \\(DC = 4\\) , compute \\(DE\\) .", "solution": "\n \n\nFrom \\(\\angle E = 60^{\\circ}\\) , we get that \\(\\angle AEF = 120^{\\circ} - \\angle CED = \\angle CDE\\) . Therefore, \\(\\triangle AEF \\sim \\triangle CDE\\) . Since \\(EF:DE = 2:1\\) , the ratio of similarity must be \\(2:1\\) , so \\(AE = 2CD = 8\\) . Recall \\(ABC\\) has side length \\(7 + 4 = 11\\) , so \\(EC = 11 - 8 = 3\\) . Law of Cosines on \\(\\triangle CDE\\) gives \\(DE^2 = \\sqrt{3^2 + 4^2 - 3\\cdot 4} = \\sqrt{13}\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n15. [9]", "solution_match": "\nSolution 1: \n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "15", "problem_type": null, "exam": "HMMT", "problem": "Right triangle \\(\\triangle DEF\\) with \\(\\angle D = 90^{\\circ}\\) and \\(\\angle F = 30^{\\circ}\\) is inscribed in equilateral triangle \\(\\triangle ABC\\) such that \\(D\\) , \\(E\\) , and \\(F\\) lie on segments \\(\\overline{BC}\\) , \\(\\overline{CA}\\) , and \\(\\overline{AB}\\) , respectively. Given that \\(BD = 7\\) and \\(DC = 4\\) , compute \\(DE\\) .", "solution": "\n \n\nLet \\(\\odot (DEF)\\) meet \\(AC\\) again at point \\(X\\) . Then, \\(\\angle FXA = 180^{\\circ} - \\angle FXE = \\angle FDE = 90^{\\circ}\\) and \\(\\angle XDC = 180^{\\circ} - \\angle DCX - \\angle DXC = 120^{\\circ} - \\angle DXE = 120^{\\circ} - \\angle DFE = 90^{\\circ}\\) . It follows that \\(CX = 2CD = 8\\) , so \\(AX = 11 - CX = 3\\) , and \\(AF = 2AX = 6\\) . Thus, Law of Cosines on \\(\\triangle AEF\\) gives \\(EF = \\sqrt{8^2 + 6^2 - 8\\cdot 6} = 2\\sqrt{13}\\) , implying that \\(DE = \\sqrt{13}\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n15. [9]", "solution_match": "\nSolution 2: \n\n"}}
{"year": "2025", "tier": "T4", "problem_label": "16", "problem_type": null, "exam": "HMMT", "problem": "The Cantor set is defined as the set of real numbers \\(x\\) such that \\(0 \\leq x < 1\\) and the digit 1 does not appear in the base-3 expansion of \\(x\\) . Two numbers are uniformly and independently selected at random from the Cantor set. Compute the expected value of their absolute difference. \n\n(Formally, one can pick a number \\(x\\) uniformly at random from the Cantor set by first picking a real number \\(y\\) uniformly at random from the interval \\([0,1)\\) , writing it out in binary, reading its digits as if they were in base- 3, and setting \\(x\\) to 2 times the result.)", "solution": "Answer: \\(\\boxed{\\frac{2}{5}}\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n16. [9]", "solution_match": "\nProposed by: Derek Liu \n\n"}}
{"year": "2025", "tier": "T4", "problem_label": "16", "problem_type": null, "exam": "HMMT", "problem": "The Cantor set is defined as the set of real numbers \\(x\\) such that \\(0 \\leq x < 1\\) and the digit 1 does not appear in the base-3 expansion of \\(x\\) . Two numbers are uniformly and independently selected at random from the Cantor set. Compute the expected value of their absolute difference. \n\n(Formally, one can pick a number \\(x\\) uniformly at random from the Cantor set by first picking a real number \\(y\\) uniformly at random from the interval \\([0,1)\\) , writing it out in binary, reading its digits as if they were in base- 3, and setting \\(x\\) to 2 times the result.)", "solution": "Let \\(d\\) be the expected value of the absolute difference. Observe that the Cantor set is made up of two smaller copies of itself, each scaled down by a factor of 3. There is a \\(\\frac{1}{2}\\) chance that the two selected numbers are in the same copy, in which case the expected value of their absolute difference is \\(\\frac{1}{3} d\\) . Otherwise, we can write them as \\(\\frac{2 + x}{3}\\) and \\(\\frac{y}{3}\\) for independently and uniformly randomly selected \\(x\\) and \\(y\\) in the Cantor set. Their difference is \\(\\frac{2 + (x - y)}{3}\\) , which by symmetry has expected value \\(\\frac{2}{3}\\) . Thus \n\n\\[d = \\frac{1}{2}\\cdot \\frac{1}{3} d + \\frac{1}{2}\\cdot \\frac{2}{3}\\Rightarrow d = \\left[\\frac{2}{5}\\right].\\]", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n16. [9]", "solution_match": "\nSolution: "}}
{"year": "2025", "tier": "T4", "problem_label": "17", "problem_type": null, "exam": "HMMT", "problem": "Let \\(f\\) be a quadratic polynomial with real coefficients, and let \\(g_{1}\\) , \\(g_{2}\\) , \\(g_{3}\\) , ... be a geometric progression of real numbers. Define \\(a_{n} = f(n) + g_{n}\\) . Given that \\(a_{1}\\) , \\(a_{2}\\) , \\(a_{3}\\) , \\(a_{4}\\) , and \\(a_{5}\\) are equal to 1, 2, 3, 14, and 16, respectively, compute \\(\\frac{g_{2}}{g_{1}}\\) .", "solution": "Answer: \\(\\boxed{\\frac{19}{10}}\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n17. [11]", "solution_match": "\nProposed by: Pitchayut Saengrungkonkha \n\n"}}
@@ -44,13 +44,13 @@
{"year": "2025", "tier": "T4", "problem_label": "22", "problem_type": null, "exam": "HMMT", "problem": "Let \\(a\\) , \\(b\\) , and \\(c\\) be real numbers such that \\(a^2(b + c) = 1\\) , \\(b^2(c + a) = 2\\) , and \\(c^2(a + b) = 5\\) . Given that there are three possible values for \\(abc\\) , compute the minimum possible value of \\(abc\\) .", "solution": "Answer: \\(\\frac{- 5 - \\sqrt{5}}{2}\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n22. [12]", "solution_match": "\nProposed by: Pitchayut Saengrungkonkha \n\n"}}
{"year": "2025", "tier": "T4", "problem_label": "22", "problem_type": null, "exam": "HMMT", "problem": "Let \\(a\\) , \\(b\\) , and \\(c\\) be real numbers such that \\(a^2(b + c) = 1\\) , \\(b^2(c + a) = 2\\) , and \\(c^2(a + b) = 5\\) . Given that there are three possible values for \\(abc\\) , compute the minimum possible value of \\(abc\\) .", "solution": "Let \\(x = abc\\) . Multiplying all equations together and simplifying gives \n\n\\[(abc)^2 (a + b)(b + c)(c + a) = 10,\\] \n\n\\[(abc)^2 \\left(a^2(b + c) + b^2(c + a) + c^2(a + b) + 2abc\\right) = 10,\\] \n\n\\[x^2 (1 + 2 + 5 + 2x) = 10,\\] \n\n\\[x^2 (x + 4) = 5.\\] \n\nThe resulting cubic factors as \\((x - 1)(x^2 + 5x + 5) = 0\\) . Therefore, the smallest possible value of \\(abc\\) is \\(\\frac{- 5 - \\sqrt{5^2 - 4\\cdot 5}}{2} = \\frac{- 5 - \\sqrt{5}}{2}\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n22. [12]", "solution_match": "\nSolution: "}}
{"year": "2025", "tier": "T4", "problem_label": "23", "problem_type": null, "exam": "HMMT", "problem": "Regular hexagon \\(ABCDEF\\) has side length 2. Circle \\(\\omega\\) lies inside the hexagon and is tangent to segments \\(\\overline{AB}\\) and \\(\\overline{AF}\\) . There exist two perpendicular lines tangent to \\(\\omega\\) that pass through \\(C\\) and \\(E\\) , respectively. Given that these two lines do not intersect on line \\(AD\\) , compute the radius of \\(\\omega\\) .", "solution": "Answer: \\(\\frac{3\\sqrt{3} - 3}{2} = \\frac{3}{2} (\\sqrt{3} - 1)\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n23. [12]", "solution_match": "\nProposed by: Karthik Venkata Vedula \n\n"}}
-{"year": "2025", "tier": "T4", "problem_label": "23", "problem_type": null, "exam": "HMMT", "problem": "Regular hexagon \\(ABCDEF\\) has side length 2. Circle \\(\\omega\\) lies inside the hexagon and is tangent to segments \\(\\overline{AB}\\) and \\(\\overline{AF}\\) . There exist two perpendicular lines tangent to \\(\\omega\\) that pass through \\(C\\) and \\(E\\) , respectively. Given that these two lines do not intersect on line \\(AD\\) , compute the radius of \\(\\omega\\) .", "solution": "\n \n\nLet \\(O\\) be the center of \\(\\omega\\) , and let the two tangent lines intersect at \\(P\\) . Note that \\(O\\) lies on the external angle bisector of \\(\\angle CPE\\) because the tangents are symmetric about line \\(PO\\) . Additionally, \\(O\\) lies on the perpendicular bisector of \\(CE\\) by symmetry. By Fact 5, \\(COPE\\) is cyclic and \\(\\angle COE = 90^{\\circ}\\) . To finish, observe that \\(\\angle COD = 45^{\\circ}\\) . Dropping the altitude \\(CH\\) down to \\(AD\\) gives \\(OH = CH = \\sqrt{3}\\) . So, \\(AO = AH - OH = 3 - \\sqrt{3}\\) . The desired answer is then \\(\\frac{\\sqrt{3}}{2} \\cdot AO = \\left[\\frac{3\\sqrt{3} - 3}{2}\\right]\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n23. [12]", "solution_match": "\nSolution 1: \n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "23", "problem_type": null, "exam": "HMMT", "problem": "Regular hexagon \\(ABCDEF\\) has side length 2. Circle \\(\\omega\\) lies inside the hexagon and is tangent to segments \\(\\overline{AB}\\) and \\(\\overline{AF}\\) . There exist two perpendicular lines tangent to \\(\\omega\\) that pass through \\(C\\) and \\(E\\) , respectively. Given that these two lines do not intersect on line \\(AD\\) , compute the radius of \\(\\omega\\) .", "solution": "\n \n\nLet \\(O\\) be the center of \\(\\omega\\) , and let the two tangent lines intersect at \\(P\\) . Note that \\(O\\) lies on the external angle bisector of \\(\\angle CPE\\) because the tangents are symmetric about line \\(PO\\) . Additionally, \\(O\\) lies on the perpendicular bisector of \\(CE\\) by symmetry. By Fact 5, \\(COPE\\) is cyclic and \\(\\angle COE = 90^{\\circ}\\) . To finish, observe that \\(\\angle COD = 45^{\\circ}\\) . Dropping the altitude \\(CH\\) down to \\(AD\\) gives \\(OH = CH = \\sqrt{3}\\) . So, \\(AO = AH - OH = 3 - \\sqrt{3}\\) . The desired answer is then \\(\\frac{\\sqrt{3}}{2} \\cdot AO = \\left[\\frac{3\\sqrt{3} - 3}{2}\\right]\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n23. [12]", "solution_match": "\nSolution 1: \n\n"}}
{"year": "2025", "tier": "T4", "problem_label": "23", "problem_type": null, "exam": "HMMT", "problem": "Regular hexagon \\(ABCDEF\\) has side length 2. Circle \\(\\omega\\) lies inside the hexagon and is tangent to segments \\(\\overline{AB}\\) and \\(\\overline{AF}\\) . There exist two perpendicular lines tangent to \\(\\omega\\) that pass through \\(C\\) and \\(E\\) , respectively. Given that these two lines do not intersect on line \\(AD\\) , compute the radius of \\(\\omega\\) .", "solution": "Another way to get \\(\\angle COE = 90^{\\circ}\\) is as follows. \n\nLet \\(\\omega\\) meet the tangents from \\(C\\) and \\(E\\) at \\(Q\\) and \\(R\\) , respectively. Observe \\(OC = OE\\) (as \\(O\\) lies on the perpendicular bisector of \\(CE\\) ) and \\(OQ = OR\\) , so \\(\\triangle OCQ \\stackrel{\\triangle}{=} \\triangle OER\\) . Then \\(\\angle COE = \\angle QOR = 90^{\\circ}\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n23. [12]", "solution_match": "\nSolution 2: "}}
{"year": "2025", "tier": "T4", "problem_label": "24", "problem_type": null, "exam": "HMMT", "problem": "For any integer \\(x\\) , let \n\n\\[f(x) = 100! \\left(1 + x + \\frac{x^{2}}{2!} + \\frac{x^{3}}{3!} + \\dots + \\frac{x^{100}}{100!}\\right).\\] \n\nA positive integer \\(a\\) is chosen such that \\(f(a) - 20\\) is divisible by \\(101^{2}\\) . Compute the remainder when \\(f(a + 101)\\) is divided by \\(101^{2}\\) .", "solution": "Answer: [1939]", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n24. [12]", "solution_match": "\nProposed by: Pitchayut Saengrungkonkha \n\n"}}
{"year": "2025", "tier": "T4", "problem_label": "24", "problem_type": null, "exam": "HMMT", "problem": "For any integer \\(x\\) , let \n\n\\[f(x) = 100! \\left(1 + x + \\frac{x^{2}}{2!} + \\frac{x^{3}}{3!} + \\dots + \\frac{x^{100}}{100!}\\right).\\] \n\nA positive integer \\(a\\) is chosen such that \\(f(a) - 20\\) is divisible by \\(101^{2}\\) . Compute the remainder when \\(f(a + 101)\\) is divided by \\(101^{2}\\) .", "solution": "By the binomial theorem, \n\n\\[(a + 101)^{n} \\equiv a^{n} + \\binom{n}{1} a^{n - 1}101 = a^{n} + 101na^{n - 1} \\pmod {101^{2}}.\\]\n\n\n\nUsing this gives (all congruences are modulo \\(101^{2}\\) ) \n\n\\[f(a + 101) = 100!\\sum_{n = 0}^{100}\\frac{(a + 101)^{n}}{n!}\\] \\[\\equiv 100!\\sum_{n = 0}^{100}\\left(\\frac{a^{n}}{n!} +\\frac{101n a^{n - 1}}{n!}\\right)\\] \\[\\equiv f(a) + 100!\\cdot 101\\sum_{n = 1}^{100}\\frac{a^{n - 1}}{(n - 1)!}\\] \\[\\equiv f(a) + 101f(a) - 100!\\cdot 101\\frac{a^{100}}{100!}\\] \\[\\equiv f(a) + 101(f(a) - 1)\\] \\[\\equiv 20 + 101(20 - 1) = \\boxed {1939}\\quad (\\mathrm{mod}101^{2}).\\]", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n24. [12]", "solution_match": "\nSolution 1: "}}
{"year": "2025", "tier": "T4", "problem_label": "24", "problem_type": null, "exam": "HMMT", "problem": "For any integer \\(x\\) , let \n\n\\[f(x) = 100! \\left(1 + x + \\frac{x^{2}}{2!} + \\frac{x^{3}}{3!} + \\dots + \\frac{x^{100}}{100!}\\right).\\] \n\nA positive integer \\(a\\) is chosen such that \\(f(a) - 20\\) is divisible by \\(101^{2}\\) . Compute the remainder when \\(f(a + 101)\\) is divided by \\(101^{2}\\) .", "solution": "The above solution can be viewed as a consequence of Hensel's lemma as follows. Because 101 is prime, for any integer \\(x\\) not divisible by 101, we have that \n\n\\[f^{\\prime}(x) = 100!\\left(1 + x + \\frac{x^{2}}{2!} +\\dots +\\frac{x^{99}}{99!}\\right) = f(x) - x^{100}\\equiv f(x) - 1\\pmod {101}.\\] \n\nClearly \\(101 \\nmid a\\) . Hence, by Hensel's lemma, we get that \n\n\\[f(a + 101) \\equiv f(a) + 101f^{\\prime}(a) \\equiv 20 + 101 \\cdot 19 \\equiv \\boxed {1939} \\pmod {101^{2}}.\\] \n\nRemark. All possible \\(a\\) 's are \\(a \\equiv 1012, 6670\\) , and 9885 (mod \\(101^{2}\\) ). They all lead to the same answer.", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n24. [12]", "solution_match": "\nSolution 2: "}}
{"year": "2025", "tier": "T4", "problem_label": "25", "problem_type": null, "exam": "HMMT", "problem": "Let \\(A B C D\\) be a trapezoid such that \\(A B \\parallel C D\\) , \\(A D = 13\\) , \\(B C = 15\\) , \\(A B = 20\\) , and \\(C D = 34\\) . Point \\(X\\) lies inside the trapezoid such that \\(\\angle X A B = 2\\angle X B A\\) and \\(\\angle X D C = 2\\angle X C D\\) . Compute \\(X D - X A\\) .", "solution": "Answer: 4", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n25. [14]", "solution_match": "\nProposed by: Pitchayut Saengrungkonkha \n\n"}}
-{"year": "2025", "tier": "T4", "problem_label": "25", "problem_type": null, "exam": "HMMT", "problem": "Let \\(A B C D\\) be a trapezoid such that \\(A B \\parallel C D\\) , \\(A D = 13\\) , \\(B C = 15\\) , \\(A B = 20\\) , and \\(C D = 34\\) . Point \\(X\\) lies inside the trapezoid such that \\(\\angle X A B = 2\\angle X B A\\) and \\(\\angle X D C = 2\\angle X C D\\) . Compute \\(X D - X A\\) .", "solution": "\n \n\nConstruct point \\(P\\) on \\(A B\\) such that \\(X A = X P\\) and point \\(Q\\) on \\(C D\\) such that \\(X D = X Q\\) . The angle condition gives \\(Q C = X Q = X D\\) and \\(P B = X P = X A\\) . Moreover, \\(A D Q P\\) is an isosceles trapezoid.\n\n\n\nLet \\(S\\) be the projection of \\(A\\) onto \\(CD\\) , and let \\(T\\) be on \\(CD\\) such that \\(AT \\parallel BC\\) . Then \\(ADT\\) is a 13- 14- 15 triangle, so \\(DS = 5\\) . Therefore, \\(QD - PA = 10\\) . Finally, we get \n\n\\[XD - XA = QC - PB = (34 - QD) - (20 - PA) = 14 - 10 = \\boxed{4}.\\]", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n25. [14]", "solution_match": "\nSolution: \n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "25", "problem_type": null, "exam": "HMMT", "problem": "Let \\(A B C D\\) be a trapezoid such that \\(A B \\parallel C D\\) , \\(A D = 13\\) , \\(B C = 15\\) , \\(A B = 20\\) , and \\(C D = 34\\) . Point \\(X\\) lies inside the trapezoid such that \\(\\angle X A B = 2\\angle X B A\\) and \\(\\angle X D C = 2\\angle X C D\\) . Compute \\(X D - X A\\) .", "solution": "\n \n\nConstruct point \\(P\\) on \\(A B\\) such that \\(X A = X P\\) and point \\(Q\\) on \\(C D\\) such that \\(X D = X Q\\) . The angle condition gives \\(Q C = X Q = X D\\) and \\(P B = X P = X A\\) . Moreover, \\(A D Q P\\) is an isosceles trapezoid.\n\n\n\nLet \\(S\\) be the projection of \\(A\\) onto \\(CD\\) , and let \\(T\\) be on \\(CD\\) such that \\(AT \\parallel BC\\) . Then \\(ADT\\) is a 13- 14- 15 triangle, so \\(DS = 5\\) . Therefore, \\(QD - PA = 10\\) . Finally, we get \n\n\\[XD - XA = QC - PB = (34 - QD) - (20 - PA) = 14 - 10 = \\boxed{4}.\\]", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n25. [14]", "solution_match": "\nSolution: \n\n"}}
{"year": "2025", "tier": "T4", "problem_label": "26", "problem_type": null, "exam": "HMMT", "problem": "Isabella has a bag with 20 blue diamonds and 25 purple diamonds. She repeats the following process 44 times: she removes a diamond from the bag uniformly at random, then puts one blue diamond and one purple diamond into the bag. Compute the expected number of blue diamonds in the bag after all 44 repetitions.", "solution": "Answer: \\(\\boxed{\\frac{173}{4}}\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n26. [14]", "solution_match": "\nProposed by: Henrick Rabinovitz, Srinivas Arun \n\n"}}
{"year": "2025", "tier": "T4", "problem_label": "26", "problem_type": null, "exam": "HMMT", "problem": "Isabella has a bag with 20 blue diamonds and 25 purple diamonds. She repeats the following process 44 times: she removes a diamond from the bag uniformly at random, then puts one blue diamond and one purple diamond into the bag. Compute the expected number of blue diamonds in the bag after all 44 repetitions.", "solution": "Let \\(a = 20\\) and \\(b = 25\\) be the initial numbers of blue and purple diamonds, respectively, and let \\(c = 44\\) be the number of times Isabella performs the operation. Suppose that at some point, the bag contains \\(x\\) blue diamonds and \\(y\\) purple diamonds, for \\(x + y = z\\) total diamonds. After one step, the bag will have \\(z + 1\\) diamonds. The expected change in the number of blue diamonds in this step is \\((- x / z) + 1 = y / z\\) , and likewise this quantity for purple diamonds is \\(x / z\\) . Thus, the expected change in the difference between the number of blue and purple diamonds is \\((y - x) / z\\) . Since this difference was initially \\(x - y\\) , the expected value of this difference is multiplied by \\((z - 1) / z\\) at each step (regardless of \\(x - y\\) ). Since \\(z\\) starts at \\(a + b\\) and ends at \\(a + b + c\\) , the expected difference after \\(c\\) operations is \n\n\\[(a - b)\\cdot \\prod_{z = a + b}^{a + b + c - 1}\\frac{z - 1}{z} = \\frac{(a + b - 1)(a - b)}{(a + b + c - 1)},\\] \n\nand as the total number of diamonds is \\(a + b + c\\) , the expected number of blue diamonds at the end is \n\n\\[\\frac{a + b + c}{2} +\\frac{(a + b - 1)(a - b)}{2(a + b + c - 1)}.\\] \n\nPlugging in \\(a = 20\\) , \\(b = 25\\) , and \\(c = 44\\) gives us the answer, \\(\\boxed{\\frac{173}{4}}\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n26. [14]", "solution_match": "\nSolution: "}}
{"year": "2025", "tier": "T4", "problem_label": "27", "problem_type": null, "exam": "HMMT", "problem": "Compute the number of ordered pairs \\((m, n)\\) of odd positive integers both less than 80 such that \n\n\\[\\gcd (4^{m} + 2^{m} + 1,4^{n} + 2^{n} + 1) > 1.\\]", "solution": "Answer: \\(\\boxed{820}\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n27. [14]", "solution_match": "\nProposed by: Pitchayut Saengrungkonkha \n\n"}}
@@ -59,13 +59,13 @@
{"year": "2025", "tier": "T4", "problem_label": "28", "problem_type": null, "exam": "HMMT", "problem": "Let \\(f\\) be a function from nonnegative integers to nonnegative integers such that \\(f(0) = 0\\) and \n\n\\[f(m) = f\\left(\\left\\lfloor \\frac{m}{2}\\right\\rfloor\\right) + \\left\\lceil \\frac{m}{2}\\right\\rceil^{2}\\] \n\nfor all positive integers \\(m\\) . Compute \n\n\\[\\frac{f(1)}{1\\cdot2} +\\frac{f(2)}{2\\cdot3} +\\frac{f(3)}{3\\cdot4} +\\dots +\\frac{f(31)}{31\\cdot32}.\\] \n\n(Here, \\(\\lfloor z\\rfloor\\) is the greatest integer less than or equal to \\(z\\) , and \\(\\lceil z\\rceil\\) is the least positive integer greater than or equal to \\(z\\) .)", "solution": "For all positive integers \\(n\\) , let \\(\\omega (n) = f(n) - f(n - 1)\\) . We claim that \\(\\omega (n)\\) is the largest odd divisor of \\(n\\) for all \\(n > 0\\) . Indeed, for all positive integers \\(k\\) , we have \n\n\\[\\omega (2k) = f(2k) - f(2k - 1) = f(k) + k^{2} - (f(k - 1) + k^{2}) = f(k) - f(k - 1) = \\omega (k)\\] \n\nand \n\n\\[\\omega (2k + 1) = f(2k + 1) - f(2k) = f(k) + (k + 1)^{2} - (f(k) + k^{2}) = 2k + 1.\\] \n\nInducting on the positive integers implies that \\(\\omega (n)\\) is indeed the largest odd divisor of \\(n\\) . \n\nWe can now rewrite the sum as \n\n\\[\\sum_{n = 1}^{31}\\frac{f(n)}{n(n + 1)} = \\sum_{n = 1}^{31}\\left(\\frac{f(n)}{n} -\\frac{f(n)}{n + 1}\\right) = \\left(\\sum_{n = 1}^{30}\\frac{f(n) - f(n - 1)}{n}\\right) - \\frac{f(31)}{32}.\\] \n\nNote that by using the original recursive definition, we can compute \n\n\\[f(31) = 16^{2} + 8^{2} + 4^{2} + 2^{2} + 1^{2} = 341.\\] \n\nMoreover, we also see that \\(\\frac{f(n) - f(n - 1)}{n} = \\frac{\\omega(n)}{n} = 2^{-\\nu_{2}(n)}\\) , where \\(2^{\\nu_{2}(n)}\\) is the largest power of 2 dividing \\(n\\) . Thus, our desired sum is \n\n\\[\\left(\\sum_{n = 1}^{31}2^{-\\nu_{2}(n)}\\right) - \\frac{341}{32} = 16\\cdot 2^{-0} + 8\\cdot 2^{-1} + 4\\cdot 2^{-2} + 2\\cdot 2^{-3} + 1\\cdot 2^{-4} - \\frac{341}{32} = \\left\\lfloor \\frac{341}{32}\\right\\rfloor\\]", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n28. [14]", "solution_match": "\nSolution 1: "}}
{"year": "2025", "tier": "T4", "problem_label": "28", "problem_type": null, "exam": "HMMT", "problem": "Let \\(f\\) be a function from nonnegative integers to nonnegative integers such that \\(f(0) = 0\\) and \n\n\\[f(m) = f\\left(\\left\\lfloor \\frac{m}{2}\\right\\rfloor\\right) + \\left\\lceil \\frac{m}{2}\\right\\rceil^{2}\\] \n\nfor all positive integers \\(m\\) . Compute \n\n\\[\\frac{f(1)}{1\\cdot2} +\\frac{f(2)}{2\\cdot3} +\\frac{f(3)}{3\\cdot4} +\\dots +\\frac{f(31)}{31\\cdot32}.\\] \n\n(Here, \\(\\lfloor z\\rfloor\\) is the greatest integer less than or equal to \\(z\\) , and \\(\\lceil z\\rceil\\) is the least positive integer greater than or equal to \\(z\\) .)", "solution": "From the original recursion, for all positive integers \\(n\\) , we have \n\n\\[\\frac{f(2n)}{2n(2n + 1)} = \\frac{f(n) + n^{2}}{2n(2n + 1)} = \\frac{f(n)}{2n(2n + 1)} +\\frac{n}{2(2n + 1)}\\]\n\n\n\nand \n\n\\[\\frac{f(2n + 1)}{(2n + 1)(2n + 2)} = \\frac{f(n) + (n + 1)^{2}}{(2n + 1)(2n + 2)} = \\frac{f(n)}{(2n + 1)(2n + 2)} +\\frac{n + 1}{2(2n + 1)}.\\] \n\nAdding these two equations gives \n\n\\[\\frac{f(2n)}{2n(2n + 1)} +\\frac{f(2n + 1)}{(2n + 1)(2n + 2)} = \\frac{f(n)}{2n(n + 1)} +\\frac{1}{2}.\\] \n\nThus, if \\(T(n) = \\sum_{k = 1}^{n} \\frac{f(k)}{k(k + 1)}\\) , we have \n\n\\[T(2n + 1) = \\sum_{k = 1}^{2n + 1}\\frac{f(k)}{k(k + 1)}\\] \\[\\qquad = \\frac{f(1)}{1(2)} +\\sum_{m = 1}^{n}\\left(\\frac{f(2m)}{2m(2m + 1)} +\\frac{f(2m + 1)}{(2m + 1)(2m + 2)}\\right)\\] \\[\\qquad = \\frac{1}{2} +\\sum_{m = 1}^{n}\\left(\\frac{f(m)}{2m(m + 1)} +\\frac{1}{2}\\right)\\] \\[\\qquad = \\frac{n + 1}{2} +\\frac{1}{2} T(n).\\] \n\nStarting from \\(T(1) = 1\\) , we can compute \\(T(3) = \\frac{5}{4}\\) , \\(T(7) = \\frac{21}{8}\\) , \\(T(15) = \\frac{85}{16}\\) , and finally, \\(T(31) = \\left\\lfloor \\frac{341}{32} \\right\\rfloor\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n28. [14]", "solution_match": "\nSolution 2: "}}
{"year": "2025", "tier": "T4", "problem_label": "29", "problem_type": null, "exam": "HMMT", "problem": "Points \\(A\\) and \\(B\\) lie on circle \\(\\omega\\) with center \\(O\\) . Let \\(X\\) be a point inside \\(\\omega\\) . Suppose that \\(XO = 2\\sqrt{2}\\) , \\(XA = 1\\) , \\(XB = 3\\) , and \\(\\angle AXB = 90^{\\circ}\\) . Points \\(Y\\) and \\(Z\\) are on \\(\\omega\\) such that \\(Y \\neq A\\) and triangles \\(\\triangle AXB\\) and \\(\\triangle YXZ\\) are similar with the same orientation. Compute \\(XY\\) .", "solution": "Answer: \\(\\frac{11}{5}\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n29. [16]", "solution_match": "\nProposed by: Ethan Liu \n\n"}}
-{"year": "2025", "tier": "T4", "problem_label": "29", "problem_type": null, "exam": "HMMT", "problem": "Points \\(A\\) and \\(B\\) lie on circle \\(\\omega\\) with center \\(O\\) . Let \\(X\\) be a point inside \\(\\omega\\) . Suppose that \\(XO = 2\\sqrt{2}\\) , \\(XA = 1\\) , \\(XB = 3\\) , and \\(\\angle AXB = 90^{\\circ}\\) . Points \\(Y\\) and \\(Z\\) are on \\(\\omega\\) such that \\(Y \\neq A\\) and triangles \\(\\triangle AXB\\) and \\(\\triangle YXZ\\) are similar with the same orientation. Compute \\(XY\\) .", "solution": "Consider a rotation about \\(X\\) by \\(90^{\\circ}\\) followed by a homothety with ratio \\(\\frac{1}{3}\\) that sends \\(B\\) to \\(A\\) . This sends \\(\\omega\\) to \\(\\omega^{\\prime}\\) with radius \\(\\frac{1}{3}\\) of the radius of \\(\\omega\\) and center \\(O^{\\prime}\\) . Since \\(A\\) is the image of \\(B\\) under this rotation, we know \\(A\\) lies on both circles; the same argument shows \\(Y\\) must lie on both circles. Thus, \\(Y\\) is the reflection of \\(A\\) over \\(OO^{\\prime}\\) . In particular, this means that \\(XY = AX^{\\prime}\\) , where \\(X^{\\prime}\\) is the reflection of \\(X\\) over \\(OO^{\\prime}\\) . \n\nLet \\(M\\) be the midpoint of \\(AB\\) . Note that because \\(\\triangle OXO^{\\prime} \\sim \\triangle BXA\\) , we also have \\(\\triangle XOX^{\\prime} \\sim \\triangle XMA\\) , as they are both isosceles and \\(\\angle XOX^{\\prime} = 2\\angle XOO^{\\prime} = 2\\angle XBA = \\angle XMA\\) . This implies that \\(\\triangle XOM \\sim \\triangle XX^{\\prime}A\\) . Thus, we know that \\(AX^{\\prime} = OM \\cdot \\frac{XA}{XM} = \\frac{2}{\\sqrt{10}} OM\\) . It remains to compute \\(OM\\) ; noting that the distance between \\(O\\) and the foot from \\(X\\) to \\(AB\\) is \\(\\frac{2}{5}\\sqrt{10}\\) , and that the altitude of \\(\\triangle AXY\\) has length \\(\\frac{3}{10}\\sqrt{10}\\) , we get that the distance from \\(O\\) to \\(AB\\) is \n\n\\[\\frac{3}{10}\\sqrt{10} +\\sqrt{\\left(2\\sqrt{2}\\right)^{2} - \\left(\\frac{2}{5}\\sqrt{10}\\right)^{2}} = \\frac{11}{10}\\sqrt{10}\\] \n\nby the Pythagorean theorem, which means that \\(XY = AX^{\\prime} = \\left\\lfloor \\frac{11}{5} \\right\\rfloor\\) .\n\n\n", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n29. [16]", "solution_match": "\nSolution 1: "}}
-{"year": "2025", "tier": "T4", "problem_label": "29", "problem_type": null, "exam": "HMMT", "problem": "Points \\(A\\) and \\(B\\) lie on circle \\(\\omega\\) with center \\(O\\) . Let \\(X\\) be a point inside \\(\\omega\\) . Suppose that \\(XO = 2\\sqrt{2}\\) , \\(XA = 1\\) , \\(XB = 3\\) , and \\(\\angle AXB = 90^{\\circ}\\) . Points \\(Y\\) and \\(Z\\) are on \\(\\omega\\) such that \\(Y \\neq A\\) and triangles \\(\\triangle AXB\\) and \\(\\triangle YXZ\\) are similar with the same orientation. Compute \\(XY\\) .", "solution": "\n \n\nLet \\(M\\) be the midpoint of \\(AB\\) . We will find \\(MO\\) first. \n\nLet the internal bisector of \\(\\angle A X B\\) intersect \\(\\odot (A X B)\\) at \\(P\\) . From Ptolemy, \\(X P = 2\\sqrt{2} = X O\\) . Let \\(X^{\\prime}\\) be the foot of altitude from \\(X\\) to \\(M O\\) . Observe that \\(O\\) is the reflection of \\(P\\) across \\(X X^{\\prime}\\) . By the\n\n\n\narea of \\(\\triangle A X B\\) , we have \\(X^{\\prime}M = \\frac{3}{\\sqrt{10}}\\) . Therefore, \n\n\\[M O = O X^{\\prime} + X^{\\prime}M = X^{\\prime}P + X^{\\prime}M = 2X^{\\prime}M + M P = \\frac{11\\sqrt{10}}{10}.\\] \n\nBy the spiral similarity \\(\\triangle X A B\\mapsto \\triangle X Y Z\\) , we have that \\(A Y\\perp B Z\\) and \\(\\angle A O B + \\angle Y O Z = 90^{\\circ}\\) Therefore, \\(\\triangle X A M\\sim \\triangle X Y N\\) where \\(N\\) is the midpoint of \\(Y Z\\) . Thus, \\(Y Z = \\frac{11\\sqrt{10}}{5}\\) and \\(X Y = \\left[\\frac{11}{5}\\right]\\) \n\nNote: if \\(\\triangle X A B\\widetilde{\\sim}\\triangle X Y Z\\) , just consider another \\(\\triangle X Y^{\\prime}Z^{\\prime}\\) caused by reflecting \\(\\triangle X Y Z\\) across \\(X O\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n29. [16]", "solution_match": "\nSolution 2: \n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "29", "problem_type": null, "exam": "HMMT", "problem": "Points \\(A\\) and \\(B\\) lie on circle \\(\\omega\\) with center \\(O\\) . Let \\(X\\) be a point inside \\(\\omega\\) . Suppose that \\(XO = 2\\sqrt{2}\\) , \\(XA = 1\\) , \\(XB = 3\\) , and \\(\\angle AXB = 90^{\\circ}\\) . Points \\(Y\\) and \\(Z\\) are on \\(\\omega\\) such that \\(Y \\neq A\\) and triangles \\(\\triangle AXB\\) and \\(\\triangle YXZ\\) are similar with the same orientation. Compute \\(XY\\) .", "solution": "Consider a rotation about \\(X\\) by \\(90^{\\circ}\\) followed by a homothety with ratio \\(\\frac{1}{3}\\) that sends \\(B\\) to \\(A\\) . This sends \\(\\omega\\) to \\(\\omega^{\\prime}\\) with radius \\(\\frac{1}{3}\\) of the radius of \\(\\omega\\) and center \\(O^{\\prime}\\) . Since \\(A\\) is the image of \\(B\\) under this rotation, we know \\(A\\) lies on both circles; the same argument shows \\(Y\\) must lie on both circles. Thus, \\(Y\\) is the reflection of \\(A\\) over \\(OO^{\\prime}\\) . In particular, this means that \\(XY = AX^{\\prime}\\) , where \\(X^{\\prime}\\) is the reflection of \\(X\\) over \\(OO^{\\prime}\\) . \n\nLet \\(M\\) be the midpoint of \\(AB\\) . Note that because \\(\\triangle OXO^{\\prime} \\sim \\triangle BXA\\) , we also have \\(\\triangle XOX^{\\prime} \\sim \\triangle XMA\\) , as they are both isosceles and \\(\\angle XOX^{\\prime} = 2\\angle XOO^{\\prime} = 2\\angle XBA = \\angle XMA\\) . This implies that \\(\\triangle XOM \\sim \\triangle XX^{\\prime}A\\) . Thus, we know that \\(AX^{\\prime} = OM \\cdot \\frac{XA}{XM} = \\frac{2}{\\sqrt{10}} OM\\) . It remains to compute \\(OM\\) ; noting that the distance between \\(O\\) and the foot from \\(X\\) to \\(AB\\) is \\(\\frac{2}{5}\\sqrt{10}\\) , and that the altitude of \\(\\triangle AXY\\) has length \\(\\frac{3}{10}\\sqrt{10}\\) , we get that the distance from \\(O\\) to \\(AB\\) is \n\n\\[\\frac{3}{10}\\sqrt{10} +\\sqrt{\\left(2\\sqrt{2}\\right)^{2} - \\left(\\frac{2}{5}\\sqrt{10}\\right)^{2}} = \\frac{11}{10}\\sqrt{10}\\] \n\nby the Pythagorean theorem, which means that \\(XY = AX^{\\prime} = \\left\\lfloor \\frac{11}{5} \\right\\rfloor\\) .\n\n\n", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n29. [16]", "solution_match": "\nSolution 1: "}}
+{"year": "2025", "tier": "T4", "problem_label": "29", "problem_type": null, "exam": "HMMT", "problem": "Points \\(A\\) and \\(B\\) lie on circle \\(\\omega\\) with center \\(O\\) . Let \\(X\\) be a point inside \\(\\omega\\) . Suppose that \\(XO = 2\\sqrt{2}\\) , \\(XA = 1\\) , \\(XB = 3\\) , and \\(\\angle AXB = 90^{\\circ}\\) . Points \\(Y\\) and \\(Z\\) are on \\(\\omega\\) such that \\(Y \\neq A\\) and triangles \\(\\triangle AXB\\) and \\(\\triangle YXZ\\) are similar with the same orientation. Compute \\(XY\\) .", "solution": "\n \n\nLet \\(M\\) be the midpoint of \\(AB\\) . We will find \\(MO\\) first. \n\nLet the internal bisector of \\(\\angle A X B\\) intersect \\(\\odot (A X B)\\) at \\(P\\) . From Ptolemy, \\(X P = 2\\sqrt{2} = X O\\) . Let \\(X^{\\prime}\\) be the foot of altitude from \\(X\\) to \\(M O\\) . Observe that \\(O\\) is the reflection of \\(P\\) across \\(X X^{\\prime}\\) . By the\n\n\n\narea of \\(\\triangle A X B\\) , we have \\(X^{\\prime}M = \\frac{3}{\\sqrt{10}}\\) . Therefore, \n\n\\[M O = O X^{\\prime} + X^{\\prime}M = X^{\\prime}P + X^{\\prime}M = 2X^{\\prime}M + M P = \\frac{11\\sqrt{10}}{10}.\\] \n\nBy the spiral similarity \\(\\triangle X A B\\mapsto \\triangle X Y Z\\) , we have that \\(A Y\\perp B Z\\) and \\(\\angle A O B + \\angle Y O Z = 90^{\\circ}\\) Therefore, \\(\\triangle X A M\\sim \\triangle X Y N\\) where \\(N\\) is the midpoint of \\(Y Z\\) . Thus, \\(Y Z = \\frac{11\\sqrt{10}}{5}\\) and \\(X Y = \\left[\\frac{11}{5}\\right]\\) \n\nNote: if \\(\\triangle X A B\\widetilde{\\sim}\\triangle X Y Z\\) , just consider another \\(\\triangle X Y^{\\prime}Z^{\\prime}\\) caused by reflecting \\(\\triangle X Y Z\\) across \\(X O\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n29. [16]", "solution_match": "\nSolution 2: \n\n"}}
{"year": "2025", "tier": "T4", "problem_label": "30", "problem_type": null, "exam": "HMMT", "problem": "Let \\(a\\) , \\(b\\) , and \\(c\\) be real numbers satisfying the system of equations \n\n\\[a\\sqrt{1 + b^{2}} +b\\sqrt{1 + a^{2}} = \\frac{3}{4},\\] \\[b\\sqrt{1 + c^{2}} +c\\sqrt{1 + b^{2}} = \\frac{5}{12},\\mathrm{and}\\] \\[c\\sqrt{1 + a^{2}} +a\\sqrt{1 + c^{2}} = \\frac{21}{20}.\\] \n\nCompute \\(a\\) .", "solution": "Answer: \\(\\frac{7}{2\\sqrt{30}}\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n30. [16]", "solution_match": "\nProposed by: Pitchayut Saengrungkonga \n\n"}}
{"year": "2025", "tier": "T4", "problem_label": "30", "problem_type": null, "exam": "HMMT", "problem": "Let \\(a\\) , \\(b\\) , and \\(c\\) be real numbers satisfying the system of equations \n\n\\[a\\sqrt{1 + b^{2}} +b\\sqrt{1 + a^{2}} = \\frac{3}{4},\\] \\[b\\sqrt{1 + c^{2}} +c\\sqrt{1 + b^{2}} = \\frac{5}{12},\\mathrm{and}\\] \\[c\\sqrt{1 + a^{2}} +a\\sqrt{1 + c^{2}} = \\frac{21}{20}.\\] \n\nCompute \\(a\\) .", "solution": "Recall that the functions \\(\\sinh (x) = \\frac{e^{x} - e^{- x}}{2}\\) and \\(\\cosh (x) = \\frac{e^{- x} + e^{- x}}{2}\\) satisfy the relation \n\n\\[\\sinh (x + y) = \\sinh (x)\\cosh (y) + \\cosh (x)\\sinh (y) = \\sinh (x)\\sqrt{1 + \\sinh (y)^{2}} +\\sinh (y)\\sqrt{1 + \\sinh (x)^{2}}.\\] \n\nSince \\(\\sinh\\) is surjective, we can perform the substitution \\(a = \\sinh (x)\\) , \\(b = \\sinh (y)\\) , and \\(c = \\sinh (z)\\) , which turns the equations into \n\n\\[\\sinh (x + y) = \\frac{2 - \\frac{1}{2}}{2},\\] \\[\\sinh (y + z) = \\frac{\\frac{3}{2} - \\frac{2}{3}}{2},\\] \\[\\sinh (z + x) = \\frac{\\frac{5}{2} - \\frac{2}{5}}{2}.\\] \n\nThus, \\(x + y = \\log (2)\\) , \\(y + z = \\log (3 / 2)\\) , and \\(z + x = \\log (5 / 2)\\) . Solving these equations gives \\(x = \\log (\\sqrt{10 / 3})\\) , so \\(a = \\frac{1}{2}\\left(\\sqrt{\\frac{10}{3}} - \\sqrt{\\frac{3}{10}}\\right) = \\left[\\frac{7}{2\\sqrt{30}}\\right]\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n30. [16]", "solution_match": "\nSolution 1: "}}
{"year": "2025", "tier": "T4", "problem_label": "30", "problem_type": null, "exam": "HMMT", "problem": "Let \\(a\\) , \\(b\\) , and \\(c\\) be real numbers satisfying the system of equations \n\n\\[a\\sqrt{1 + b^{2}} +b\\sqrt{1 + a^{2}} = \\frac{3}{4},\\] \\[b\\sqrt{1 + c^{2}} +c\\sqrt{1 + b^{2}} = \\frac{5}{12},\\mathrm{and}\\] \\[c\\sqrt{1 + a^{2}} +a\\sqrt{1 + c^{2}} = \\frac{21}{20}.\\] \n\nCompute \\(a\\) .", "solution": "We can find positive real numbers \\(x\\) , \\(y\\) , and \\(z\\) such that \\(a = \\frac{x^{2} - 1}{2x}\\) , \\(b = \\frac{y^{2} - 1}{2y}\\) , and \\(c = \\frac{z^{2} - 1}{2z}\\) . Then, the first equation becomes \n\n\\[\\frac{x^{2} - 1}{2x} \\cdot \\frac{y^{2} + 1}{2y} + \\frac{y^{2} - 1}{2y} \\cdot \\frac{x^{2} + 1}{2x} = \\frac{3}{4},\\] \n\nwhich simplifies to \n\n\\[x y - \\frac{1}{x y} = \\frac{3}{2},\\] \n\nfrom which it follows that \\(x y = 2\\) . Similarly, \\(y z - \\frac{1}{y z} = \\frac{5}{6}\\) and \\(z x - \\frac{1}{z x} = \\frac{21}{10}\\) , so \\(y z = \\frac{3}{2}\\) and \\(z x = \\frac{5}{2}\\) . Thus \\(x = \\sqrt{(2 \\cdot \\frac{5}{2}) / (\\frac{3}{2})} = \\sqrt{\\frac{10}{3}}\\) , and \\(a = \\frac{(10 / 3) - 1}{2\\sqrt{10 / 3}} = \\left[\\frac{7}{2\\sqrt{30}}\\right]\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n30. [16]", "solution_match": "\nSolution 2: "}}
{"year": "2025", "tier": "T4", "problem_label": "31", "problem_type": null, "exam": "HMMT", "problem": "There exists a unique circle that is both tangent to the parabola \\(y = x^{2}\\) at two points and tangent to the curve \\(x = \\sqrt{\\frac{y^{3}}{1 - y}}\\) . Compute the radius of this circle.", "solution": "Answer: \\(\\frac{\\sqrt{5}}{2}\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n31. [16]", "solution_match": "\nProposed by: Karthik Venkata Vedula \n\n"}}
-{"year": "2025", "tier": "T4", "problem_label": "31", "problem_type": null, "exam": "HMMT", "problem": "There exists a unique circle that is both tangent to the parabola \\(y = x^{2}\\) at two points and tangent to the curve \\(x = \\sqrt{\\frac{y^{3}}{1 - y}}\\) . Compute the radius of this circle.", "solution": "\n \n\nWe can square both sides of the second curve to get \\(x^{2} = \\frac{y^{3}}{1 - y}\\) , which further rearranges to \n\n\\[\\frac{x^{2}}{(x^{2} + y^{2})^{2}} = \\frac{y}{x^{2} + y^{2}}.\\] \n\nThis relation implies that curves \\(y = x^{2}\\) and \\(x^{2} = \\frac{y^{3}}{1 - y}\\) map to each other under inversion about the unit circle \\(x^{2} + y^{2} = 1\\) . Therefore, the unique circle we seek must be invariant under inversion about \\(x^{2} + y^{2} = 1\\) . \n\nSince the circle is tangent to \\(y = x^{2}\\) , we know that the circle is of the form \n\n\\[x^{2} + (y - y_{0})^{2} = r^{2}.\\] \n\nWe know that the length of the tangent from \\((0,0)\\) to this circle is 1. Since the distance from \\((0,0)\\) to the center of the circle is \\(y_{0}\\) , using the Pythagorean Theorem gives \\(r^{2} + 1 = y_{0}^{2}\\) . Because the parabola \\(y = x^{2}\\) is tangent to this circle at two distinct points, the equation \\(x^{2} + (x^{2} - y_{0})^{2} = r^{2} = y_{0}^{2} - 1\\) must have two double roots. Therefore, \n\n\\[x^{2} + (x^{2} - y_{0})^{2} - (y_{0}^{2} - 1) = x^{4} - (2y_{0} - 1)x^{2} + 1\\] \n\nmust be a perfect square, so \\(2y_{0} - 1 = 2 \\implies y_{0} = \\frac{3}{2}\\) . This means \\(r^{2} = y_{0}^{2} - 1 = \\frac{5}{4}\\) , so the radius of the circle is \\(\\left[\\frac{\\sqrt{5}}{2}\\right]\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n31. [16]", "solution_match": "\nSolution: \n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "31", "problem_type": null, "exam": "HMMT", "problem": "There exists a unique circle that is both tangent to the parabola \\(y = x^{2}\\) at two points and tangent to the curve \\(x = \\sqrt{\\frac{y^{3}}{1 - y}}\\) . Compute the radius of this circle.", "solution": "\n \n\nWe can square both sides of the second curve to get \\(x^{2} = \\frac{y^{3}}{1 - y}\\) , which further rearranges to \n\n\\[\\frac{x^{2}}{(x^{2} + y^{2})^{2}} = \\frac{y}{x^{2} + y^{2}}.\\] \n\nThis relation implies that curves \\(y = x^{2}\\) and \\(x^{2} = \\frac{y^{3}}{1 - y}\\) map to each other under inversion about the unit circle \\(x^{2} + y^{2} = 1\\) . Therefore, the unique circle we seek must be invariant under inversion about \\(x^{2} + y^{2} = 1\\) . \n\nSince the circle is tangent to \\(y = x^{2}\\) , we know that the circle is of the form \n\n\\[x^{2} + (y - y_{0})^{2} = r^{2}.\\] \n\nWe know that the length of the tangent from \\((0,0)\\) to this circle is 1. Since the distance from \\((0,0)\\) to the center of the circle is \\(y_{0}\\) , using the Pythagorean Theorem gives \\(r^{2} + 1 = y_{0}^{2}\\) . Because the parabola \\(y = x^{2}\\) is tangent to this circle at two distinct points, the equation \\(x^{2} + (x^{2} - y_{0})^{2} = r^{2} = y_{0}^{2} - 1\\) must have two double roots. Therefore, \n\n\\[x^{2} + (x^{2} - y_{0})^{2} - (y_{0}^{2} - 1) = x^{4} - (2y_{0} - 1)x^{2} + 1\\] \n\nmust be a perfect square, so \\(2y_{0} - 1 = 2 \\implies y_{0} = \\frac{3}{2}\\) . This means \\(r^{2} = y_{0}^{2} - 1 = \\frac{5}{4}\\) , so the radius of the circle is \\(\\left[\\frac{\\sqrt{5}}{2}\\right]\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n31. [16]", "solution_match": "\nSolution: \n\n"}}
{"year": "2025", "tier": "T4", "problem_label": "32", "problem_type": null, "exam": "HMMT", "problem": "In the coordinate plane, a closed lattice loop of length \\(2n\\) is a sequence of lattice points \\(P_{0}, P_{1}, P_{2}, \\ldots , P_{2n}\\) such that \\(P_{0}\\) and \\(P_{2n}\\) are both the origin and \\(P_{i}P_{i + 1} = 1\\) for each \\(i\\) . A closed lattice loop of length 2026 is chosen uniformly at random from all such loops. Let \\(k\\) be the maximum integer such that the line \\(\\ell\\) with equation \\(x + y = k\\) passes through at least one point of the loop. Compute the expected number of indices \\(i\\) such that \\(0 \\leq i \\leq 2025\\) and \\(P_{i}\\) lies on \\(\\ell\\) .\n\n\n\n(A lattice point is a point with integer coordinates.)", "solution": "Answer: \\(\\frac{1013}{507}\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n32. [16]", "solution_match": "\nProposed by: Carlos Rodriguez, Jordan Lefkowitz \n\n"}}
{"year": "2025", "tier": "T4", "problem_label": "32", "problem_type": null, "exam": "HMMT", "problem": "In the coordinate plane, a closed lattice loop of length \\(2n\\) is a sequence of lattice points \\(P_{0}, P_{1}, P_{2}, \\ldots , P_{2n}\\) such that \\(P_{0}\\) and \\(P_{2n}\\) are both the origin and \\(P_{i}P_{i + 1} = 1\\) for each \\(i\\) . A closed lattice loop of length 2026 is chosen uniformly at random from all such loops. Let \\(k\\) be the maximum integer such that the line \\(\\ell\\) with equation \\(x + y = k\\) passes through at least one point of the loop. Compute the expected number of indices \\(i\\) such that \\(0 \\leq i \\leq 2025\\) and \\(P_{i}\\) lies on \\(\\ell\\) .\n\n\n\n(A lattice point is a point with integer coordinates.)", "solution": "We claim that if 2026 is replaced with \\(2n\\) , the answer is \\(\\frac{2n}{n + 1}\\) . \n\nWrite the path as a sequence of \\(U\\) , \\(D\\) , \\(L\\) , and \\(R\\) moves. The possible sequences that can result are precisely those with an equal number of \\(U\\) 's and \\(D\\) 's, and an equal number of \\(R\\) 's and \\(L\\) 's. We first project this sequence onto a single dimension by converting each \\(U\\) and \\(R\\) to a 1, and each \\(D\\) and \\(L\\) to a \\(- 1\\) . The resulting sequence will have an equal number of 1's and \\(- 1\\) 's. \n\nWe claim that every such sequence of \\(n\\) 1's and \\(n - 1\\) 's corresponds to the same number of closed lattice loops. Indeed, given such a sequence, a corresponding lattice loop can be made by replacing all 1's with \\(U\\) 's and \\(R\\) 's, and all \\(- 1\\) 's with \\(D\\) 's and \\(L\\) 's, so that there are an equal number of \\(U\\) 's and \\(D\\) 's. Nothing in this replacement process is order- dependent, so the number of ways to create this loop does not depend on the initial sequence. \n\nWe can see that each of the 1's move the path towards \\(\\ell\\) and the \\(- 1\\) 's move the path away. Hence, we can think of the path as a one- dimensional walk, starting and ending in the same place, where the points on \\(\\ell\\) correspond exactly to the maxima of this one- dimensional walk. \n\nUsing Catalan numbers, the probability that any given point is at the maxima is \n\n\\[\\frac{\\frac{1}{n + 1}\\binom{2n}{n}}{\\binom{2n}{n}} = \\frac{1}{n + 1},\\] \n\nand thus by linearity, the expected number of \\(i \\in [0, 2025]\\) such that \\(P_{i}\\) is on \\(\\ell\\) is \\(\\frac{2n}{n + 1}\\) . Plugging in \\(n = 1013\\) gives a final answer of \\(\\left[\\frac{1013}{507}\\right]\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n32. [16]", "solution_match": "\nSolution: "}}
{"year": "2025", "tier": "T4", "problem_label": "33", "problem_type": null, "exam": "HMMT", "problem": "Estimate the total number of pages that teams submitted to the Team Round this year. (All pages associated to at least one problem number count as submitted pages, even blank cover sheets for a problem.) \n\nSubmit a positive integer \\(E\\) . If the correct answer is \\(A\\) , you will receive \\(\\max \\left(0, \\left[20 \\left(1 - \\left(\\frac{|E - A|}{100}\\right)^{2 / 3}\\right)\\right]\\right)\\) points.", "solution": "Answer: \\(\\boxed{1003}\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n33. [20]", "solution_match": "\nProposed by: Derek Liu \n\n"}}
diff --git a/HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl b/HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl
index 5015c03e73b6167b8304e80ad21bdc40e71abd94..204a21cacbd2fc9093b502eeaeab16d1c5d59740 100644
--- a/HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl
+++ b/HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl
@@ -2,22 +2,22 @@
{"year": "2025", "tier": "T4", "problem_label": "1", "problem_type": null, "exam": "HMMT", "problem": "Let \\(a\\) , \\(b\\) , and \\(c\\) be pairwise distinct positive integers such that \\(\\frac{1}{a}\\) , \\(\\frac{1}{b}\\) , \\(\\frac{1}{c}\\) is an increasing arithmetic sequence in that order. Prove that \\(\\gcd (a,b) > 1\\) .", "solution": "Observe that \\(\\frac{2}{b} -\\frac{1}{a} = \\frac{1}{c}\\) , so \\((2a - b)c = ab\\) and thus \\(2a - b \\mid ab\\) . If we assume that \\(\\gcd (a,b) = 1\\) , then \\(\\gcd (2a - b,a) = 1\\) , so \\(2a - b \\mid b\\) . Then \\(2a - b \\mid (2a - b) + b = 2a\\) , so \\(2a - b \\mid \\gcd (2a,b) \\leq 2\\) . Thus \\(2a - b \\leq 2\\) . But \\(a > b\\) , contradiction. Thus, \\(\\gcd (a,b) > 1\\) , as desired.", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n1. [20]", "solution_match": "\nSolution 2: "}}
{"year": "2025", "tier": "T4", "problem_label": "2", "problem_type": null, "exam": "HMMT", "problem": "A polyomino is a connected figure constructed by joining one or more unit squares edge-to-edge. Determine, with proof, the number of non-congruent polyominoes with no holes, perimeter 180, and area 2024.", "solution": "Answer: 2", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n2. [25]", "solution_match": "\nProposed by: Albert Wang \n\n"}}
{"year": "2025", "tier": "T4", "problem_label": "2", "problem_type": null, "exam": "HMMT", "problem": "A polyomino is a connected figure constructed by joining one or more unit squares edge-to-edge. Determine, with proof, the number of non-congruent polyominoes with no holes, perimeter 180, and area 2024.", "solution": "Define the bounding box of a polyomino to be the smallest axis-aligned rectangle that contains the entire polyomino. Suppose a polyomino satisfying the given conditions has a bounding box with dimensions \\(w \\times h\\) . \n\nClaim 1. \\(w + h \\leq 90\\) . \n\nProof. The polyomino has at least \\(2w\\) horizontal edges and at least \\(2h\\) vertical edges. Moreover, it has a perimeter of 180. Therefore, \\(2w + 2h \\leq 180\\) , so \\(w + h \\leq 90\\) . \\(\\square\\) \n\nClaim 2. The dimensions of the bounding box are either \\(44 \\times 46\\) , \\(45 \\times 45\\) , or \\(46 \\times 44\\) . \n\nProof. Note that \\(hw \\geq 2024\\) since it contains the polyomino with area 2024. Suppose for sake of contradiction that \\(h + w \\leq 89\\) . Then, \n\n\\[(h - w)^{2} = (h + w)^{2} - 4hw \\leq 89^{2} - 4 \\cdot 2024 = -175,\\] \n\ncontradiction. Therefore, \\(h + w = 90\\) , so we can let \\((h,w) = (45 + x,45 - x)\\) . Then, \\(2025 - x^{2} = hw \\geq 2024\\) implies that \\(x \\in \\{- 1,0,1\\}\\) , as desired. \\(\\square\\) \n\nIn the first and third cases, the bounding box has area 2024, so it must be the entire polyomino, giving us the \\(44 \\times 46\\) rectangle (and its rotation) as a possible answer. In the second case, the bounding box has area 2025, so one cell must be removed to form the polyomino. Removing the corner cell yields a polyomino with perimeter 180, and removing any other cells yields a polyomino with perimeter greater than 180. Therefore, the only other possibility is a \\(45 \\times 45\\) square missing a corner. Thus the answer is \\(\\boxed{2}\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n2. [25]", "solution_match": "\nSolution: "}}
-{"year": "2025", "tier": "T4", "problem_label": "3", "problem_type": null, "exam": "HMMT", "problem": "Let \\(\\omega_{1}\\) and \\(\\omega_{2}\\) be two circles intersecting at distinct points \\(A\\) and \\(B\\) . Point \\(X\\) varies along \\(\\omega_{1}\\) , and point \\(Y\\) on \\(\\omega_{2}\\) is chosen such that \\(AB\\) bisects the angle \\(\\angle XAY\\) . Prove that as \\(X\\) varies along \\(\\omega_{1}\\) , the circumcenter of \\(\\triangle AXY\\) (if it exists) varies along a fixed line.", "solution": "Solution 1:\n\n\n\n \n\nLet \\(O_{1}\\) , \\(O_{2}\\) , and \\(O\\) be the centers of \\(\\omega_{1}\\) , \\(\\omega_{2}\\) , and the circumcircle of \\(\\triangle AXY\\) , respectively. \n\nWe claim that triangle \\(O O_{1}O_{2}\\) is isosceles with \\(O O_{1} = O O_{2}\\) , and thus in particular \\(O\\) always lies on the perpendicular bisector of \\(O_{1}O_{2}\\) . \n\nTo this end, observe that \\(O O_{1} \\perp A X\\) and \\(O_{1}O_{2} \\perp A B\\) , so \\(\\angle O O_{1}O_{2} = \\angle X A B\\) . Analogously, \\(\\angle O O_{2}O_{1} = \\angle Y A B\\) . So indeed \\(O O_{1}O_{2}\\) is isosceles, and we are done. \n\nRemark. One may also consider the antipodes of \\(A\\) on \\(\\omega_{1}\\) and \\(\\omega_{2}\\) for an equivalent but more natural angle- chase.", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n3. [30]", "solution_match": "\nProposed by: Pitchayut Saengrungkonkka \n\n"}}
-{"year": "2025", "tier": "T4", "problem_label": "3", "problem_type": null, "exam": "HMMT", "problem": "Let \\(\\omega_{1}\\) and \\(\\omega_{2}\\) be two circles intersecting at distinct points \\(A\\) and \\(B\\) . Point \\(X\\) varies along \\(\\omega_{1}\\) , and point \\(Y\\) on \\(\\omega_{2}\\) is chosen such that \\(AB\\) bisects the angle \\(\\angle XAY\\) . Prove that as \\(X\\) varies along \\(\\omega_{1}\\) , the circumcenter of \\(\\triangle AXY\\) (if it exists) varies along a fixed line.", "solution": "\n \n\nLet \\(A^{\\prime}\\) be the \\(A\\) - antipode in circle \\((A X Y)\\) . It suffices to show that \\(A^{\\prime}\\) lies on a fixed line. We will show that this line is one that is parallel to \\(A B\\) . \n\nLet \\(M\\) be the second intersection of line \\(A B\\) with circle \\((A X Y)\\) , and let \\(N\\) be the antipode of \\(M\\) on this circle. Since \\(A M A^{\\prime}N\\) is a rectangle with \\(N\\) lying on the line through \\(A\\) perpendicular to \\(A B\\) , it suffices to show that \\(N\\) is fixed (independent of \\(X\\) and \\(Y\\) ). \n\nTo this end, take an inversion at \\(A\\) with arbitrary radius, denoting images with \\(\\bullet \\mapsto \\bullet^{*}\\) . \n\nObserve that \\(X^{*}\\) and \\(Y^{*}\\) lie on the fixed lines \\(\\ell_{1} = \\omega_{1}^{*}\\) and \\(\\ell_{2} = \\omega_{2}^{*}\\) . Let \\(\\ell\\) be the line through \\(A\\) perpendicular to \\(A B^{*}\\) , and suppose that \\(\\ell_{1}\\) and \\(\\ell_{2}\\) intersect \\(\\ell\\) at \\(P\\) and \\(Q\\) , respectively. \n\nSince \\(\\angle X A B = \\angle B A Y\\) , we have \\(\\angle X^{*}A B^{*} = \\angle B^{*}A Y^{*}\\) . Circles \\(\\omega_{1}\\) , \\(\\omega_{2}\\) , and \\((A X Y)\\) are mapped to lines \\(B^{*}X^{*}\\) , \\(B^{*}Y^{*}\\) , and \\(X^{*}Y^{*}\\) . As \\(A N \\perp A B\\) , it follows that \\(N^{*}\\) is the intersection of \\(X^{*}Y^{*}\\) and \\(\\ell\\) .\n\n\n\nFinally, observe that \\((N^{*},A;P,Q)\\stackrel {B^{*}}{=}(N^{*},M^{*};X,Y)\\) is a harmonic bundle, as \\(A M^{*}\\) bisects \\(\\angle X^{*}A Y^{*}\\) and \\(\\angle M^{*}A N^{*} = 90^{\\circ}\\) . Since \\(A\\) , \\(P\\) , and \\(Q\\) are fixed, so is \\(N^{*}\\) . Thus \\(N\\) is fixed, and \\(O\\) lies on the perpendicular bisector of \\(A N\\) , which is also fixed. \n\nRemark. An alternative approach to the last paragraph is to recall Blanchet's theorem, which states that \\(P Y^{*}\\) , \\(Q X^{*}\\) , and \\(A B^{*}\\) are concurrent. By Ceva and Menelaus, we get that \\((N^{*},A;P,Q) = - 1\\) . \n\nRemark. If one projects the kite/harmonic quadrilateral \\(N X M Y\\) from \\(A^{\\prime}\\) onto the line through the antipodes defined in the previous remark, we obtain that \\(A^{\\prime}N\\) (a line parallel to \\(A B\\) ) passes through the midpoint of the two antipodes, directly finishing.", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n3. [30]", "solution_match": "\nSolution 2: \n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "3", "problem_type": null, "exam": "HMMT", "problem": "Let \\(\\omega_{1}\\) and \\(\\omega_{2}\\) be two circles intersecting at distinct points \\(A\\) and \\(B\\) . Point \\(X\\) varies along \\(\\omega_{1}\\) , and point \\(Y\\) on \\(\\omega_{2}\\) is chosen such that \\(AB\\) bisects the angle \\(\\angle XAY\\) . Prove that as \\(X\\) varies along \\(\\omega_{1}\\) , the circumcenter of \\(\\triangle AXY\\) (if it exists) varies along a fixed line.", "solution": "Solution 1:\n\n\n\n \n\nLet \\(O_{1}\\) , \\(O_{2}\\) , and \\(O\\) be the centers of \\(\\omega_{1}\\) , \\(\\omega_{2}\\) , and the circumcircle of \\(\\triangle AXY\\) , respectively. \n\nWe claim that triangle \\(O O_{1}O_{2}\\) is isosceles with \\(O O_{1} = O O_{2}\\) , and thus in particular \\(O\\) always lies on the perpendicular bisector of \\(O_{1}O_{2}\\) . \n\nTo this end, observe that \\(O O_{1} \\perp A X\\) and \\(O_{1}O_{2} \\perp A B\\) , so \\(\\angle O O_{1}O_{2} = \\angle X A B\\) . Analogously, \\(\\angle O O_{2}O_{1} = \\angle Y A B\\) . So indeed \\(O O_{1}O_{2}\\) is isosceles, and we are done. \n\nRemark. One may also consider the antipodes of \\(A\\) on \\(\\omega_{1}\\) and \\(\\omega_{2}\\) for an equivalent but more natural angle- chase.", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n3. [30]", "solution_match": "\nProposed by: Pitchayut Saengrungkonkka \n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "3", "problem_type": null, "exam": "HMMT", "problem": "Let \\(\\omega_{1}\\) and \\(\\omega_{2}\\) be two circles intersecting at distinct points \\(A\\) and \\(B\\) . Point \\(X\\) varies along \\(\\omega_{1}\\) , and point \\(Y\\) on \\(\\omega_{2}\\) is chosen such that \\(AB\\) bisects the angle \\(\\angle XAY\\) . Prove that as \\(X\\) varies along \\(\\omega_{1}\\) , the circumcenter of \\(\\triangle AXY\\) (if it exists) varies along a fixed line.", "solution": "\n \n\nLet \\(A^{\\prime}\\) be the \\(A\\) - antipode in circle \\((A X Y)\\) . It suffices to show that \\(A^{\\prime}\\) lies on a fixed line. We will show that this line is one that is parallel to \\(A B\\) . \n\nLet \\(M\\) be the second intersection of line \\(A B\\) with circle \\((A X Y)\\) , and let \\(N\\) be the antipode of \\(M\\) on this circle. Since \\(A M A^{\\prime}N\\) is a rectangle with \\(N\\) lying on the line through \\(A\\) perpendicular to \\(A B\\) , it suffices to show that \\(N\\) is fixed (independent of \\(X\\) and \\(Y\\) ). \n\nTo this end, take an inversion at \\(A\\) with arbitrary radius, denoting images with \\(\\bullet \\mapsto \\bullet^{*}\\) . \n\nObserve that \\(X^{*}\\) and \\(Y^{*}\\) lie on the fixed lines \\(\\ell_{1} = \\omega_{1}^{*}\\) and \\(\\ell_{2} = \\omega_{2}^{*}\\) . Let \\(\\ell\\) be the line through \\(A\\) perpendicular to \\(A B^{*}\\) , and suppose that \\(\\ell_{1}\\) and \\(\\ell_{2}\\) intersect \\(\\ell\\) at \\(P\\) and \\(Q\\) , respectively. \n\nSince \\(\\angle X A B = \\angle B A Y\\) , we have \\(\\angle X^{*}A B^{*} = \\angle B^{*}A Y^{*}\\) . Circles \\(\\omega_{1}\\) , \\(\\omega_{2}\\) , and \\((A X Y)\\) are mapped to lines \\(B^{*}X^{*}\\) , \\(B^{*}Y^{*}\\) , and \\(X^{*}Y^{*}\\) . As \\(A N \\perp A B\\) , it follows that \\(N^{*}\\) is the intersection of \\(X^{*}Y^{*}\\) and \\(\\ell\\) .\n\n\n\nFinally, observe that \\((N^{*},A;P,Q)\\stackrel {B^{*}}{=}(N^{*},M^{*};X,Y)\\) is a harmonic bundle, as \\(A M^{*}\\) bisects \\(\\angle X^{*}A Y^{*}\\) and \\(\\angle M^{*}A N^{*} = 90^{\\circ}\\) . Since \\(A\\) , \\(P\\) , and \\(Q\\) are fixed, so is \\(N^{*}\\) . Thus \\(N\\) is fixed, and \\(O\\) lies on the perpendicular bisector of \\(A N\\) , which is also fixed. \n\nRemark. An alternative approach to the last paragraph is to recall Blanchet's theorem, which states that \\(P Y^{*}\\) , \\(Q X^{*}\\) , and \\(A B^{*}\\) are concurrent. By Ceva and Menelaus, we get that \\((N^{*},A;P,Q) = - 1\\) . \n\nRemark. If one projects the kite/harmonic quadrilateral \\(N X M Y\\) from \\(A^{\\prime}\\) onto the line through the antipodes defined in the previous remark, we obtain that \\(A^{\\prime}N\\) (a line parallel to \\(A B\\) ) passes through the midpoint of the two antipodes, directly finishing.", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n3. [30]", "solution_match": "\nSolution 2: \n\n"}}
{"year": "2025", "tier": "T4", "problem_label": "4", "problem_type": null, "exam": "HMMT", "problem": "Jerry places at most one rook in each cell of a \\(2025 \\times 2025\\) grid of cells. A rook attacks another rook if the two rooks are in the same row or column and there are no other rooks between them. \n\nDetermine, with proof, the maximum number of rooks Jerry can place on the grid such that no rook attacks 4 other rooks.", "solution": "Answer: 8096", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n4. [35]", "solution_match": "\nProposed by: Arul Kolla \n\n"}}
{"year": "2025", "tier": "T4", "problem_label": "4", "problem_type": null, "exam": "HMMT", "problem": "Jerry places at most one rook in each cell of a \\(2025 \\times 2025\\) grid of cells. A rook attacks another rook if the two rooks are in the same row or column and there are no other rooks between them. \n\nDetermine, with proof, the maximum number of rooks Jerry can place on the grid such that no rook attacks 4 other rooks.", "solution": "The answer is \\(2024 \\times 4 = 8096\\) . More generally, for an \\(n \\times n\\) grid, the answer is \\(4n - 4\\) . Call a rook that attacks at most 3 other rooks good. \n\nWe use the following observation in both parts of the solution: a rook on the border of the grid must be good. \n\nLower Bound: Place rooks on all \\(4n - 4\\) border cells of the grid. By the above observation, every rook is good. \n\nUpper Bound: Consider any valid placement of rooks, and assume there exists a rook that is not on the border. We can move this rook to the border via a usual rook move, since this rook is good and thus the path to one of the four border cells in its row or column must be empty. \n\nAfter this move, we claim the placement of rooks is still valid. Indeed: \n\n- the moved rook is now on the border, so by the observation above, it must be good; \n\n- any rook that used to attack this rook cannot attack more rooks after the move, so such rooks must still be good; \n\n- any rook attacked in the final position must be either: \n\n- opposite the direction moved, in which case it attacked the moved rook both before and after the move (so is still good), or \n- perpendicular to the direction moved, in which case it is a border rook and must always be good. \n\nBy repeating the above process, we can always move from any good position to one where all rooks are on the border. This implies that the number of rooks in any good position is at most \\(4n - 4\\) . When \\(n = 2024\\) , the answer is \\(4n - 4 = \\left\\lfloor 8096 \\right\\rfloor\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n4. [35]", "solution_match": "\nSolution 1: "}}
{"year": "2025", "tier": "T4", "problem_label": "4", "problem_type": null, "exam": "HMMT", "problem": "Jerry places at most one rook in each cell of a \\(2025 \\times 2025\\) grid of cells. A rook attacks another rook if the two rooks are in the same row or column and there are no other rooks between them. \n\nDetermine, with proof, the maximum number of rooks Jerry can place on the grid such that no rook attacks 4 other rooks.", "solution": "Consider the set of all rooks which are either the leftmost or rightmost in their row, or the topmost or bottommost in their column. Note that this set must include every rook, as any rook not in this set attacks a rook in all 4 directions. \n\nEach column contributes at most 2 rooks to this set, and each row contributes at most 2 rooks. We can safely ignore the top and bottom rows in this count, as any rook in the top or bottom row is already the topmost or bottommost rook in its column. Thus the number of rooks in the set is at most \\(2 \\cdot (2025 + 2025 - 2) = \\left\\lfloor 8096 \\right\\rfloor\\) , which can be constructed as seen before.", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n4. [35]", "solution_match": "\nSolution 2: "}}
-{"year": "2025", "tier": "T4", "problem_label": "5", "problem_type": null, "exam": "HMMT", "problem": "Let \\(\\triangle ABC\\) be an acute triangle with orthocenter \\(H\\) . Points \\(E\\) and \\(F\\) are on segments \\(\\overline{AC}\\) and \\(\\overline{AB}\\) , respectively, such that \\(\\angle EHF = 90^{\\circ}\\) . Let \\(X\\) be the foot of the altitude from \\(H\\) to \\(\\overline{EF}\\) . Prove that \\(\\angle BXC = 90^{\\circ}\\) .", "solution": "Solution 1: \n\n\n \n\nWe use \\(\\angle\\) to denote directed angles. Let \\(Y\\) and \\(Z\\) be the feet of the altitudes from \\(B\\) and \\(C\\) to \\(AC\\) and \\(AB\\) , respectively. Then \\(\\angle HZF = \\angle HXF = 90^{\\circ}\\) , so \\(HZFX\\) is cyclic. Similarly, \\(HYEX\\) is cyclic. Therefore, \n\n\\[\\angle BYX = \\angle HYX = \\angle HEX = \\angle FHX = \\angle FZX = \\angle BZX.\\] \n\nHence, \\(BZXY\\) is cyclic. A symmetric argument shows \\(C\\) lies on this circle as well. It follows that \\(\\angle BXC = \\angle BYC = 90^{\\circ}\\) , as desired.", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n5. [35]", "solution_match": "\nProposed by: Pitchayut Saengrungkonka \n\n"}}
-{"year": "2025", "tier": "T4", "problem_label": "5", "problem_type": null, "exam": "HMMT", "problem": "Let \\(\\triangle ABC\\) be an acute triangle with orthocenter \\(H\\) . Points \\(E\\) and \\(F\\) are on segments \\(\\overline{AC}\\) and \\(\\overline{AB}\\) , respectively, such that \\(\\angle EHF = 90^{\\circ}\\) . Let \\(X\\) be the foot of the altitude from \\(H\\) to \\(\\overline{EF}\\) . Prove that \\(\\angle BXC = 90^{\\circ}\\) .", "solution": "Let \\(T\\) be the foot of altitude from \\(A\\) to \\(BC\\) . For any point \\(X\\) , let \\(X'\\) denote the image of \\(X\\) under the negative inversion at \\(H\\) with radius \\(\\sqrt{HA \\cdot HT}\\) . Then \\(B'\\) and \\(C'\\) are the feet of the altitudes from \\(B\\) and \\(C\\) to sides \\(AC\\) and \\(AB\\) , respectively. \n\nClaim 1. \\(\\angle BX'C = 90^{\\circ}\\) . \n\nProof. Because \\(HX \\perp EF\\) and \\(HE \\perp HF\\) , the quadrilateral \\(HE'X'F'\\) is a rectangle. Note that \\(\\angle BE'H = \\angle EB'H = 90^{\\circ}\\) and \\(\\angle X'E'H = 90^{\\circ}\\) . Consequently, \\(X', B\\) , and \\(E'\\) are collinear. Similarly, \\(X', C\\) , and \\(F'\\) are collinear. Then, \\(\\angle BX'C = \\angle E'X'F' = 90^{\\circ}\\) , as desired. \\(\\square\\) \n\nFrom the claim, \\(X'\\) lies on the circle with diameter \\(BC\\) (which \\(B'\\) and \\(C'\\) also lie on). Since this circle is invariant under the inversion, \\(X\\) lies on the circle with diameter \\(BC\\) as well, and \\(\\angle BXC = 90^{\\circ}\\) .\n\n\n", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n5. [35]", "solution_match": "\nSolution 2: "}}
-{"year": "2025", "tier": "T4", "problem_label": "5", "problem_type": null, "exam": "HMMT", "problem": "Let \\(\\triangle ABC\\) be an acute triangle with orthocenter \\(H\\) . Points \\(E\\) and \\(F\\) are on segments \\(\\overline{AC}\\) and \\(\\overline{AB}\\) , respectively, such that \\(\\angle EHF = 90^{\\circ}\\) . Let \\(X\\) be the foot of the altitude from \\(H\\) to \\(\\overline{EF}\\) . Prove that \\(\\angle BXC = 90^{\\circ}\\) .", "solution": "We begin by proving the following lemma. \n\nLemma 2. Let \\(A B C D\\) be a quadrilateral and \\(P\\) be a point such that \\(\\angle A P B + \\angle C P D = 180^{\\circ}\\) . Then, the feet of the altitudes from \\(P\\) to each side of \\(A B C D\\) are concyclic. \n\nProof. Let \\(P_{A},P_{B},P_{C},P_{D}\\) the feet of the altitudes from \\(P\\) to \\(A B\\) , \\(B C\\) , \\(C D\\) , and \\(D A\\) respectively. Note that quadrilateral \\(P_{A}P P_{B}B\\) is cyclic. By angle chasing, \n\n\\[\\angle P_{D}P_{A}P_{B} + \\angle P_{B}P_{C}P_{D} = \\angle P_{D}P_{A}P + \\angle P P_{A}P_{B} + \\angle P_{B}P_{C}P + \\angle P P_{C}P_{D}\\] \\[\\qquad = \\angle P_{D}A P + \\angle P B P_{B} + \\angle P_{B}C P + \\angle P D P_{D}\\] \\[\\qquad = (180^{\\circ} - \\angle A P D) + (180^{\\circ} - \\angle B P C)\\] \\[\\qquad = \\angle B P A + \\angle D P C\\] \\[\\qquad = 180^{\\circ}.\\] \n\nTherefore, \\(P_{A}P_{B}P_{C}P_{D}\\) is cyclic as desired. \n\n\n\n\n\n\nLet \\(P_{\\infty}\\) be the point at infinity on line \\(A C\\) . Let \\(Y\\) and \\(Z\\) be the feet of the altitudes from \\(H\\) to \\(A B\\) and \\(A C\\) , respectively. Note that \\(\\angle E H F + \\angle B H P_{\\infty} = 90^{\\circ} + 90^{\\circ} = 180^{\\circ}\\) . Thus, the feet of the altitudes from \\(H\\) to \\(E F\\) , \\(E B\\) , \\(B P_{\\infty}\\) , and \\(C P_{\\infty}\\) are concyclic. In other words, \\(X Y Z B\\) is cyclic. Since \\(B C Y Z\\) is a cyclic quadrilateral, we conclude \\(X\\) lies on this circle, giving us that \\(\\angle B X C = 90^{\\circ}\\) as desired. \n\nRemark. Here's another way to prove the lemma. \n\nIt is well known that, with the provided condition, there is a point \\(P^{\\prime}\\) that is the isogonal conjugate of \\(P\\) with respect to quadrilateral \\(A B C D\\) . Let \\(P_{A}\\) , \\(P_{B}\\) , \\(P_{C}\\) , and \\(P_{D}\\) be the feet of the altitudes from \\(P\\) to \\(A B\\) , \\(B C\\) , \\(C D\\) , and \\(D A\\) , respectively, and let \\(Q\\) be the foot of the altitude from \\(P^{\\prime}\\) to \\(A B\\) . Because \\(P\\) and \\(P^{\\prime}\\) are isogonal conjugates with respect to the triangle formed by lines \\(A B\\) , \\(B C\\) , and \\(C D\\) , we have \\(P_{A}P_{B}P_{C}Q\\) is cyclic. Similarly, because \\(P\\) and \\(P^{\\prime}\\) are also isogonal conjugate with respect to the triangle formed by lines \\(D A\\) , \\(A B\\) , and \\(B C\\) , we have \\(P_{D}P_{A}P_{B}Q\\) is cyclic. Consequently, \\(P_{A}P_{B}P_{C}P_{C}\\) is cyclic as desired.", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n5. [35]", "solution_match": "\nSolution 3: "}}
+{"year": "2025", "tier": "T4", "problem_label": "5", "problem_type": null, "exam": "HMMT", "problem": "Let \\(\\triangle ABC\\) be an acute triangle with orthocenter \\(H\\) . Points \\(E\\) and \\(F\\) are on segments \\(\\overline{AC}\\) and \\(\\overline{AB}\\) , respectively, such that \\(\\angle EHF = 90^{\\circ}\\) . Let \\(X\\) be the foot of the altitude from \\(H\\) to \\(\\overline{EF}\\) . Prove that \\(\\angle BXC = 90^{\\circ}\\) .", "solution": "Solution 1: \n\n\n \n\nWe use \\(\\angle\\) to denote directed angles. Let \\(Y\\) and \\(Z\\) be the feet of the altitudes from \\(B\\) and \\(C\\) to \\(AC\\) and \\(AB\\) , respectively. Then \\(\\angle HZF = \\angle HXF = 90^{\\circ}\\) , so \\(HZFX\\) is cyclic. Similarly, \\(HYEX\\) is cyclic. Therefore, \n\n\\[\\angle BYX = \\angle HYX = \\angle HEX = \\angle FHX = \\angle FZX = \\angle BZX.\\] \n\nHence, \\(BZXY\\) is cyclic. A symmetric argument shows \\(C\\) lies on this circle as well. It follows that \\(\\angle BXC = \\angle BYC = 90^{\\circ}\\) , as desired.", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n5. [35]", "solution_match": "\nProposed by: Pitchayut Saengrungkonka \n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "5", "problem_type": null, "exam": "HMMT", "problem": "Let \\(\\triangle ABC\\) be an acute triangle with orthocenter \\(H\\) . Points \\(E\\) and \\(F\\) are on segments \\(\\overline{AC}\\) and \\(\\overline{AB}\\) , respectively, such that \\(\\angle EHF = 90^{\\circ}\\) . Let \\(X\\) be the foot of the altitude from \\(H\\) to \\(\\overline{EF}\\) . Prove that \\(\\angle BXC = 90^{\\circ}\\) .", "solution": "Let \\(T\\) be the foot of altitude from \\(A\\) to \\(BC\\) . For any point \\(X\\) , let \\(X'\\) denote the image of \\(X\\) under the negative inversion at \\(H\\) with radius \\(\\sqrt{HA \\cdot HT}\\) . Then \\(B'\\) and \\(C'\\) are the feet of the altitudes from \\(B\\) and \\(C\\) to sides \\(AC\\) and \\(AB\\) , respectively. \n\nClaim 1. \\(\\angle BX'C = 90^{\\circ}\\) . \n\nProof. Because \\(HX \\perp EF\\) and \\(HE \\perp HF\\) , the quadrilateral \\(HE'X'F'\\) is a rectangle. Note that \\(\\angle BE'H = \\angle EB'H = 90^{\\circ}\\) and \\(\\angle X'E'H = 90^{\\circ}\\) . Consequently, \\(X', B\\) , and \\(E'\\) are collinear. Similarly, \\(X', C\\) , and \\(F'\\) are collinear. Then, \\(\\angle BX'C = \\angle E'X'F' = 90^{\\circ}\\) , as desired. \\(\\square\\) \n\nFrom the claim, \\(X'\\) lies on the circle with diameter \\(BC\\) (which \\(B'\\) and \\(C'\\) also lie on). Since this circle is invariant under the inversion, \\(X\\) lies on the circle with diameter \\(BC\\) as well, and \\(\\angle BXC = 90^{\\circ}\\) .\n\n\n", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n5. [35]", "solution_match": "\nSolution 2: "}}
+{"year": "2025", "tier": "T4", "problem_label": "5", "problem_type": null, "exam": "HMMT", "problem": "Let \\(\\triangle ABC\\) be an acute triangle with orthocenter \\(H\\) . Points \\(E\\) and \\(F\\) are on segments \\(\\overline{AC}\\) and \\(\\overline{AB}\\) , respectively, such that \\(\\angle EHF = 90^{\\circ}\\) . Let \\(X\\) be the foot of the altitude from \\(H\\) to \\(\\overline{EF}\\) . Prove that \\(\\angle BXC = 90^{\\circ}\\) .", "solution": "We begin by proving the following lemma. \n\nLemma 2. Let \\(A B C D\\) be a quadrilateral and \\(P\\) be a point such that \\(\\angle A P B + \\angle C P D = 180^{\\circ}\\) . Then, the feet of the altitudes from \\(P\\) to each side of \\(A B C D\\) are concyclic. \n\nProof. Let \\(P_{A},P_{B},P_{C},P_{D}\\) the feet of the altitudes from \\(P\\) to \\(A B\\) , \\(B C\\) , \\(C D\\) , and \\(D A\\) respectively. Note that quadrilateral \\(P_{A}P P_{B}B\\) is cyclic. By angle chasing, \n\n\\[\\angle P_{D}P_{A}P_{B} + \\angle P_{B}P_{C}P_{D} = \\angle P_{D}P_{A}P + \\angle P P_{A}P_{B} + \\angle P_{B}P_{C}P + \\angle P P_{C}P_{D}\\] \\[\\qquad = \\angle P_{D}A P + \\angle P B P_{B} + \\angle P_{B}C P + \\angle P D P_{D}\\] \\[\\qquad = (180^{\\circ} - \\angle A P D) + (180^{\\circ} - \\angle B P C)\\] \\[\\qquad = \\angle B P A + \\angle D P C\\] \\[\\qquad = 180^{\\circ}.\\] \n\nTherefore, \\(P_{A}P_{B}P_{C}P_{D}\\) is cyclic as desired. \n\n\n\n\n\n\nLet \\(P_{\\infty}\\) be the point at infinity on line \\(A C\\) . Let \\(Y\\) and \\(Z\\) be the feet of the altitudes from \\(H\\) to \\(A B\\) and \\(A C\\) , respectively. Note that \\(\\angle E H F + \\angle B H P_{\\infty} = 90^{\\circ} + 90^{\\circ} = 180^{\\circ}\\) . Thus, the feet of the altitudes from \\(H\\) to \\(E F\\) , \\(E B\\) , \\(B P_{\\infty}\\) , and \\(C P_{\\infty}\\) are concyclic. In other words, \\(X Y Z B\\) is cyclic. Since \\(B C Y Z\\) is a cyclic quadrilateral, we conclude \\(X\\) lies on this circle, giving us that \\(\\angle B X C = 90^{\\circ}\\) as desired. \n\nRemark. Here's another way to prove the lemma. \n\nIt is well known that, with the provided condition, there is a point \\(P^{\\prime}\\) that is the isogonal conjugate of \\(P\\) with respect to quadrilateral \\(A B C D\\) . Let \\(P_{A}\\) , \\(P_{B}\\) , \\(P_{C}\\) , and \\(P_{D}\\) be the feet of the altitudes from \\(P\\) to \\(A B\\) , \\(B C\\) , \\(C D\\) , and \\(D A\\) , respectively, and let \\(Q\\) be the foot of the altitude from \\(P^{\\prime}\\) to \\(A B\\) . Because \\(P\\) and \\(P^{\\prime}\\) are isogonal conjugates with respect to the triangle formed by lines \\(A B\\) , \\(B C\\) , and \\(C D\\) , we have \\(P_{A}P_{B}P_{C}Q\\) is cyclic. Similarly, because \\(P\\) and \\(P^{\\prime}\\) are also isogonal conjugate with respect to the triangle formed by lines \\(D A\\) , \\(A B\\) , and \\(B C\\) , we have \\(P_{D}P_{A}P_{B}Q\\) is cyclic. Consequently, \\(P_{A}P_{B}P_{C}P_{C}\\) is cyclic as desired.", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n5. [35]", "solution_match": "\nSolution 3: "}}
{"year": "2025", "tier": "T4", "problem_label": "6", "problem_type": null, "exam": "HMMT", "problem": "Complex numbers \\(\\omega_{1}\\) , ..., \\(\\omega_{n}\\) each have magnitude 1. Let \\(z\\) be a complex number distinct from \\(\\omega_{1}\\) , ..., \\(\\omega_{n}\\) such that \n\n\\[\\frac{z + \\omega_{1}}{z - \\omega_{1}} +\\dots +\\frac{z + \\omega_{n}}{z - \\omega_{n}} = 0.\\] \n\nProve that \\(|z| = 1\\)", "solution": "Solution 1: We show that no solutions \\(z\\) not on the unit circle can exist. First, we eliminate \\(|z| > 1\\) \n\nClaim 1. For all \\(j\\) and \\(|z| > 1\\) , the real part of \\(\\frac{z + \\omega_{j}}{z - \\omega_{j}}\\) is positive. \n\nProof. We use geometry. Note that \\(\\omega_{j}\\) and \\(- \\omega_{j}\\) are antipodes on the unit circle. Since \\(z\\) lies outside the unit circle, it follows that \\(\\angle \\omega_{j}z(- \\omega_{j})\\) is acute. But this means the complex number \\(\\frac{z + \\omega_{j}}{z - \\omega_{j}}\\) lies strictly in the first or fourth quadrant of the complex plane and thus has positive real part, as desired. \\(\\square\\) \n\nIt is then clear that whenever \\(|z| > 1\\) , the sum \\(\\sum_{j = 1}^{n} \\frac{z + \\omega_{j}}{z - \\omega_{j}}\\) has positive real part and thus cannot be 0. The case where \\(|z| < 1\\) is analogous, except \\(\\angle \\omega_{j}z(- \\omega_{j})\\) is obtuse instead, so \\(\\frac{z + \\omega_{j}}{z - \\omega_{j}}\\) has negative real part for all \\(j\\) . Therefore, all solutions \\(z\\) to the original equation must satisfy \\(|z| = 1\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n6. [40]", "solution_match": "\nProposed by: Karthik Venkata Vedula \n\n"}}
{"year": "2025", "tier": "T4", "problem_label": "6", "problem_type": null, "exam": "HMMT", "problem": "Complex numbers \\(\\omega_{1}\\) , ..., \\(\\omega_{n}\\) each have magnitude 1. Let \\(z\\) be a complex number distinct from \\(\\omega_{1}\\) , ..., \\(\\omega_{n}\\) such that \n\n\\[\\frac{z + \\omega_{1}}{z - \\omega_{1}} +\\dots +\\frac{z + \\omega_{n}}{z - \\omega_{n}} = 0.\\] \n\nProve that \\(|z| = 1\\)", "solution": "We show more generally that for any positive integers \\(k\\) , \\(a_{1}\\) , ..., \\(a_{k}\\) , and distinct \\(\\omega_{j}\\) on the unit circle, the equation \n\n\\[\\sum_{j = 1}^{k} a_{j} \\left(\\frac{z + \\omega_{i}}{z - \\omega_{j}}\\right) = 0\\] \n\nhas \\(k\\) distinct solutions on the unit circle. The original problem then follows upon consolidating duplicate \\(\\omega_{j}\\) 's. Without loss of generality, assume that \\(\\omega_{1}\\) , ..., \\(\\omega_{k}\\) are in this order going clockwise around the unit circle. \n\nClaim 2. There is a solution on the (clockwise) arc from \\(\\omega_{j}\\) to \\(\\omega_{j + 1}\\) for all \\(j\\) (where \\(\\omega_{k + 1} = \\omega_{1}\\) ). \n\nProof. First, \\(\\omega_{j}\\) and \\(- \\omega_{j}\\) are antipodes on the unit circle, so if \\(z\\) is on the unit circle, \\(\\angle \\omega_{j}z(- \\omega_{j}) = 90^{\\circ}\\) This means \\(\\frac{z + \\omega_{j}}{z - \\omega_{j}}\\) is purely imaginary. Now consider the imaginary part of the left hand side of the equation, which is a real and continuous function on the arc strictly between \\(\\omega_{j}\\) and \\(\\omega_{j + 1}\\) for each \\(j\\) . In particular, as \\(z\\) approaches \\(\\omega_{j + 1}\\) from the clockwise direction, this function approaches \\(\\infty\\) . On the other hand, as \\(z\\) approaches \\(\\omega_{j}\\) from the counterclockwise direction, this function approaches \\(-\\infty\\) . By the Intermediate Value Theorem, there must be a solution on this arc, as desired. \\(\\square\\)\n\n\n\nIt follows that there are at least \\(k\\) solutions on the unit circle. But the equation is equivalent to a polynomial of degree \\(k\\) . Hence, there are exactly \\(k\\) solutions, all of which lie on the unit circle. \n\nRemark. The coefficients \\(a_{j}\\) 's are introduced to handle the case where some of \\(\\omega_{1},\\ldots ,\\omega_{n}\\) are equal. Another way to get around this case is to utilize the fact that roots of polynomials are continuous, so we can take the limit where several \\(\\omega_{j}\\) 's approach each other. \n\nRemark. The Möbius transformation \\(z\\mapsto i\\frac{z - 1}{z + 1}\\) sends the unit circle to a real line. One can rephrase both solutions as working on the real line instead of the unit circle.", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n6. [40]", "solution_match": "\nSolution 2: "}}
{"year": "2025", "tier": "T4", "problem_label": "7", "problem_type": null, "exam": "HMMT", "problem": "Determine, with proof, whether a square can be dissected into finitely many (not necessarily congruent) triangles, each of which has interior angles \\(30^{\\circ}\\) , \\(75^{\\circ}\\) , and \\(75^{\\circ}\\) .", "solution": "Answer: No", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n7. [45]", "solution_match": "\nProposed by: Derek Liu \n\n"}}
{"year": "2025", "tier": "T4", "problem_label": "7", "problem_type": null, "exam": "HMMT", "problem": "Determine, with proof, whether a square can be dissected into finitely many (not necessarily congruent) triangles, each of which has interior angles \\(30^{\\circ}\\) , \\(75^{\\circ}\\) , and \\(75^{\\circ}\\) .", "solution": "Assume for sake of contradiction that such a dissection exists. It has exactly half as many \\(30^{\\circ}\\) angles as \\(75^{\\circ}\\) angles. \n\nAround any intersection point except the square's vertices, the only angles that can appear are \\(30^{\\circ}\\) , \\(75^{\\circ}\\) , and \\(180^{\\circ}\\) . The only combinations of these that sum to \\(180^{\\circ}\\) or \\(360^{\\circ}\\) are \n\n\\[6\\cdot 30^{\\circ} = 180^{\\circ},\\] \\[30^{\\circ} + 2\\cdot 75^{\\circ} = 180^{\\circ},\\] \\[180^{\\circ} = 180^{\\circ},\\] \\[12\\cdot 30^{\\circ} = 360^{\\circ},\\] \\[7\\cdot 30^{\\circ} + 2\\cdot 75^{\\circ} = 360^{\\circ},\\] \\[2\\cdot 30^{\\circ} + 4\\cdot 75^{\\circ} = 360^{\\circ},\\] \\[6\\cdot 30^{\\circ} + 180^{\\circ} = 360^{\\circ},\\] \\[30^{\\circ} + 2\\cdot 75^{\\circ} + 180^{\\circ} = 360^{\\circ},\\] \\[180^{\\circ} + 180^{\\circ} = 360^{\\circ}.\\] \n\nIn particular, around any such point, there are at least half as many \\(30^{\\circ}\\) angles as \\(75^{\\circ}\\) angles. \n\nHowever, the square's vertices must each be surrounded by three \\(30^{\\circ}\\) angles and zero \\(75^{\\circ}\\) angles, as there is no other way to get a sum of \\(90^{\\circ}\\) . Thus the total number of \\(30^{\\circ}\\) angles in the dissection must be at least 12 more than half the number of \\(75^{\\circ}\\) angles, contradiction. \n\nThus no such dissection exists.", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n7. [45]", "solution_match": "\nSolution 1: "}}
{"year": "2025", "tier": "T4", "problem_label": "7", "problem_type": null, "exam": "HMMT", "problem": "Determine, with proof, whether a square can be dissected into finitely many (not necessarily congruent) triangles, each of which has interior angles \\(30^{\\circ}\\) , \\(75^{\\circ}\\) , and \\(75^{\\circ}\\) .", "solution": "Again assume for sake of contradiction that a dissection exists. Interpret the dissection as a graph \\(G\\) , where the vertices of the graph are the vertices of all the triangles, and edges connect each pair of consecutive vertices along a line segment. \n\nCall a vertex flat if it is on either the boundary of the square (including its corners) or the interior of an edge of any triangle. Let \\(X\\) be the number of flat vertices and \\(Y\\) be the number of non- flat vertices in \\(G\\) . Let \\(E\\) and \\(F\\) be the number of edges and faces (triangles) in the dissection, respectively. Then \\((X + Y) - E + F = 1\\) . \n\nObserving the angle combinations in the first solution, we see that any non- flat vertex must have at least 6 incident edges, and any flat vertex must have at least 4. Thus \\(2E \\geq 6Y + 4X\\) , so \\(E \\geq 3Y + 2X\\) . \n\nThe sum of the angles of all \\(F\\) triangles is \\(\\pi F\\) . Around any non- flat vertex, such angles sum to \\(2\\pi\\) . Around any flat vertex, the angles sum to \\(\\pi\\) , with the exception of the four corners of the square, where they sum to \\(\\pi /2\\) instead. Thus \n\n\\[F\\pi = (X - 4)\\pi +4(\\pi /2) + Y(2\\pi) = (X + 2Y - 2)\\pi ,\\]\n\n\n\nso \\(F = X + 2Y - 2\\) . This means \n\n\\[X + Y - E + F\\leq (X + Y) - (3Y + 2X) + (X + 2Y - 2) = -2,\\] \n\ncontradiction. Thus no dissection exists.", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n7. [45]", "solution_match": "\nSolution 2: "}}
-{"year": "2025", "tier": "T4", "problem_label": "8", "problem_type": null, "exam": "HMMT", "problem": "Let \\(\\triangle ABC\\) be a triangle with incenter \\(I\\) . The incircle of triangle \\(\\triangle ABC\\) touches \\(\\overline{BC}\\) at \\(D\\) . Let \\(M\\) be the midpoint of \\(\\overline{BC}\\) , and let line \\(AI\\) meet the circumcircle of triangle \\(\\triangle ABC\\) again at \\(L \\neq A\\) . Let \\(\\omega\\) be the circle centered at \\(L\\) tangent to \\(AB\\) and \\(AC\\) . If \\(\\omega\\) intersects segment \\(\\overline{AD}\\) at point \\(P\\) , prove that \\(\\angle IPM = 90^{\\circ}\\) .", "solution": "Solution 1: Let \\(X\\) and \\(Y\\) be the bottom and top point on \\(\\omega\\) (i.e., the tangents of \\(X\\) and \\(Y\\) to \\(\\omega\\) are parallel to \\(BC\\) , and \\(Y\\) and \\(A\\) lie on the same side of \\(BC\\) ). Note that \\(A\\) , \\(P\\) , \\(D\\) , and \\(X\\) are collinear by homothety between the incircle and \\(\\omega\\) . The key claim is the following. \n\nClaim 1. Line \\(IY\\) is tangent to \\(\\omega\\) . \n\nProof. Let the line through \\(I\\) parallel to \\(BC\\) meet \\(AB\\) and \\(AC\\) at \\(B^{\\prime}\\) and \\(C^{\\prime}\\) , respectively. Notice that \\(B^{\\prime}L\\) is the perpendicular bisector of \\(BI\\) , so \\(B^{\\prime}L\\) externally bisects \\(\\angle AB^{\\prime}C^{\\prime}\\) . Similarly, \\(C^{\\prime}L\\) externally bisects \\(\\angle AC^{\\prime}B^{\\prime}\\) . Hence, \\(L\\) is the excenter of \\(\\triangle AB^{\\prime}C^{\\prime}\\) , which means that \\(B^{\\prime}C^{\\prime}\\) is tangent to \\(\\omega\\) . \\(\\square\\) \n\n\n \n\nNow, we note that \\(LY \\perp BC\\) , so \\(L\\) , \\(Y\\) , and \\(M\\) are collinear (on the perpendicular bisector of \\(BC\\) ). Since \\(\\angle YPX = 90^{\\circ}\\) and \\(\\angle YMD = 90^{\\circ}\\) , \\(PDMY\\) is cyclic. However, \\(IYMD\\) is a rectangle, so \\(IPDMY\\) is a cyclic pentagon. Hence, \\(\\angle IPM = \\angle IDM = 90^{\\circ}\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n8. [50]", "solution_match": "\nProposed by: Pitchayut Saengrungkongka \n\n"}}
-{"year": "2025", "tier": "T4", "problem_label": "8", "problem_type": null, "exam": "HMMT", "problem": "Let \\(\\triangle ABC\\) be a triangle with incenter \\(I\\) . The incircle of triangle \\(\\triangle ABC\\) touches \\(\\overline{BC}\\) at \\(D\\) . Let \\(M\\) be the midpoint of \\(\\overline{BC}\\) , and let line \\(AI\\) meet the circumcircle of triangle \\(\\triangle ABC\\) again at \\(L \\neq A\\) . Let \\(\\omega\\) be the circle centered at \\(L\\) tangent to \\(AB\\) and \\(AC\\) . If \\(\\omega\\) intersects segment \\(\\overline{AD}\\) at point \\(P\\) , prove that \\(\\angle IPM = 90^{\\circ}\\) .", "solution": "Let the incircle touch \\(AC\\) and \\(AB\\) at \\(E\\) and \\(F\\) , respectively. Let \\(DI\\) intersect \\(EF\\) at \\(X\\) . Let \\(D'\\) be the other intersection of \\(AD\\) and the incircle.\n\n\n\nClaim 2. \\(P M \\parallel D^{\\prime}X\\) . \n\nProof. Consider the homothety at \\(A\\) that sends \\(\\omega\\) to the incircle. It sends \\(L\\) to \\(I\\) and \\(P\\) to \\(D^{\\prime}\\) . Furthermore, it's well- known that \\(X\\) lies on \\(A M\\) . Because \\(I X \\parallel L M\\) , we also have that the homothety sends \\(M\\) to \\(X\\) . These facts imply that \\(D^{\\prime}X \\parallel P M\\) . \\(\\square\\) \n\n\n \n\nLet \\(T\\) be the antipode of \\(D\\) on the incircle. Let \\(A T\\) intersect the incircle again at \\(T^{\\prime}\\) . Since \\(X\\) lies on the polar of \\(A\\) with respect to the incircle, by Brocard's theorem, we have \\(D^{\\prime}\\) , \\(X\\) , and \\(T^{\\prime}\\) are collinear. It is well- known that \\(A T \\parallel I M\\) . Therefore, \\(\\angle D P M = \\angle D D^{\\prime}X = \\angle D D^{\\prime}T^{\\prime} = \\angle D T T^{\\prime} = \\angle D I M\\) . Consequently, \\(I M D P\\) is cyclic, and \\(\\angle I P M = \\angle I D M = 90^{\\circ}\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n8. [50]", "solution_match": "\nSolution 2: "}}
-{"year": "2025", "tier": "T4", "problem_label": "8", "problem_type": null, "exam": "HMMT", "problem": "Let \\(\\triangle ABC\\) be a triangle with incenter \\(I\\) . The incircle of triangle \\(\\triangle ABC\\) touches \\(\\overline{BC}\\) at \\(D\\) . Let \\(M\\) be the midpoint of \\(\\overline{BC}\\) , and let line \\(AI\\) meet the circumcircle of triangle \\(\\triangle ABC\\) again at \\(L \\neq A\\) . Let \\(\\omega\\) be the circle centered at \\(L\\) tangent to \\(AB\\) and \\(AC\\) . If \\(\\omega\\) intersects segment \\(\\overline{AD}\\) at point \\(P\\) , prove that \\(\\angle IPM = 90^{\\circ}\\) .", "solution": "Let \\(\\omega\\) be tangent to \\(A B\\) and \\(A C\\) at \\(E\\) and \\(F\\) , respectively. Note that these are the feet of the altitudes from \\(L\\) to \\(A B\\) and \\(A C\\) , and \\(L\\) lies on the circumcircle of \\(\\triangle A B C\\) by Fact 5. As \\(M\\) is clearly the foot from \\(L\\) to \\(B C\\) , it follows that \\(E\\) , \\(F\\) , and \\(M\\) are collinear on the Simson Line of \\(L\\) with respect to \\(\\triangle A B C\\) . \n\nLastly, we want \\(P\\) to be on the circle with diameter \\(I M\\) . This circle intersects \\(E F\\) again at the foot from \\(M\\) to \\(A I\\) , which is the midpoint of \\(E F\\) . Let this point be \\(M^{\\prime}\\) . Consider the homothety sending the incircle to \\(\\omega\\) . This clearly sends \\(D\\) to the second intersection of \\(A D\\) and \\(\\omega\\) , which is \\(P^{\\prime}\\) , and it sends \\(I\\) to \\(L\\) . Note that \\(A P\\cdot A P^{\\prime} = A E^{2} = A M\\cdot A L\\) , as the circle with diameter \\(L E\\) is tangent to \\(A E\\) . Thus, \\(P P^{\\prime}M^{\\prime}L\\) is cyclic. Since \\(I D \\parallel L P^{\\prime}\\) , \\(I\\) lies on \\(M^{\\prime}L\\) , and \\(D\\) lies on \\(P P^{\\prime}\\) . By Reim's, we also have \\(P D M^{\\prime}I\\) is cyclic. As \\(I M\\) is a diameter of \\((D M^{\\prime}I)\\) , we have \\(\\angle I P M = 90^{\\circ}\\) .\n\n\n", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n8. [50]", "solution_match": "\nSolution 3: "}}
+{"year": "2025", "tier": "T4", "problem_label": "8", "problem_type": null, "exam": "HMMT", "problem": "Let \\(\\triangle ABC\\) be a triangle with incenter \\(I\\) . The incircle of triangle \\(\\triangle ABC\\) touches \\(\\overline{BC}\\) at \\(D\\) . Let \\(M\\) be the midpoint of \\(\\overline{BC}\\) , and let line \\(AI\\) meet the circumcircle of triangle \\(\\triangle ABC\\) again at \\(L \\neq A\\) . Let \\(\\omega\\) be the circle centered at \\(L\\) tangent to \\(AB\\) and \\(AC\\) . If \\(\\omega\\) intersects segment \\(\\overline{AD}\\) at point \\(P\\) , prove that \\(\\angle IPM = 90^{\\circ}\\) .", "solution": "Solution 1: Let \\(X\\) and \\(Y\\) be the bottom and top point on \\(\\omega\\) (i.e., the tangents of \\(X\\) and \\(Y\\) to \\(\\omega\\) are parallel to \\(BC\\) , and \\(Y\\) and \\(A\\) lie on the same side of \\(BC\\) ). Note that \\(A\\) , \\(P\\) , \\(D\\) , and \\(X\\) are collinear by homothety between the incircle and \\(\\omega\\) . The key claim is the following. \n\nClaim 1. Line \\(IY\\) is tangent to \\(\\omega\\) . \n\nProof. Let the line through \\(I\\) parallel to \\(BC\\) meet \\(AB\\) and \\(AC\\) at \\(B^{\\prime}\\) and \\(C^{\\prime}\\) , respectively. Notice that \\(B^{\\prime}L\\) is the perpendicular bisector of \\(BI\\) , so \\(B^{\\prime}L\\) externally bisects \\(\\angle AB^{\\prime}C^{\\prime}\\) . Similarly, \\(C^{\\prime}L\\) externally bisects \\(\\angle AC^{\\prime}B^{\\prime}\\) . Hence, \\(L\\) is the excenter of \\(\\triangle AB^{\\prime}C^{\\prime}\\) , which means that \\(B^{\\prime}C^{\\prime}\\) is tangent to \\(\\omega\\) . \\(\\square\\) \n\n\n \n\nNow, we note that \\(LY \\perp BC\\) , so \\(L\\) , \\(Y\\) , and \\(M\\) are collinear (on the perpendicular bisector of \\(BC\\) ). Since \\(\\angle YPX = 90^{\\circ}\\) and \\(\\angle YMD = 90^{\\circ}\\) , \\(PDMY\\) is cyclic. However, \\(IYMD\\) is a rectangle, so \\(IPDMY\\) is a cyclic pentagon. Hence, \\(\\angle IPM = \\angle IDM = 90^{\\circ}\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n8. [50]", "solution_match": "\nProposed by: Pitchayut Saengrungkongka \n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "8", "problem_type": null, "exam": "HMMT", "problem": "Let \\(\\triangle ABC\\) be a triangle with incenter \\(I\\) . The incircle of triangle \\(\\triangle ABC\\) touches \\(\\overline{BC}\\) at \\(D\\) . Let \\(M\\) be the midpoint of \\(\\overline{BC}\\) , and let line \\(AI\\) meet the circumcircle of triangle \\(\\triangle ABC\\) again at \\(L \\neq A\\) . Let \\(\\omega\\) be the circle centered at \\(L\\) tangent to \\(AB\\) and \\(AC\\) . If \\(\\omega\\) intersects segment \\(\\overline{AD}\\) at point \\(P\\) , prove that \\(\\angle IPM = 90^{\\circ}\\) .", "solution": "Let the incircle touch \\(AC\\) and \\(AB\\) at \\(E\\) and \\(F\\) , respectively. Let \\(DI\\) intersect \\(EF\\) at \\(X\\) . Let \\(D'\\) be the other intersection of \\(AD\\) and the incircle.\n\n\n\nClaim 2. \\(P M \\parallel D^{\\prime}X\\) . \n\nProof. Consider the homothety at \\(A\\) that sends \\(\\omega\\) to the incircle. It sends \\(L\\) to \\(I\\) and \\(P\\) to \\(D^{\\prime}\\) . Furthermore, it's well- known that \\(X\\) lies on \\(A M\\) . Because \\(I X \\parallel L M\\) , we also have that the homothety sends \\(M\\) to \\(X\\) . These facts imply that \\(D^{\\prime}X \\parallel P M\\) . \\(\\square\\) \n\n\n \n\nLet \\(T\\) be the antipode of \\(D\\) on the incircle. Let \\(A T\\) intersect the incircle again at \\(T^{\\prime}\\) . Since \\(X\\) lies on the polar of \\(A\\) with respect to the incircle, by Brocard's theorem, we have \\(D^{\\prime}\\) , \\(X\\) , and \\(T^{\\prime}\\) are collinear. It is well- known that \\(A T \\parallel I M\\) . Therefore, \\(\\angle D P M = \\angle D D^{\\prime}X = \\angle D D^{\\prime}T^{\\prime} = \\angle D T T^{\\prime} = \\angle D I M\\) . Consequently, \\(I M D P\\) is cyclic, and \\(\\angle I P M = \\angle I D M = 90^{\\circ}\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n8. [50]", "solution_match": "\nSolution 2: "}}
+{"year": "2025", "tier": "T4", "problem_label": "8", "problem_type": null, "exam": "HMMT", "problem": "Let \\(\\triangle ABC\\) be a triangle with incenter \\(I\\) . The incircle of triangle \\(\\triangle ABC\\) touches \\(\\overline{BC}\\) at \\(D\\) . Let \\(M\\) be the midpoint of \\(\\overline{BC}\\) , and let line \\(AI\\) meet the circumcircle of triangle \\(\\triangle ABC\\) again at \\(L \\neq A\\) . Let \\(\\omega\\) be the circle centered at \\(L\\) tangent to \\(AB\\) and \\(AC\\) . If \\(\\omega\\) intersects segment \\(\\overline{AD}\\) at point \\(P\\) , prove that \\(\\angle IPM = 90^{\\circ}\\) .", "solution": "Let \\(\\omega\\) be tangent to \\(A B\\) and \\(A C\\) at \\(E\\) and \\(F\\) , respectively. Note that these are the feet of the altitudes from \\(L\\) to \\(A B\\) and \\(A C\\) , and \\(L\\) lies on the circumcircle of \\(\\triangle A B C\\) by Fact 5. As \\(M\\) is clearly the foot from \\(L\\) to \\(B C\\) , it follows that \\(E\\) , \\(F\\) , and \\(M\\) are collinear on the Simson Line of \\(L\\) with respect to \\(\\triangle A B C\\) . \n\nLastly, we want \\(P\\) to be on the circle with diameter \\(I M\\) . This circle intersects \\(E F\\) again at the foot from \\(M\\) to \\(A I\\) , which is the midpoint of \\(E F\\) . Let this point be \\(M^{\\prime}\\) . Consider the homothety sending the incircle to \\(\\omega\\) . This clearly sends \\(D\\) to the second intersection of \\(A D\\) and \\(\\omega\\) , which is \\(P^{\\prime}\\) , and it sends \\(I\\) to \\(L\\) . Note that \\(A P\\cdot A P^{\\prime} = A E^{2} = A M\\cdot A L\\) , as the circle with diameter \\(L E\\) is tangent to \\(A E\\) . Thus, \\(P P^{\\prime}M^{\\prime}L\\) is cyclic. Since \\(I D \\parallel L P^{\\prime}\\) , \\(I\\) lies on \\(M^{\\prime}L\\) , and \\(D\\) lies on \\(P P^{\\prime}\\) . By Reim's, we also have \\(P D M^{\\prime}I\\) is cyclic. As \\(I M\\) is a diameter of \\((D M^{\\prime}I)\\) , we have \\(\\angle I P M = 90^{\\circ}\\) .\n\n\n", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n8. [50]", "solution_match": "\nSolution 3: "}}
{"year": "2025", "tier": "T4", "problem_label": "9", "problem_type": null, "exam": "HMMT", "problem": "Let \\(\\mathbb{Z}\\) be the set of integers. Determine, with proof, all primes \\(p\\) for which there exists a function \\(f\\colon \\mathbb{Z}\\to \\mathbb{Z}\\) such that for any integer \\(x\\) , \n\n\\(f(x + p) = f(x)\\) and \\(p\\) divides \\(f(x + f(x)) - x\\) .", "solution": "Answer: \\(\\boxed{p = 5}\\) and all primes \\(p\\equiv \\pm 1\\) (mod 5)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n9. [60]", "solution_match": "\nProposed by: Marin Hristov Hristov \n\n"}}
{"year": "2025", "tier": "T4", "problem_label": "9", "problem_type": null, "exam": "HMMT", "problem": "Let \\(\\mathbb{Z}\\) be the set of integers. Determine, with proof, all primes \\(p\\) for which there exists a function \\(f\\colon \\mathbb{Z}\\to \\mathbb{Z}\\) such that for any integer \\(x\\) , \n\n\\(f(x + p) = f(x)\\) and \\(p\\) divides \\(f(x + f(x)) - x\\) .", "solution": "We work in \\(\\mathbb{F}_{p}\\) , treating \\(f\\) as a map from \\(\\mathbb{F}_{p}\\) to itself. Clearly, \\(p = 2\\) doesn't work. For \\(p > 2\\) such that 5 is a quadratic residue mod \\(p\\) , as well as \\(p = 5\\) itself, there exists some \\(\\alpha\\) such that \\((2\\alpha +1)^{2}\\equiv 5\\) (mod \\(p\\) ). Taking \\(f(x) = \\alpha x\\) then works because \n\n\\[f(x + f(x)) - x = (\\alpha^{2} + \\alpha -1)x = \\frac{1}{4}\\left((2\\alpha +1)^{2} - 5\\right)x\\equiv 0\\pmod {p}.\\] \n\nTo prove no other primes satisfy the conditions in the problem statement, note that \\(f\\) is surjective, as for any \\(x\\) , \\(f(x + f(x)) = x\\) . As \\(\\mathbb{F}_{p}\\) is finite, \\(f\\) is bijective. Plugging in \\(x = f(y)\\) yields \n\n\\[f(f(y) + f(f(y))) = f(y)\\implies f(y) + f(f(y)) = y.\\] \n\nSince \\(f\\) is bijective, there exists \\(z\\in \\mathbb{F}_{p}\\) such that \\(f(z) = 0\\) , then \\(z = f(z + f(z)) = f(z) = 0\\) . Therefore, \\(f(0) = 0\\) . This is the only fixed point, as any fixed point \\(d\\) would satisfy \\(d = f(d) + f(f(d)) = 2d\\) ,\n\n\n\nwhich is impossible if \\(d \\neq 0\\) . Hence the remaining residues form nontrivial cycles \\(y\\) , \\(f(y)\\) , \\(f(f(y))\\) , etc. If the cycle containing \\(y\\) is of length \\(n\\) , then \n\n\\[\\begin{array}{r l} & {f^{-1}(y) = y + f(y),}\\\\ & {f^{-2}(y) = f^{-1}(y) + y = 2y + f(y),}\\\\ & {\\qquad \\vdots}\\\\ & {f^{-(n - 1)}(y) = f(y) = F_{n}y + F_{n - 1}f(y),}\\\\ & {\\qquad y = F_{n + 1}y + F_{n}f(y),} \\end{array} \\quad (by induction)\\] \n\nwhere \\(F_{k}\\) is the \\(k\\) - th Fibonacci number. As \\(y \\neq 0\\) and \\(f(y) \\neq 0\\) , the last two equations tell us \n\n\\[F_{n}^{2} \\equiv \\left(\\frac{(1 - F_{n - 1})f(y)}{y}\\right) \\left(\\frac{(1 - F_{n + 1})y}{f(y)}\\right) \\equiv (F_{n + 1} - 1)(F_{n - 1} - 1) \\pmod {p}.\\] \n\nLet \\(A = F_{n + 1} - 1\\) and \\(B = F_{n - 1} - 1\\) for brevity. The last equation becomes \n\n\\[(A - B)^{2} \\equiv AB \\equiv \\frac{1}{4} ((A + B)^{2} - (A - B)^{2}) \\pmod {p} \\Longrightarrow (A + B)^{2} \\equiv 5(A - B)^{2} \\pmod {p}.\\] \n\nAs 5 is not a quadratic residue, this implies \\(F_{n + 1} \\equiv F_{n - 1} \\equiv 1 \\pmod {p}\\) . Hence, if \\(d\\) is the smallest positive integer such that \\(F_{d} \\equiv 0 \\pmod {p}\\) and \\(F_{d + 1} \\equiv 1 \\pmod {p}\\) , then the Fibonacci sequence is periodic modulo \\(p\\) with period \\(d\\) , so \\(d \\mid n\\) . The sum of all cycle lengths (excluding the fixed point 0) is \\(p - 1\\) , so \\(d \\mid p - 1\\) . The following well- known lemma will give us a contradiction. \n\nLemma 1. If 5 is not a quadratic residue modulo a prime \\(p\\) , then \\(p \\nmid F_{p - 1}\\) . \n\nProof 1. Recall Binet's formula, \n\n\\[F_{p - 1} = \\frac{1}{\\sqrt{5}} \\left(\\left(\\frac{1 + \\sqrt{5}}{2}\\right)^{p - 1} - \\left(\\frac{1 - \\sqrt{5}}{2}\\right)^{p - 1}\\right).\\] \n\nMultiplying both sides by \\(2^{p - 1}\\) and expanding via the binomial theorem, we have \n\n\\[2^{p - 1}F_{p - 1} = 2\\sum_{k = 0}^{\\frac{p - 3}{2}}5^{k}\\binom{p - 1}{2k + 1}.\\] \n\nHowever, \\(\\binom{p- 1}{2k+1} \\equiv(- 1)^{2k+1} \\equiv- 1 \\pmod{p}\\) for all \\(k\\) , so \n\n\\[p \\mid F_{p - 1} \\quad \\text{if and only if} \\quad p \\left| \\sum_{k = 0}^{\\frac{p - 3}{2}} 5^{k} = \\frac{5^{\\frac{p - 1}{2}} - 1}{5 - 1} \\right.\\] \n\nTherefore \\(p \\mid F_{p - 1}\\) if and only if \\(5^{\\frac{p - 1}{2}} \\equiv 1 \\pmod {p}\\) , which doesn't hold if 5 is not a quadratic residue modulo \\(p\\) , as desired. \\(\\square\\) \n\nProof 2. Work in \\(\\mathbb{F}_{p^{2}} = \\mathbb{F}_{p}[\\sqrt{5}]\\) . Since 5 is not a quadratic residue, we get that \\((\\sqrt{5})^{p} = - \\sqrt{5}\\) . Using the fact that \\((a + b)^{p} = a^{p} + b^{p}\\) (because all other terms have coefficient divisible by \\(p\\) ), we get that \n\n\\[\\left(\\frac{1 + \\sqrt{5}}{2}\\right)^{p} = \\frac{1 - \\sqrt{5}}{2} \\implies \\left(\\frac{1 + \\sqrt{5}}{2}\\right)^{p - 1} = \\left(\\frac{1 - \\sqrt{5}}{2}\\right)^{2} = \\frac{3 - \\sqrt{5}}{2}.\\]\n\n\n\nSimilarly, \\(\\left(\\frac{1 - \\sqrt{5}}{2}\\right)^{p - 1} = \\frac{3 + \\sqrt{5}}{2}\\) . Hence, by Binet's formula, \n\n\\[F_{p - 1} = \\frac{1}{\\sqrt{5}}\\left(\\left(\\frac{1 + \\sqrt{5}}{2}\\right)^{p - 1} - \\left(\\frac{1 - \\sqrt{5}}{2}\\right)^{p - 1}\\right)\\] \\[= \\frac{1}{\\sqrt{5}}\\left(\\frac{3 - \\sqrt{5}}{2} -\\frac{3 + \\sqrt{5}}{2}\\right) = -1,\\] \n\nso it is not divisible by \\(p\\) . \n\nHence, \\(F_{p - 1} \\neq 0\\) (mod \\(p\\) ) if 5 is not a quadratic residue modulo \\(p\\) , which contradicts \\(d \\mid p - 1\\) above. This completes the solution.", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n9. [60]", "solution_match": "\nSolution: "}}
{"year": "2025", "tier": "T4", "problem_label": "10", "problem_type": null, "exam": "HMMT", "problem": "Determine, with proof, all possible values of \\(\\gcd (a^{2} + b^{2} + c^{2}, abc)\\) across all triples of positive integers \\((a, b, c)\\) .", "solution": "Answer: All positive integers \\(n\\) such that \\(\\nu_{p}(n) \\neq 1\\) for all prime \\(p \\equiv 3\\) (mod 4)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n10. [60]", "solution_match": "\nProposed by: Henrik Rabinovitz \n\n"}}
diff --git a/HarvardMIT/segmented/en-284-tournaments-2025-hmic-solutions.jsonl b/HarvardMIT/segmented/en-284-tournaments-2025-hmic-solutions.jsonl
index 066f73fc7446abe88c7e4ec7bbd18e634a39f417..b2dfa2b593b80899d8d9fbd1ea8a06e4b5f6cadc 100644
--- a/HarvardMIT/segmented/en-284-tournaments-2025-hmic-solutions.jsonl
+++ b/HarvardMIT/segmented/en-284-tournaments-2025-hmic-solutions.jsonl
@@ -1,10 +1,10 @@
-{"year": "2025", "tier": "T4", "problem_label": "1", "problem_type": null, "exam": "HMMT", "problem": "Let \\(A B C D\\) be a convex quadrilateral. Define parabolas \\(\\mathcal{P}_{A}\\) \\(\\mathcal{P}_{B}\\) \\(\\mathcal{P}_{C}\\) , and \\(\\mathcal{P}_{D}\\) to have directrices \\(B D\\) \\(C A\\) \\(D B\\) , and \\(A C\\) , and foci \\(A\\) \\(B\\) \\(C\\) , and \\(D\\) , respectively. Prove that no two of these parabolas intersect more than once. \n\n(A parabola with directrix \\(\\ell\\) and focus \\(P\\) consists of all points \\(X\\) for which \\(P X\\) equals the distance from \\(P\\) to \\(\\ell\\) .)", "solution": "", "metadata": {"resource_path": "HarvardMIT/segmented/en-284-tournaments-2025-hmic-solutions.jsonl", "problem_match": "\n1. [5]", "solution_match": "\nProposed by: Albert Wang \n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "1", "problem_type": null, "exam": "HMMT", "problem": "Let \\(A B C D\\) be a convex quadrilateral. Define parabolas \\(\\mathcal{P}_{A}\\) \\(\\mathcal{P}_{B}\\) \\(\\mathcal{P}_{C}\\) , and \\(\\mathcal{P}_{D}\\) to have directrices \\(B D\\) \\(C A\\) \\(D B\\) , and \\(A C\\) , and foci \\(A\\) \\(B\\) \\(C\\) , and \\(D\\) , respectively. Prove that no two of these parabolas intersect more than once. \n\n(A parabola with directrix \\(\\ell\\) and focus \\(P\\) consists of all points \\(X\\) for which \\(P X\\) equals the distance from \\(P\\) to \\(\\ell\\) .)", "solution": "", "metadata": {"resource_path": "HarvardMIT/segmented/en-284-tournaments-2025-hmic-solutions.jsonl", "problem_match": "\n1. [5]", "solution_match": "\nProposed by: Albert Wang \n\n"}}
{"year": "2025", "tier": "T4", "problem_label": "1", "problem_type": null, "exam": "HMMT", "problem": "Let \\(A B C D\\) be a convex quadrilateral. Define parabolas \\(\\mathcal{P}_{A}\\) \\(\\mathcal{P}_{B}\\) \\(\\mathcal{P}_{C}\\) , and \\(\\mathcal{P}_{D}\\) to have directrices \\(B D\\) \\(C A\\) \\(D B\\) , and \\(A C\\) , and foci \\(A\\) \\(B\\) \\(C\\) , and \\(D\\) , respectively. Prove that no two of these parabolas intersect more than once. \n\n(A parabola with directrix \\(\\ell\\) and focus \\(P\\) consists of all points \\(X\\) for which \\(P X\\) equals the distance from \\(P\\) to \\(\\ell\\) .)", "solution": "Let \\(d(P, X Y)\\) be the distance from \\(P\\) to line \\(X Y\\) . We will first prove \\(\\mathcal{P}_{A}\\) and \\(\\mathcal{P}_{B}\\) intersect at most once. \n\nClaim 1. Let \\(\\ell_{A B}\\) be the perpendicular bisector of \\(A B\\) . Then, \\(\\mathcal{P}_{A}\\) is tangent to \\(\\ell_{A B}\\) . \n\nProof. Consider any point \\(X\\) on \\(\\mathcal{P}_{A}\\) . Then, \n\n\\[X A = d(X,B D)\\leq X B,\\] \n\nwith equality only holding at the unique point \\(X\\) for which \\(X B\\perp B D\\) . Thus, \\(\\mathcal{P}_{A}\\) lies entirely on one side of \\(\\ell_{A B}\\) , touching it once at this point \\(X\\) . \\(\\square\\) \n\nIt follows that \\(\\mathcal{P}_{A}\\) and \\(\\mathcal{P}_{B}\\) are on different sides of \\(\\ell_{A B}\\) and hence can intersect at most once (possibly at a common tangency point to \\(\\ell_{A B}\\) ). \n\nSince \\(A B C D\\) is convex, \\(A\\) and \\(C\\) lie on opposite sides of line \\(B D\\) , the common directrix of \\(\\mathcal{P}_{A}\\) and \\(\\mathcal{P}_{C}\\) . Thus, \\(\\mathcal{P}_{A}\\) and \\(\\mathcal{P}_{C}\\) lie on opposite sides of line \\(B D\\) and cannot intersect at all. \n\nThe remaining pairs of parabolas are handled similarly.", "metadata": {"resource_path": "HarvardMIT/segmented/en-284-tournaments-2025-hmic-solutions.jsonl", "problem_match": "\n1. [5]", "solution_match": "\nSolution 1: "}}
{"year": "2025", "tier": "T4", "problem_label": "1", "problem_type": null, "exam": "HMMT", "problem": "Let \\(A B C D\\) be a convex quadrilateral. Define parabolas \\(\\mathcal{P}_{A}\\) \\(\\mathcal{P}_{B}\\) \\(\\mathcal{P}_{C}\\) , and \\(\\mathcal{P}_{D}\\) to have directrices \\(B D\\) \\(C A\\) \\(D B\\) , and \\(A C\\) , and foci \\(A\\) \\(B\\) \\(C\\) , and \\(D\\) , respectively. Prove that no two of these parabolas intersect more than once. \n\n(A parabola with directrix \\(\\ell\\) and focus \\(P\\) consists of all points \\(X\\) for which \\(P X\\) equals the distance from \\(P\\) to \\(\\ell\\) .)", "solution": "Let \\(d(P, X Y)\\) be the distance from \\(P\\) to line \\(X Y\\) . \n\nClaim 2. \\(\\mathcal{P}_{A}\\) and \\(\\mathcal{P}_{B}\\) intersect at most once. \n\nProof. Let \\(P\\) be a common point of both parabolas. Then, \\(d(P, B D) = P A\\) and \\(d(P, A C) = P B\\) , which combined imply \n\n\\[P A = d(P,B D)\\leq P B = d(P,A C)\\leq P A.\\] \n\nThus, the inequalities above are equalities, i.e., \\(P A = P B\\) , \\(P A\\perp A C\\) , and \\(P B\\perp B D\\) . Such \\(P\\) , if it exists, is unique. \\(\\square\\)\n\n\n\nThe remaining pairs of parabolas are handled similarly. As in solution 1, \\(\\mathcal{P}_{A}\\) and \\(\\mathcal{P}_{C}\\) cannot intersect, nor can \\(\\mathcal{P}_{B}\\) and \\(\\mathcal{P}_{D}\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-284-tournaments-2025-hmic-solutions.jsonl", "problem_match": "\n1. [5]", "solution_match": "\nSolution 2: "}}
{"year": "2025", "tier": "T4", "problem_label": "2", "problem_type": null, "exam": "HMMT", "problem": "Find all polynomials \\(P\\) with real coefficients for which there exists a polynomial \\(Q\\) with real coefficients such that for all real \\(t\\) , \n\n\\[\\cos (P(t)) = Q(\\cos t).\\]", "solution": "Answer: All constant functions and \\(P(x) = a x + b\\pi\\) for all nonzero integers \\(a\\) and integers \\(b\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-284-tournaments-2025-hmic-solutions.jsonl", "problem_match": "\n2. [7]", "solution_match": "\nProposed by: Karthik Venkata Vedula \n\n"}}
{"year": "2025", "tier": "T4", "problem_label": "2", "problem_type": null, "exam": "HMMT", "problem": "Find all polynomials \\(P\\) with real coefficients for which there exists a polynomial \\(Q\\) with real coefficients such that for all real \\(t\\) , \n\n\\[\\cos (P(t)) = Q(\\cos t).\\]", "solution": "It is well- known that these polynomials work by taking \\(Q\\) to be a Chebyshev polynomial (if \\(P\\) is linear) or a constant (if \\(P\\) is constant). \n\nSuppose that \\(\\deg P \\geq 2\\) . Now consider the density of the roots of \\(\\cos (P(t))\\) , i.e. \n\n\\[\\lim_{n \\to \\infty} \\frac{\\text{number of roots in the interval} [-n, n]}{n}.\\] \n\nSince \\(\\cos (P(t)) = Q(\\cos t)\\) , the density is finite, because for each interval of length \\(2\\pi\\) , there can only be a finite number of roots (i.e. twice the degree of \\(Q\\) ). However, we claim that \\(\\cos (P(t))\\) has an infinite density of roots. In particular, consider the solutions to \\(P(x) = \\pm (2k - 1)\\pi /2\\) over positive integers \\(k\\) . Asymptotically, for large \\(k\\) , such \\(x\\) will always exist and be \\(\\Theta (k^{1 / \\deg P})\\) . As \\(k \\to \\infty\\) , such \\(x\\) become infinitely dense, contradicting the finite density of roots of \\(Q(\\cos t)\\) . Therefore, \\(\\deg P \\leq 1\\) . \n\nIf \\(\\deg P = 1\\) , let \\(P(t) = at + b\\) . Observe that \\(Q(\\cos t)\\) is periodic with period \\(2\\pi\\) , so \\(\\cos (P(t))\\) must be as well. This is only the case when \\(a\\) is an integer. Furthermore, \n\n\\[Q(\\cos t) = \\cos (at + b) = \\cos (at)\\cos (b) - \\sin (at)\\sin (b)\\] \n\nmust be an even function. Note that \\(\\cos (at)\\cos (b)\\) is even and \\(\\sin (at)\\sin (b)\\) is odd, so \\(\\sin (at)\\sin (b) = 0\\) for all \\(t\\) , and hence \\(b\\) is an integer multiple of \\(\\pi\\) . \n\nThus, the only polynomials that work are the ones claimed above.", "metadata": {"resource_path": "HarvardMIT/segmented/en-284-tournaments-2025-hmic-solutions.jsonl", "problem_match": "\n2. [7]", "solution_match": "\nSolution 1: "}}
{"year": "2025", "tier": "T4", "problem_label": "2", "problem_type": null, "exam": "HMMT", "problem": "Find all polynomials \\(P\\) with real coefficients for which there exists a polynomial \\(Q\\) with real coefficients such that for all real \\(t\\) , \n\n\\[\\cos (P(t)) = Q(\\cos t).\\]", "solution": "Taking the derivative of both sides, \n\n\\[\\sin (P(t))P^{\\prime}(t) = (- \\sin t)Q^{\\prime}(\\cos t).\\] \n\nThe right- hand side is bounded in \\(t\\) , so the left- hand side must also be bounded. If \\(\\deg P \\geq 2\\) , then as \\(t\\) approaches \\(\\infty\\) , \\(P^{\\prime}(t)\\) approaches \\(\\pm \\infty\\) and \\(\\sin (P(t))\\) does not approach 0, contradiction. Thus, \\(\\deg P \\leq 1\\) , and we finish as before.", "metadata": {"resource_path": "HarvardMIT/segmented/en-284-tournaments-2025-hmic-solutions.jsonl", "problem_match": "\n2. [7]", "solution_match": "\nSolution 2: "}}
-{"year": "2025", "tier": "T4", "problem_label": "3", "problem_type": null, "exam": "HMMT", "problem": "Let \\(ABCD\\) be a parallelogram, and let \\(O\\) be a point inside \\(ABCD\\) . Suppose the circumcircles of triangles \\(OAB\\) and \\(OCD\\) intersect at \\(P \\neq O\\) , and the circumcircles of triangles \\(OBC\\) and \\(OAD\\) intersect at \\(Q \\neq O\\) . Prove \\(\\angle POQ\\) equals one of the angles of quadrilateral \\(ABCD\\) .", "solution": "", "metadata": {"resource_path": "HarvardMIT/segmented/en-284-tournaments-2025-hmic-solutions.jsonl", "problem_match": "\n3. [8]", "solution_match": "\nProposed by: Derek Liu\n\n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "3", "problem_type": null, "exam": "HMMT", "problem": "Let \\(ABCD\\) be a parallelogram, and let \\(O\\) be a point inside \\(ABCD\\) . Suppose the circumcircles of triangles \\(OAB\\) and \\(OCD\\) intersect at \\(P \\neq O\\) , and the circumcircles of triangles \\(OBC\\) and \\(OAD\\) intersect at \\(Q \\neq O\\) . Prove \\(\\angle POQ\\) equals one of the angles of quadrilateral \\(ABCD\\) .", "solution": "", "metadata": {"resource_path": "HarvardMIT/segmented/en-284-tournaments-2025-hmic-solutions.jsonl", "problem_match": "\n3. [8]", "solution_match": "\nProposed by: Derek Liu\n\n\n"}}
{"year": "2025", "tier": "T4", "problem_label": "3", "problem_type": null, "exam": "HMMT", "problem": "Let \\(ABCD\\) be a parallelogram, and let \\(O\\) be a point inside \\(ABCD\\) . Suppose the circumcircles of triangles \\(OAB\\) and \\(OCD\\) intersect at \\(P \\neq O\\) , and the circumcircles of triangles \\(OBC\\) and \\(OAD\\) intersect at \\(Q \\neq O\\) . Prove \\(\\angle POQ\\) equals one of the angles of quadrilateral \\(ABCD\\) .", "solution": "In what follows, all angles are directed. \n\nClaim 1. The points \\(P\\) and \\(Q\\) are symmetric over the center of \\(ABCD\\) . \n\nProof. Note that \n\n\\[\\angle A P B = \\angle A O B = \\angle O A D + \\angle C B O = \\angle O Q D + \\angle C Q O = \\angle C Q D.\\] \n\nSimilar equalities hold for each pair of opposite sides, so \\(P\\) and \\(Q\\) are symmetric across the parallelogram's center. \\(\\square\\) \n\nConsequently, \\(AP\\) and \\(CQ\\) are parallel, so \n\n\\[\\angle P O Q = \\angle P O B + \\angle B O Q = \\angle P A B + \\angle B C Q = \\angle C B A,\\] \n\nas desired. (Once we undirect the angles, \\(\\angle P O Q\\) is either \\(\\angle B\\) or \\(\\pi - \\angle B = \\angle A\\) .)", "metadata": {"resource_path": "HarvardMIT/segmented/en-284-tournaments-2025-hmic-solutions.jsonl", "problem_match": "\n3. [8]", "solution_match": "\nSolution: "}}
{"year": "2025", "tier": "T4", "problem_label": "4", "problem_type": null, "exam": "HMMT", "problem": "Determine whether there exist infinitely many pairs of distinct positive integers \\(m\\) and \\(n\\) such that \\(2^{m} + n\\) divides \\(2^{n} + m\\) .", "solution": "Answer: Yes", "metadata": {"resource_path": "HarvardMIT/segmented/en-284-tournaments-2025-hmic-solutions.jsonl", "problem_match": "\n4. [9]", "solution_match": "\nProposed by: Carlos Rodriguez, Jordan Lefkowitz \n\n"}}
{"year": "2025", "tier": "T4", "problem_label": "4", "problem_type": null, "exam": "HMMT", "problem": "Determine whether there exist infinitely many pairs of distinct positive integers \\(m\\) and \\(n\\) such that \\(2^{m} + n\\) divides \\(2^{n} + m\\) .", "solution": "Let \\(k\\) be a positive integer, and set \\(m = 2^{k}\\) and \\(n = p - 2^{2^{k}}\\) for prime \\(p\\) to be chosen later. We want \\(2^{m} + n = p\\) to divide \\(2^{p - 2^{2^{k}}} + 2^{k}\\) , which is equivalent to having \n\n\\[0\\equiv 2^{p - 2^{2^{k}}} + 2^{k}\\equiv 2^{1 - 2^{2^{k}}} + 2^{k}\\equiv 2^{1-2^{2^{k}}}\\left(2^{2^{2^{k}} + k - 1} + 1\\right)\\pmod {p}.\\] \n\nLet \\(r = 2^{2^{k}} + k - 1\\) . Since \\(r \\neq 3\\) , by Zsigmondy, we can pick a prime \\(p\\) that divides \\(2^{r} + 1\\) but not \\(2^{s} + 1\\) for any nonnegative integer \\(s < r\\) . Let \\(d = \\operatorname{ord}_{p}(2)\\) . Then, \\(2^{|d - r|} \\equiv -1 \\pmod {p}\\) , so by definition of \\(p\\) , we have \\(|d - r| \\geq r\\) . Hence, \\(d \\geq 2r\\) . As \\(d \\mid p - 1\\) , we conclude \n\n\\[p > 2r = 2\\left(2^{2^{k}} + k - 1\\right) > 2^{2^{k}} + 2^{k},\\] \n\nso \\(n = p - 2^{2^{k}} > m\\) . Since \\(p \\mid 2^{r} + 1\\) by definition, \\((m, n)\\) is a pair of distinct positive integers with \\(2^{m} + n \\mid 2^{n} + m\\) . \n\nAs \\(k\\) was arbitrary (and \\(m = 2^{k}\\) ), there exist infinitely many such pairs.", "metadata": {"resource_path": "HarvardMIT/segmented/en-284-tournaments-2025-hmic-solutions.jsonl", "problem_match": "\n4. [9]", "solution_match": "\nSolution: "}}
diff --git a/IMO/md/en-IMO-2005-notes.md b/IMO/md/en-IMO-2005-notes.md
index e8593e637002948c70098fad16b6cbc9b0672e08..621e236fc7a0e7fc3f10fb98d281acbfdb0a2fd3 100644
--- a/IMO/md/en-IMO-2005-notes.md
+++ b/IMO/md/en-IMO-2005-notes.md
@@ -73,7 +73,7 @@ Thus
and since three unit vectors with vanishing sum must be rotations of each other by \(120^{\circ}\) , it follows they must also form an equilateral triangle.
-
+
Consequently, triangles \(A_1A_2B_1\) , \(B_1B_2C_1\) , \(C_1C_2A_1\) are congruent, as \(\angle A_2 = \angle B_2 = \angle C_2\) . So triangle \(A_1B_1C_1\) is equilateral and the diagonals are concurrent at the center.
@@ -108,7 +108,7 @@ From this we deduce \(a_{n + 1} \in \{k, k + n + 1\}\) as desired.
This gives us actually a complete description of all possible sequences satisfying the hypothesis: choose any value of \(a_{1}\) to start. Then, for the \(n\) th term, the set \(S = \{a_{1}, \ldots , a_{n - 1}\}\) is (in some order) a set of \(n - 1\) consecutive integers. We then let \(a_{n} = \max S + 1\) or \(a_{n} = \min S - 1\) . A picture of six possible starting terms is shown below.
-
+
@@ -193,7 +193,7 @@ Let \(A B C D\) be a fixed convex quadrilateral with \(B C = D A\) and \(\overli
Let \(M\) be the Miquel point of complete quadrilateral \(A D B C\) ; in other words, let \(M\) be the second intersection point of the circumcircles of \(\triangle A P D\) and \(\triangle B P C\) . (A good diagram should betray this secret; all the points are given in the picture.) This makes lots of sense since we know \(E\) and \(F\) will be sent to each other under the spiral similarity too.
-
+
Thus \(M\) is the Miquel point of complete quadrilateral \(F A C E\) . As \(R = \overline{{F E}}\cap \overline{{A C}}\) we deduce \(F A R M\) is a cyclic quadrilateral (among many others, but we'll only need one).
diff --git a/IMO/md/en-IMO-2006-notes.md b/IMO/md/en-IMO-2006-notes.md
index eb899038604d70bfc04750b6468e3a62cb210c71..97bba764ec3317b6073a02376eeae5d3a8639826 100644
--- a/IMO/md/en-IMO-2006-notes.md
+++ b/IMO/md/en-IMO-2006-notes.md
@@ -251,7 +251,7 @@ The main work is the proof of the lemma.
Label the polygon \(P_{0}P_{1}\ldots P_{N - 1}\) . Consider the \(N / 2\) major diagonals of the almost convex \(N\) - gon, \(P_{0}P_{N / 2}\) , \(P_{1}P_{N / 2 + 1}\) , et cetera. A butterfly refers to a self- intersecting quadrilateral \(P_{i}P_{i + 1}P_{i + 1 + N / 2}P_{i + N / 2}\) . An example of a butterfly is shown below for \(N = 8\) .
-
+
Claim — Every point \(X\) in the polygon is contained in the wingspan of some butterfly.
diff --git a/IMO/md/en-IMO-2007-notes.md b/IMO/md/en-IMO-2007-notes.md
index 38afbf84a7fe70e9c443aa5d7aefbff55af9735b..8d1f1d33237f200cf13296e4a8231b76e95f7c9b 100644
--- a/IMO/md/en-IMO-2007-notes.md
+++ b/IMO/md/en-IMO-2007-notes.md
@@ -113,7 +113,7 @@ Let \(M\) , \(N\) , \(P\) denote the midpoints of \(\overline{CF}\) , \(\overlin
By a homothety at \(C\) with ratio \(\frac{1}{2}\) , we find \(\overline{MNP}\) is the image of line \(\ell \equiv \overline{AGF}\) .
-
+
However, since we also have \(\overline{EM} \perp \overline{CF}\) and \(\overline{EN} \perp \overline{CG}\) (from \(EF = EG = EC\) ) we conclude \(\overline{PMN}\) is the Simson line of \(E\) with respect to \(\triangle BCD\) , which implies \(\overline{EP} \perp \overline{BD}\) . In other words, \(\overline{EP}\) is the perpendicular bisector of \(\overline{BD}\) , so \(E\) is the midpoint of arc \(\overline{BCD}\) .
@@ -150,7 +150,7 @@ So let's consider the situation
At this point \(A\) is a set of \(k\) red vertices, while \(B\) has the remaining \(2r - k\) red vertices (and all the green ones). An example is shown below with \(k = 4\) and \(2r = 6\) .
-
+
Now, if we can move any red vertex from \(B\) back to \(A\) without changing the clique number of \(B\) , we do so, and win.
@@ -203,7 +203,7 @@ Thus \(OP = OQ\) . Since \(OC = OR\) as well, we get the conclusion.
Denote by \(X\) and \(Y\) the feet from \(R\) to \(\overline{CA}\) and \(\overline{CB}\) , so \(\triangle CXR \cong \triangle CYR\) . Then, let \(t = \frac{CQ}{CR} = 1 - \frac{CP}{CR}\) .
-
+
Then it follows that
diff --git a/IMO/md/en-IMO-2008-notes.md b/IMO/md/en-IMO-2008-notes.md
index 0970fe8b0c8a4a3a3d7a2e81a30af85a64ca29d8..daf6f4a96456eb7c2ed54849501e735150a24ab7 100644
--- a/IMO/md/en-IMO-2008-notes.md
+++ b/IMO/md/en-IMO-2008-notes.md
@@ -73,7 +73,7 @@ We show two solutions.
We first show that \(B_{1}\) , \(B_{2}\) , \(C_{1}\) , \(C_{2}\) are concyclic. It suffices to prove that \(A\) lies on the radical axis of the circles \(\Gamma_{B}\) and \(\Gamma_{C}\) .
-
+
Let \(X\) be the second intersection of \(\Gamma_{B}\) and \(\Gamma_{C}\) . Clearly \(\overline{XH}\) is perpendicular to the line joining the centers of the circles, namely \(\overline{EF}\) . But \(\overline{EF} \parallel \overline{BC}\) , so \(\overline{XH} \perp \overline{BC}\) . Since \(\overline{AH} \perp \overline{BC}\) as well, we find that \(A\) , \(X\) , \(H\) are collinear, as needed.
@@ -240,7 +240,7 @@ Let \(\overline{{P Q}}\) and \(\overline{{S T}}\) be diameters of \(\omega_{1}\)
Now orient \(A C\) horizontally and let \(K\) be the "uppermost" point of \(\omega\) , as shown.
-
+
Consequently, a homothety at \(B\) maps \(Q\) , \(T\) , \(K\) to each other (since \(T\) is the uppermost of the excircle, \(Q\) of the incircle). Similarly, a homothety at \(D\) maps \(P\) , \(S\) , \(K\) to each other. As \(\overline{{P Q}}\) and \(\overline{{S T}}\) are parallel diameters it then follows \(K\) is the exsimilicenter of \(\omega_{1}\) and \(\omega_{2}\) .
diff --git a/IMO/md/en-IMO-2009-notes.md b/IMO/md/en-IMO-2009-notes.md
index 9237a285a4c5e6534d636cc03056f591ef73fee7..7920e6eccdd1410faaf13039ef40e9c25457643c 100644
--- a/IMO/md/en-IMO-2009-notes.md
+++ b/IMO/md/en-IMO-2009-notes.md
@@ -101,7 +101,7 @@ Let \(ABC\) be a triangle with circumcenter \(O\) . The points \(P\) and \(Q\) a
By power of a point, we have \(- AQ \cdot QB = OQ^2 - R^2\) and \(- AP \cdot PC = OP^2 - R^2\) . Therefore, it suffices to show \(AQ \cdot QB = AP \cdot PC\) .
-
+
As \(\overline{ML} \parallel \overline{AC}\) and \(\overline{MK} \parallel \overline{AB}\) we have that
@@ -192,7 +192,7 @@ Let \(I\) be the incenter of \(A B C\) , and set \(\angle D A C = 2x\) (so that
Having chased all the angles we want, we need a relationship. We can find it by considering the side ratio \(\frac{I K}{K C}\) . Using the angle bisector theorem, we can express this in terms of triangle \(I D C\) ; however we can also express it in terms of triangle \(I E C\) .
-
+
By the law of sines, we obtain
diff --git a/IMO/md/en-IMO-2010-notes.md b/IMO/md/en-IMO-2010-notes.md
index b188741cd6a9491950731245cb32627b01ef9dd5..9b002a318e58038967e211ab60d0a14d86a191db 100644
--- a/IMO/md/en-IMO-2010-notes.md
+++ b/IMO/md/en-IMO-2010-notes.md
@@ -130,7 +130,7 @@ Let \(I_{A}\) be the \(A\) - excenter, and set \(Q = \overline{{I_{A}F}}\cap \ov
where \(\infty\) is the point at infinity along \(\overline{{I P Q}}\) . Thus \(P\) is the midpoint of \(\overline{{I Q}}\) . Since \(D\) is the midpoint of \(\overline{{I I_{A}}}\) by "Fact 5", it follows that \(\overline{{D P}}\) bisects \(\overline{{I F}}\) .
-
+
@@ -187,7 +187,7 @@ We present two solutions using harmonic bundles.
\(\P\) First solution (Evan Chen). Let \(N\) be the antipode of \(M\) , and let \(NP\) meet \(\Gamma\) again at \(D\) . Focus only on \(CDMN\) for now (ignoring the condition). Then \(C\) and \(D\) are feet of altitudes in \(\triangle MNP\) ; it is well- known that the circumcircle of \(\triangle CDP\) is orthogonal to \(\Gamma\) (passing through the orthocenter of \(\triangle MPN\) ).
-
+
Now, we are given that point \(S\) is such that \(\overline{SC}\) is tangent to \(\Gamma\) , and \(SC = SP\) . It follows that \(S\) is the circumcenter of \(\triangle CDP\) , and hence \(\overline{SC}\) and \(\overline{SD}\) are tangents to \(\Gamma\) .
diff --git a/IMO/md/en-IMO-2011-notes.md b/IMO/md/en-IMO-2011-notes.md
index 6a300ea68a1ecd468929b6619d1aca4987d2a78b..afbd56a3feae33bdcf26ea392129cf1b722f9aee 100644
--- a/IMO/md/en-IMO-2011-notes.md
+++ b/IMO/md/en-IMO-2011-notes.md
@@ -253,7 +253,7 @@ Let \(A B C\) be an acute triangle with circumcircle \(\Gamma\) . Let \(\ell\) b
This is a hard problem with many beautiful solutions. The following solution is not very beautiful but not too hard to find during an olympiad, as the only major insight it requires is the construction of \(A_{2}\) , \(B_{2}\) , and \(C_{2}\) .
-
+
We apply complex numbers with \(\omega\) the unit circle and \(p = 1\) . Let \(A_{1} = \ell_{B} \cap \ell_{C}\) , and let \(a_{2} = a^{2}\) (in other words, \(A_{2}\) is the reflection of \(P\) across the diameter of \(\omega\) through \(A\) ). Define the points \(B_{1}\) , \(C_{1}\) , \(B_{2}\) , \(C_{2}\) similarly.
diff --git a/IMO/md/en-IMO-2012-notes.md b/IMO/md/en-IMO-2012-notes.md
index ace1768c6f28a3f5b27a55bdbcd16a4941c8fc10..d17adc271053e6b60a629ef1751c516f3872cfc9 100644
--- a/IMO/md/en-IMO-2012-notes.md
+++ b/IMO/md/en-IMO-2012-notes.md
@@ -242,7 +242,7 @@ Let \(A B C\) be a triangle with \(\angle B C A = 90^{\circ}\) , and let \(D\) b
Let \(\omega_{A}\) and \(\omega_{B}\) be the circles through \(C\) centered at \(A\) and \(B\) ; extend rays \(A K\) and \(B L\) to hit \(\omega_{B}\) and \(\omega_{A}\) again at \(K^{*}\) , \(L^{*}\) . By radical center \(X\) , we have \(K L K^{*}L^{*}\) is cyclic, say with circumcircle \(\omega\) .
-
+
By orthogonality of \((A)\) and \((B)\) we find that \(\overline{A L}\) , \(\overline{A L^{*}}\) , \(\overline{B K}\) , \(\overline{B K^{*}}\) are tangents to \(\omega\) (in particular, \(K L K^{*}L^{*}\) is harmonic). In particular \(\overline{M K}\) and \(\overline{M L}\) are tangents to \(\omega\) , so \(M K = M L\) .
diff --git a/IMO/md/en-IMO-2013-notes.md b/IMO/md/en-IMO-2013-notes.md
index 9b1f955580e0f7b48969baa3755853057bf163bb..88376a87c9469c2cfa9ddcadd9c79f73dc64e575 100644
--- a/IMO/md/en-IMO-2013-notes.md
+++ b/IMO/md/en-IMO-2013-notes.md
@@ -139,7 +139,7 @@ Proof. This follows from the fact that we have a spiral similarity \(\triangle X
We define the arc midpoints \(Y\) and \(Z\) similarly, which lie on the perpendicular bisectors of \(\overline{A_{1}C_{1}}\) , \(\overline{A_{1}B_{1}}\) .
-
+
We now turn to the problem condition which asserts the circumcenter \(W\) of \(\triangle A_{1}B_{1}C_{1}\) lies on \((ABC)\) .
@@ -176,7 +176,7 @@ We present two solutions, an elementary one and then an advanced one by moving p
\(\P\) First solution, classical. Let \(P\) be the second intersection of \(\omega_{1}\) and \(\omega_{2}\) ; this is the Miquel point, so \(P\) also lies on the circumcircle of \(AMN\) , which is the circle with diameter \(\overline{AH}\) .
-
+
We now contend:
@@ -273,7 +273,7 @@ Let \(n \geq 3\) be an integer, and consider a circle with \(n + 1\) equally spa
First, here are half of the beautiful labellings up to reflection for \(n = 6\) , just for concreteness.
-
+
Abbreviate "beautiful labelling of points around a circle" to ring. Moreover, throughout the solution we will allow degenerate chords that join a point to itself; this has no effect on the problem statement.
@@ -298,7 +298,7 @@ In any ring, the chords of sum \(k\) (even including degenerate ones) are pseudo
Proof. By induction on \(n\) . By shifting, we may assume that one of the chords is \(\{0, k\}\) and discard all numbers exceeding \(k\) ; that is, assume \(n = k\) . Suppose the other two chords are \(\{a, n - a\}\) and \(\{b, n - b\}\) .
-
+
We consider the chord \(\{u, v\}\) directly above \(\{0, n\}\) , drawn in blue. There are now three cases.
@@ -327,7 +327,7 @@ Proof. Because the chords of sum \(n\) are pseudo- parallel, there is at most on
Conversely, we claim that this works. The chords of sum \(n\) (and less than \(n\) ) are OK by construction, so assume for contradiction that there exists \(a, b, c \in \{1, \ldots , n - 1\}\) such that \(a + b = n + c\) . Then, we can "reflect" them using the (pseudo- parallel) chords of length \(n\) to find that \((n - a) + (n - b) = 0 + (n - c)\) , and the chords joining 0 to \(n - c\) and \(n - a\) to \(n - b\) intersect, by definition.
-
+
This is a contradiction that the original numbers on \([0, n - 1]\) form a ring.
diff --git a/IMO/md/en-IMO-2014-notes.md b/IMO/md/en-IMO-2014-notes.md
index a986d5fbaea0d80551b35ea6150c1c1556214450..18b26dcdb32915fa0adf580c16d613afd3d6d240 100644
--- a/IMO/md/en-IMO-2014-notes.md
+++ b/IMO/md/en-IMO-2014-notes.md
@@ -91,12 +91,12 @@ The answer is \(k = \lfloor \sqrt{n - 1} \rfloor\) .
\(\P\) Proof that the property holds when \(n\geq k^{2} + 1\) . First, assume \(n > k^{2}\) for some \(k\) . We will prove we can find an empty \(k\times k\) square. Indeed, let \(R\) be a rook in the uppermost column, and draw \(k\) squares of size \(k\times k\) directly below it, aligned. There are at most \(k - 1\) rooks among these squares, as desired.
-
+
\(\P\) Construction for all \(n\leq k^{2}\) . We first give an example where for \(n = k^{2}\) showing there may be no empty \(k\times k\) square. We draw the example for \(k = 3\) (with the generalization being obvious);
-
+
To show that this works, consider for each rook drawing an \(k\times k\) square of \(X\) 's whose bottom- right hand corner is the rook (these may go off the board). These indicate
@@ -131,7 +131,7 @@ Proof.
which implies conclusion.
-
+
Let the perpendicular bisector of \(T H\) meet \(A H\) at \(P\) now. It suffices to show that \(\frac{A P}{P H}\) is symmetric in \(b = A D\) and \(d = A B\) , because then \(P\) will be the circumcenter of \(\triangle T S H\) . To do this, set \(A H = \frac{b d}{2 R}\) and \(A C = 2 R\) .
@@ -174,7 +174,7 @@ Now for concreteness we will use a negative inversion at \(H\) which swaps \(B\)
Let us describe the inverted problem. We let \(M\) and \(N\) denote the midpoints of \(\overline{A^{*}B^{*}}\) and \(\overline{A^{*}D^{*}}\) , which are the centers of \((HA^{*}B^{*})\) and \((HA^{*}D^{*})\) . From \(\overline{T^{*}C^{*}} \perp (HA^{*}D^{*})\) , we know have \(C^{*}\) , \(M\) , \(T^{*}\) collinear. Similarly, \(C^{*}\) , \(N\) , \(S^{*}\) are collinear. We have that \((A^{*}HC^{*})\) is orthogonal to \((ABCD)\) which remains fixed. We wish to show \(\overline{T^{*}S^{*}}\) and \(\overline{MN}\) are parallel.
-
+
@@ -199,7 +199,7 @@ We give three solutions.
\(\P\) First solution by harmonic bundles. Let \(\overline{BM}\) intersect the circumcircle again at \(X\) .
-
+
The angle conditions imply that the tangent to \((ABC)\) at \(B\) is parallel to \(\overline{AP}\) . Let \(\infty\) be the point at infinity along line \(AP\) . Then
@@ -278,7 +278,7 @@ A set of lines in the plane is in general position if no two are parallel and no
Suppose we have colored \(k\) of the lines blue, and that it is not possible to color any additional lines. That means any of the \(n - k\) non- blue lines is the side of some finite region with an otherwise entirely blue perimeter. For each such line \(\ell\) , select one such region, and take the next counterclockwise vertex; this is the intersection of two blue lines \(v\) . We'll say \(\ell\) is the eyelid of \(v\) .
-
+
You can prove without too much difficulty that every intersection of two blue lines has at most two eyelids. Since there are \(\binom{k}{2}\) such intersections, we see that
@@ -289,7 +289,7 @@ so \(n\leq k^{2}\) , as required.
Remark. In fact, \(k = \sqrt{n}\) is "sharp for greedy algorithms", as illustrated below for \(k = 3\) :
-
+
diff --git a/IMO/md/en-IMO-2015-notes.md b/IMO/md/en-IMO-2015-notes.md
index 7bf4f46023245fd3b25b59a461497bcf58c36468..8003044441151f10d702ff1cdfe97fa521b887da 100644
--- a/IMO/md/en-IMO-2015-notes.md
+++ b/IMO/md/en-IMO-2015-notes.md
@@ -84,7 +84,7 @@ We say that a finite set \(S\) of points in the plane is balanced if, for any tw
For part (a), take a circle centered at a point \(O\) , and add \(n - 1\) additional points by adding pairs of points separated by an arc of \(60^{\circ}\) or similar triples. An example for \(n = 6\) is shown below.
-
+
For part (b), the answer is odd \(n\) , achieved by taking a regular \(n\) - gon. To show even \(n\) fail, note that some point is on the perpendicular bisector of
@@ -163,7 +163,7 @@ Let \(L\) be on the nine- point circle with \(\angle HML = 90^{\circ}\) . The ne
In the inverted statement, we want line \(ML\) to be tangent to \((AQL)\) .
-
+
Claim — \(\overline{LM} \parallel \overline{AQ}\) .
@@ -194,7 +194,7 @@ Since \(\overline{AO} \perp \overline{FG}\) for obvious reasons, we will only ne
Let line \(FG\) meet \((BDF)\) and \((CGE)\) again at \(F_{2}\) and \(G_{2}\) .
-
+
Claim — Quadrilaterals \(FBDF_{2}\) and \(G_{2}ECG\) are similar, actually homothetic through \(\overline{FG} \cap \overline{BC}\) .
diff --git a/IMO/md/en-IMO-2016-notes.md b/IMO/md/en-IMO-2016-notes.md
index b69be409df4f7876e4ec6c416dc7b3d922e2d874..f046ebb7a76a5b06edd30ad8fe52a3a7f3268f75 100644
--- a/IMO/md/en-IMO-2016-notes.md
+++ b/IMO/md/en-IMO-2016-notes.md
@@ -95,7 +95,7 @@ Proof. The proof uses three observations:
These three facts, together with \(F\) lying inside \(\triangle ABD\) , are enough to imply the result. \(\square\)
-
+
@@ -108,7 +108,7 @@ Proof. Fact 5.
Using these observations as the anchor for everything that follows, we now prove several claims about \(X\) and \(E\) in succession.
-
+
Claim — Point \(E\) is the midpoint of arc \(\overline{AD}\) in \((ABMD)\) , and hence lies on ray \(BF\) .
@@ -287,7 +287,7 @@ The following solution was communicated to me by Yang Liu.
Imagine taking a larger circle \(\omega\) encasing all \(\binom{n}{2}\) intersection points. Denote by \(P_{1}\) , \(P_{2}\) , ..., \(P_{2n}\) the order of the points on \(\omega\) in clockwise order; we imagine placing the frogs on \(P_{i}\) instead. Observe that, in order for every pair of segments to meet, each line segment must be of the form \(P_{i}P_{i + n}\) .
-
+
Then:
diff --git a/IMO/md/en-IMO-2017-notes.md b/IMO/md/en-IMO-2017-notes.md
index b2dc76f00ae7f3b0eda75cabe327d73821845ded..f0c514d3fc3663bdf7970feeb04f5dad391ed734 100644
--- a/IMO/md/en-IMO-2017-notes.md
+++ b/IMO/md/en-IMO-2017-notes.md
@@ -211,7 +211,7 @@ Now, we may assume the rabbit reveals its location \(A\) , so that all previous
The rabbit chooses two points \(X\) and \(Y\) symmetric about \(\ell\) such that \(XY = 2\) and \(AX = AY = n\) , as shown. The rabbit can then hop to either \(X\) or \(Y\) , pinging the point \(P_{n}\) on the \(\ell\) each time. This takes \(n\) hops.
-
+
Now among all points \(H\) the hunter can go to, \(\min \max \{HX, HY\}\) is clearly minimized with \(H \in \ell\) by symmetry. So the hunter moves to a point \(H\) such that \(BH = n\) as well. In that case the new distance is \(HX = HY\) .
@@ -264,7 +264,7 @@ Similarly, \(\overline{TK}\) is tangent at \(T\) iff \(\triangle ATS \sim \trian
The two similarities are equivalent because \(RS = ST\) the SAS gives \(KR \cdot TA = RS \cdot RT = TS \cdot TR\) .
-
+
Remark. The problem is actually symmetric with respect to two circles; \(\overline{RA}\) is tangent at \(R\) if and only if \(\overline{TK}\) at \(T\) .
@@ -299,7 +299,7 @@ Fix \(N \geq 1\) . A collection of \(N(N + 1)\) soccer players of distinct heigh
Some opening remarks: location and height are symmetric to each other, if one thinks about this problem as permutation pattern avoidance. So while officially there are multiple solutions, they are basically isomorphic to one another, and I am not aware of any solution otherwise.
-
+
Take a partition of \(N\) groups in order by height: \(G_{1} = \{1,\ldots ,N + 1\}\) , \(G_{2} = \{N+\) \(2,\ldots ,2N + 2\}\) , and so on. We will pick two people from each group \(G_{k}\)
diff --git a/IMO/md/en-IMO-2018-notes.md b/IMO/md/en-IMO-2018-notes.md
index 80290146e2a3967cf5a1ca4f12ef56d6c3020a48..5be9421d42c8ef26a5eeb53d8e2a36adaa8caf74 100644
--- a/IMO/md/en-IMO-2018-notes.md
+++ b/IMO/md/en-IMO-2018-notes.md
@@ -81,7 +81,7 @@ We present a synthetic solution from the IMO shortlist as well as a complex numb
\(\P\) Synthetic solution (from Shortlist). Construct parallelograms \(AXFD\) and \(AEGY\) , noting that \(X\) and \(Y\) lie on \(\Gamma\) . As \(\overline{XF} \parallel \overline{AB}\) we can let \(M\) denote the midpoint of minor arcs \(\overline{XF}\) and \(\overline{AB}\) (which coincide). Define \(N\) similarly.
-
+
Observe that \(XF = AD = AE = YG\) , so arcs \(\widehat{XF}\) and \(\widehat{YG}\) have equal measure; hence arcs \(\widehat{MF}\) and \(\widehat{NG}\) have equal measure; therefore \(\widehat{MN} \parallel \widehat{FG}\) .
@@ -121,7 +121,7 @@ Thus, \(\overline{{F G^{\prime}}}\parallel \overline{{D E}}\) , so \(G = G^{\pri
\(\P\) Synthetic solution from Derek Liu. Let lines \(F D\) and \(G E\) intersect \(\Gamma\) again at \(J\) and \(K\) , respectively.
-
+
Notice that \(\triangle BFD \sim \triangle JAD\) ; as \(FB = FD\) , it follows that \(AJ = AD\) . Likewise, \(\triangle CGE \sim \triangle KAE\) and \(GC = GE\) , so \(AK = AE\) . Hence,
@@ -210,7 +210,7 @@ Let \(n = 2018\) and \(N = 1 + 2 + \dots +n\) . For every number \(d\) not in th
Consider the directed path starting from the top vertex \(A\) . Starting from the first number, it increments by at least \(1 + 2 + \dots +n\) , since the increments at each step in the path are distinct; therefore equality must hold and thus the path from the top ends at \(N = 1 + 2 + \dots +n\) with all the numbers \(\{1,2,\ldots ,n\}\) being close by. Let \(B\) be that position.
-
+
Consider the two left/right neighbors \(X\) and \(Y\) of the endpoint \(B\) . Assume that \(B\) is to the right of the midpoint of the bottom side, and complete the equilateral triangle
@@ -281,7 +281,7 @@ Claim — Let \(p\) be a prime. Consider the sequence \(\nu_{p}(a_{N + 1})\) , \
A cartoon of the situation is drawn below.
-
+
@@ -381,7 +381,7 @@ In other words, the problem is to show (1) and (2) implies (3).
Henceforth drop apostrophes. Here is the inverted diagram (with apostrophes dropped).
-
+
Let \(Q\) denote the reflection of \(P\) and let \(Y\) denote the second intersection of \((BQC)\) and \((AQD)\) . Then
diff --git a/IMO/md/en-IMO-2019-notes.md b/IMO/md/en-IMO-2019-notes.md
index 677d8861ae2b99d1649165e0158d39e95ea25adf..a8b896b3c54e5ad9892bdaab58dc748661679736 100644
--- a/IMO/md/en-IMO-2019-notes.md
+++ b/IMO/md/en-IMO-2019-notes.md
@@ -103,7 +103,7 @@ Note the angle condition implies \(P_{1}C X A\) and \(Q_{1}C Y B\) are cyclic.
Letting \(T = \overline{{P X}}\cap \overline{{Q Y}}\) (possibly at infinity), it suffices to show that the radical axis of \(\triangle C X A\) and \(\triangle C Y B\) passes through \(T\) , because that would imply \(P_{1}X Y Q_{1}\) is cyclic (by power of a point when \(T\) is Euclidean, and because it is an isosceles trapezoid if \(T\) is at infinity).
-
+
To this end we use barycentric coordinates on \(\triangle A B C\) . We begin by writing
@@ -132,7 +132,7 @@ This completes the proof.
\(\P\) Second official solution by tricky angle chasing. Let lines \(A A_{1}\) and \(B B_{1}\) meet at the circumcircle of \(\triangle A B C\) again at points \(A_{2}\) and \(B_{2}\) . By Reim's theorem, \(P Q A_{2}B_{2}\) are cyclic.
-
+
Claim — The points \(P\) , \(Q\) , \(A_{2}\) , \(Q_{1}\) are cyclic. Similarly the points \(P\) , \(Q\) , \(B_{2}\) , \(P_{1}\) are cyclic.
@@ -247,7 +247,7 @@ The main claim is that \(G_{n}\) can be described explicitly in terms of \(G_{n
An illustration of \(G_{4}\) is given below.
-
+
To prove this claim, we need only show the arrows of this directed graph remain valid. The graph \(X\) is correct as a subgraph of \(G_{n}\) , since the extra 0 makes no difference. As for \(Y\) , note that if \(s = s_{1}\dots s_{n - 1}\) had \(k\) ones, then the modified string has \((n - 1 - k) + 1 = n - k\)
@@ -282,7 +282,7 @@ We present five solutions.
\(\P\) First solution by complex numbers (Evan Chen, with Yang Liu). We use complex numbers with \(D = x\) , \(E = y\) , \(F = z\) .
-
+
Then \(A = \frac{2y z}{y + z}\) , \(R = \frac{- y z}{x}\) and so
@@ -318,7 +318,7 @@ This gives \(\overline{PT} \perp \overline{O_{B}O_{C}}\) as needed.
\(\P\) Second solution by tethered moving points, with optimization (Evan Chen). Fix \(\triangle DEF\) and \(\omega\) , with \(B = \overline{DD} \cap \overline{FF}\) and \(C = \overline{DD} \cap \overline{EE}\) . We consider a variable point \(M\) on \(\omega\) and let \(X\) , \(Y\) be on \(\overline{EF}\) with \(\overline{CY} \cap \| \overline{ME}\) , \(\overline{BX} \cap \| \overline{MF}\) . We define \(W = \overline{CY} \cap \overline{BX}\) . Also, let line \(MW\) meet \(\omega\) again at \(V\) .
-
+
Claim (Angle chasing) — Pentagons \(C V W X E\) and \(B V W Y F\) are cyclic.
@@ -400,7 +400,7 @@ Dividing out,
We perform inversion around \(\omega\) , using \(\bullet^{*}\) for the inverse. Then \(\triangle A^{*}B^{*}C^{*}\) is the medial triangle of \(\triangle D E F\) , and \(T^{*}\) is the foot from \(A^{*}\) on to \(\overline{{D I}}\) . If we denote \(Q^{*}\) as the second intersection of \((P C^{*}E)\) and \((P B^{*}F)\) , then the goal is to show that \(Q^{*}\) lies on \((P I T^{*})\) .
-
+
Claim — Points \(Q^{*}\) , \(B^{*}\) , \(C^{*}\) are collinear.
@@ -431,7 +431,7 @@ In \(\triangle PEF\) , the \(P\) - symmedian meets \(\overline{EF}\) and \((PEF)
The proof proceeds in three steps. Suppose the line through \(L\) perpendicular to \(\overline{EF}\) meets \(\overline{EF}\) at \(W\) and \((PEF)\) at \(Z\) .
-
+
1. Since \(\angle ZEP = \angle WLP = \angle WDP\) , it follows \(\overline{ZE}\) is tangent to \((PDNE)\) . Similarly, \(\overline{ZF}\) is tangent to \((PDMF)\) .
diff --git a/IMO/md/en-IMO-2020-notes.md b/IMO/md/en-IMO-2020-notes.md
index 2302ce2d7515e6faedcb87ecd9d4545ef5979896..475e95fdccec3cc6c5a8889592f9bd52d2ebbf98 100644
--- a/IMO/md/en-IMO-2020-notes.md
+++ b/IMO/md/en-IMO-2020-notes.md
@@ -73,7 +73,7 @@ Prove that the following three lines meet in a point: the internal bisectors of
Let \(O\) denote the circumcenter of \(\triangle PAB\) . We claim it is the desired concurrency point.
-
+
Indeed, \(O\) obviously lies on the perpendicular bisector of \(AB\) . Now
@@ -144,7 +144,7 @@ The first key idea is the deep fact that
So, place all four pebbles of the same colour in a box (hence \(n\) boxes). For each \(k = 1,2,\ldots ,2n\) we tape a piece of string between pebble \(k\) and \(4n + 1 - k\) . To solve the problem, it suffices to paint each string either blue or green such that each box has two blue strings and two green strings (where a string between two pebbles in the same box counts double).
-
+
We can therefore rephrase the problem as follows, if we view boxes as vertices and strings as edges:
@@ -171,7 +171,7 @@ Answer: \(k = n^2 - n + 1\)
When \(k = n^2 - n\) , the construction for \(n = 4\) is shown below which generalizes readily. (We draw \(A\) in red and \(B\) in blue.)
-
+
To see this is sharp, view \(A\) and \(B\) as graphs whose connected components are paths (possibly with 0 edges; the direction of these edges is irrelevant). Now, if \(k = n^2 - n + 1\) it follows that \(A\) and \(B\) each have exactly \(n - 1\) connected components.
@@ -240,7 +240,7 @@ We make the following the definitions:
The figure is shown below with \(\mathcal{T}\) drawn in yellow, and points of \(S\) drawn in blue.
-
+
diff --git a/IMO/md/en-IMO-2021-notes.md b/IMO/md/en-IMO-2021-notes.md
index cb37268e35bcfe80077c1076999962c55f8d37bc..bdc9b274f74156a13b194101325a09494fcac823 100644
--- a/IMO/md/en-IMO-2021-notes.md
+++ b/IMO/md/en-IMO-2021-notes.md
@@ -164,7 +164,7 @@ Proof. Notice that \(\angle EMB = 180^{\circ} - \angle AMB - \angle EMZ = 180^{\
Let \(N\) be the other intersection of circles \((ACD)\) and \((DEX)\) and let \(R\) be the intersection of \(AC\) and \(BM\) .
-
+
Claim — Points \(B\) , \(D\) , \(M\) , \(N\) are cyclic.
@@ -201,7 +201,7 @@ Let \(\Gamma\) be a circle with center \(I\) , and \(ABCD\) a convex quadrilater
Let \(PQR S\) be the contact points of \(\Gamma\) an \(\overline{AB}\) , \(\overline{BC}\) , \(\overline{CD}\) , \(\overline{DA}\) .
-
+
Claim — We have \(\triangle IQZ \cong \triangle IRT\) . Similarly, \(\triangle IPX \cong \triangle ISY\) .
@@ -242,7 +242,7 @@ Assume for contradiction no such \(k\) exists. We will use a so- called "thresho
This process takes exactly 2021 steps. Right after the \(k\) th move, we consider a situation where we color walnut \(k\) red as well, so at the \(k\) th step there are \(k\) ones. For brevity, a non- red walnut is called black. An example is illustrated below with 2021 replaced by 6.
-
+
Claim — At each step, the walnut that becomes red is between two non- red or two red walnuts.
diff --git a/IMO/md/en-IMO-2022-notes.md b/IMO/md/en-IMO-2022-notes.md
index 21ada0491a15f282b3cb024bf70229f96aa5139c..7aeb0a8a91261bc31fbc29fd0ae522b10c3efa86 100644
--- a/IMO/md/en-IMO-2022-notes.md
+++ b/IMO/md/en-IMO-2022-notes.md
@@ -203,7 +203,7 @@ The conditions imply
Define \(K = \overline{CT} \cap \overline{AE}\) , \(L = \overline{DT} \cap \overline{AB}\) , \(X = \overline{BT} \cap \overline{AE}\) , \(Y = \overline{ET} \cap \overline{BY}\) .
-
+
Claim (Main claim) — We have
diff --git a/IMO/md/en-IMO-2023-notes.md b/IMO/md/en-IMO-2023-notes.md
index 9c58d17173453af48e914bfa321403ce6c3d16a2..87c8ff4b63b93e8944d09b45a364812cc9aac019 100644
--- a/IMO/md/en-IMO-2023-notes.md
+++ b/IMO/md/en-IMO-2023-notes.md
@@ -57,7 +57,7 @@ is an integer for every \(n = 1,2,\ldots ,2023\) . Prove that \(a_{2023}\geq 303
5. Let \(n\) be a positive integer. A Japanese triangle consists of \(1 + 2 + \dots +n\) circles arranged in an equilateral triangular shape such that for each \(1\leq i\leq n\) , the \(i^{\mathrm{th}}\) row contains exactly \(i\) circles, exactly one of which is colored red. A ninja path in a Japanese triangle is a sequence of \(n\) circles obtained by starting in the top row, then repeatedly going from a circle to one of the two circles immediately below it and finishing in the bottom row. Here is an example of a Japanese triangle with \(n = 6\) , along with a ninja path in that triangle containing two red circles.
-
+
In terms of \(n\) , find the greatest \(k\) such that in each Japanese triangle there is a ninja path containing at least \(k\) red circles.
@@ -122,7 +122,7 @@ Proof. Because \(\angle SPD = \angle LPD = \angle LBD = \angle SBE = \angle FCS
Let us define \(X = \overline{AM} \cap \overline{BS}\) and complete chord \(\overline{PXQ}\) . We aim to show that \(\overline{PXQ}\) is tangent to \((PDLB)\) .
-
+
Claim (Main projective claim) — We have \(XP = XA\) .
@@ -134,7 +134,7 @@ Proof. Introduce \(Y = \overline{PDF} \cap \overline{AM}\) . Note that
where \(\infty = \overline{AM} \cap \overline{SF}\) is at infinity (because \(AMSF\) is a rectangle). Thus, \(XY = XA\) .
-
+
Since \(\triangle APY\) is also right, we get \(XP = XA\) .
@@ -283,7 +283,7 @@ Available online at https://aops.com/community/p28104367.
Let \(n\) be a positive integer. A Japanese triangle consists of \(1 + 2 + \dots + n\) circles arranged in an equilateral triangular shape such that for each \(1\leq i\leq n\) , the \(i^{\mathrm{th}}\) row contains exactly \(i\) circles, exactly one of which is colored red. A ninja path in a Japanese triangle is a sequence of \(n\) circles obtained by starting in the top row, then repeatedly going from a circle to one of the two circles immediately below it and finishing in the bottom row. Here is an example of a Japanese triangle with \(n = 6\) , along with a ninja path in that triangle containing two red circles.
-
+
In terms of \(n\) , find the greatest \(k\) such that in each Japanese triangle there is a ninja path containing at least \(k\) red circles.
@@ -294,7 +294,7 @@ The answer is
Construction. It suffices to find a Japanese triangle for \(n = 2^{e} - 1\) with the property that at most \(e\) red circles in any ninja path. The construction shown below for \(e = 4\) obviously generalizes, and works because in each of the sets \(\{1\}\) , \(\{2,3\}\) , \(\{4,5,6,7\}\) , ..., \(\{2^{e - 1}, \ldots , 2^{e} - 1\}\) , at most one red circle can be taken. (These sets are colored in different shades of red for visual clarity).
-
+
@@ -311,14 +311,14 @@ Connect the red circle to both circles under it; White circles to the left of th
The circles in the bottom row are all leaves of this tree. For example, the \(n = 6\) construction in the beginning gives the tree shown on the left half of the figure below:
-
+
Now focus on only the red circles, as shown in the right half of the figure. We build a new rooted tree \(T_{2}\) where each red circle is joined to the red circle below it if there was a path of (zero or more) white circles in \(T_{1}\) between them. Then each red circle has at most 2 direct descendants in \(T_{2}\) . Hence the depth of the new tree \(T_{2}\) exceeds \(\log_2(n)\) , which produces the desired path.
Another recursive proof of bound, communicated by Helio Ng. We give another proof that \(\lfloor \log_2n\rfloor +1\) is always achievable. Define \(f(i,j)\) to be the maximum number of red circles contained in the portion of a ninja path from \((1,1)\) to \((i,j)\) , including the endpoints \((1,1)\) and \((i,j)\) . (If \((i,j)\) is not a valid circle in the triangle, define \(f(i,j) = 0\) for convenience.) An example is shown below with the values of \(f(i,j)\) drawn in the circles.
-
+
@@ -378,7 +378,7 @@ so that
In particular, \(\max (\alpha ,\beta ,\gamma)< 30^{\circ}\) , so it follows that \(A_{1}\) lies inside \(\triangle O B C\) , and similarly for the others. This means for example that \(C_{1}\) lies between \(B\) and \(A_{2}\) , and so on. Therefore the polygon \(A_{2}C_{1}B_{2}A_{1}C_{2}B_{1}\) is convex.
-
+
We start by providing the "interpretation" for the \(480^{\circ}\) angle in the statement:
diff --git a/IMO/md/en-IMO-2024-notes.md b/IMO/md/en-IMO-2024-notes.md
index 0405dc7368c0748f64018a6995c5d42a5ba417e2..10a57bcb861b17b2a43fa50edfad6a74df2c2d62 100644
--- a/IMO/md/en-IMO-2024-notes.md
+++ b/IMO/md/en-IMO-2024-notes.md
@@ -183,7 +183,7 @@ Then there are \(N\) initial blocks placed, colored red. The rest of the blocks
In other words, the yellow blocks \(B_{i}\) for \(i > N\) are given coordinates \(B_{i} = (a_{i},a_{i + 1})\) for \(i > N\) . Note in particular that in towers \(M + 1\) , \(M + 2\) , ..., the blocks are all yellow.
-
+
We let \(h_{\ell}\) denote the height of the \(\ell^{\mathrm{th}}\) tower at a given time \(n\) . (This is an abuse of notation and we should write \(h_{\ell}(n)\) at time \(n\) , but \(n\) will always be clear from context.)
@@ -191,7 +191,7 @@ We let \(h_{\ell}\) denote the height of the \(\ell^{\mathrm{th}}\) tower at a g
\(\P\) Up to alternating up and down. We start with two independent easy observations: the set of numbers that occur infinitely often is downwards closed, and consecutive terms cannot both be huge.
-
+
Claim — If the \((k + 1)^{\mathrm{st}}\) tower grows arbitrarily high, so does tower \(k\) . In fact, there exists a constant \(C\) such that \(h_{k} \geq h_{k + 1} - C\) at all times.
@@ -228,7 +228,7 @@ For example, in the figure below, we illustrate how
\[S(34) = (9, 11; a_{34} = 1) \longrightarrow S(36) = (9, 12; a_{36} = 2)\]
-
+
The final element \(a_{n}\) simply reminds us which tower was most recently incremented. At this point we can give a complete description of how to move from \(S(n)\) to \(S(n + 2)\) :
@@ -289,7 +289,7 @@ Proof. The first part is true since \(\triangle BTC\) is the image of \(\triangl
We thus delete \(K\) and \(L\) from the picture altogether; they aren't needed anymore.
-
+
@@ -326,7 +326,7 @@ Surprisingly the answer is \(n = 3\) for any grid size \(s\times (s - 1)\) when
Case where \(M_{1}\) is not on the edge. In the first case, if that monster \(M_{1}\) is not on the edge of the row, then Turbo can trace two paths below it as shown below. At least one of these paths works, hence three attempts is sufficient.
-
+
Case where \(M_{1}\) is on the edge. WLOG, \(M_{1}\) is in the leftmost cell. Then Turbo follows the green staircase pattern shown in the left figure below. If the staircase is free
@@ -335,7 +335,7 @@ Case where \(M_{1}\) is on the edge. WLOG, \(M_{1}\) is in the leftmost cell. Th
of monsters, then Turbo wins on the second attempt. Otherwise, if a monster \(M_{2}\) is encountered on the staircase, Turbo has found a safe path to the left of \(M_{2}\) ; then Turbo can use this to reach the column \(M_{1}\) is in, and escape from there. This is shown in purple in the center and right figure (there are two slightly different cases depending on whether \(M_{2}\) was encountered going east or south).
-
+
Thus the problem is solved in three attempts, as promised.
@@ -348,7 +348,7 @@ All \(s - 1\) cells in the first row (which has no monsters). The center \(s - 3
For \(s = 12\) , the happy triangle is the region shaded in the thick border below.
-
+
@@ -359,12 +359,12 @@ Then solutions roughly must distinguish between these two cases:
- Inside happy triangle: If the first monster \(M_{1}\) is found in the happy triangle, and there is a safe path found by Turbo to the two shoulders (marked \(\bigstar\) in the figure), then one can finish in two more moves by considering the two paths from \(\bigstar\) that cut under the monster \(M_{1}\) ; one of them must work. This slightly generalizes the easier case in the solution above (which focuses only on the case where \(M_{1}\) is in the first row).
-
+
- Outside happy triangle: Now suppose the first monster \(M_{1}\) is outside the happy triangle. Of the two shoulders, take the one closer to the center (if in the center column, either one works; if only one shoulder, use it). If there is a safe path to that shoulder, then one can take a staircase pattern towards the center, as shown in the figure. In that case, the choice of shoulder and position guarantees the staircase reaches the bottom row, so that if no monster is along this path, the algorithm ends. Otherwise, if one encounters a second monster along the staircase, then one can use the third trial to cut under the monster \(M_{1}\) .
-
+
We now prove the following proposition: in any valid strategy for Turbo, in the case where Turbo first encounters a monster upon leaving the happy triangle, the second path must outline the same staircase shape.
@@ -373,12 +373,12 @@ We now prove the following proposition: in any valid strategy for Turbo, in the
The monsters pre- commit to choosing their pattern to be either a NW- SE diagonal or NE- SW diagonal, with a single one- column gap; see figure below for an example. Note that this forces any valid path for Turbo to pass through the particular gap.
-
+
We may assume without loss of generality that Turbo first encounters a monster \(M_{1}\) when Turbo first leaves the happy triangle, and that this forces an NW- SE configuration.
-
+
Then the following is true:
diff --git a/IMO/md/en-IMO-2025-notes.md b/IMO/md/en-IMO-2025-notes.md
index 6285da04f40f2f8ccddae8991399143a73f737ad..ec650d450a41aec4c9c282f937dd43b48b533a1d 100644
--- a/IMO/md/en-IMO-2025-notes.md
+++ b/IMO/md/en-IMO-2025-notes.md
@@ -105,11 +105,11 @@ Hence, by induction we may repeatedly delete a long line without changing the nu
We now classify all the ways to cover the \(1 + 2 + 3 = 6\) points in an \(n = 3\) grid with 3 lines.
-
+
Long line present
-
+
No long line
@@ -150,7 +150,7 @@ Proof. \(\angle AEC = \angle ABC = \angle CAB = 90^{\circ} - \alpha\) .
Hence, if we let \(A' := \overline{CE} \cap \overline{DF}\) , we have a parallelogram \(ACA'D\) . Note in particular that \(\overline{BA'} \parallel \overline{CD}\) .
-
+
Next, let \(T\) denote the circumcenter of \(\triangle A^{\prime}EF\) . (This will be the tangency point later in the problem.)
@@ -435,7 +435,7 @@ Construction. We show a general construction when \(n = k^{2}\) illustrated belo
tiles as promised.
-
+
@@ -457,7 +457,7 @@ To apply this to the present problem, take the \(n\) uncovered squares which we
The figure below shows two examples of the process, each for a board with \(n = 9\) , for two choices of LIS and LDS. The cells in the LIS and LDS have been marked with green circles, and the boundaries of the quadrants are drawn in green lines. In the left example, the LIS and LDS have a black square in common (that cell has all four directions labeled). In the right example, the LIS and do not have common squares.
-
+
We observe that:
diff --git a/IMO/segmented/en-IMO-2005-notes.jsonl b/IMO/segmented/en-IMO-2005-notes.jsonl
index cbba0fe2c3680669b46ec9c57d8e4199092eb174..93ab4a683762411aa30d5da992023fb3a8a3d673 100644
--- a/IMO/segmented/en-IMO-2005-notes.jsonl
+++ b/IMO/segmented/en-IMO-2005-notes.jsonl
@@ -1,6 +1,6 @@
-{"year": "2005", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "Six points are chosen on the sides of an equilateral triangle \\(A B C\\) .. \\(A_{1}\\) \\(A_{2}\\) on \\(B C\\) \\(B_{1}\\) \\(B_{2}\\) on \\(C A\\) and \\(C_{1}\\) \\(C_{2}\\) on \\(A B\\) , such that they are the vertices of a convex hexagon \\(A_{1}A_{2}B_{1}B_{2}C_{1}C_{2}\\) with equal side lengths. Prove that the lines \\(A_{1}B_{2}\\) \\(B_{1}C_{2}\\) and \\(C_{1}A_{2}\\) are concurrent.", "solution": "The six sides of the hexagon, when oriented, comprise six vectors with vanishing sum. However note that \n\n\\[\\overrightarrow{A_1A_2} +\\overrightarrow{B_1B_2} +\\overrightarrow{C_1C_2} = 0.\\] \n\nThus \n\n\\[\\overrightarrow{A_2B_1} +\\overrightarrow{B_2C_1} +\\overrightarrow{C_2A_1} = 0\\] \n\nand since three unit vectors with vanishing sum must be rotations of each other by \\(120^{\\circ}\\) , it follows they must also form an equilateral triangle. \n\n\n \n\nConsequently, triangles \\(A_1A_2B_1\\) , \\(B_1B_2C_1\\) , \\(C_1C_2A_1\\) are congruent, as \\(\\angle A_2 = \\angle B_2 = \\angle C_2\\) . So triangle \\(A_1B_1C_1\\) is equilateral and the diagonals are concurrent at the center.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2005-notes.jsonl", "problem_match": "1. ", "solution_match": "## \\(\\S 1.1\\) IMO 2005/1, proposed by Bogdan Enescu (ROU) \n"}}
-{"year": "2005", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "Let \\(a_{1}\\) , \\(a_{2}\\) , ... be a sequence of integers with infinitely many positive and negative terms. Suppose that for every positive integer \\(n\\) the numbers \\(a_{1}\\) , \\(a_{2}\\) , ..., \\(a_{n}\\) leave \\(n\\) different remainders upon division by \\(n\\) . Prove that every integer occurs exactly once in the sequence.", "solution": "Obviously every integer appears at most once (otherwise take \\(n\\) much larger). So we will prove every integer appears at least once. \n\nClaim — For any \\(i < j\\) we have \\(|a_{i} - a_{j}| < j\\) . \n\nProof. Otherwise, let \\(n = |a_{i} - a_{j}| \\neq 0\\) . Then \\(i, j \\in [1, n]\\) and \\(a_{i} \\equiv a_{j} \\pmod{n}\\) , contradiction. \n\nClaim — For any \\(n\\) , the set \\(\\{a_{1}, \\ldots , a_{n}\\}\\) is of the form \\(\\{k + 1, \\ldots , k + n\\}\\) for some integer \\(k\\) . \n\nProof. By induction, with the base case \\(n = 1\\) being vacuous. For the inductive step, suppose \\(\\{a_{1}, \\ldots , a_{n}\\} = \\{k + 1, \\ldots , k + n\\}\\) are determined. Then \n\n\\[a_{n + 1} \\equiv k \\pmod{n + 1}.\\] \n\nMoreover by the earlier claim we have \n\n\\[|a_{n + 1} - a_{1}| < n + 1.\\] \n\nFrom this we deduce \\(a_{n + 1} \\in \\{k, k + n + 1\\}\\) as desired. \n\nThis gives us actually a complete description of all possible sequences satisfying the hypothesis: choose any value of \\(a_{1}\\) to start. Then, for the \\(n\\) th term, the set \\(S = \\{a_{1}, \\ldots , a_{n - 1}\\}\\) is (in some order) a set of \\(n - 1\\) consecutive integers. We then let \\(a_{n} = \\max S + 1\\) or \\(a_{n} = \\min S - 1\\) . A picture of six possible starting terms is shown below. \n\n\n\n\n\n\nFinally, we observe that the condition that the sequence has infinitely many positive and negative terms (which we have not used until now) implies it is unbounded above and below. Thus it must contain every integer.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2005-notes.jsonl", "problem_match": "2. ", "solution_match": "## \\(\\S 1.2\\) IMO 2005/2, proposed by Nicholas de Bruijn (NLD) \n"}}
+{"year": "2005", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "Six points are chosen on the sides of an equilateral triangle \\(A B C\\) .. \\(A_{1}\\) \\(A_{2}\\) on \\(B C\\) \\(B_{1}\\) \\(B_{2}\\) on \\(C A\\) and \\(C_{1}\\) \\(C_{2}\\) on \\(A B\\) , such that they are the vertices of a convex hexagon \\(A_{1}A_{2}B_{1}B_{2}C_{1}C_{2}\\) with equal side lengths. Prove that the lines \\(A_{1}B_{2}\\) \\(B_{1}C_{2}\\) and \\(C_{1}A_{2}\\) are concurrent.", "solution": "The six sides of the hexagon, when oriented, comprise six vectors with vanishing sum. However note that \n\n\\[\\overrightarrow{A_1A_2} +\\overrightarrow{B_1B_2} +\\overrightarrow{C_1C_2} = 0.\\] \n\nThus \n\n\\[\\overrightarrow{A_2B_1} +\\overrightarrow{B_2C_1} +\\overrightarrow{C_2A_1} = 0\\] \n\nand since three unit vectors with vanishing sum must be rotations of each other by \\(120^{\\circ}\\) , it follows they must also form an equilateral triangle. \n\n\n \n\nConsequently, triangles \\(A_1A_2B_1\\) , \\(B_1B_2C_1\\) , \\(C_1C_2A_1\\) are congruent, as \\(\\angle A_2 = \\angle B_2 = \\angle C_2\\) . So triangle \\(A_1B_1C_1\\) is equilateral and the diagonals are concurrent at the center.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2005-notes.jsonl", "problem_match": "1. ", "solution_match": "## \\(\\S 1.1\\) IMO 2005/1, proposed by Bogdan Enescu (ROU) \n"}}
+{"year": "2005", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "Let \\(a_{1}\\) , \\(a_{2}\\) , ... be a sequence of integers with infinitely many positive and negative terms. Suppose that for every positive integer \\(n\\) the numbers \\(a_{1}\\) , \\(a_{2}\\) , ..., \\(a_{n}\\) leave \\(n\\) different remainders upon division by \\(n\\) . Prove that every integer occurs exactly once in the sequence.", "solution": "Obviously every integer appears at most once (otherwise take \\(n\\) much larger). So we will prove every integer appears at least once. \n\nClaim — For any \\(i < j\\) we have \\(|a_{i} - a_{j}| < j\\) . \n\nProof. Otherwise, let \\(n = |a_{i} - a_{j}| \\neq 0\\) . Then \\(i, j \\in [1, n]\\) and \\(a_{i} \\equiv a_{j} \\pmod{n}\\) , contradiction. \n\nClaim — For any \\(n\\) , the set \\(\\{a_{1}, \\ldots , a_{n}\\}\\) is of the form \\(\\{k + 1, \\ldots , k + n\\}\\) for some integer \\(k\\) . \n\nProof. By induction, with the base case \\(n = 1\\) being vacuous. For the inductive step, suppose \\(\\{a_{1}, \\ldots , a_{n}\\} = \\{k + 1, \\ldots , k + n\\}\\) are determined. Then \n\n\\[a_{n + 1} \\equiv k \\pmod{n + 1}.\\] \n\nMoreover by the earlier claim we have \n\n\\[|a_{n + 1} - a_{1}| < n + 1.\\] \n\nFrom this we deduce \\(a_{n + 1} \\in \\{k, k + n + 1\\}\\) as desired. \n\nThis gives us actually a complete description of all possible sequences satisfying the hypothesis: choose any value of \\(a_{1}\\) to start. Then, for the \\(n\\) th term, the set \\(S = \\{a_{1}, \\ldots , a_{n - 1}\\}\\) is (in some order) a set of \\(n - 1\\) consecutive integers. We then let \\(a_{n} = \\max S + 1\\) or \\(a_{n} = \\min S - 1\\) . A picture of six possible starting terms is shown below. \n\n\n\n\n\n\nFinally, we observe that the condition that the sequence has infinitely many positive and negative terms (which we have not used until now) implies it is unbounded above and below. Thus it must contain every integer.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2005-notes.jsonl", "problem_match": "2. ", "solution_match": "## \\(\\S 1.2\\) IMO 2005/2, proposed by Nicholas de Bruijn (NLD) \n"}}
{"year": "2005", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "Let \\(x,y,z > 0\\) satisfy \\(x y z\\geq 1\\) . Prove that \n\n\\[\\frac{x^{5} - x^{2}}{x^{5} + y^{2} + z^{2}} +\\frac{y^{5} - y^{2}}{x^{2} + y^{5} + z^{2}} +\\frac{z^{5} - z^{2}}{x^{2} + y^{2} + z^{5}}\\geq 0.\\]", "solution": "Negating both sides and adding 3 eliminates the minus signs: \n\n\\[\\sum_{\\mathrm{cyc}}\\frac{1}{x^{5} + y^{2} + z^{2}}\\leq \\frac{3}{x^{2} + y^{2} + z^{2}}.\\] \n\nThus we only need to consider the case \\(x y z = 1\\) \n\nDirect expansion and Muirhead works now! As advertised, once we show it suffices to analyze if \\(x y z = 1\\) the inequality becomes more economically written as \n\n\\[S = \\sum_{\\mathrm{cyc}}x^{2}(x^{2} - y z)(y^{4} + x^{3}z + x z^{3})(z^{4} + x^{3}y + x y^{3})\\stackrel {?}{\\geq}0.\\] \n\nSo, clearing all the denominators gives \n\n\\[S = \\sum_{\\mathrm{cyc}}x^{2}(x^{2} - y z)\\left[y^{4}z^{4} + x^{3}y^{5} + x y^{7} + x^{3}z^{5} + x^{6}y z + x^{4}y^{3}z + x z^{7} + x^{4}y z^{3} + x^{2}y^{3}z^{3}\\right]\\] \\[\\quad = \\sum_{\\mathrm{cyc}}\\left[x^{4}y^{4}z^{4} + x^{7}y^{5} + x^{5}y^{7} + x^{7}z^{5} + x^{10}y z + x^{8}y^{3}z + x^{5}z^{7} + x^{8}y z^{3} + x^{6}y^{3}z^{3}\\right]\\] \\[\\quad -\\sum_{\\mathrm{cyc}}\\left[x^{2}y^{5}z^{5} + x^{5}y^{6}z + x^{3}y^{8}z + x^{5}y z^{6} + x^{8}y^{2}z^{2} + x^{6}y^{4}z^{2} + x^{3}y z^{8} + x^{6}y^{2}z^{4} + x^{4}y^{4}z^{4}\\right]\\] \\[\\quad = \\sum_{\\mathrm{cyc}}\\left[x^{7}y^{5} + x^{5}y^{7} + x^{7}z^{5} + x^{10}y z + x^{5}z^{7} + x^{6}y^{3}z^{3}\\right]\\] \\[\\quad -\\sum_{\\mathrm{cyc}}\\left[x^{2}y^{5}z^{5} + x^{5}y^{6}z + x^{5}y z^{6} + x^{8}y^{2}z^{2} + x^{6}y^{4}z^{2} + x^{6}y^{2}z^{4}\\right]\\] \n\nIn other words we need to show \n\n\\[\\sum_{\\mathrm{sym}}\\left(2x^{7}y^{5} + \\frac{1}{2} x^{10}y z + \\frac{1}{2} x^{6}y^{3}z^{3}\\right)\\geq \\sum_{\\mathrm{sym}}\\left(\\frac{1}{2} x^{8}y^{2}z^{2} + \\frac{1}{2} x^{5}y^{5}z^{2} + x^{6}y^{4}z^{2} + x^{6}y^{5}z\\right).\\] \n\nwhich follows by summing \n\n\\[\\sum_{\\mathrm{sym}}\\frac{x^{10}y z + x^{6}y^{3}z^{3}}{2}\\geq \\sum_{\\mathrm{sym}}x^{8}y^{2}z^{2}\\] \\[\\frac{1}{2}\\sum_{\\mathrm{sym}}x^{8}y^{2}z^{2}\\geq \\frac{1}{2}\\sum_{\\mathrm{sym}}x^{6}y^{4}z^{2}\\] \\[\\frac{1}{2}\\sum_{\\mathrm{sym}}x^{7}y^{5}\\geq \\frac{1}{2}\\sum_{\\mathrm{sym}}x^{5}y^{5}z^{2}\\]\n\n\n\n\\[{\\frac{1}{2}}\\sum_{\\mathrm{sym}}x^{7}y^{5}\\geq{\\frac{1}{2}}\\sum_{\\mathrm{sym}}x^{6}y^{4}z^{2}\\] \\[\\sum_{\\mathrm{sym}}x^{7}y^{5}\\geq\\sum_{\\mathrm{sym}}x^{6}y^{5}z.\\] \n\nThe first line here comes from AM- GM, the rest come from Muirhead. \n\nRemark. More elegant approach is to use Cauchy in the form \n\n\\[\\frac{1}{x^{5} + y^{2} + z^{2}} \\leq \\frac{x^{-1} + y^{2} + z^{2}}{(x^{2} + y^{2} + z^{2})^{2}}.\\]\n\n\n\n## \\(\\S 2\\) Solutions to Day 2", "metadata": {"resource_path": "IMO/segmented/en-IMO-2005-notes.jsonl", "problem_match": "3. ", "solution_match": "## \\(\\S 1.3\\) IMO 2005/3, proposed by Hojoo Lee (KOR) \n"}}
{"year": "2005", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "Determine all positive integers relatively prime to all the terms of the infinite sequence \n\n\\[a_{n} = 2^{n} + 3^{n} + 6^{n} - 1,\\quad n\\geq 1.\\]", "solution": "The answer is 1 only (which works). \n\nIt suffices to show there are no primes. For the primes \\(p = 2\\) and \\(p = 3\\) , take \\(a_{2} = 48\\) . For any prime \\(p \\geq 5\\) notice that \n\n\\[a_{p - 2} = 2^{p - 2} + 3^{p - 2} + 6^{p - 2} - 1\\] \\[\\equiv \\frac{1}{2} +\\frac{1}{3} +\\frac{1}{6} -1\\pmod {p\\] \\[\\equiv 0\\pmod {p\\] \n\nso no other larger prime works.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2005-notes.jsonl", "problem_match": "4. ", "solution_match": "## \\(\\S 2.1\\) IMO 2005/4, proposed by Mariusz Skalba (POL) \n"}}
-{"year": "2005", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "Let \\(A B C D\\) be a fixed convex quadrilateral with \\(B C = D A\\) and \\(\\overline{{B C}}\\not\\parallel\\overline{{D A}}\\) . Let two variable points \\(E\\) and \\(F\\) lie on the sides \\(B C\\) and \\(D A\\) , respectively, and satisfy \\(B E = D F\\) . The lines \\(A C\\) and \\(B D\\) meet at \\(P\\) , the lines \\(B D\\) and \\(E F\\) meet at \\(Q\\) , the lines \\(E F\\) and \\(A C\\) meet at \\(R\\) . Prove that the circumcircles of the triangles \\(P Q R\\) , as \\(E\\) and \\(F\\) vary, have a common point other than \\(P\\) .", "solution": "Let \\(M\\) be the Miquel point of complete quadrilateral \\(A D B C\\) ; in other words, let \\(M\\) be the second intersection point of the circumcircles of \\(\\triangle A P D\\) and \\(\\triangle B P C\\) . (A good diagram should betray this secret; all the points are given in the picture.) This makes lots of sense since we know \\(E\\) and \\(F\\) will be sent to each other under the spiral similarity too. \n\n\n \n\nThus \\(M\\) is the Miquel point of complete quadrilateral \\(F A C E\\) . As \\(R = \\overline{{F E}}\\cap \\overline{{A C}}\\) we deduce \\(F A R M\\) is a cyclic quadrilateral (among many others, but we'll only need one). \n\nNow look at complete quadrilateral \\(A F Q P\\) . Since \\(M\\) lies on \\((D F Q)\\) and \\((R A F)\\) , it follows that \\(M\\) is in fact the Miquel point of \\(A F Q P\\) as well. So \\(M\\) lies on \\((P Q R)\\) . \n\nThus \\(M\\) is the fixed point that we wanted. \n\nRemark. Naturally, the congruent length condition can be relaxed to \\(D F / D A = B E / B C\\) .", "metadata": {"resource_path": "IMO/segmented/en-IMO-2005-notes.jsonl", "problem_match": "5. ", "solution_match": "## \\(\\S 2.2\\) IMO 2005/5, proposed by Waldemar Pompe (POL) \n"}}
+{"year": "2005", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "Let \\(A B C D\\) be a fixed convex quadrilateral with \\(B C = D A\\) and \\(\\overline{{B C}}\\not\\parallel\\overline{{D A}}\\) . Let two variable points \\(E\\) and \\(F\\) lie on the sides \\(B C\\) and \\(D A\\) , respectively, and satisfy \\(B E = D F\\) . The lines \\(A C\\) and \\(B D\\) meet at \\(P\\) , the lines \\(B D\\) and \\(E F\\) meet at \\(Q\\) , the lines \\(E F\\) and \\(A C\\) meet at \\(R\\) . Prove that the circumcircles of the triangles \\(P Q R\\) , as \\(E\\) and \\(F\\) vary, have a common point other than \\(P\\) .", "solution": "Let \\(M\\) be the Miquel point of complete quadrilateral \\(A D B C\\) ; in other words, let \\(M\\) be the second intersection point of the circumcircles of \\(\\triangle A P D\\) and \\(\\triangle B P C\\) . (A good diagram should betray this secret; all the points are given in the picture.) This makes lots of sense since we know \\(E\\) and \\(F\\) will be sent to each other under the spiral similarity too. \n\n\n \n\nThus \\(M\\) is the Miquel point of complete quadrilateral \\(F A C E\\) . As \\(R = \\overline{{F E}}\\cap \\overline{{A C}}\\) we deduce \\(F A R M\\) is a cyclic quadrilateral (among many others, but we'll only need one). \n\nNow look at complete quadrilateral \\(A F Q P\\) . Since \\(M\\) lies on \\((D F Q)\\) and \\((R A F)\\) , it follows that \\(M\\) is in fact the Miquel point of \\(A F Q P\\) as well. So \\(M\\) lies on \\((P Q R)\\) . \n\nThus \\(M\\) is the fixed point that we wanted. \n\nRemark. Naturally, the congruent length condition can be relaxed to \\(D F / D A = B E / B C\\) .", "metadata": {"resource_path": "IMO/segmented/en-IMO-2005-notes.jsonl", "problem_match": "5. ", "solution_match": "## \\(\\S 2.2\\) IMO 2005/5, proposed by Waldemar Pompe (POL) \n"}}
{"year": "2005", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "In a mathematical competition 6 problems were posed to the contestants. Each pair of problems was solved by more than \\(\\frac{2}{5}\\) of the contestants. Nobody solved all 6 problems. Show that there were at least 2 contestants who each solved exactly 5 problems.", "solution": "Assume not and at most one contestant solved five problems. By adding in solves, we can assume WLOG that one contestant solved problems one through five, and every other contestant solved four of the six problems. \n\nWe split the remaining contestants based on whether they solved P6. Let \\(a_{i}\\) denote the number of contestants who solved \\(\\{1,2,\\ldots ,5\\} \\setminus \\{i\\}\\) (and missed P6). Let \\(b_{ij}\\) denote the number of contestants who solved \\(\\{1,2,\\ldots ,5,6\\} \\setminus \\{i,j\\}\\) , for \\(1\\leq i< j\\leq 5\\) (thus in particular they solved P6). Thus \n\n\\[n = 1 + \\sum_{1\\leq i\\leq 5}a_{i} + \\sum_{1\\leq i< j\\leq 5}b_{ij}\\] \n\ndenotes the total number of contestants. \n\nConsidering contestants who solved P1/P6 we have \n\n\\[t_{1}:= b_{23} + b_{24} + b_{25} + b_{34} + b_{35} + b_{45}\\geq \\frac{2}{5} n + \\frac{1}{5}\\] \n\nand we similarly define \\(t_{2}\\) , \\(t_{3}\\) , \\(t_{4}\\) , \\(t_{5}\\) . (We have written \\(\\frac{2}{5} n + \\frac{1}{5}\\) since we know the left- hand side is an integer strictly larger than \\(\\frac{2}{5} n\\) .) Also, by considering contestants who solved P1/P2 we have \n\n\\[t_{12} = 1 + a_{3} + a_{4} + a_{5} + b_{34} + b_{35} + b_{45}\\geq \\frac{2}{5} n + \\frac{1}{5}\\] \n\nand we similarly define \\(t_{ij}\\) for \\(1\\leq i< j\\leq 5\\) \n\nClaim — The number \\(\\frac{2n + 1}{5}\\) is equal to some integer \\(k\\) , fourteen of the \\(t\\) 's are equal to \\(k\\) , and the last one is equal to \\(k + 1\\) . \n\nProof. First, summing all fifteen equations gives \n\n\\[6n + 4 = 10 + 6(n - 1) = 10 + \\sum_{1\\leq i\\leq 5}6a_{i} + \\sum_{1\\leq i< j\\leq 5}6b_{ij}\\] \\[\\qquad = \\sum_{1\\leq i\\leq 5}t_{i} + \\sum_{1\\leq i< j\\leq 5}t_{ij}.\\] \n\nThus the sum of the 15 \\(t\\) 's is \\(6n + 4\\) . But since all the \\(t\\) 's are integers at least \\(\\frac{2n + 1}{5} = \\frac{6n + 3}{15}\\) , the conclusion follows. \\(\\square\\) \n\nHowever, we will also manipulate the equations to get the following.\n\n\n\nClaim — We have \n\n\\[t_{45} \\equiv 1 + t_{1} + t_{2} + t_{3} + t_{12} + t_{23} + t_{31} \\pmod {3}.\\] \n\nProof. This follows directly by computing the coefficient of the \\(a\\) 's and \\(b\\) 's. We will nonetheless write out a derivation of this equation, to motivate it, but the proof stands without it. \n\nLet \\(B = \\sum_{1 \\leq i < j \\leq 5} b_{ij}\\) be the sum of all \\(b\\) 's. First, note that \n\n\\[t_{1} + t_{2} = B + b_{34} + b_{45} + b_{35} - b_{12}\\] \\[\\qquad = B + (t_{12} - 1 - a_{3} - a_{4} - a_{5}) - b_{12}\\] \\[\\Rightarrow b_{12} = B - (t_{1} + t_{2}) + t_{12} - 1 - (a_{3} + a_{4} + a_{5}).\\] \n\nThis means we have more or less solved for each \\(b_{ij}\\) in terms of only \\(t\\) and \\(a\\) variables. Now \n\n\\[t_{45} = 1 + a_{1} + a_{2} + a_{3} + b_{12} + b_{23} + b_{31}\\] \\[\\qquad = 1 + a_{1} + a_{2} + a_{3}\\] \\[\\qquad +[B - (t_{1} + t_{2}) + t_{12} - 1 - (a_{3} + a_{4} + a_{5})]\\] \\[\\qquad +[B - (t_{2} + t_{3}) + t_{23} - 1 - (a_{1} + a_{4} + a_{5})]\\] \\[\\qquad +[B - (t_{3} + t_{1}) + t_{13} - 1 - (a_{2} + a_{4} + a_{5})]\\] \\[\\qquad \\equiv 1 + t_{1} + t_{2} + t_{3} + t_{12} + t_{23} + t_{31} \\pmod {3}\\] \n\nas desired. \n\nHowever, we now show the two claims are incompatible (and this is easy, many ways to do this). There are two cases. \n\n- Say \\(t_{5} = k + 1\\) and the others are \\(k\\) . Then the equation for \\(t_{45}\\) gives that \\(k \\equiv 6k + 1 \\pmod {3}\\) . But now the equation for \\(t_{12}\\) give \\(k \\equiv 6k \\pmod {3}\\) . \n\n- Say \\(t_{45} = k + 1\\) and the others are \\(k\\) . Then the equation for \\(t_{45}\\) gives that \\(k + 1 \\equiv 6k \\pmod {3}\\) . But now the equation for \\(t_{12}\\) give \\(k \\equiv 6k + 1 \\pmod {3}\\) . \n\nRemark. It is significantly easier to prove that there is at least one contestant who solved five problems. One can see it by dropping the \\(+10\\) in the proof of the claim, and arrives at a contradiction. In this situation it is not even necessary to set up the many \\(a\\) and \\(b\\) variables; just note that the expected number of contestants solving any particular pair of problems is \\(\\frac{\\binom{4}{2}n}{\\binom{6}{2}} = \\frac{2}{5} n\\) . \n\nThe fact that \\(\\frac{2n + 1}{5}\\) should be an integer also follows quickly, since if not one can improve the bound to \\(\\frac{2n + 5}{5}\\) and quickly run into a contradiction. Again one can get here without setting up \\(a\\) and \\(b\\) . \n\nThe main difficulty seems to be the precision required in order to nail down the second 5- problem solve. \n\nRemark. The second claim may look miraculous, but the proof shows that it is not too unnatural to consider \\(t_{1} + t_{2} - t_{12}\\) to isolate \\(b_{12}\\) in terms of \\(a\\) 's and \\(t\\) 's. The main trick is: why mod 3? \n\nThe reason is that if one looks closely, for a fixed \\(k\\) we have a system of 15 equations in 15 variables. Unless the determinant \\(D\\) of that system happens to be zero, this means there will be a rational solution in \\(a\\) and \\(b\\) , whose denominators are bounded by \\(D\\) . However if\n\n\n\n\\(p\\mid D\\) then we may conceivably run into mod \\(p\\) issues. \n\nThis motivates the choice \\(p = 3\\) , since it is easy to see the determinant is divisible by 3, since constant shifts of \\(\\vec{a}\\) and \\(\\vec{b}\\) are also solutions mod 3. (The choice \\(p = 2\\) is a possible guess as well for this reason, but the problem seems to have better 3- symmetry.)", "metadata": {"resource_path": "IMO/segmented/en-IMO-2005-notes.jsonl", "problem_match": "6. ", "solution_match": "## \\(\\S 2.3\\) IMO 2005/6, proposed by Radu Gologan, Dan Schwartz (ROU) \n"}}
diff --git a/IMO/segmented/en-IMO-2006-notes.jsonl b/IMO/segmented/en-IMO-2006-notes.jsonl
index 4051410b33d799cb8cb554746f3449620b13dd2c..21fc88280f24e22a2f47fca0cd9ea51d5d77ee23 100644
--- a/IMO/segmented/en-IMO-2006-notes.jsonl
+++ b/IMO/segmented/en-IMO-2006-notes.jsonl
@@ -3,4 +3,4 @@
{"year": "2006", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "Determine the least real number \\(M\\) such that the inequality \n\n\\[\\left|a b(a^{2} - b^{2}) + b c(b^{2} - c^{2}) + c a(c^{2} - a^{2})\\right|\\leq M\\left(a^{2} + b^{2} + c^{2}\\right)^{2}\\] \n\nholds for all real numbers \\(a\\) , \\(b\\) and \\(c\\) .", "solution": "It's the same as \n\n\\[|(a - b)(b - c)(c - a)(a + b + c)|\\leq M\\left(a^{2} + b^{2} + c^{2}\\right)^{2}.\\] \n\nLet \\(x = a - b\\) , \\(y = b - c\\) , \\(z = c - a\\) , \\(s = a + b + c\\) . Then we want to have \n\n\\[|xyz s|\\leq \\frac{M}{9} (x^{2} + y^{2} + z^{2} + s^{2})^{2}.\\] \n\nHere \\(x + y + z = 0\\) . \n\nNow if \\(x\\) and \\(y\\) have the same sign, we can replace them with the average (this increases the LHS and decreases RHS). So we can have \\(x = y\\) , \\(z = - 2x\\) . Now WLOG \\(x > 0\\) to get \n\n\\[2x^{3}\\cdot s\\leq \\frac{M}{9}\\left(6x^{2} + s^{2}\\right)^{2}.\\] \n\nAfter this routine calculation gives \\(M = \\frac{9}{12}\\sqrt{2}\\) works and is optimal (by \\(6x^{2} + s^{2} = 2x^{2} + 2x^{2} + 2x^{2} + s^{2}\\) and AM- GM).\n\n\n\n## \\(\\S 2\\) Solutions to Day 2", "metadata": {"resource_path": "IMO/segmented/en-IMO-2006-notes.jsonl", "problem_match": "3. ", "solution_match": "## \\(\\S 1.3\\) IMO 2006/3, proposed by Finbarr Holland (IRL) \n"}}
{"year": "2006", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "Determine all pairs \\((x,y)\\) of integers such that \n\n\\[1 + 2^{x} + 2^{2x + 1} = y^{2}.\\]", "solution": "Answers: \\((0,\\pm 2)\\) , \\((4,\\pm 23)\\) , which work. \n\nAssume \\(x \\geq 4\\) . \n\n\\[2^{x}\\left(1 + 2^{x + 1}\\right) = 2^{x} + 2^{2x + 1} = y^{2} - 1 = (y - 1)(y + 1).\\] \n\nSo either: \n\n- \\(y = 2^{x - 1}m + 1\\) for some odd \\(m\\) , so \n\n\\[1 + 2^{x + 1} = m\\left(2^{x - 2}m + 1\\right)\\Longrightarrow 2^{x} = \\frac{4(1 - m)}{m^{2} - 8}.\\] \n\n- \\(y = 2^{x - 1}m - 1\\) for some odd \\(m\\) , so \n\n\\[1 + 2^{x + 1} = m\\left(2^{x - 2}m - 1\\right)\\Longrightarrow 2^{x} = \\frac{4(1 + m)}{m^{2} - 8}.\\] \n\nIn particular we need \\(4|1\\pm m|\\geq 2^{4}|m^{2} - 8|\\) , which is enough to imply \\(m< 5\\) . From here easily recover \\(x = 4\\) , \\(m = 3\\) as the last solution (in the second case).", "metadata": {"resource_path": "IMO/segmented/en-IMO-2006-notes.jsonl", "problem_match": "4. ", "solution_match": "## \\(\\S 2.1\\) IMO 2006/4, proposed by Zuming Feng (USA) \n"}}
{"year": "2006", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "Let \\(P(x)\\) be a polynomial of degree \\(n > 1\\) with integer coefficients and let \\(k\\) be a positive integer. Consider the polynomial \n\n\\[Q(x) = P(P(\\ldots P(P(x))\\ldots))\\] \n\nwhere \\(P\\) occurs \\(k\\) times. Prove that there are at most \\(n\\) integers \\(t\\) such that \\(Q(t) = t\\) .", "solution": "First, we prove that: \n\nClaim (Putnam 2000 et al) — If a number is periodic under \\(P\\) then in fact it's fixed by \\(P \\circ P\\) . \n\nProof. Let \\(x_{1}\\) , \\(x_{2}\\) , ..., \\(x_{n}\\) be a minimal orbit. Then \n\n\\[x_{i} - x_{i + 1} \\mid P(x_{i}) - P(x_{i + 1}) = x_{i + 1} - x_{i + 2}\\] \n\nand so on cyclically. \n\nIf any of the quantities are zero we are done. Else, we must eventually have \\(x_{i} - x_{i + 1} = -(x_{i + 1} - x_{i + 2})\\) , so \\(x_{i} = x_{i + 2}\\) and we get 2- periodicity. \n\nThe tricky part is to study the 2- orbits. Suppose there exists a fixed pair \\(u \\neq v\\) with \\(P(u) = v\\) , \\(P(v) = u\\) . (If no such pair exists, we are already done.) Let \\((a, b)\\) be any other pair with \\(P(a) = b\\) , \\(P(b) = a\\) , possibly even \\(a = b\\) , but \\(\\{a, b\\} \\cap \\{u, v\\} = \\emptyset\\) . Then we should have \n\n\\[u - a \\mid P(u) - P(a) = v - b \\mid P(v) - P(b) = u - a\\] \n\nand so \\(u - a\\) and \\(v - b\\) divide each other (and are nonzero). Similarly, \\(u - b\\) and \\(v - a\\) divide each other. \n\nHence \\(u - a = \\pm (v - b)\\) and \\(u - b = \\pm (v - a)\\) . We consider all four cases: \n\n- If \\(u - a = v - b\\) and \\(u - b = v - a\\) then \\(u - v = b - a = b\\) , contradiction. \n\n- If \\(u - a = -(v - b)\\) and \\(u - b = -(v - a)\\) then \\(u + v = u - v = a + b\\) . \n\n- If \\(u - a = -(v - b)\\) and \\(u - b = v - a\\) , we get \\(a + b = u + v\\) from the first one (discarding the second). \n\n- If \\(u - a = v - b\\) and \\(u - b = -(v - a)\\) , we get \\(a + b = u + v\\) from the second one (discarding the first one). \n\nThus in all possible situations we have \n\n\\[a + b = c := u + v\\] \n\na fixed constant. \n\nTherefore, any pair \\((a, b)\\) with \\(P(a) = b\\) and \\(P(b) = a\\) actually satisfies \\(P(a) = c - a\\) . And since \\(\\deg P > 1\\) , this means there are at most \\(n\\) roots to \\(a + P(a) = c\\) , as needed.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2006-notes.jsonl", "problem_match": "5. ", "solution_match": "## \\(\\S 2.2\\) IMO 2006/5, proposed by Dan Schwarz (ROU) \n"}}
-{"year": "2006", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "Assign to each side \\(b\\) of a convex polygon \\(P\\) the maximum area of a triangle that has \\(b\\) as a side and is contained in \\(P\\) . Show that the sum of the areas assigned to the sides of \\(P\\) is at least twice the area of \\(P\\) .", "solution": "We say a polygon in almost convex if all its angles are at most \\(180^{\\circ}\\) . \n\nNote that given any convex or almost convex polygon, we can take any side \\(b\\) and add another vertex on it, and the sum of the labels doesn't change (since the label of a side is the length of the side times the distance of the farthest point). \n\n## Lemma \n\nLet \\(N\\) be an even integer. Then any almost convex \\(N\\) - gon with area \\(S\\) should have an inscribed triangle with area at least \\(2S / N\\) . \n\nThe main work is the proof of the lemma. \n\nLabel the polygon \\(P_{0}P_{1}\\ldots P_{N - 1}\\) . Consider the \\(N / 2\\) major diagonals of the almost convex \\(N\\) - gon, \\(P_{0}P_{N / 2}\\) , \\(P_{1}P_{N / 2 + 1}\\) , et cetera. A butterfly refers to a self- intersecting quadrilateral \\(P_{i}P_{i + 1}P_{i + 1 + N / 2}P_{i + N / 2}\\) . An example of a butterfly is shown below for \\(N = 8\\) . \n\n\n \n\nClaim — Every point \\(X\\) in the polygon is contained in the wingspan of some butterfly. \n\nProof. Consider a windmill- like process which \n\n- starts from some oriented red line \\(P_{0}P_{N / 2}\\) , oriented to face \\(P_{0}P_{N / 2}\\)\n\n\n\n- rotates through \\(P_{0}P_{N / 2} \\cap P_{1}P_{N / 2 + 1}\\) to get line \\(P_{1}P_{N / 2 + 1}\\) , \n\n- rotates through \\(P_{1}P_{N / 2 + 1} \\cap P_{2}P_{N / 2 + 2}\\) to get line \\(P_{2}P_{N / 2 + 2}\\) , \n\n- ...et cetera, until returning to line \\(P_{N / 2}P_{0}\\) , but in the reverse orientation. \n\nAt the end of the process, every point in the plane has switched sides with our moving line. The moment that \\(X\\) crosses the moving red line, we get it contained in a butterfly, as needed. \\(\\square\\) \n\nClaim — If \\(A B D C = P_{i}P_{i + 1}P_{i + 1 + N / 2}P_{i + N / 2}\\) is a butterfly, one of the triangles \\(A B C\\) , \\(B C D\\) , \\(C D A\\) , \\(D A B\\) has area at least that of the butterfly. \n\nProof. Let the diagonals of the butterfly meet at \\(O\\) , and let \\(a = A O\\) , \\(b = B O\\) , \\(c = C O\\) , \\(d = D O\\) . If we assume WLOG \\(d = \\min (a,b,c,d)\\) then it follows \\([A B C] = [A O B] + [B O C] \\geq [A O B] + [C O D]\\) , as needed. \\(\\square\\) \n\nNow, since the \\(N / 2\\) butterflies cover an area of \\(S\\) , it follows that one of the butterflies has area at least \\(S / (N / 2) = 2S / N\\) , and so that butterfly gives a triangle with area at least \\(2S / N\\) , completing the proof of the lemma. \n\nMain proof. Let \\(a_{1}\\) , ..., \\(a_{n}\\) be the numbers assigned to the sides. Assume for contradiction \\(a_{1} + \\dots +a_{n}< 2S\\) . We pick even integers \\(m_{1}\\) , \\(m_{2}\\) , ..., \\(m_{n}\\) such that \n\n\\[\\frac{a_{1}}{S} < \\frac{2m_{1}}{m_{1} + \\dots +m_{n}}\\] \\[\\frac{a_{2}}{S} < \\frac{2m_{2}}{m_{1} + \\dots +m_{n}}\\] \\[\\vdots\\] \\[\\frac{a_{n}}{S} < \\frac{2m_{n}}{m_{1} + \\dots +m_{n}}.\\] \n\nwhich is possible by rational approximation, since the right- hand sides sum to 2 and the left- hand sides sum to strictly less than 2. \n\nNow we break every side of \\(P\\) into \\(m_{i}\\) equal parts to get an almost convex \\(N\\) - gon, where \\(N = m_{1} + \\dots +m_{n}\\) . \n\nThe main lemma then gives us a triangle \\(\\Delta\\) of the almost convex \\(N\\) - gon which has area at least \\(\\frac{2S}{N}\\) . If \\(\\Delta\\) used the \\(i\\) th side then it then follows the label \\(a_{i}\\) on that side should be at least \\(m_{i} \\cdot \\frac{2S}{N}\\) , contradiction.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2006-notes.jsonl", "problem_match": "6. ", "solution_match": "## \\(\\S 2.3\\) IMO 2006/6, proposed by Dušan Djukić (SRB) \n"}}
+{"year": "2006", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "Assign to each side \\(b\\) of a convex polygon \\(P\\) the maximum area of a triangle that has \\(b\\) as a side and is contained in \\(P\\) . Show that the sum of the areas assigned to the sides of \\(P\\) is at least twice the area of \\(P\\) .", "solution": "We say a polygon in almost convex if all its angles are at most \\(180^{\\circ}\\) . \n\nNote that given any convex or almost convex polygon, we can take any side \\(b\\) and add another vertex on it, and the sum of the labels doesn't change (since the label of a side is the length of the side times the distance of the farthest point). \n\n## Lemma \n\nLet \\(N\\) be an even integer. Then any almost convex \\(N\\) - gon with area \\(S\\) should have an inscribed triangle with area at least \\(2S / N\\) . \n\nThe main work is the proof of the lemma. \n\nLabel the polygon \\(P_{0}P_{1}\\ldots P_{N - 1}\\) . Consider the \\(N / 2\\) major diagonals of the almost convex \\(N\\) - gon, \\(P_{0}P_{N / 2}\\) , \\(P_{1}P_{N / 2 + 1}\\) , et cetera. A butterfly refers to a self- intersecting quadrilateral \\(P_{i}P_{i + 1}P_{i + 1 + N / 2}P_{i + N / 2}\\) . An example of a butterfly is shown below for \\(N = 8\\) . \n\n\n \n\nClaim — Every point \\(X\\) in the polygon is contained in the wingspan of some butterfly. \n\nProof. Consider a windmill- like process which \n\n- starts from some oriented red line \\(P_{0}P_{N / 2}\\) , oriented to face \\(P_{0}P_{N / 2}\\)\n\n\n\n- rotates through \\(P_{0}P_{N / 2} \\cap P_{1}P_{N / 2 + 1}\\) to get line \\(P_{1}P_{N / 2 + 1}\\) , \n\n- rotates through \\(P_{1}P_{N / 2 + 1} \\cap P_{2}P_{N / 2 + 2}\\) to get line \\(P_{2}P_{N / 2 + 2}\\) , \n\n- ...et cetera, until returning to line \\(P_{N / 2}P_{0}\\) , but in the reverse orientation. \n\nAt the end of the process, every point in the plane has switched sides with our moving line. The moment that \\(X\\) crosses the moving red line, we get it contained in a butterfly, as needed. \\(\\square\\) \n\nClaim — If \\(A B D C = P_{i}P_{i + 1}P_{i + 1 + N / 2}P_{i + N / 2}\\) is a butterfly, one of the triangles \\(A B C\\) , \\(B C D\\) , \\(C D A\\) , \\(D A B\\) has area at least that of the butterfly. \n\nProof. Let the diagonals of the butterfly meet at \\(O\\) , and let \\(a = A O\\) , \\(b = B O\\) , \\(c = C O\\) , \\(d = D O\\) . If we assume WLOG \\(d = \\min (a,b,c,d)\\) then it follows \\([A B C] = [A O B] + [B O C] \\geq [A O B] + [C O D]\\) , as needed. \\(\\square\\) \n\nNow, since the \\(N / 2\\) butterflies cover an area of \\(S\\) , it follows that one of the butterflies has area at least \\(S / (N / 2) = 2S / N\\) , and so that butterfly gives a triangle with area at least \\(2S / N\\) , completing the proof of the lemma. \n\nMain proof. Let \\(a_{1}\\) , ..., \\(a_{n}\\) be the numbers assigned to the sides. Assume for contradiction \\(a_{1} + \\dots +a_{n}< 2S\\) . We pick even integers \\(m_{1}\\) , \\(m_{2}\\) , ..., \\(m_{n}\\) such that \n\n\\[\\frac{a_{1}}{S} < \\frac{2m_{1}}{m_{1} + \\dots +m_{n}}\\] \\[\\frac{a_{2}}{S} < \\frac{2m_{2}}{m_{1} + \\dots +m_{n}}\\] \\[\\vdots\\] \\[\\frac{a_{n}}{S} < \\frac{2m_{n}}{m_{1} + \\dots +m_{n}}.\\] \n\nwhich is possible by rational approximation, since the right- hand sides sum to 2 and the left- hand sides sum to strictly less than 2. \n\nNow we break every side of \\(P\\) into \\(m_{i}\\) equal parts to get an almost convex \\(N\\) - gon, where \\(N = m_{1} + \\dots +m_{n}\\) . \n\nThe main lemma then gives us a triangle \\(\\Delta\\) of the almost convex \\(N\\) - gon which has area at least \\(\\frac{2S}{N}\\) . If \\(\\Delta\\) used the \\(i\\) th side then it then follows the label \\(a_{i}\\) on that side should be at least \\(m_{i} \\cdot \\frac{2S}{N}\\) , contradiction.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2006-notes.jsonl", "problem_match": "6. ", "solution_match": "## \\(\\S 2.3\\) IMO 2006/6, proposed by Dušan Djukić (SRB) \n"}}
diff --git a/IMO/segmented/en-IMO-2007-notes.jsonl b/IMO/segmented/en-IMO-2007-notes.jsonl
index e411f0021f9f6b799636d7315b586a59d6c651a9..5acfe249ce4beb48abf7828a17144437612bbe14 100644
--- a/IMO/segmented/en-IMO-2007-notes.jsonl
+++ b/IMO/segmented/en-IMO-2007-notes.jsonl
@@ -1,6 +1,6 @@
{"year": "2007", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "Real numbers \\(a_{1}\\) , \\(a_{2}\\) , ..., \\(a_{n}\\) are fixed. For each \\(1 \\leq i \\leq n\\) we let \\(d_{i} = \\max \\{a_{j} : 1 \\leq j \\leq i\\} - \\min \\{a_{j} : i \\leq j \\leq n\\}\\) and let \\(d = \\max \\{d_{i} : 1 \\leq i \\leq n\\}\\) . \n\n(a) Prove that for any real numbers \\(x_{1} \\leq \\dots \\leq x_{n}\\) we have \n\n\\[\\max \\left\\{|x_{i} - a_{i}|:1\\leq i\\leq n\\right\\} \\geq \\frac{1}{2} d.\\] \n\n(b) Moreover, show that there exists some choice of \\(x_{1} \\leq \\dots \\leq x_{n}\\) which achieves equality.", "solution": "Note that we can dispense of \\(d_{i}\\) immediately by realizing that the definition of \\(d\\) just says \n\n\\[d = \\max_{1 \\leq i \\leq j \\leq n} (a_{i} - a_{j}).\\] \n\nIf \\(a_{1} \\leq \\dots \\leq a_{n}\\) are already nondecreasing then \\(d = 0\\) and there is nothing to prove (for the equality case, just let \\(x_{i} = a_{i}\\) ), so we will no longer consider this case. \n\nOtherwise, consider any indices \\(i < j\\) with \\(a_{i} > a_{j}\\) . We first prove (a) by applying the following claim with \\(p = a_{i}\\) and \\(q = a_{j}\\) : \n\nClaim — For any \\(p \\leq q\\) , we have either \\(|p - a_{i}| \\geq \\frac{1}{2} (a_{i} - a_{j})\\) or \\(|q - a_{j}| \\geq \\frac{1}{2} (a_{i} - a_{j}) \n\nProof. Assume for contradiction both are false. Then \\(p > a_{i} - \\frac{1}{2} (a_{i} - a_{j}) = a_{j} + \\frac{1}{2} (a_{i} - a_{j}) > q\\) , contradiction. \\(\\square\\) \n\nAs for (b), we let \\(i < j\\) be any indices for which \\(a_{i} - a_{j} = d > 0\\) achieves the maximal difference. We then define \\(x_{\\bullet}\\) in three steps: \n\nWe set \\(x_{k} = \\frac{a_{i} + a_{j}}{2}\\) for \\(k = i, \\ldots , j\\) . We recursively set \\(x_{k} = \\max (x_{k - 1}, a_{k})\\) for \\(k = j + 1, j + 2, \\ldots\\) . We recursively set \\(x_{k} = \\min (x_{k + 1}, a_{k})\\) for \\(k = i - 1, i - 2, \\ldots\\) . \n\nBy definition, these \\(x_{\\bullet}\\) are weakly increasing. To prove this satisfies (b) we only need to check that \n\n\\[|x_{k} - a_{k}| \\leq \\frac{a_{i} - a_{j}}{2} \\qquad (\\star)\\] \n\nfor any index \\(k\\) (as equality holds for \\(k = i\\) or \\(k = j\\) ). \n\nWe note \\((\\star)\\) holds for \\(i < k < j\\) by construction. For \\(k > j\\) , note that \\(x_{k} \\in \\{a_{j}, a_{j + 1}, \\ldots , a_{k}\\}\\) by construction, so \\((\\star)\\) follows from our choice of \\(i\\) and \\(j\\) giving the largest possible difference; the case \\(k < i\\) is similar.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2007-notes.jsonl", "problem_match": "1. ", "solution_match": "## \\(\\S 1.1\\) IMO 2007/1, proposed by Michael Albert (NZL) \n"}}
-{"year": "2007", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "Consider five points \\(A\\) , \\(B\\) , \\(C\\) , \\(D\\) and \\(E\\) such that \\(ABCD\\) is a parallelogram and \\(BCED\\) is a cyclic quadrilateral. Let \\(\\ell\\) be a line passing through \\(A\\) . Suppose that \\(\\ell\\) intersects the interior of the segment \\(DC\\) at \\(F\\) and intersects line \\(BC\\) at \\(G\\) . Suppose also that \\(EF = EG = EC\\) . Prove that \\(\\ell\\) is the bisector of angle \\(DAB\\) .", "solution": "Let \\(M\\) , \\(N\\) , \\(P\\) denote the midpoints of \\(\\overline{CF}\\) , \\(\\overline{CG}\\) , \\(\\overline{AC}\\) (noting \\(P\\) is also the midpoint of \\(\\overline{BD}\\) ). \n\nBy a homothety at \\(C\\) with ratio \\(\\frac{1}{2}\\) , we find \\(\\overline{MNP}\\) is the image of line \\(\\ell \\equiv \\overline{AGF}\\) . \n\n\n \n\nHowever, since we also have \\(\\overline{EM} \\perp \\overline{CF}\\) and \\(\\overline{EN} \\perp \\overline{CG}\\) (from \\(EF = EG = EC\\) ) we conclude \\(\\overline{PMN}\\) is the Simson line of \\(E\\) with respect to \\(\\triangle BCD\\) , which implies \\(\\overline{EP} \\perp \\overline{BD}\\) . In other words, \\(\\overline{EP}\\) is the perpendicular bisector of \\(\\overline{BD}\\) , so \\(E\\) is the midpoint of arc \\(\\overline{BCD}\\) . \n\nFinally, \n\n\\[\\angle (\\overline{AB}, \\ell) = \\angle (\\overline{CD}, \\overline{MNP}) = \\angle CMN = \\angle CEN\\] \\[\\qquad = 90^{\\circ} - \\angle NCE = 90^{\\circ} + \\angle ECB\\] \n\nwhich means that \\(\\ell\\) is parallel to a bisector of \\(\\angle BCD\\) , and hence to one of \\(\\angle BAD\\) . (Moreover since \\(F\\) lies on the interior of \\(\\overline{CD}\\) , it is actually the internal bisector.)", "metadata": {"resource_path": "IMO/segmented/en-IMO-2007-notes.jsonl", "problem_match": "2. ", "solution_match": "## \\(\\S 1.2\\) IMO 2007/2, proposed by Charles Leytem (LUX) \n"}}
-{"year": "2007", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "In a mathematical competition some competitors are (mutual) friends. Call a group of competitors a clique if each two of them are friends. Given that the largest size of a clique is even, prove that the competitors can be arranged into two rooms such that the largest size of a clique contained in one room is the same as the largest size of a clique contained in the other room.", "solution": "Take the obvious graph interpretation \\(G\\) . We paint red any vertices in one of the maximal cliques \\(K\\) , which we assume has \\(2r\\) vertices, and paint the remaining vertices green. We let \\(\\alpha (\\bullet)\\) denote the clique number. \n\nInitially, let the two rooms \\(A = K\\) , \\(B = G - K\\) . \n\nClaim — We can move at most \\(r\\) vertices of \\(A\\) into \\(B\\) to arrive at \\(\\alpha (A) \\leq \\alpha (B) \\leq \\alpha (A) + 1\\) . \n\nProof. This is actually obvious by discrete continuity. We move one vertex at a time, noting \\(\\alpha (A)\\) decreases by one at each step, while \\(\\alpha (B)\\) increases by either zero or one at each step. \n\nWe stop once \\(\\alpha (B) \\geq \\alpha (A)\\) , which happens before we have moved \\(r\\) vertices (since then we have \\(\\alpha (B) \\geq r = \\alpha (A)\\) ). The conclusion follows. \\(\\square\\) \n\nSo let's consider the situation \n\n\\[\\alpha (A) = k \\geq r \\qquad \\text{and} \\qquad \\alpha (B) = k + 1.\\] \n\nAt this point \\(A\\) is a set of \\(k\\) red vertices, while \\(B\\) has the remaining \\(2r - k\\) red vertices (and all the green ones). An example is shown below with \\(k = 4\\) and \\(2r = 6\\) . \n\n\n \n\nNow, if we can move any red vertex from \\(B\\) back to \\(A\\) without changing the clique number of \\(B\\) , we do so, and win. \n\nOtherwise, it must be the case that every \\((k + 1)\\) - clique in \\(B\\) uses every red vertex in \\(B\\) . For each \\((k + 1)\\) - clique in \\(B\\) (in arbitrary order), we do the following procedure.\n\n\n\n- If all \\(k + 1\\) vertices are still green, pick one and re-color it blue. This is possible since \\(k + 1 > 2r - k\\) . \n\n- Otherwise, do nothing. \n\nThen we move all the blue vertices from \\(B\\) to \\(A\\) , one at a time, in the same order we re- colored them. This forcibly decreases the clique number of \\(B\\) to \\(k\\) , since the clique number is \\(k + 1\\) just before the last blue vertex is moved, and strictly less than \\(k + 1\\) (hence equal to \\(k\\) ) immediately after that. \n\nClaim — After this, \\(\\alpha (A) = k\\) still holds. \n\nProof. Assume not, and we have a \\((k + 1)\\) - clique which uses \\(b\\) blue vertices and \\((k + 1) - b\\) red vertices in \\(A\\) . Together with the \\(2r - k\\) red vertices already in \\(B\\) we then get a clique of size \n\n\\[b + ((k + 1 - b)) + (2r - k) = 2r + 1\\] \n\nwhich is a contradiction. \n\nRemark. Dragomir Grozev posted the following motivation on his blog: \n\nI think, it's a natural idea to place all students in one room and begin moving them one by one into the other one. Then the max size of the cliques in the first and second room increase (resp. decrease) at most with one. So, there would be a moment both sizes are almost the same. At that moment we may adjust something. \n\nTrying the idea, I had some difficulties keeping track of the maximal cliques in the both rooms. It seemed easier all the students in one of the rooms to comprise a clique. It could be achieved by moving only the members of the maximal clique. Following this path the remaining obstacles can be overcome naturally.\n\n\n\n## \\(\\S 2\\) Solutions to Day 2", "metadata": {"resource_path": "IMO/segmented/en-IMO-2007-notes.jsonl", "problem_match": "3. ", "solution_match": "## \\(\\S 1.3\\) IMO 2007/3, proposed by Vasily Astakhov (RUS) \n"}}
-{"year": "2007", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "In triangle \\(ABC\\) the bisector of \\(\\angle BCA\\) meets the circumcircle again at \\(R\\) , the perpendicular bisector of \\(\\overline{BC}\\) at \\(P\\) , and the perpendicular bisector of \\(\\overline{AC}\\) at \\(Q\\) . The midpoint of \\(\\overline{BC}\\) is \\(K\\) and the midpoint of \\(\\overline{AC}\\) is \\(L\\) . Prove that the triangles \\(RPK\\) and \\(RQL\\) have the same area.", "solution": "We first begin by proving the following claim. \n\nClaim — We have \\(CQ = PR\\) (equivalently, \\(CP = QR\\) ). \n\nProof. Let \\(O = \\overline{LQ} \\cap \\overline{KP}\\) be the circumcenter. Then \n\n\\[\\angle OPQ = \\angle KPC = 90^{\\circ} - \\angle PCK = 90^{\\circ} - \\angle LCQ = \\angle \\angle CQL = \\angle PQO.\\] \n\nThus \\(OP = OQ\\) . Since \\(OC = OR\\) as well, we get the conclusion. \n\nDenote by \\(X\\) and \\(Y\\) the feet from \\(R\\) to \\(\\overline{CA}\\) and \\(\\overline{CB}\\) , so \\(\\triangle CXR \\cong \\triangle CYR\\) . Then, let \\(t = \\frac{CQ}{CR} = 1 - \\frac{CP}{CR}\\) . \n\n\n \n\nThen it follows that \n\n\\[[RQ L] = [X Q L] = t(1 - t)\\cdot [X R C] = t(1 - t)\\cdot [Y C R] = [Y K P] = [R K P]\\] \n\nas needed. \n\nRemark. Trigonometric approaches are very possible (and easier to find) as well: both areas work out to be \\(\\frac{1}{8} ab \\tan \\frac{1}{2} C\\) .", "metadata": {"resource_path": "IMO/segmented/en-IMO-2007-notes.jsonl", "problem_match": "4. ", "solution_match": "## \\(\\S 2.1\\) IMO 2007/4, proposed by Marek Pechal (CZE) \n"}}
+{"year": "2007", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "Consider five points \\(A\\) , \\(B\\) , \\(C\\) , \\(D\\) and \\(E\\) such that \\(ABCD\\) is a parallelogram and \\(BCED\\) is a cyclic quadrilateral. Let \\(\\ell\\) be a line passing through \\(A\\) . Suppose that \\(\\ell\\) intersects the interior of the segment \\(DC\\) at \\(F\\) and intersects line \\(BC\\) at \\(G\\) . Suppose also that \\(EF = EG = EC\\) . Prove that \\(\\ell\\) is the bisector of angle \\(DAB\\) .", "solution": "Let \\(M\\) , \\(N\\) , \\(P\\) denote the midpoints of \\(\\overline{CF}\\) , \\(\\overline{CG}\\) , \\(\\overline{AC}\\) (noting \\(P\\) is also the midpoint of \\(\\overline{BD}\\) ). \n\nBy a homothety at \\(C\\) with ratio \\(\\frac{1}{2}\\) , we find \\(\\overline{MNP}\\) is the image of line \\(\\ell \\equiv \\overline{AGF}\\) . \n\n\n \n\nHowever, since we also have \\(\\overline{EM} \\perp \\overline{CF}\\) and \\(\\overline{EN} \\perp \\overline{CG}\\) (from \\(EF = EG = EC\\) ) we conclude \\(\\overline{PMN}\\) is the Simson line of \\(E\\) with respect to \\(\\triangle BCD\\) , which implies \\(\\overline{EP} \\perp \\overline{BD}\\) . In other words, \\(\\overline{EP}\\) is the perpendicular bisector of \\(\\overline{BD}\\) , so \\(E\\) is the midpoint of arc \\(\\overline{BCD}\\) . \n\nFinally, \n\n\\[\\angle (\\overline{AB}, \\ell) = \\angle (\\overline{CD}, \\overline{MNP}) = \\angle CMN = \\angle CEN\\] \\[\\qquad = 90^{\\circ} - \\angle NCE = 90^{\\circ} + \\angle ECB\\] \n\nwhich means that \\(\\ell\\) is parallel to a bisector of \\(\\angle BCD\\) , and hence to one of \\(\\angle BAD\\) . (Moreover since \\(F\\) lies on the interior of \\(\\overline{CD}\\) , it is actually the internal bisector.)", "metadata": {"resource_path": "IMO/segmented/en-IMO-2007-notes.jsonl", "problem_match": "2. ", "solution_match": "## \\(\\S 1.2\\) IMO 2007/2, proposed by Charles Leytem (LUX) \n"}}
+{"year": "2007", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "In a mathematical competition some competitors are (mutual) friends. Call a group of competitors a clique if each two of them are friends. Given that the largest size of a clique is even, prove that the competitors can be arranged into two rooms such that the largest size of a clique contained in one room is the same as the largest size of a clique contained in the other room.", "solution": "Take the obvious graph interpretation \\(G\\) . We paint red any vertices in one of the maximal cliques \\(K\\) , which we assume has \\(2r\\) vertices, and paint the remaining vertices green. We let \\(\\alpha (\\bullet)\\) denote the clique number. \n\nInitially, let the two rooms \\(A = K\\) , \\(B = G - K\\) . \n\nClaim — We can move at most \\(r\\) vertices of \\(A\\) into \\(B\\) to arrive at \\(\\alpha (A) \\leq \\alpha (B) \\leq \\alpha (A) + 1\\) . \n\nProof. This is actually obvious by discrete continuity. We move one vertex at a time, noting \\(\\alpha (A)\\) decreases by one at each step, while \\(\\alpha (B)\\) increases by either zero or one at each step. \n\nWe stop once \\(\\alpha (B) \\geq \\alpha (A)\\) , which happens before we have moved \\(r\\) vertices (since then we have \\(\\alpha (B) \\geq r = \\alpha (A)\\) ). The conclusion follows. \\(\\square\\) \n\nSo let's consider the situation \n\n\\[\\alpha (A) = k \\geq r \\qquad \\text{and} \\qquad \\alpha (B) = k + 1.\\] \n\nAt this point \\(A\\) is a set of \\(k\\) red vertices, while \\(B\\) has the remaining \\(2r - k\\) red vertices (and all the green ones). An example is shown below with \\(k = 4\\) and \\(2r = 6\\) . \n\n\n \n\nNow, if we can move any red vertex from \\(B\\) back to \\(A\\) without changing the clique number of \\(B\\) , we do so, and win. \n\nOtherwise, it must be the case that every \\((k + 1)\\) - clique in \\(B\\) uses every red vertex in \\(B\\) . For each \\((k + 1)\\) - clique in \\(B\\) (in arbitrary order), we do the following procedure.\n\n\n\n- If all \\(k + 1\\) vertices are still green, pick one and re-color it blue. This is possible since \\(k + 1 > 2r - k\\) . \n\n- Otherwise, do nothing. \n\nThen we move all the blue vertices from \\(B\\) to \\(A\\) , one at a time, in the same order we re- colored them. This forcibly decreases the clique number of \\(B\\) to \\(k\\) , since the clique number is \\(k + 1\\) just before the last blue vertex is moved, and strictly less than \\(k + 1\\) (hence equal to \\(k\\) ) immediately after that. \n\nClaim — After this, \\(\\alpha (A) = k\\) still holds. \n\nProof. Assume not, and we have a \\((k + 1)\\) - clique which uses \\(b\\) blue vertices and \\((k + 1) - b\\) red vertices in \\(A\\) . Together with the \\(2r - k\\) red vertices already in \\(B\\) we then get a clique of size \n\n\\[b + ((k + 1 - b)) + (2r - k) = 2r + 1\\] \n\nwhich is a contradiction. \n\nRemark. Dragomir Grozev posted the following motivation on his blog: \n\nI think, it's a natural idea to place all students in one room and begin moving them one by one into the other one. Then the max size of the cliques in the first and second room increase (resp. decrease) at most with one. So, there would be a moment both sizes are almost the same. At that moment we may adjust something. \n\nTrying the idea, I had some difficulties keeping track of the maximal cliques in the both rooms. It seemed easier all the students in one of the rooms to comprise a clique. It could be achieved by moving only the members of the maximal clique. Following this path the remaining obstacles can be overcome naturally.\n\n\n\n## \\(\\S 2\\) Solutions to Day 2", "metadata": {"resource_path": "IMO/segmented/en-IMO-2007-notes.jsonl", "problem_match": "3. ", "solution_match": "## \\(\\S 1.3\\) IMO 2007/3, proposed by Vasily Astakhov (RUS) \n"}}
+{"year": "2007", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "In triangle \\(ABC\\) the bisector of \\(\\angle BCA\\) meets the circumcircle again at \\(R\\) , the perpendicular bisector of \\(\\overline{BC}\\) at \\(P\\) , and the perpendicular bisector of \\(\\overline{AC}\\) at \\(Q\\) . The midpoint of \\(\\overline{BC}\\) is \\(K\\) and the midpoint of \\(\\overline{AC}\\) is \\(L\\) . Prove that the triangles \\(RPK\\) and \\(RQL\\) have the same area.", "solution": "We first begin by proving the following claim. \n\nClaim — We have \\(CQ = PR\\) (equivalently, \\(CP = QR\\) ). \n\nProof. Let \\(O = \\overline{LQ} \\cap \\overline{KP}\\) be the circumcenter. Then \n\n\\[\\angle OPQ = \\angle KPC = 90^{\\circ} - \\angle PCK = 90^{\\circ} - \\angle LCQ = \\angle \\angle CQL = \\angle PQO.\\] \n\nThus \\(OP = OQ\\) . Since \\(OC = OR\\) as well, we get the conclusion. \n\nDenote by \\(X\\) and \\(Y\\) the feet from \\(R\\) to \\(\\overline{CA}\\) and \\(\\overline{CB}\\) , so \\(\\triangle CXR \\cong \\triangle CYR\\) . Then, let \\(t = \\frac{CQ}{CR} = 1 - \\frac{CP}{CR}\\) . \n\n\n \n\nThen it follows that \n\n\\[[RQ L] = [X Q L] = t(1 - t)\\cdot [X R C] = t(1 - t)\\cdot [Y C R] = [Y K P] = [R K P]\\] \n\nas needed. \n\nRemark. Trigonometric approaches are very possible (and easier to find) as well: both areas work out to be \\(\\frac{1}{8} ab \\tan \\frac{1}{2} C\\) .", "metadata": {"resource_path": "IMO/segmented/en-IMO-2007-notes.jsonl", "problem_match": "4. ", "solution_match": "## \\(\\S 2.1\\) IMO 2007/4, proposed by Marek Pechal (CZE) \n"}}
{"year": "2007", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "Let \\(a\\) and \\(b\\) be positive integers. Show that if \\(4ab - 1\\) divides \\((4a^{2} - 1)^{2}\\) , then \\(a = b\\) .", "solution": "As usual, \n\n\\[4ab - 1\\mid (4a^{2} - 1)^{2}\\iff 4ab - 1\\mid (4ab\\cdot a - b)^{2}\\iff 4ab - 1\\mid (a - b)^{2}.\\] \n\nThen we use a typical Vieta jumping argument. Define \n\n\\[k = \\frac{(a - b)^{2}}{4ab - 1}.\\] \n\nNote that \\(k = 0 \\iff a = b\\) . So we will prove that \\(k > 0\\) leads to a contradiction. \n\nIndeed, suppose \\((a,b)\\) is a minimal solution with \\(a > b\\) (we have \\(a\\neq b\\) since \\(k\\neq 0\\) - By Vieta jumping, \\((b,\\frac{b^{2} + k}{a})\\) is also such a solution. But now \n\n\\[\\frac{b^{2} + k}{a}\\geq a\\Rightarrow k\\geq a^{2} - b^{2}\\] \\[\\Rightarrow \\frac{(a - b)^{2}}{4ab - 1}\\geq a^{2} - b^{2}\\] \\[\\Rightarrow a - b\\geq (4ab - 1)(a + b)\\] \n\nwhich is absurd for \\(a,b\\in \\mathbb{Z}_{>0}\\) . (In the last step we divided by \\(a - b > 0\\) -", "metadata": {"resource_path": "IMO/segmented/en-IMO-2007-notes.jsonl", "problem_match": "5. ", "solution_match": "## \\(\\S 2.2\\) IMO 2007/5, proposed by Kevin Buzzard, Edward Crane (UNK) \n"}}
{"year": "2007", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "Let \\(n\\) be a positive integer. Consider \n\n\\[S = \\{(x,y,z) \\mid x,y,z \\in \\{0,1,\\ldots ,n\\} , x + y + z > 0\\}\\] \n\nas a set of \\((n + 1)^{3} - 1\\) points in the three- dimensional space. Determine the smallest possible number of planes, the union of which contains \\(S\\) but does not include \\((0,0,0)\\) .", "solution": "The answer is \\(3n\\) . Here are two examples of constructions with \\(3n\\) planes: \n\n\\(\\cdot x + y + z = i\\) for \\(i = 1,\\ldots ,3n\\) \n\n\\(\\cdot x = i\\) \\(y = i\\) \\(z = i\\) for \\(i = 1,\\ldots ,n\\) \n\nSuppose for contradiction we have \\(N< 3n\\) planes. Let them be \\(a_{i}x + b_{i}y + c_{i}z + 1 = 0\\) for \\(i = 1,\\ldots ,N\\) . Define the polynomials \n\n\\[A(x,y,z) = \\prod_{i = 1}^{n}(x - i)\\prod_{i = 1}^{n}(y - i)\\prod_{i = 1}^{n}(z - i)\\] \\[B(x,y,z) = \\prod_{i = 1}^{N}(a_{i}x + b_{i}y + c_{i}z + 1).\\] \n\nNote that \\(A(0,0,0) = (- 1)^{n}(n!)^{3}\\neq 0\\) and \\(B(0,0,0) = 1\\neq 0\\) , but \\(A(x,y,z) =\\) \\(B(x,y,z) = 0\\) for any \\((x,y,z)\\in S\\) . Also, the coefficient of \\(x^{n}y^{n}z^{n}\\) in \\(A\\) is 1, while the coefficient of \\(x^{n}y^{n}z^{n}\\) in \\(B\\) is 0. \n\nNow, define \n\n\\[P(x,y,z):= A(x,y,z) - \\lambda B(x,y,z).\\] \n\nwhere \\(\\lambda = \\frac{A(0,0,0)}{B(0,0,0)} = (- 1)^{n}(n!)^{3}\\) . We now have that \n\n\\(\\cdot P(x,y,z) = 0\\) for any \\(x,y,z\\in \\{0,1,\\ldots ,n\\}^{3}\\) \n\nBut the coefficient of \\(x^{n}y^{n}z^{n}\\) is 1. \n\nThis is a contradiction to Alon's combinatorial nullstellensatz.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2007-notes.jsonl", "problem_match": "6. ", "solution_match": "## \\(\\S 2.3\\) IMO 2007/6, proposed by Gerhard Woeginger (NLD) \n"}}
diff --git a/IMO/segmented/en-IMO-2008-notes.jsonl b/IMO/segmented/en-IMO-2008-notes.jsonl
index 59a696f232f99806d4dea35f472b4a721384a375..93479925ca3bcec68a6444703db4d3a43be54398 100644
--- a/IMO/segmented/en-IMO-2008-notes.jsonl
+++ b/IMO/segmented/en-IMO-2008-notes.jsonl
@@ -1,6 +1,6 @@
-{"year": "2008", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "Let \\(H\\) be the orthocenter of an acute-angled triangle \\(ABC\\) . The circle \\(\\Gamma_{A}\\) centered at the midpoint of \\(\\overline{BC}\\) and passing through \\(H\\) intersects the sideline \\(BC\\) at points \\(A_{1}\\) and \\(A_{2}\\) . Similarly, define the points \\(B_{1}\\) , \\(B_{2}\\) , \\(C_{1}\\) , and \\(C_{2}\\) . Prove that six points \\(A_{1}\\) , \\(A_{2}\\) , \\(B_{1}\\) , \\(B_{2}\\) , \\(C_{1}\\) , \\(C_{2}\\) are concyclic.", "solution": ". \n\nWe show two solutions. \n\n\\(\\P\\) First solution using power of a point. Let \\(D\\) , \\(E\\) , \\(F\\) be the centers of \\(\\Gamma_{A}\\) , \\(\\Gamma_{B}\\) , \\(\\Gamma_{C}\\) (in other words, the midpoints of the sides). \n\nWe first show that \\(B_{1}\\) , \\(B_{2}\\) , \\(C_{1}\\) , \\(C_{2}\\) are concyclic. It suffices to prove that \\(A\\) lies on the radical axis of the circles \\(\\Gamma_{B}\\) and \\(\\Gamma_{C}\\) . \n\n\n \n\nLet \\(X\\) be the second intersection of \\(\\Gamma_{B}\\) and \\(\\Gamma_{C}\\) . Clearly \\(\\overline{XH}\\) is perpendicular to the line joining the centers of the circles, namely \\(\\overline{EF}\\) . But \\(\\overline{EF} \\parallel \\overline{BC}\\) , so \\(\\overline{XH} \\perp \\overline{BC}\\) . Since \\(\\overline{AH} \\perp \\overline{BC}\\) as well, we find that \\(A\\) , \\(X\\) , \\(H\\) are collinear, as needed. \n\nThus, \\(B_{1}\\) , \\(B_{2}\\) , \\(C_{1}\\) , \\(C_{2}\\) are concyclic. Similarly, \\(C_{1}\\) , \\(C_{2}\\) , \\(A_{1}\\) , \\(A_{2}\\) are concyclic, as are \\(A_{1}\\) , \\(A_{2}\\) , \\(B_{1}\\) , \\(B_{2}\\) . Now if any two of these three circles coincide, we are done; else the pairwise radical axii are not concurrent, contradiction. (Alternatively, one can argue directly that \\(O\\) is the center of all three circles, by taking the perpendicular bisectors.) \n\n\\(\\P\\) Second solution using length chase (Ritwin Narra). We claim the circumcenter \\(O\\) of \\(\\triangle ABC\\) is in fact the center of \\((A_{1}A_{2}B_{1}B_{2}C_{1}C_{2})\\) . \n\nDefine \\(D\\) , \\(E\\) , \\(F\\) as before. Then since \\(\\overline{OD} \\perp \\overline{A_{1}A_{2}}\\) and \\(DA_{1} = DA_{2}\\) , which means \\(OA_{1} = OA_{2}\\) . Similarly, we have \\(OB_{1} = OB_{2}\\) and \\(OC_{1} = OC_{2}\\) . \n\nNow since \\(DA_{1} = DA_{2} = DH\\) , we have \\(OA_{1}^{2} = OD^{2} + HD^{2}\\) . We seek to show \n\n\\[OD^{2} + HD^{2} = OE^{2} + HE^{2} = OF^{2} + HF^{2}.\\]\n\n\n\nThis is clear by Appollonius's Theorem since \\(D\\) , \\(E\\) , and \\(F\\) lie on the nine- point circle, which is centered at the midpoint of \\(\\overline{OH}\\) .", "metadata": {"resource_path": "IMO/segmented/en-IMO-2008-notes.jsonl", "problem_match": "1. ", "solution_match": "## \\(\\S 1.1\\) IMO 2008/1, proposed by Andrey Gavrilyuk (RUS) \n"}}
+{"year": "2008", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "Let \\(H\\) be the orthocenter of an acute-angled triangle \\(ABC\\) . The circle \\(\\Gamma_{A}\\) centered at the midpoint of \\(\\overline{BC}\\) and passing through \\(H\\) intersects the sideline \\(BC\\) at points \\(A_{1}\\) and \\(A_{2}\\) . Similarly, define the points \\(B_{1}\\) , \\(B_{2}\\) , \\(C_{1}\\) , and \\(C_{2}\\) . Prove that six points \\(A_{1}\\) , \\(A_{2}\\) , \\(B_{1}\\) , \\(B_{2}\\) , \\(C_{1}\\) , \\(C_{2}\\) are concyclic.", "solution": ". \n\nWe show two solutions. \n\n\\(\\P\\) First solution using power of a point. Let \\(D\\) , \\(E\\) , \\(F\\) be the centers of \\(\\Gamma_{A}\\) , \\(\\Gamma_{B}\\) , \\(\\Gamma_{C}\\) (in other words, the midpoints of the sides). \n\nWe first show that \\(B_{1}\\) , \\(B_{2}\\) , \\(C_{1}\\) , \\(C_{2}\\) are concyclic. It suffices to prove that \\(A\\) lies on the radical axis of the circles \\(\\Gamma_{B}\\) and \\(\\Gamma_{C}\\) . \n\n\n \n\nLet \\(X\\) be the second intersection of \\(\\Gamma_{B}\\) and \\(\\Gamma_{C}\\) . Clearly \\(\\overline{XH}\\) is perpendicular to the line joining the centers of the circles, namely \\(\\overline{EF}\\) . But \\(\\overline{EF} \\parallel \\overline{BC}\\) , so \\(\\overline{XH} \\perp \\overline{BC}\\) . Since \\(\\overline{AH} \\perp \\overline{BC}\\) as well, we find that \\(A\\) , \\(X\\) , \\(H\\) are collinear, as needed. \n\nThus, \\(B_{1}\\) , \\(B_{2}\\) , \\(C_{1}\\) , \\(C_{2}\\) are concyclic. Similarly, \\(C_{1}\\) , \\(C_{2}\\) , \\(A_{1}\\) , \\(A_{2}\\) are concyclic, as are \\(A_{1}\\) , \\(A_{2}\\) , \\(B_{1}\\) , \\(B_{2}\\) . Now if any two of these three circles coincide, we are done; else the pairwise radical axii are not concurrent, contradiction. (Alternatively, one can argue directly that \\(O\\) is the center of all three circles, by taking the perpendicular bisectors.) \n\n\\(\\P\\) Second solution using length chase (Ritwin Narra). We claim the circumcenter \\(O\\) of \\(\\triangle ABC\\) is in fact the center of \\((A_{1}A_{2}B_{1}B_{2}C_{1}C_{2})\\) . \n\nDefine \\(D\\) , \\(E\\) , \\(F\\) as before. Then since \\(\\overline{OD} \\perp \\overline{A_{1}A_{2}}\\) and \\(DA_{1} = DA_{2}\\) , which means \\(OA_{1} = OA_{2}\\) . Similarly, we have \\(OB_{1} = OB_{2}\\) and \\(OC_{1} = OC_{2}\\) . \n\nNow since \\(DA_{1} = DA_{2} = DH\\) , we have \\(OA_{1}^{2} = OD^{2} + HD^{2}\\) . We seek to show \n\n\\[OD^{2} + HD^{2} = OE^{2} + HE^{2} = OF^{2} + HF^{2}.\\]\n\n\n\nThis is clear by Appollonius's Theorem since \\(D\\) , \\(E\\) , and \\(F\\) lie on the nine- point circle, which is centered at the midpoint of \\(\\overline{OH}\\) .", "metadata": {"resource_path": "IMO/segmented/en-IMO-2008-notes.jsonl", "problem_match": "1. ", "solution_match": "## \\(\\S 1.1\\) IMO 2008/1, proposed by Andrey Gavrilyuk (RUS) \n"}}
{"year": "2008", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "Let \\(x\\) , \\(y\\) , \\(z\\) be real numbers with \\(xyz = 1\\) , all different from 1. Prove that \n\n\\[\\frac{x^{2}}{(x - 1)^{2}} +\\frac{y^{2}}{(y - 1)^{2}} +\\frac{z^{2}}{(z - 1)^{2}}\\geq 1\\] \n\nand show that equality holds for infinitely many choices of rational numbers \\(x\\) , \\(y\\) , \\(z\\) .", "solution": "Let \\(x = a / b\\) , \\(y = b / c\\) , \\(z = c / a\\) , so we want to show \n\n\\[\\left(\\frac{a}{a - b}\\right)^{2} + \\left(\\frac{b}{b - c}\\right)^{2} + \\left(\\frac{c}{c - a}\\right)^{2}\\geq 1.\\] \n\nA boring computation shows this is equivalent to \n\n\\[\\frac{(a^{2}b + b^{2}c + c^{2}a - 3abc)^{2}}{(a - b)^{2}(b - c)^{2}(c - a)^{2}}\\geq 0\\] \n\nwhich proves the inequality (and it is unsurprising we are in such a situation, given that there is an infinite curve of rationals). \n\nFor equality, it suffices to show there are infinitely many integer solutions to \n\n\\[a^{2}b + b^{2}c + c^{2}a = 3abc\\iff \\frac{a}{c} +\\frac{b}{a} +\\frac{c}{a} = 3\\] \n\nor equivalently that there are infinitely many rational solutions to \n\n\\[u + v + \\frac{1}{uv} = 3.\\] \n\nFor any \\(0\\neq u\\in \\mathbb{Q}\\) the real solution for \\(u\\) is \n\n\\[v = \\frac{-u + (u - 1)\\sqrt{1 - 4 / u} + 3}{2}\\] \n\nand there are certainly infinitely many rational numbers \\(u\\) for which \\(1 - 4 / u\\) is a rational square (say, \\(u = \\frac{- 4}{q^{2} - 1}\\) for \\(q\\neq \\pm 1\\) a rational number).", "metadata": {"resource_path": "IMO/segmented/en-IMO-2008-notes.jsonl", "problem_match": "2. ", "solution_match": "## \\(\\S 1.2\\) IMO 2008/2, proposed by Walther Janous (AUT) \n"}}
{"year": "2008", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "Prove that there are infinitely many positive integers \\(n\\) such that \\(n^{2} + 1\\) has a prime factor greater than \\(2n + \\sqrt{2n}\\) .", "solution": "Problem statement \n\nThe idea is to pick the prime \\(p\\) first! \n\nSelect any large prime \\(p \\geq 2013\\) , and let \\(h = \\left\\lceil \\sqrt{p}\\right\\rceil\\) . We will try to find an \\(n\\) such that \n\n\\[n \\leq \\frac{1}{2} (p - h) \\quad \\text{and} \\quad p \\mid n^{2} + 1.\\] \n\nThis implies \\(p \\geq 2n + \\sqrt{p}\\) which is enough to ensure \\(p \\geq 2n + \\sqrt{2n}\\) . \n\nAssume \\(p \\equiv 1\\) (mod 8) henceforth. Then there exists some \\(\\frac{1}{2} p < x < p\\) such that \\(x^{2} \\equiv - 1\\) (mod \\(p\\) ), and we set \n\n\\[x = \\frac{p + 1}{2} +t.\\] \n\nClaim — We have \\(t \\geq \\frac{h - 1}{2}\\) and hence may take \\(n = p - x\\) . \n\nProof. Assume for contradiction this is false; then \n\n\\[0\\equiv 4(x^{2} + 1)\\pmod{p\\] \\[\\quad = (p + 1 + 2t)^{2} + 4\\] \\[\\quad \\equiv (2t + 1)^{2} + 4\\pmod{p\\] \\[\\quad < h^{2} + 4\\] \n\nSo we have that \\((2t + 1)^{2} + 4\\) is positive and divisible by \\(p\\) , yet at most \\(\\left\\lceil \\sqrt{p}\\right\\rceil^{2} + 4 < 2p\\) . So it must be the case that \\((2t + 1)^{2} + 4 = p\\) , but this has no solutions modulo 8. \\(\\square\\)\n\n\n\n## \\(\\S 2\\) Solutions to Day 2", "metadata": {"resource_path": "IMO/segmented/en-IMO-2008-notes.jsonl", "problem_match": "3. ", "solution_match": "## \\(\\S 1.3\\) IMO 2008/3, proposed by Kestutis Česnavičius (LTU) \n"}}
{"year": "2008", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "Find all functions \\(f\\) from the positive reals to the positive reals such that \n\n\\[\\frac{f(w)^{2} + f(x)^{2}}{f(y^{2}) + f(z^{2})} = \\frac{w^{2} + x^{2}}{y^{2} + z^{2}}\\] \n\nfor all positive real numbers \\(w\\) , \\(x\\) , \\(y\\) , \\(z\\) satisfying \\(wx = yz\\) .", "solution": "The answers are \\(f(x) \\equiv x\\) and \\(f(x) \\equiv 1 / x\\) . These work, so we show they are the only ones. \n\nFirst, setting \\((t, t, t, t)\\) gives \\(f(t^{2}) = f(t)^{2}\\) . In particular, \\(f(1) = 1\\) . Next, setting \\((t, 1, \\sqrt{t}, \\sqrt{t})\\) gives \n\n\\[\\frac{f(t)^{2} + 1}{2f(t)} = \\frac{t^{2} + 1}{2t}\\] \n\nwhich as a quadratic implies \\(f(t) \\in \\{t, 1 / t\\}\\) . \n\nNow assume \\(f(a) = a\\) and \\(f(b) = 1 / b\\) . Setting \\((\\sqrt{a}, \\sqrt{b}, 1, \\sqrt{ab})\\) gives \n\n\\[\\frac{a + 1 / b}{f(ab) + 1} = \\frac{a + b}{ab + 1}.\\] \n\nOne can check the two cases on \\(f(ab)\\) each imply \\(a = 1\\) and \\(b = 1\\) respectively. Hence the only answers are those claimed.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2008-notes.jsonl", "problem_match": "4. ", "solution_match": "## \\(\\S 2.1\\) IMO 2008/4, proposed by Hojoo Lee (KOR) \n"}}
{"year": "2008", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "Let \\(n\\) and \\(k\\) be positive integers with \\(k \\geq n\\) and \\(k - n\\) an even number. There are \\(2n\\) lamps labelled 1, 2, ..., \\(2n\\) each of which can be either on or off. Initially all the lamps are off. We consider sequences of steps: at each step one of the lamps is switched (from on to off or from off to on). Let \\(N\\) be the number of such sequences consisting of \\(k\\) steps and resulting in the state where lamps 1 through \\(n\\) are all on, and lamps \\(n + 1\\) through \\(2n\\) are all off. Let \\(M\\) be number of such sequences consisting of \\(k\\) steps, resulting in the state where lamps 1 through \\(n\\) are all on, and lamps \\(n + 1\\) through \\(2n\\) are all off, but where none of the lamps \\(n + 1\\) through \\(2n\\) is ever switched on. Determine \\(\\frac{N}{M}\\) .", "solution": "The answer is \\(2^{k - n}\\) . \n\nConsider the following map \\(\\Psi\\) from \\(N\\) - sequences to \\(M\\) - sequences: \n\n- change every instance of \\(n + 1\\) to 1; \n\n- change every instance of \\(n + 2\\) to 2; \n\n: \n\n- change every instance of \\(2n\\) to \\(n\\) . \n\n(For example, suppose \\(k = 9\\) , \\(n = 3\\) ; then \\(144225253 \\mapsto 111222223\\) .) \n\nClearly this is map is well- defined and surjective. So all that remains is: \n\nClaim — Every \\(M\\) - sequence has exactly \\(2^{k - n}\\) pre- images under \\(\\Psi\\) . \n\nProof. Indeed, suppose that there are \\(c_{1}\\) instances of lamp 1. Then we want to pick an odd subset of the 1's to change to \\(n + 1\\) 's, so \\(2^{c_{1} - 1}\\) ways to do this. And so on. Hence the number of pre- images is \n\n\\[\\prod_{i}2^{c_{i} - 1} = 2^{k - n}.\\]", "metadata": {"resource_path": "IMO/segmented/en-IMO-2008-notes.jsonl", "problem_match": "5. ", "solution_match": "## \\(\\S 2.2\\) IMO 2008/5, proposed by Bruno Le Floch and Ilia Smilga (FRA) \n"}}
-{"year": "2008", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "Let \\(ABCD\\) be a convex quadrilateral with \\(BA \\neq BC\\) . Denote the incircles of triangles \\(ABC\\) and \\(ADC\\) by \\(\\omega_{1}\\) and \\(\\omega_{2}\\) respectively. Suppose that there exists a circle \\(\\omega\\) tangent to ray \\(BA\\) beyond \\(A\\) and to the ray \\(BC\\) beyond \\(C\\) , which is also tangent to the lines \\(AD\\) and \\(CD\\) . Prove that the common external tangents to \\(\\omega_{1}\\) and \\(\\omega_{2}\\) intersect on \\(\\omega\\) .", "solution": "Let \\(A B C D \n\nBy the external version of Pitot theorem, the existence of \\(\\omega\\) implies that \n\n\\[B A + A D = C B + C D.\\] \n\nLet \\(\\overline{{P Q}}\\) and \\(\\overline{{S T}}\\) be diameters of \\(\\omega_{1}\\) and \\(\\omega_{2}\\) with \\(P,T\\in \\overline{{A C}}\\) . Then the length relation on \\(A B C D\\) implies that \\(P\\) and \\(T\\) are reflections about the midpoint of \\(\\overline{{A C}}\\) \n\nNow orient \\(A C\\) horizontally and let \\(K\\) be the \"uppermost\" point of \\(\\omega\\) , as shown. \n\n\n \n\nConsequently, a homothety at \\(B\\) maps \\(Q\\) , \\(T\\) , \\(K\\) to each other (since \\(T\\) is the uppermost of the excircle, \\(Q\\) of the incircle). Similarly, a homothety at \\(D\\) maps \\(P\\) , \\(S\\) , \\(K\\) to each other. As \\(\\overline{{P Q}}\\) and \\(\\overline{{S T}}\\) are parallel diameters it then follows \\(K\\) is the exsimilicenter of \\(\\omega_{1}\\) and \\(\\omega_{2}\\) .", "metadata": {"resource_path": "IMO/segmented/en-IMO-2008-notes.jsonl", "problem_match": "6. ", "solution_match": "## \\(\\S 2.3\\) IMO 2008/6, proposed by Vladimir Shmarov (RUS) \n"}}
+{"year": "2008", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "Let \\(ABCD\\) be a convex quadrilateral with \\(BA \\neq BC\\) . Denote the incircles of triangles \\(ABC\\) and \\(ADC\\) by \\(\\omega_{1}\\) and \\(\\omega_{2}\\) respectively. Suppose that there exists a circle \\(\\omega\\) tangent to ray \\(BA\\) beyond \\(A\\) and to the ray \\(BC\\) beyond \\(C\\) , which is also tangent to the lines \\(AD\\) and \\(CD\\) . Prove that the common external tangents to \\(\\omega_{1}\\) and \\(\\omega_{2}\\) intersect on \\(\\omega\\) .", "solution": "Let \\(A B C D \n\nBy the external version of Pitot theorem, the existence of \\(\\omega\\) implies that \n\n\\[B A + A D = C B + C D.\\] \n\nLet \\(\\overline{{P Q}}\\) and \\(\\overline{{S T}}\\) be diameters of \\(\\omega_{1}\\) and \\(\\omega_{2}\\) with \\(P,T\\in \\overline{{A C}}\\) . Then the length relation on \\(A B C D\\) implies that \\(P\\) and \\(T\\) are reflections about the midpoint of \\(\\overline{{A C}}\\) \n\nNow orient \\(A C\\) horizontally and let \\(K\\) be the \"uppermost\" point of \\(\\omega\\) , as shown. \n\n\n \n\nConsequently, a homothety at \\(B\\) maps \\(Q\\) , \\(T\\) , \\(K\\) to each other (since \\(T\\) is the uppermost of the excircle, \\(Q\\) of the incircle). Similarly, a homothety at \\(D\\) maps \\(P\\) , \\(S\\) , \\(K\\) to each other. As \\(\\overline{{P Q}}\\) and \\(\\overline{{S T}}\\) are parallel diameters it then follows \\(K\\) is the exsimilicenter of \\(\\omega_{1}\\) and \\(\\omega_{2}\\) .", "metadata": {"resource_path": "IMO/segmented/en-IMO-2008-notes.jsonl", "problem_match": "6. ", "solution_match": "## \\(\\S 2.3\\) IMO 2008/6, proposed by Vladimir Shmarov (RUS) \n"}}
diff --git a/IMO/segmented/en-IMO-2009-notes.jsonl b/IMO/segmented/en-IMO-2009-notes.jsonl
index e2a49f89e0bf20a4c1168f12c7176419ef4d69a2..1ea65baa228b731b8c593b8f2d1d4cf3d8048a48 100644
--- a/IMO/segmented/en-IMO-2009-notes.jsonl
+++ b/IMO/segmented/en-IMO-2009-notes.jsonl
@@ -1,6 +1,6 @@
{"year": "2009", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "Let \\(n,k\\geq 2\\) be positive integers and let \\(a_{1}\\) , \\(a_{2}\\) , \\(a_{3}\\) , ..., \\(a_{k}\\) be distinct integers in the set \\(\\{1,2,\\ldots ,n\\}\\) such that \\(n\\) divides \\(a_{i}(a_{i + 1} - 1)\\) for \\(i = 1,2,\\ldots ,k - 1\\) . Prove that \\(n\\) does not divide \\(a_{k}(a_{1} - 1)\\) .", "solution": "We proceed indirectly and assume that \n\n\\[a_{i}(a_{i + 1} - 1)\\equiv 0\\pmod {n}\\] \n\nfor \\(i = 1,\\ldots ,k\\) (indices taken modulo \\(k\\) ). We claim that this implies all the \\(a_{i}\\) are equal modulo \\(n\\) . \n\nLet \\(q = p^{e}\\) be any prime power dividing \\(n\\) . Then, \\(a_{1}(a_{2} - 1)\\equiv 0\\) (mod \\(q\\) ), so \\(p\\) divides either \\(a_{1}\\) or \\(a_{2} - 1\\) . \n\n- If \\(p \\mid a_{1}\\) , then \\(p \\nmid a_{1} - 1\\) . Then \n\n\\[a_{k}(a_{1} - 1)\\equiv 0\\pmod {q}\\implies a_{k}\\equiv 0\\pmod {q}.\\] \n\nIn particular, \\(p \\mid a_{k}\\) . So repeating this argument, we get \\(a_{k - 1} \\equiv 0\\) (mod \\(q\\) ), \\(a_{k - 2} \\equiv 0\\) (mod \\(q\\) ), and so on. \n\n- Similarly, if \\(p \\mid a_{2} - 1\\) then \\(p \\nmid a_{2}\\) , and \n\n\\[a_{2}(a_{3} - 1)\\equiv 0\\pmod {q}\\implies a_{3}\\equiv 1\\pmod {q}.\\] \n\nIn particular, \\(p \\mid a_{3} - 1\\) . So repeating this argument, we get \\(a_{4} \\equiv 0\\) (mod \\(q\\) ), \\(a_{5} \\equiv 0\\) (mod \\(q\\) ), and so on. \n\nEither way, we find \\(a_{i}\\) (mod \\(q\\) ) is constant (and either 0 or 1). \n\nSince \\(q\\) was an arbitrary prime power dividing \\(n\\) , by Chinese remainder theorem we conclude that \\(a_{i}\\) (mod \\(n\\) ) is constant as well. But this contradicts the assumption of distinctness.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2009-notes.jsonl", "problem_match": "1. ", "solution_match": "## \\(\\S 1.1\\) IMO 2009/1, proposed by Ross Atkins (AUS) \n"}}
-{"year": "2009", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "Let \\(A B C\\) be a triangle with circumcenter \\(o\\) . The points \\(P\\) and \\(Q\\) are interior points of the sides \\(C A\\) and \\(A B\\) respectively. Let \\(K\\) , \\(L\\) , \\(M\\) be the midpoints of \\(\\overline{{B P}}\\) \\(\\overline{{C Q}}\\) , \\(\\overline{{P Q}}\\) . Suppose that \\(\\overline{{P Q}}\\) is tangent to the circumcircle of \\(\\triangle K L M\\) . Prove that \\(O P = O Q\\) .", "solution": "a point, we have \\(- AQ \\cdot QB = OQ^2 - R^2\\) and \\(- AP \\cdot PC = OP^2 - R^2\\) . Therefore, it suffices to show \\(AQ \\cdot QB = AP \\cdot PC\\) . \n\n\n \n\nAs \\(\\overline{ML} \\parallel \\overline{AC}\\) and \\(\\overline{MK} \\parallel \\overline{AB}\\) we have that \n\n\\[\\angle APQ = \\angle LMP = \\angle LKM\\] \\[\\angle PQA = \\angle KMQ = \\angle MLK\\] \n\nand consequently we have the (opposite orientation) similarity \n\n\\[\\triangle APQ \\sim \\triangle MKL.\\] \n\nTherefore \n\n\\[\\frac{AQ}{AP} = \\frac{ML}{MK} = \\frac{2ML}{2MK} = \\frac{PC}{QB}\\] \n\nid est \\(AQ \\cdot QB = AP \\cdot PC\\) , which is what we wanted to prove.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2009-notes.jsonl", "problem_match": "2. ", "solution_match": "## \\(\\S 1.2\\) IMO 2009/2, proposed by Sergei Berlov (RUS) \n"}}
+{"year": "2009", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "Let \\(A B C\\) be a triangle with circumcenter \\(o\\) . The points \\(P\\) and \\(Q\\) are interior points of the sides \\(C A\\) and \\(A B\\) respectively. Let \\(K\\) , \\(L\\) , \\(M\\) be the midpoints of \\(\\overline{{B P}}\\) \\(\\overline{{C Q}}\\) , \\(\\overline{{P Q}}\\) . Suppose that \\(\\overline{{P Q}}\\) is tangent to the circumcircle of \\(\\triangle K L M\\) . Prove that \\(O P = O Q\\) .", "solution": "a point, we have \\(- AQ \\cdot QB = OQ^2 - R^2\\) and \\(- AP \\cdot PC = OP^2 - R^2\\) . Therefore, it suffices to show \\(AQ \\cdot QB = AP \\cdot PC\\) . \n\n\n \n\nAs \\(\\overline{ML} \\parallel \\overline{AC}\\) and \\(\\overline{MK} \\parallel \\overline{AB}\\) we have that \n\n\\[\\angle APQ = \\angle LMP = \\angle LKM\\] \\[\\angle PQA = \\angle KMQ = \\angle MLK\\] \n\nand consequently we have the (opposite orientation) similarity \n\n\\[\\triangle APQ \\sim \\triangle MKL.\\] \n\nTherefore \n\n\\[\\frac{AQ}{AP} = \\frac{ML}{MK} = \\frac{2ML}{2MK} = \\frac{PC}{QB}\\] \n\nid est \\(AQ \\cdot QB = AP \\cdot PC\\) , which is what we wanted to prove.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2009-notes.jsonl", "problem_match": "2. ", "solution_match": "## \\(\\S 1.2\\) IMO 2009/2, proposed by Sergei Berlov (RUS) \n"}}
{"year": "2009", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "Suppose that \\(s_{1},s_{2},s_{3},\\ldots\\) is a strictly increasing sequence of positive integers such that the sub-sequences \\(s_{s_{1}}\\) , \\(s_{s_{2}}\\) , \\(s_{s_{3}}\\) , ...and \\(s_{s_{1} + 1}\\) , \\(s_{s_{2} + 1}\\) , \\(s_{s_{3} + 1}\\) , ...are both arithmetic progressions. Prove that the sequence \\(s_{1}\\) , \\(s_{2}\\) , \\(s_{3}\\) , ...is itself an arithmetic progression.", "solution": "We present two solutions. \n\n\\(\\P\\) First solution (Alex Zhai). Let \\(s(n):= s_{n}\\) and write \n\n\\[s(s(n)) = Dn + A\\] \\[s(s(n) + 1) = D^{\\prime}n + B.\\] \n\nIn light of the bounds \\(s(s(n))\\leq s(s(n) + 1)\\leq s(s(n + 1))\\) we right away recover \\(D = D^{\\prime}\\) and \\(A\\leq B\\) \n\nLet \\(d_{n} = s(n + 1) - s(n)\\) . Note that \\(\\sup d_{n}< \\infty\\) since \\(d_{n}\\) is bounded above by \\(A\\) \n\nThen we let \n\n\\[m:= \\min d_{n},\\qquad M:= \\max d_{n}.\\] \n\nNow suppose \\(a\\) achieves the maximum, meaning \\(s(a + 1) - s(a) = M\\) . Then \n\n\\[\\underbrace{d_{s(s(a))} + \\cdot\\cdot\\cdot + d_{s(s(a + 1)) - 1}}_{D\\mathrm{~terms}} = \\underbrace{\\left[s(s(s(a + 1))) - s(s(s(a)))\\right]}_{D\\mathrm{~terms}}\\] \\[\\qquad = (D\\cdot s(a + 1) + A) - (D\\cdot s(a) + A) = DM.\\] \n\nNow \\(M\\) was maximal hence \\(M = d_{s(s(a))} = \\dots = d_{s(s(a + 1)) - 1}\\) . But \\(d_{s(s(a))} = B - A\\) is a constant. Hence \\(M = B - A\\) . In the same way \\(m = B - A\\) as desired. \n\n\\(\\P\\) Second solution. We retain the notation \\(D\\) , \\(A\\) , \\(B\\) above, as well as \\(m = \\min_{n}s(n+\\) \\(1) - s(n)\\geq B - A\\) . We do the involution trick first as: \n\n\\[D = \\left[s(s(s(n) + 1)) - s(s(s(n)))\\right] = s(Dn + B) - s(Dn + A)\\] \n\nand hence we recover \\(D\\geq m(B - A)\\) \n\nThe edge case \\(D = B - A\\) is easy since then \\(m = 1\\) and \\(D = s(Dn + B) - s(Dn + A)\\) forces \\(s\\) to be a constant shift. So henceforth assume \\(D > B - A\\) \n\nThe idea is that right now the \\(B\\) terms are \"too big\", so we want to use the involution trick in a way that makes as many \" \\(A\\) minus \\(B\\) \" shape expressions as possible. This motivates considering \\(s(s(s(n + 1))) - s(s(s(n) + 1) + 1) > 0\\) , since the first expression will have all \\(A\\) 's and the second expression will have all \\(B\\) 's. Calculation gives: \n\n\\[s(D(n + 1) + A) - s(Dn + B + 1) = \\left[s(s(s(n + 1))) - s(s(s(n) + 1) + 1)\\right]\\] \\[\\qquad = (D s(n + 1) + A) - (D(s(n) + 1) + B)\\] \\[\\qquad = D(s(n + 1) - s(n)) + A - B - D.\\]\n\n\n\nThen by picking \\(n\\) achieving the minimum \\(m\\) , \n\n\\[\\underbrace{m(D + A - B - 1)}_{>0}\\leq s(s(s(n + 1))) - s(s(s(n) + 1) + 1)\\leq Dm + A - B - D\\] \n\nwhich becomes \n\n\\[(D - m(B - A)) + ((B - A) - m)\\leq 0.\\] \n\nSince both of these quantities were supposed to be nonnegative, we conclude \\(m = B - A\\) and \\(D = m^{2}\\) . Now the estimate \\(D = s(Dn + B) - s(Dn + A)\\geq m(B - A)\\) is actually sharp, so it follows that \\(s(n)\\) is arithmetic.\n\n\n\n## \\(\\S 2\\) Solutions to Day 2", "metadata": {"resource_path": "IMO/segmented/en-IMO-2009-notes.jsonl", "problem_match": "3. ", "solution_match": "## \\(\\S 1.3\\) IMO 2009/3, proposed by Gabriel Carroll (USA) \n"}}
-{"year": "2009", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "Let \\(A B C\\) be a triangle with \\(A B = A C\\) . The angle bisectors of \\(\\angle C A B\\) and \\(\\angle A B C\\) meet the sides \\(B C\\) and \\(C A\\) at \\(D\\) and \\(E\\) , respectively. Let \\(K\\) be the incenter of triangle \\(A D C\\) . Suppose that \\(\\angle B E K = 45^{\\circ}\\) . Find all possible values of \\(\\angle C A B\\) .", "solution": "Here is the solution presented in my book EGM0. \n\nLet \\(I\\) be the incenter of \\(A B C\\) , and set \\(\\angle D A C = 2x\\) (so that \\(0^{\\circ}< x< 45^{\\circ}\\) ). From \\(\\angle A I E = \\angle D I C\\) , it is easy to compute \n\n\\[\\angle K I E = 90^{\\circ} - 2x, \\angle E C I = 45^{\\circ} - x, \\angle I E K = 45^{\\circ}, \\angle K E C = 3x.\\] \n\nHaving chased all the angles we want, we need a relationship. We can find it by considering the side ratio \\(\\frac{I K}{K C}\\) . Using the angle bisector theorem, we can express this in terms of triangle \\(I D C\\) ; however we can also express it in terms of triangle \\(I E C\\) . \n\n\n \n\nBy the law of sines, we obtain \n\n\\[\\frac{I K}{K C} = \\frac{\\sin 45^{\\circ}\\cdot\\frac{E K}{\\sin(90^{\\circ} - 2x)}}{\\sin(3x)\\cdot\\frac{E K}{\\sin(45^{\\circ} - x)}} = \\frac{\\sin 45^{\\circ}\\sin(45^{\\circ} - x)}{\\sin(3x)\\sin(90^{\\circ} - 2x)}.\\] \n\nAlso, by the angle bisector theorem on \\(\\triangle I D C\\) , we have \n\n\\[\\frac{I K}{K C} = \\frac{I D}{D C} = \\frac{\\sin(45^{\\circ} - x)}{\\sin(45^{\\circ} + x)}.\\]\n\n\n\nEquating these and cancelling \\(\\sin \\left(45^{\\circ} - x\\right) \\neq 0\\) gives \n\n\\[\\sin 45^{\\circ}\\sin \\left(45^{\\circ} + x\\right) = \\sin 3x\\sin \\left(90^{\\circ} - 2x\\right).\\] \n\nApplying the product- sum formula (again, we are just trying to break down things as much as possible), this just becomes \n\n\\[\\cos \\left(x\\right) - \\cos \\left(90^{\\circ} + x\\right) = \\cos \\left(5x - 90^{\\circ}\\right) - \\cos \\left(90^{\\circ} + x\\right)\\] \n\nor \\(\\cos x = \\cos \\left(5x - 90^{\\circ}\\right)\\) \n\nAt this point we are basically done; the rest is making sure we do not miss any solutions and write up the completion nicely. One nice way to do this is by using product- sum in reverse as \n\n\\[0 = \\cos \\left(5x - 90^{\\circ}\\right) - \\cos x = 2\\sin \\left(3x - 45^{\\circ}\\right)\\sin \\left(2x - 45^{\\circ}\\right).\\] \n\nThis way we merely consider the two cases \n\n\\[\\sin \\left(3x - 45^{\\circ}\\right) = 0 \\mathrm{and} \\sin \\left(2x - 45^{\\circ}\\right) = 0.\\] \n\nNotice that \\(\\sin \\theta = 0\\) if and only \\(\\theta\\) is an integer multiple of \\(180^{\\circ}\\) . Using the bound \\(0^{\\circ} < x < 45^{\\circ}\\) , it is easy to see that the permissible values of \\(x\\) are \\(x = 15^{\\circ}\\) and \\(x = \\frac{45^{\\circ}}{2}\\) . As \\(\\angle A = 4x\\) , this corresponds to \\(\\angle A = 60^{\\circ}\\) and \\(\\angle A = 90^{\\circ}\\) , which can be seen to work.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2009-notes.jsonl", "problem_match": "4. ", "solution_match": "## \\(\\S 2.1\\) IMO 2009/4, proposed by Hojoo Lee, Peter Vandendriessche, Jan Vonk (BEL) \n"}}
+{"year": "2009", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "Let \\(A B C\\) be a triangle with \\(A B = A C\\) . The angle bisectors of \\(\\angle C A B\\) and \\(\\angle A B C\\) meet the sides \\(B C\\) and \\(C A\\) at \\(D\\) and \\(E\\) , respectively. Let \\(K\\) be the incenter of triangle \\(A D C\\) . Suppose that \\(\\angle B E K = 45^{\\circ}\\) . Find all possible values of \\(\\angle C A B\\) .", "solution": "Here is the solution presented in my book EGM0. \n\nLet \\(I\\) be the incenter of \\(A B C\\) , and set \\(\\angle D A C = 2x\\) (so that \\(0^{\\circ}< x< 45^{\\circ}\\) ). From \\(\\angle A I E = \\angle D I C\\) , it is easy to compute \n\n\\[\\angle K I E = 90^{\\circ} - 2x, \\angle E C I = 45^{\\circ} - x, \\angle I E K = 45^{\\circ}, \\angle K E C = 3x.\\] \n\nHaving chased all the angles we want, we need a relationship. We can find it by considering the side ratio \\(\\frac{I K}{K C}\\) . Using the angle bisector theorem, we can express this in terms of triangle \\(I D C\\) ; however we can also express it in terms of triangle \\(I E C\\) . \n\n\n \n\nBy the law of sines, we obtain \n\n\\[\\frac{I K}{K C} = \\frac{\\sin 45^{\\circ}\\cdot\\frac{E K}{\\sin(90^{\\circ} - 2x)}}{\\sin(3x)\\cdot\\frac{E K}{\\sin(45^{\\circ} - x)}} = \\frac{\\sin 45^{\\circ}\\sin(45^{\\circ} - x)}{\\sin(3x)\\sin(90^{\\circ} - 2x)}.\\] \n\nAlso, by the angle bisector theorem on \\(\\triangle I D C\\) , we have \n\n\\[\\frac{I K}{K C} = \\frac{I D}{D C} = \\frac{\\sin(45^{\\circ} - x)}{\\sin(45^{\\circ} + x)}.\\]\n\n\n\nEquating these and cancelling \\(\\sin \\left(45^{\\circ} - x\\right) \\neq 0\\) gives \n\n\\[\\sin 45^{\\circ}\\sin \\left(45^{\\circ} + x\\right) = \\sin 3x\\sin \\left(90^{\\circ} - 2x\\right).\\] \n\nApplying the product- sum formula (again, we are just trying to break down things as much as possible), this just becomes \n\n\\[\\cos \\left(x\\right) - \\cos \\left(90^{\\circ} + x\\right) = \\cos \\left(5x - 90^{\\circ}\\right) - \\cos \\left(90^{\\circ} + x\\right)\\] \n\nor \\(\\cos x = \\cos \\left(5x - 90^{\\circ}\\right)\\) \n\nAt this point we are basically done; the rest is making sure we do not miss any solutions and write up the completion nicely. One nice way to do this is by using product- sum in reverse as \n\n\\[0 = \\cos \\left(5x - 90^{\\circ}\\right) - \\cos x = 2\\sin \\left(3x - 45^{\\circ}\\right)\\sin \\left(2x - 45^{\\circ}\\right).\\] \n\nThis way we merely consider the two cases \n\n\\[\\sin \\left(3x - 45^{\\circ}\\right) = 0 \\mathrm{and} \\sin \\left(2x - 45^{\\circ}\\right) = 0.\\] \n\nNotice that \\(\\sin \\theta = 0\\) if and only \\(\\theta\\) is an integer multiple of \\(180^{\\circ}\\) . Using the bound \\(0^{\\circ} < x < 45^{\\circ}\\) , it is easy to see that the permissible values of \\(x\\) are \\(x = 15^{\\circ}\\) and \\(x = \\frac{45^{\\circ}}{2}\\) . As \\(\\angle A = 4x\\) , this corresponds to \\(\\angle A = 60^{\\circ}\\) and \\(\\angle A = 90^{\\circ}\\) , which can be seen to work.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2009-notes.jsonl", "problem_match": "4. ", "solution_match": "## \\(\\S 2.1\\) IMO 2009/4, proposed by Hojoo Lee, Peter Vandendriessche, Jan Vonk (BEL) \n"}}
{"year": "2009", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "Find all functions \\(f\\colon \\mathbb{Z}_{>0}\\to \\mathbb{Z}_{>0}\\) such that for positive integers \\(a\\) and \\(b\\) , the numbers \n\n\\[a, f(b), f(b + f(a) - 1)\\] \n\nare the sides of a non-degenerate triangle.", "solution": ". \n\nThe only function is the identity function (which works). We prove it is the only one. Let \\(P(a,b)\\) denote the given statement. \n\nClaim — We have \\(f(1) = 1\\) , and \\(f(f(n)) = n\\) . (In particular \\(f\\) is a bijection.) \n\nProof. Note that \n\n\\[P(1,b)\\Rightarrow f(b) = f(b + f(1) - 1).\\] \n\nOtherwise, the function \\(f\\) is periodic modulo \\(N = f(1) - 1\\geq 1\\) . This is impossible since we can fix \\(b\\) and let \\(a\\) be arbitrarily large in some residue class modulo \\(N\\) . \n\nHence \\(f(1) = 1\\) , so taking \\(P(n,1)\\) gives \\(f(f(n)) = n\\) . \n\nClaim — Let \\(\\delta = f(2) - 1 > 0\\) . Then for every \\(n\\) , \n\n\\[f(n + 1) = f(n) + \\delta \\quad \\mathrm{or}\\quad f(n - 1) = f(n) + \\delta\\] \n\nProof. Use \n\n\\[P(2,f(n))\\Rightarrow n - 2< f(f(n) + \\delta)< n + 2.\\] \n\nLet \\(y = f(f(n) + \\delta)\\) , hence \\(n - 2< y< n + 2\\) and \\(f(y) = f(n) + \\delta\\) . But, remark that if \\(y = n\\) , we get \\(\\delta = 0\\) , contradiction. So \\(y\\in \\{n + 1,n - 1\\}\\) and that is all. \\(\\square\\) \n\nWe now show \\(f\\) is an arithmetic progression with common difference \\(+\\delta\\) . Indeed we already know \\(f(1) = 1\\) and \\(f(2) = 1 + \\delta\\) . Now suppose \\(f(1) = 1,\\ldots ,f(n) = 1 + (n - 1)\\delta\\) . Then by induction for any \\(n\\geq 2\\) , the second case can't hold, so we have \\(f(n + 1) = f(n) + \\delta\\) , as desired. \n\nCombined with \\(f(f(n)) = n\\) , we recover that \\(f\\) is the identity.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2009-notes.jsonl", "problem_match": "5. ", "solution_match": "## \\(\\S 2.2\\) IMO 2009/5, proposed by Bruno Le Floch (FRA) \n"}}
{"year": "2009", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "Let \\(a_{1}\\) , \\(a_{2}\\) , ..., \\(a_{n}\\) be distinct positive integers and let \\(M\\) be a set of \\(n - 1\\) positive integers not containing \\(s = a_{1} + \\dots +a_{n}\\) . A grasshopper is to jump along the real axis, starting at the point 0 and making \\(n\\) jumps to the right with lengths \\(a_{1}\\) , \\(a_{2}\\) , ..., \\(a_{n}\\) in some order. Prove that the order can be chosen in such a way that the grasshopper never lands on any point in \\(M\\) .", "solution": "The proof is by induction on \\(n\\) . Assume \\(a_{1} < \\dots < a_{n}\\) and call each element of \\(M\\) a mine. Let \\(x = s - a_{n}\\) . We consider four cases, based on whether \\(x\\) has a mine and whether there is a mine past \\(x\\) . \n\n- If \\(x\\) has no mine, and there is a mine past \\(x\\) , then at most \\(n - 2\\) mines in \\([0, x]\\) and so we use induction to reach \\(x\\) , then leap from \\(x\\) to \\(s\\) and win. \n\n- If \\(x\\) has no mine but there is also no mine to the right of \\(x\\) , then let \\(m\\) be the maximal mine. By induction hypothesis on \\(M \\setminus \\{m\\}\\) , there is a path to \\(x\\) using \\(\\{a_{1}, \\ldots , a_{n - 1}\\}\\) which avoids mines except possibly \\(m\\) . If the path hits the mine \\(m\\) on the hop of length \\(a_{k}\\) , we then swap that hop with \\(a_{n}\\) , and finish. \n\n- If \\(x\\) has a mine, but there are no mines to the right of \\(x\\) , we can repeat the previous case with \\(m = x\\) . \n\n- Now suppose \\(x\\) has a mine, and there is a mine past \\(x\\) . There should exist an integer \\(1 \\leq i \\leq n - 1\\) such that \\(s - a_{i}\\) and \\(y = s - a_{i} - a_{n}\\) both have no mine. By induction hypothesis, we can then reach \\(y\\) in \\(n - 2\\) steps (as there are two mines to the right of \\(y\\) ); then \\(y \\to s - a_{i} \\to s\\) finishes. \n\nRemark. It seems much of the difficulty of the problem is realizing induction will actually work. Attempts at induction are, indeed, a total minefield (ha!), and given the position P6 of the problem, it is expected that many contestants will abandon induction after some cursory attempts fail.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2009-notes.jsonl", "problem_match": "6. ", "solution_match": "## \\(\\S 2.3\\) IMO 2009/6, proposed by Dmitry Khramtsov (RUS) \n"}}
diff --git a/IMO/segmented/en-IMO-2010-notes.jsonl b/IMO/segmented/en-IMO-2010-notes.jsonl
index e92bafb4da4063cce24e192e2e4620290bb9950e..6afb550329a8b713f8828e0631049ab9d71f01bd 100644
--- a/IMO/segmented/en-IMO-2010-notes.jsonl
+++ b/IMO/segmented/en-IMO-2010-notes.jsonl
@@ -1,6 +1,6 @@
{"year": "2010", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "Find all functions \\(f\\colon \\mathbb{R}\\to \\mathbb{R}\\) such that for all \\(x,y\\in \\mathbb{R}\\) \n\n\\[f(\\lfloor x\\rfloor y) = f(x)\\lfloor f(y)\\rfloor .\\]", "solution": "The only solutions are \\(f(x)\\equiv c\\) , where \\(c = 0\\) or \\(1\\leq c< 2\\) . It's easy to see these work. \n\nPlug in \\(x = 0\\) to get \\(f(0) = f(0)\\lfloor f(y)\\rfloor\\) , so either \n\n\\[1\\leq f(y)< 2\\quad \\forall y\\qquad \\mathrm{or}\\qquad f(0) = 0\\] \n\nIn the first situation, plug in \\(y = 0\\) to get \\(f(x)\\lfloor f(0)\\rfloor = f(0)\\) , thus \\(f\\) is constant. Thus assume henceforth \\(f(0) = 0\\) \n\nNow set \\(x = y = 1\\) to get \n\n\\[f(1) = f(1)\\lfloor f(1)\\rfloor\\] \n\nso either \\(f(1) = 0\\) or \\(1\\leq f(1)< 2\\) . We split into cases: \n\n- If \\(f(1) = 0\\) , pick \\(x = 1\\) to get \\(f(y)\\equiv 0\\) . \n\n- If \\(1\\leq f(1)< 2\\) , then \\(y = 1\\) gives \n\n\\[f(\\lfloor x\\rfloor) = f(x)\\] \n\nfrom \\(y = 1\\) , in particular \\(f(x) = 0\\) for \\(0\\leq x< 1\\) . Choose \\((x,y) = \\left(2,\\frac{1}{2}\\right)\\) to get \\(f(1) = f(2)\\left\\lfloor f\\left(\\frac{1}{2}\\right)\\right\\rfloor = 0\\) .", "metadata": {"resource_path": "IMO/segmented/en-IMO-2010-notes.jsonl", "problem_match": "1. ", "solution_match": "## \\(\\S 1.1\\) IMO 2010/1, proposed by Pierre Bornsztein (FRA) \n"}}
-{"year": "2010", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "Let \\(I\\) be the incenter of a triangle \\(A B C\\) and let \\(\\Gamma\\) be its circumcircle. Let line \\(A I\\) intersect \\(\\Gamma\\) again at \\(D\\) . Let \\(E\\) be a point on arc \\(\\overrightarrow{B D C}\\) and \\(F\\) a point on side \\(B C\\) such that \n\n\\[\\angle B A F = \\angle C A E< \\frac{1}{2}\\angle B A C.\\] \n\nFinally, let \\(G\\) be the midpoint of \\(\\overline{{I F}}\\) . Prove that \\(\\overline{{D G}}\\) and \\(\\overline{{E I}}\\) intersect on \\(\\Gamma\\)", "solution": "Let \\(\\overline{{E I}}\\) meet \\(\\Gamma\\) again at \\(K\\) . Then it suffices to show that \\(\\overline{{K D}}\\) bisects \\(\\overline{{I F}}\\) . Let \\(\\overline{{A F}}\\) meet \\(\\Gamma\\) again at \\(H\\) , so \\(\\overline{{H E}}\\parallel \\overline{{B C}}\\) . By Pascal theorem on \n\n\\[A H E K D D\\] \n\nwe then obtain that \\(P = \\overline{{A H}}\\cap \\overline{{K D}}\\) lies on a line through \\(I\\) parallel to \\(\\overline{{B C}}\\) . \n\nLet \\(I_{A}\\) be the \\(A\\) - excenter, and set \\(Q = \\overline{{I_{A}F}}\\cap \\overline{{I P}}\\) , and \\(T = \\overline{{A I D I_{A}}}\\cap \\overline{{B F C}}\\) . Then \n\n\\[-1 = (A I;T I_{A})\\stackrel {E}{=}(I Q;\\infty P)\\] \n\nwhere \\(\\infty\\) is the point at infinity along \\(\\overline{{I P Q}}\\) . Thus \\(P\\) is the midpoint of \\(\\overline{{I Q}}\\) . Since \\(D\\) is the midpoint of \\(\\overline{{I I_{A}}}\\) by \"Fact 5\", it follows that \\(\\overline{{D P}}\\) bisects \\(\\overline{{I F}}\\) . \n\n", "metadata": {"resource_path": "IMO/segmented/en-IMO-2010-notes.jsonl", "problem_match": "2. ", "solution_match": "## §1.2 IMO 2010/2, proposed by Tai Wai Ming and Wang Chongli (HKG) \n"}}
+{"year": "2010", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "Let \\(I\\) be the incenter of a triangle \\(A B C\\) and let \\(\\Gamma\\) be its circumcircle. Let line \\(A I\\) intersect \\(\\Gamma\\) again at \\(D\\) . Let \\(E\\) be a point on arc \\(\\overrightarrow{B D C}\\) and \\(F\\) a point on side \\(B C\\) such that \n\n\\[\\angle B A F = \\angle C A E< \\frac{1}{2}\\angle B A C.\\] \n\nFinally, let \\(G\\) be the midpoint of \\(\\overline{{I F}}\\) . Prove that \\(\\overline{{D G}}\\) and \\(\\overline{{E I}}\\) intersect on \\(\\Gamma\\)", "solution": "Let \\(\\overline{{E I}}\\) meet \\(\\Gamma\\) again at \\(K\\) . Then it suffices to show that \\(\\overline{{K D}}\\) bisects \\(\\overline{{I F}}\\) . Let \\(\\overline{{A F}}\\) meet \\(\\Gamma\\) again at \\(H\\) , so \\(\\overline{{H E}}\\parallel \\overline{{B C}}\\) . By Pascal theorem on \n\n\\[A H E K D D\\] \n\nwe then obtain that \\(P = \\overline{{A H}}\\cap \\overline{{K D}}\\) lies on a line through \\(I\\) parallel to \\(\\overline{{B C}}\\) . \n\nLet \\(I_{A}\\) be the \\(A\\) - excenter, and set \\(Q = \\overline{{I_{A}F}}\\cap \\overline{{I P}}\\) , and \\(T = \\overline{{A I D I_{A}}}\\cap \\overline{{B F C}}\\) . Then \n\n\\[-1 = (A I;T I_{A})\\stackrel {E}{=}(I Q;\\infty P)\\] \n\nwhere \\(\\infty\\) is the point at infinity along \\(\\overline{{I P Q}}\\) . Thus \\(P\\) is the midpoint of \\(\\overline{{I Q}}\\) . Since \\(D\\) is the midpoint of \\(\\overline{{I I_{A}}}\\) by \"Fact 5\", it follows that \\(\\overline{{D P}}\\) bisects \\(\\overline{{I F}}\\) . \n\n", "metadata": {"resource_path": "IMO/segmented/en-IMO-2010-notes.jsonl", "problem_match": "2. ", "solution_match": "## §1.2 IMO 2010/2, proposed by Tai Wai Ming and Wang Chongli (HKG) \n"}}
{"year": "2010", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "Find all functions \\(g\\colon \\mathbb{Z}_{>0}\\to \\mathbb{Z}_{>0}\\) such that \n\n\\[(g(m) + n)(g(n) + m)\\] \n\nis always a perfect square.", "solution": "For \\(c\\geq 0\\) , the function \\(g(n) = n + c\\) works; we prove this is the only possibility. \n\nFirst, the main point of the problem is that: \n\nClaim — We have \\(g(n)\\equiv g(n^{\\prime})\\) (mod \\(p\\) ) \\(\\Longrightarrow n\\equiv n^{\\prime}\\) (mod \\(p\\) ). \n\nProof. Pick a large integer \\(M\\) such that \n\n\\[\\nu_{p}(M + g(n)),\\quad \\nu_{p}(M + g(n^{\\prime}))\\quad \\mathrm{are~both~odd}.\\] \n\n(It's not hard to see this is always possible.) Now, since each of \n\n\\[(M + g(n))(n + g(M))\\] \\[(M + g(n^{\\prime}))(n^{\\prime} + g(M))\\] \n\nis a square, we get \\(n\\equiv n^{\\prime}\\equiv - g(M)\\) (mod \\(p\\) ). \n\nThis claim implies that \n\n- The numbers \\(g(n)\\) and \\(g(n + 1)\\) differ by \\(\\pm 1\\) for any \\(n\\) , and \n\n- The function \\(g\\) is injective. \n\nIt follows \\(g\\) is a linear function with slope \\(\\pm 1\\) , hence done.\n\n\n\n## \\(\\S 2\\) Solutions to Day 2", "metadata": {"resource_path": "IMO/segmented/en-IMO-2010-notes.jsonl", "problem_match": "3. ", "solution_match": "## \\(\\S 1.3\\) IMO 2010/3, proposed by Gabriel Carroll (USA) \n"}}
-{"year": "2010", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "Let \\(P\\) be a point interior to triangle \\(A B C\\) (with \\(C A\\neq C B\\) ). The lines \\(A P\\) , \\(B P\\) and \\(C P\\) meet again its circumcircle \\(\\Gamma\\) at \\(K\\) , \\(L\\) , \\(M\\) , respectively. The tangent line at \\(C\\) to \\(\\Gamma\\) meets the line \\(A B\\) at \\(S\\) . Show that from \\(S C = S P\\) follows \\(M K = M L\\) .", "solution": "## Problem sta \n\nWe present two solutions using harmonic bundles. \n\n\\(\\P\\) First solution (Evan Chen). Let \\(N\\) be the antipode of \\(M\\) , and let \\(NP\\) meet \\(\\Gamma\\) again at \\(D\\) . Focus only on \\(CDMN\\) for now (ignoring the condition). Then \\(C\\) and \\(D\\) are feet of altitudes in \\(\\triangle MNP\\) ; it is well- known that the circumcircle of \\(\\triangle CDP\\) is orthogonal to \\(\\Gamma\\) (passing through the orthocenter of \\(\\triangle MPN\\) ). \n\n\n \n\nNow, we are given that point \\(S\\) is such that \\(\\overline{SC}\\) is tangent to \\(\\Gamma\\) , and \\(SC = SP\\) . It follows that \\(S\\) is the circumcenter of \\(\\triangle CDP\\) , and hence \\(\\overline{SC}\\) and \\(\\overline{SD}\\) are tangents to \\(\\Gamma\\) . \n\nThen \\(- 1 = (AB;CD) \\stackrel{P}{=} (KL;MN)\\) . Since \\(\\overline{MN}\\) is a diameter, this implies \\(MK = ML\\) . \n\nRemark. I think it's more natural to come up with this solution in reverse. Namely, suppose we define the points the other way: let \\(\\overline{SD}\\) be the other tangent, so \\((AB;CD) = - 1\\) . Then project through \\(P\\) to get \\((KL;MN) = - 1\\) , where \\(N\\) is the second intersection of \\(\\overline{DP}\\) . However, if \\(ML = MK\\) then \\(KMLN\\) must be a kite. Thus one can recover the solution in reverse. \n\n## \\(\\P\\) Second solution (Sebastian Jeon). We have \n\n\\[SP^{2} = SC^{2} = SA\\cdot SB\\Rightarrow \\angle SPA = \\angle PBA = \\angle LBA = \\angle LKA = \\angle LKP\\] \n\n(the latter half is Reim's theorem). Therefore \\(\\overline{SP}\\) and \\(\\overline{LK}\\) are parallel.\n\n\n\nNow, let \\(\\overline{SP}\\) meet \\(\\Gamma\\) again at \\(X\\) and \\(Y\\) , and let \\(Q\\) be the antipode of \\(P\\) on \\((S)\\) . Then \n\n\\[SP^{2} = SQ^{2} = SX\\cdot SY\\Longrightarrow (PQ;XY) = -1\\Longrightarrow \\angle QCP = 90^{\\circ}\\] \n\nthat \\(\\overline{CP}\\) bisects \\(\\angle XCY\\) . Since \\(\\overline{XY} \\parallel \\overline{KL}\\) , it follows \\(\\overline{CP}\\) bisects to \\(\\angle LCK\\) too.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2010-notes.jsonl", "problem_match": "4. ", "solution_match": "## \\(\\S 2.1\\) IMO 2010/4, proposed by Marcin Kuczma (POL) \n"}}
+{"year": "2010", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "Let \\(P\\) be a point interior to triangle \\(A B C\\) (with \\(C A\\neq C B\\) ). The lines \\(A P\\) , \\(B P\\) and \\(C P\\) meet again its circumcircle \\(\\Gamma\\) at \\(K\\) , \\(L\\) , \\(M\\) , respectively. The tangent line at \\(C\\) to \\(\\Gamma\\) meets the line \\(A B\\) at \\(S\\) . Show that from \\(S C = S P\\) follows \\(M K = M L\\) .", "solution": "## Problem sta \n\nWe present two solutions using harmonic bundles. \n\n\\(\\P\\) First solution (Evan Chen). Let \\(N\\) be the antipode of \\(M\\) , and let \\(NP\\) meet \\(\\Gamma\\) again at \\(D\\) . Focus only on \\(CDMN\\) for now (ignoring the condition). Then \\(C\\) and \\(D\\) are feet of altitudes in \\(\\triangle MNP\\) ; it is well- known that the circumcircle of \\(\\triangle CDP\\) is orthogonal to \\(\\Gamma\\) (passing through the orthocenter of \\(\\triangle MPN\\) ). \n\n\n \n\nNow, we are given that point \\(S\\) is such that \\(\\overline{SC}\\) is tangent to \\(\\Gamma\\) , and \\(SC = SP\\) . It follows that \\(S\\) is the circumcenter of \\(\\triangle CDP\\) , and hence \\(\\overline{SC}\\) and \\(\\overline{SD}\\) are tangents to \\(\\Gamma\\) . \n\nThen \\(- 1 = (AB;CD) \\stackrel{P}{=} (KL;MN)\\) . Since \\(\\overline{MN}\\) is a diameter, this implies \\(MK = ML\\) . \n\nRemark. I think it's more natural to come up with this solution in reverse. Namely, suppose we define the points the other way: let \\(\\overline{SD}\\) be the other tangent, so \\((AB;CD) = - 1\\) . Then project through \\(P\\) to get \\((KL;MN) = - 1\\) , where \\(N\\) is the second intersection of \\(\\overline{DP}\\) . However, if \\(ML = MK\\) then \\(KMLN\\) must be a kite. Thus one can recover the solution in reverse. \n\n## \\(\\P\\) Second solution (Sebastian Jeon). We have \n\n\\[SP^{2} = SC^{2} = SA\\cdot SB\\Rightarrow \\angle SPA = \\angle PBA = \\angle LBA = \\angle LKA = \\angle LKP\\] \n\n(the latter half is Reim's theorem). Therefore \\(\\overline{SP}\\) and \\(\\overline{LK}\\) are parallel.\n\n\n\nNow, let \\(\\overline{SP}\\) meet \\(\\Gamma\\) again at \\(X\\) and \\(Y\\) , and let \\(Q\\) be the antipode of \\(P\\) on \\((S)\\) . Then \n\n\\[SP^{2} = SQ^{2} = SX\\cdot SY\\Longrightarrow (PQ;XY) = -1\\Longrightarrow \\angle QCP = 90^{\\circ}\\] \n\nthat \\(\\overline{CP}\\) bisects \\(\\angle XCY\\) . Since \\(\\overline{XY} \\parallel \\overline{KL}\\) , it follows \\(\\overline{CP}\\) bisects to \\(\\angle LCK\\) too.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2010-notes.jsonl", "problem_match": "4. ", "solution_match": "## \\(\\S 2.1\\) IMO 2010/4, proposed by Marcin Kuczma (POL) \n"}}
{"year": "2010", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "Each of the six boxes \\(B_{1}\\) , \\(B_{2}\\) , \\(B_{3}\\) , \\(B_{4}\\) , \\(B_{5}\\) , \\(B_{6}\\) initially contains one coin. The following two types of operations are allowed: \n\na) Choose a non-empty box \\(B_{j}\\) , \\(1\\leq j\\leq 5\\) , remove one coin from \\(B_{j}\\) and add two coins to \\(B_{j + 1}\\) ; \nb) Choose a non-empty box \\(B_{k}\\) , \\(1\\leq k\\leq 4\\) , remove one coin from \\(B_{k}\\) and swap the contents (possibly empty) of the boxes \\(B_{k + 1}\\) and \\(B_{k + 2}\\) . \n\nDetermine if there exists a finite sequence of operations of the allowed types, such that the five boxes \\(B_{1}\\) , \\(B_{2}\\) , \\(B_{3}\\) , \\(B_{4}\\) , \\(B_{5}\\) become empty, while box \\(B_{6}\\) contains exactly \\(2010^{2010^{2010}}\\) coins.", "solution": "Each o \n\nFirst, \n\n\\[(1,1,1,1,1,1) \\to (0,3,1,0,3,1) \\to (0,0,7,0,0,7)\\] \\[\\to (0,0,6,2,0,7) \\to (0,0,6,1,2,7) \\to (0,0,6,1,0,11)\\] \\[\\to (0,0,6,0,11,0) \\to (0,0,5,11,0,0).\\] \n\nand henceforth we ignore boxes \\(B_{1}\\) and \\(B_{2}\\) , looking at just the last four boxes; so we write the current position as \\((5,11,0,0)\\) . \n\nWe prove a lemma: \n\nClaim — Let \\(k \\geq 0\\) and \\(n > 0\\) . From \\((k, n, 0, 0)\\) we may reach \\((k - 1, 2^{n}, 0, 0)\\) . \n\nProof. Working with only the last three boxes for now, \n\n\\[(n,0,0)\\to (n - 1,2,0)\\to (n - 1,0,4)\\] \\[\\to (n - 2,4,0)\\to (n - 2,0,8)\\] \\[\\to (n - 3,8,0)\\to (n - 3,0,16)\\] \\[\\to \\dots \\to (1,2^{n - 1},0)\\to (1,0,2^{n})\\to (0,2^{n},0).\\] \n\nFinally we have \\((k, n, 0, 0) \\to (k, 0, 2^{n}, 0) \\to (k - 1, 2^{n}, 0, 0)\\) . \n\nNow from \\((5,11,0,0)\\) we go as follows: \n\n\\[(5,11,0,0)\\to (4,2^{11},0,0)\\to \\left(3,2^{21},0,0\\right)\\to \\left(2,2^{22^{11}},0,0\\right)\\] \\[\\to \\left(1,2^{22^{11}},0,0\\right)\\to \\left(0,2^{22^{21^{11}}},0,0\\right).\\] \n\nLet \\(A = 2^{22^{22^{11}}} > 2010^{2010^{2010}} = B\\) . Then by using move 2 repeatedly on the fourth box (i.e., throwing away several coins by swapping the empty \\(B_{5}\\) and \\(B_{6}\\) ), we go from \\((0, A, 0, 0)\\) to \\((0, B / 4, 0, 0)\\) . From there we reach \\((0, 0, 0, B)\\) .", "metadata": {"resource_path": "IMO/segmented/en-IMO-2010-notes.jsonl", "problem_match": "5. ", "solution_match": "## \\(\\S 2.2\\) IMO 2010/5, proposed by Netherlands \n"}}
{"year": "2010", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "Let \\(a_{1},a_{2},a_{3},\\ldots\\) be a sequence of positive real numbers, and \\(s\\) be a positive integer, such that \n\n\\[a_{n} = \\max \\{a_{k} + a_{n - k}\\mid 1\\leq k\\leq n - 1\\} \\mathrm{~for~all~}n > s.\\] \n\nProve there exist positive integers \\(\\ell \\leq s\\) and \\(N\\) , such that \n\n\\[a_{n} = a_{\\ell} + a_{n - \\ell}\\mathrm{~for~all~}n\\geq N.\\]", "solution": "Let \n\n\\[w_{1} = \\frac{a_{1}}{1},\\quad w_{2} = \\frac{a_{2}}{2},\\quad \\ldots ,\\quad w_{s} = \\frac{a_{s}}{s}.\\] \n\n(The choice of the letter \\(w\\) is for \"weight\".) We claim the right choice of \\(\\ell\\) is the one maximizing \\(w_{\\ell}\\) . \n\nOur plan is to view each \\(a_{n}\\) as a linear combination of the weights \\(w_{1},\\ldots ,w_{s}\\) and track their coefficients. \n\nTo this end, let's define an \\(n\\) - type to be a vector \\(T = \\langle t_{1},\\ldots ,t_{s}\\rangle\\) of nonnegative integers such that \n\n\\(n = t_{1} + \\dots +t_{s}\\) ; and \n\n\\(t_{i}\\) is divisible by \\(i\\) for every \\(i\\) . \n\nWe then define its valuation as \\(v(T) = \\sum_{i = 1}^{s}w_{i}t_{i}\\) . \n\nNow we define a \\(n\\) - type to be valid according to the following recursive rule. For \\(1\\leq n\\leq s\\) the only valid \\(n\\) - types are \n\n\\[T_{1} = \\langle 1,0,0,\\ldots ,0\\rangle\\] \\[T_{2} = \\langle 0,2,0,\\ldots ,0\\rangle\\] \\[T_{3} = \\langle 0,0,3,\\ldots ,0\\rangle\\] \\[\\vdots\\] \\[T_{s} = \\langle 0,0,0,\\ldots ,s\\rangle\\] \n\nfor \\(n = 1,\\ldots ,s\\) , respectively. Then for any \\(n > s\\) , an \\(n\\) - type is valid if it can be written as the sum of a valid \\(k\\) - type and a valid \\((n - k)\\) - type, componentwise. These represent the linear combinations possible in the recursion; in other words the recursion in the problem is phrased as \n\n\\[a_{n} = \\max_{T\\mathrm{~is~a~valid~}n\\mathrm{-type}}v(T).\\] \n\nIn fact, we have the following description of valid \\(n\\) - types: \n\nClaim — Assume \\(n > s\\) . Then an \\(n\\) - type \\(\\langle t_{1},\\ldots ,t_{s}\\rangle\\) is valid if and only if either \n\nthere exist indices \\(i< j\\) with \\(i + j > s\\) , \\(t_{i}\\geq i\\) and \\(t_{j}\\geq j\\) ; or \n\nthere exists an index \\(i > s / 2\\) with \\(t_{i}\\geq 2i\\) .\n\n\n\nProof. Immediate by forwards induction on \\(n > s\\) that all \\(n\\) - types have this property. \n\nThe reverse direction is by downwards induction on \\(n\\) . Indeed if \\(\\sum_{i} \\frac{t_{i}}{i} > 2\\) , then we may subtract off on of \\(\\{T_{1}, \\ldots , T_{s}\\}\\) while preserving the condition; and the case \\(\\sum_{i} \\frac{t_{i}}{i} = 2\\) is essentially by definition. \\(\\square\\) \n\nRemark. The claim is a bit confusingly stated in its two cases; really the latter case should be thought of as the situation \\(i = j\\) but requiring that \\(t_{i} / i\\) is counted with multiplicity. \n\nNow, for each \\(n > s\\) we pick a valid \\(n\\) - type \\(T_{n}\\) with \\(a_{n} = v(T_{n})\\) ; if there are ties, we pick one for which the \\(\\ell\\) th entry is as large as possible. \n\nClaim — For any \\(n > s\\) and index \\(i \\neq \\ell\\) , the \\(i\\) th entry of \\(T_{n}\\) is at most \\(2s + \\ell i\\) . \n\nProof. If not, we can go back \\(i\\ell\\) steps to get a valid \\((n - i\\ell)\\) - type \\(T\\) achieved by decreasing the \\(i\\) th entry of \\(T_{n}\\) by \\(i\\ell\\) . But then we can add \\(\\ell\\) to the \\(\\ell\\) th entry \\(i\\) times to get another \\(n\\) - type \\(T^{\\prime}\\) which obviously has valuation at least as large, but with larger \\(\\ell\\) th entry. \\(\\square\\) \n\nNow since all other entries in \\(T_{n}\\) are bounded, eventually the sequence \\((T_{n})_{n > s}\\) just consists of repeatedly adding 1 to the \\(\\ell\\) th entry, as required. \n\nRemark. One big step is to consider \\(w_{k} = a_{k} / k\\) . You can get this using wishful thinking or by examining small cases. (In addition this normalization makes it easier to see why the largest \\(w\\) plays an important role, since then in the definition of type, the \\(n\\) - types all have a sum of \\(n\\) . Unfortunately, it makes the characterization of valid \\(n\\) - types somewhat clumsier too.)", "metadata": {"resource_path": "IMO/segmented/en-IMO-2010-notes.jsonl", "problem_match": "6. ", "solution_match": "## \\(\\S 2.3\\) IMO 2010/6, proposed by Morteza Saghafiyan (IRN) \n"}}
diff --git a/IMO/segmented/en-IMO-2011-notes.jsonl b/IMO/segmented/en-IMO-2011-notes.jsonl
index ee6d5e9534cf9a9501957e4d0a127f93eac79d25..7994396529b080b8c940f946ad0c7e88fa7b033f 100644
--- a/IMO/segmented/en-IMO-2011-notes.jsonl
+++ b/IMO/segmented/en-IMO-2011-notes.jsonl
@@ -3,4 +3,4 @@
{"year": "2011", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "Let \\(f\\colon \\mathbb{R}\\to \\mathbb{R}\\) be a real-valued function defined on the set of real numbers that satisfies \n\n\\[f(x + y)\\leq yf(x) + f(f(x))\\] \n\nfor all real numbers \\(x\\) and \\(y\\) . Prove that \\(f(x) = 0\\) for all \\(x\\leq 0\\)", "solution": ") \n\nWe begin by rewriting the given as \n\n\\[f(z)\\leq (z - x)f(x) + f(f(x))\\quad \\forall x,z\\in \\mathbb{R}\\qquad (\\heartsuit)\\] \n\n(which is better anyways since control over inputs to \\(f\\) is more valuable). We start by eliminating the double \\(f\\) : let \\(z = f(w)\\) to get \n\n\\[f(f(w))\\leq (f(w) - x)f(x) + f(f(x))\\] \n\nand then use the symmetry trick to write \n\n\\[f(f(x))\\leq (f(x) - w)f(w) + f(f(w))\\] \n\nso that when we sum we get \n\n\\[w f(w) + x f(x)\\leq 2f(x)f(w).\\] \n\nNext we use cancellation trick: set \\(w = 2f(x)\\) in the above to get \n\n\\[x f(x)\\leq 0\\quad \\forall x\\in \\mathbb{R}.\\qquad (\\spadesuit)\\] \n\nClaim — For every \\(p\\in \\mathbb{R}\\) , we have \\(f(p)\\leq 0\\) \n\nProof. Assume \\(f(p) > 0\\) for some \\(p\\in \\mathbb{R}\\) . Then for any negative number \\(z\\) \n\n\\[0\\leq f(z)\\leq \\stackrel {(\\heartsuit)}{(z - p)}f(p) + f(f(p)).\\] \n\nwhich is false if we let \\(z\\to - \\infty\\) \n\nTogether with \\((\\spadesuit)\\) we derive \\(f(x) = 0\\) for \\(x< 0\\) . Finally, letting \\(x\\) and \\(z\\) be any negative numbers in \\((\\heartsuit)\\) , we get \\(f(0)\\geq 0\\) , so \\(f(0) = 0\\) too. \n\nRemark. As another corollary of the claim, \\(f(f(x)) = 0\\) for all \\(x\\) . \n\nRemark. A nontrivial example of a working \\(f\\) is to take \n\n \n\nor some other negative function growing rapidly in absolute value for \\(x > 0\\) .\n\n\n\n■\n\n\n\n## \\(\\S 2\\) Solutions to Day 2", "metadata": {"resource_path": "IMO/segmented/en-IMO-2011-notes.jsonl", "problem_match": "3. ", "solution_match": "## \\(\\S 1.3\\) IMO 2011/3, proposed by Igor Voronovich, BLR \n"}}
{"year": "2011", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "Let \\(n > 0\\) be an integer. We are given a balance and \\(n\\) weights of weight \\(2^{0}\\) , \\(2^{1}\\) , ..., \\(2^{n - 1}\\) . We are to place each of the \\(n\\) weights on the balance, one after another, in such a way that the right pan is never heavier than the left pan. At each step we choose one of the weights that has not yet been placed on the balance, and place it on either the left pan or the right pan, until all of the weights have been placed. Determine the number of ways in which this can be done.", "solution": "The answer is \\(a_{n} = (2n - 1)!!\\) \n\nWe refer to what we're counting as a valid \\(n\\) - sequence: an order of which weights to place, and whether to place them on the left or right pan. \n\nWe use induction, with \\(n = 1\\) being obvious. Now consider the weight \\(2^{0} = 1\\) \n\n- If we delete it from any valid \\(n\\) -sequence, we get a valid \\((n - 1)\\) -sequence with all weights doubled. \n\n- Given a valid \\((n - 1)\\) -sequence with all weights doubled, we may insert \\(2^{0} = 1\\) it into \\(2n - 1\\) ways. Indeed, we may insert it anywhere, and designate it either left or right, except we may not designate right if we choose to insert \\(2^{0} = 1\\) at the very beginning. \n\nConsequently, we have that \n\n\\[a_{n} = (2n - 1)\\cdot a_{n - 1}.\\] \n\nSince \\(a_{1} = 1\\) , the conclusion follows. \n\nRemark (Gabriel Levin). An alternate approach can be done by considering the heaviest weight \\(2^{n - 1}\\) instead of the lightest weight \\(2^{0} = 1\\) . This gives a more complicated recursion: \n\n\\[a_{n} = \\sum_{k = 0}^{n - 1}\\binom{n - 1}{k}2^{n - k - 1}(n - k - 1)!a_{k}\\] \n\nby summing over the index \\(k\\) corresponding to the number of weights placed before the heaviest weight \\(2^{n - 1}\\) is placed. \n\nThis recursion can then be rewritten as \n\n\\[a_{n} = \\sum_{k = 0}^{n - 1}\\binom{n - 1}{k}2^{n - k - 1}(n - k - 1)!a_{k}\\] \\[\\qquad = (n - 1)!\\sum_{k = 0}^{n - 1}\\frac{2^{n - k - 1}a_{k}}{k!}\\] \\[\\qquad = 2(n - 1)!\\sum_{k = 0}^{n - 1}\\frac{2^{(n - 1) - k - 1}a_{k}}{k!}\\]\n\n\n\n\\[= 2(n - 1)(n - 2)!\\left(\\sum_{k = 0}^{(n - 1) - 1}\\frac{2^{(n - 1) - k - 1}a_k}{k!}\\right) + 2(n - 1)!\\frac{\\frac{1}{2}a_{n - 1}}{(n - 1)!}\\] \\[= 2(n - 1)a_{n - 1} + a_{n - 1}\\] \\[= (2n - 1)a_{n - 1}\\] \n\nand since \\(a_{1} = 1\\) , we have \\(a_{n} = (2n - 1)!!\\) by induction on \\(n\\) .", "metadata": {"resource_path": "IMO/segmented/en-IMO-2011-notes.jsonl", "problem_match": "4. ", "solution_match": "## \\(\\S 2.1\\) IMO 2011/4, proposed by Morteza Saghafian (IRN) \n"}}
{"year": "2011", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "Let \\(f\\colon \\mathbb{Z}\\to \\mathbb{Z}_{>0}\\) be a function such that \\(f(m - n)\\mid f(m) - f(n)\\) for \\(m,n\\in \\mathbb{Z}\\) Prove that if \\(m,n\\in \\mathbb{Z}\\) satisfy \\(f(m)\\leq f(n)\\) then \\(f(m)\\mid f(n)\\)", "solution": "\\) \n\nLet \\(P(m,n)\\) denote the given assertion. First, we claim \\(f\\) is even. This is straight calculation: \n\n\\[\\bullet P(x,0)\\Rightarrow f(x)\\mid f(x) - f(0)\\Rightarrow f(x)\\mid M:= f(0).\\] \\[\\bullet P(0,x)\\Rightarrow f(-x)\\mid M - f(x)\\Rightarrow f(-x)\\mid f(x).\\mathrm{Analogously},f(x)\\mid f(-x).\\mathrm{So\\] \\[f(x) = f(-x)\\mathrm{and} f\\mathrm{~is~even.}\\] \n\nClaim — Let \\(x,y,z\\) be integers with \\(x + y + z = 0\\) . Then among \\(f(x),f(y),f(z)\\) two of them are equal and divide the third. \n\nProof. Let \\(a = f(\\pm x)\\) , \\(b = f(\\pm y)\\) , \\(c = f(\\pm z)\\) be positive integers. Note that \n\n\\[a\\mid b - c\\] \\[b\\mid c - a\\] \\[c\\mid a - b\\] \n\nfrom \\(P(y, - z)\\) and similarly. WLOG \\(c = \\max (a,b,c)\\) ; then \\(c > |a - b|\\) so \\(a = b\\) . Thus \\(a = b\\mid c\\) from the first two. \\(\\square\\) \n\nThis implies the problem, by taking \\(x\\) and \\(y\\) in the previous claim to be the integers \\(m\\) and \\(n\\) . \n\nRemark. At https://aops.com/community/c6h418981p2381909, Davi Medeiros gives the following characterization of functions \\(f\\) satisfying the hypothesis. \n\nPick \\(f(0)\\) , \\(k\\) positive integers, a chain \\(d_{1}\\mid d_{2}\\mid \\dots \\mid d_{k}\\) of divisors of \\(f(0)\\) , and positive integers \\(a_{1},a_{2},\\ldots ,a_{k - 1}\\) , greater than 1 (if \\(k = 1\\) , \\(a_{i}\\) doesn't exist, for every \\(i\\) ). We'll define \\(f\\) as follows: \n\n\\(f(n) = d_{1}\\) , for every integer \\(n\\) that is not divisible by \\(a_{1}\\) ; \\(f(a_{1}n) = d_{2}\\) , for every integer \\(n\\) that is not divisible by \\(a_{2}\\) ; \\(f(a_{1}a_{2}n) = d_{3}\\) , for every integer \\(n\\) that is not divisible by \\(a_{3}\\) ; \\(f(a_{1}a_{2}a_{3}n) = d_{4}\\) , for every integer \\(n\\) that is not divisible by \\(a_{4}\\) ; \\(\\dots\\) \\(f(a_{1}a_{2}\\ldots a_{k - 1}n) = d_{k}\\) , for every integer \\(n\\) ;", "metadata": {"resource_path": "IMO/segmented/en-IMO-2011-notes.jsonl", "problem_match": "5. ", "solution_match": "## \\(\\S 2.2\\) IMO 2011/5, proposed by Mahyar Sefidgaran (IRN) \n"}}
-{"year": "2011", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "Let \\(A B C\\) be an acute triangle with circumcircle \\(\\Gamma\\) . Let \\(\\ell\\) be a tangent line to \\(\\Gamma\\) and let \\(\\ell_{a}\\) , \\(\\ell_{b}\\) , \\(\\ell_{c}\\) be the lines obtained by reflecting \\(\\ell\\) in the lines \\(B C\\) , \\(C A\\) , and \\(A B\\) , respectively. Show that the circumcircle of the triangle determined by the lines \\(\\ell_{a}\\) , \\(\\ell_{b}\\) , and \\(\\ell_{c}\\) is tangent to the circle \\(\\Gamma\\) .", "solution": "This is a hard problem with many beautiful solutions. The following solution is not very beautiful but not too hard to find during an olympiad, as the only major insight it requires is the construction of \\(A_{2}\\) , \\(B_{2}\\) , and \\(C_{2}\\) . \n\n\n \n\nWe apply complex numbers with \\(\\omega\\) the unit circle and \\(p = 1\\) . Let \\(A_{1} = \\ell_{B} \\cap \\ell_{C}\\) , and let \\(a_{2} = a^{2}\\) (in other words, \\(A_{2}\\) is the reflection of \\(P\\) across the diameter of \\(\\omega\\) through \\(A\\) ). Define the points \\(B_{1}\\) , \\(C_{1}\\) , \\(B_{2}\\) , \\(C_{2}\\) similarly. \n\nWe claim that \\(\\overline{A_{1}A_{2}}\\) , \\(\\overline{B_{1}B_{2}}\\) , \\(\\overline{C_{1}C_{2}}\\) concur at a point on \\(\\Gamma\\) . \n\nWe begin by finding \\(A_{1}\\) . If we reflect the points \\(1 + i\\) and \\(1 - i\\) over \\(\\overline{AB}\\) , then we get two points \\(Z_{1}\\) , \\(Z_{2}\\) with \n\n\\[z_{1} = a + b - a b(1 - i) = a + b - a b + a b i\\] \\[z_{2} = a + b - a b(1 + i) = a + b - a b - a b i.\\] \n\nTherefore, \n\n\\[z_{1} - z_{2} = 2a b i\\] \\[\\overline{{z_{1}z_{2}}} -\\overline{{z_{2}}} z_{1} = -2i\\left(a + b + \\frac{1}{a} +\\frac{1}{b} -2\\right).\\]\n\n\n\nNow \\(\\ell_{C}\\) is the line \\(\\overline{Z_{1}Z_{2}}\\) , so with the analogous equation \\(\\ell_{B}\\) we obtain: \n\n\\[a_{1} = \\frac{-2i\\left(a + b + \\frac{1}{a} +\\frac{1}{b} - 2\\right)(2a c i) + 2i\\left(a + c + \\frac{1}{a} +\\frac{1}{c} - 2\\right)(2a b i)}{\\left(-\\frac{2}{a b} i\\right)(2a c i) - \\left(-\\frac{2}{a c} i\\right)(2a b i)}\\] \\[\\quad = \\frac{\\left[c - b\\right]a^{2} + \\left[\\frac{c}{b} -\\frac{b}{c} -2c + 2b\\right]a + (c - b)}{\\frac{c}{b} -\\frac{b}{c}}\\] \\[\\quad = a + \\frac{(c - b)\\left[a^{2} - 2a + 1\\right]}{(c - b)(c + b) / b c}\\] \\[\\quad = a + \\frac{b c}{b + c} (a - 1)^{2}.\\] \n\nThen the second intersection of \\(\\overline{A_{1}A_{2}}\\) with \\(\\omega\\) is given by \n\n\\[\\frac{a_{1} - a_{2}}{1 - a_{2}\\overline{a_{1}}} = \\frac{a + \\frac{b c}{b + c}(a - 1)^{2} - a^{2}}{1 - a - a^{2}\\cdot\\frac{(1 - 1 / a)^{2}}{b + c}}\\] \\[\\qquad = \\frac{a + \\frac{b c}{b + c}(1 - a)}{1 - \\frac{1}{b + c}(1 - a)}\\] \\[\\qquad = \\frac{a b + b c + c a - a b c}{a + b + c - 1}.\\] \n\nThus, the claim is proved. \n\nFinally, it suffices to show \\(\\overline{A_{1}B_{1}}\\parallel \\overline{A_{2}B_{2}}\\) . One can also do this with complex numbers; it amounts to showing \\(a^{2} - b^{2}\\) , \\(a - b\\) , \\(i\\) (corresponding to \\(\\overline{A_{2}B_{2}}\\) , \\(\\overline{A_{1}B_{1}}\\) , \\(\\overline{PP}\\) ) have their arguments an arithmetic progression, equivalently \n\n\\[\\frac{(a - b)^{2}}{i(a^{2} - b^{2})}\\in \\mathbb{R}\\iff \\frac{(a - b)^{2}}{i(a^{2} - b^{2})} = \\frac{\\left(\\frac{1}{a} - \\frac{1}{b}\\right)^{2}}{\\frac{1}{i}\\left(\\frac{1}{a^{2}} - \\frac{1}{b^{2}}\\right)}\\] \n\nwhich is obvious. \n\nRemark. One can use directed angle chasing for this last part too. Let \\(\\overline{BC}\\) meet \\(\\ell\\) at \\(K\\) and \\(\\overline{B_{2}C_{2}}\\) meet \\(\\ell\\) at \\(L\\) . Evidently \n\n\\[-\\angle B_{2}L P = \\angle L P B_{2} + \\angle P B_{2}L\\] \\[\\qquad = 2\\angle K P B + \\angle P B_{2}C_{2}\\] \\[\\qquad = 2\\angle K P B + 2\\angle P B C\\] \\[\\qquad = -2\\angle P K B\\] \\[\\qquad = \\angle P K B_{1}\\] \n\nas required.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2011-notes.jsonl", "problem_match": "6. ", "solution_match": "## \\(\\S 2.3\\) IMO 2011/6, proposed by Japan \n"}}
+{"year": "2011", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "Let \\(A B C\\) be an acute triangle with circumcircle \\(\\Gamma\\) . Let \\(\\ell\\) be a tangent line to \\(\\Gamma\\) and let \\(\\ell_{a}\\) , \\(\\ell_{b}\\) , \\(\\ell_{c}\\) be the lines obtained by reflecting \\(\\ell\\) in the lines \\(B C\\) , \\(C A\\) , and \\(A B\\) , respectively. Show that the circumcircle of the triangle determined by the lines \\(\\ell_{a}\\) , \\(\\ell_{b}\\) , and \\(\\ell_{c}\\) is tangent to the circle \\(\\Gamma\\) .", "solution": "This is a hard problem with many beautiful solutions. The following solution is not very beautiful but not too hard to find during an olympiad, as the only major insight it requires is the construction of \\(A_{2}\\) , \\(B_{2}\\) , and \\(C_{2}\\) . \n\n\n \n\nWe apply complex numbers with \\(\\omega\\) the unit circle and \\(p = 1\\) . Let \\(A_{1} = \\ell_{B} \\cap \\ell_{C}\\) , and let \\(a_{2} = a^{2}\\) (in other words, \\(A_{2}\\) is the reflection of \\(P\\) across the diameter of \\(\\omega\\) through \\(A\\) ). Define the points \\(B_{1}\\) , \\(C_{1}\\) , \\(B_{2}\\) , \\(C_{2}\\) similarly. \n\nWe claim that \\(\\overline{A_{1}A_{2}}\\) , \\(\\overline{B_{1}B_{2}}\\) , \\(\\overline{C_{1}C_{2}}\\) concur at a point on \\(\\Gamma\\) . \n\nWe begin by finding \\(A_{1}\\) . If we reflect the points \\(1 + i\\) and \\(1 - i\\) over \\(\\overline{AB}\\) , then we get two points \\(Z_{1}\\) , \\(Z_{2}\\) with \n\n\\[z_{1} = a + b - a b(1 - i) = a + b - a b + a b i\\] \\[z_{2} = a + b - a b(1 + i) = a + b - a b - a b i.\\] \n\nTherefore, \n\n\\[z_{1} - z_{2} = 2a b i\\] \\[\\overline{{z_{1}z_{2}}} -\\overline{{z_{2}}} z_{1} = -2i\\left(a + b + \\frac{1}{a} +\\frac{1}{b} -2\\right).\\]\n\n\n\nNow \\(\\ell_{C}\\) is the line \\(\\overline{Z_{1}Z_{2}}\\) , so with the analogous equation \\(\\ell_{B}\\) we obtain: \n\n\\[a_{1} = \\frac{-2i\\left(a + b + \\frac{1}{a} +\\frac{1}{b} - 2\\right)(2a c i) + 2i\\left(a + c + \\frac{1}{a} +\\frac{1}{c} - 2\\right)(2a b i)}{\\left(-\\frac{2}{a b} i\\right)(2a c i) - \\left(-\\frac{2}{a c} i\\right)(2a b i)}\\] \\[\\quad = \\frac{\\left[c - b\\right]a^{2} + \\left[\\frac{c}{b} -\\frac{b}{c} -2c + 2b\\right]a + (c - b)}{\\frac{c}{b} -\\frac{b}{c}}\\] \\[\\quad = a + \\frac{(c - b)\\left[a^{2} - 2a + 1\\right]}{(c - b)(c + b) / b c}\\] \\[\\quad = a + \\frac{b c}{b + c} (a - 1)^{2}.\\] \n\nThen the second intersection of \\(\\overline{A_{1}A_{2}}\\) with \\(\\omega\\) is given by \n\n\\[\\frac{a_{1} - a_{2}}{1 - a_{2}\\overline{a_{1}}} = \\frac{a + \\frac{b c}{b + c}(a - 1)^{2} - a^{2}}{1 - a - a^{2}\\cdot\\frac{(1 - 1 / a)^{2}}{b + c}}\\] \\[\\qquad = \\frac{a + \\frac{b c}{b + c}(1 - a)}{1 - \\frac{1}{b + c}(1 - a)}\\] \\[\\qquad = \\frac{a b + b c + c a - a b c}{a + b + c - 1}.\\] \n\nThus, the claim is proved. \n\nFinally, it suffices to show \\(\\overline{A_{1}B_{1}}\\parallel \\overline{A_{2}B_{2}}\\) . One can also do this with complex numbers; it amounts to showing \\(a^{2} - b^{2}\\) , \\(a - b\\) , \\(i\\) (corresponding to \\(\\overline{A_{2}B_{2}}\\) , \\(\\overline{A_{1}B_{1}}\\) , \\(\\overline{PP}\\) ) have their arguments an arithmetic progression, equivalently \n\n\\[\\frac{(a - b)^{2}}{i(a^{2} - b^{2})}\\in \\mathbb{R}\\iff \\frac{(a - b)^{2}}{i(a^{2} - b^{2})} = \\frac{\\left(\\frac{1}{a} - \\frac{1}{b}\\right)^{2}}{\\frac{1}{i}\\left(\\frac{1}{a^{2}} - \\frac{1}{b^{2}}\\right)}\\] \n\nwhich is obvious. \n\nRemark. One can use directed angle chasing for this last part too. Let \\(\\overline{BC}\\) meet \\(\\ell\\) at \\(K\\) and \\(\\overline{B_{2}C_{2}}\\) meet \\(\\ell\\) at \\(L\\) . Evidently \n\n\\[-\\angle B_{2}L P = \\angle L P B_{2} + \\angle P B_{2}L\\] \\[\\qquad = 2\\angle K P B + \\angle P B_{2}C_{2}\\] \\[\\qquad = 2\\angle K P B + 2\\angle P B C\\] \\[\\qquad = -2\\angle P K B\\] \\[\\qquad = \\angle P K B_{1}\\] \n\nas required.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2011-notes.jsonl", "problem_match": "6. ", "solution_match": "## \\(\\S 2.3\\) IMO 2011/6, proposed by Japan \n"}}
diff --git a/IMO/segmented/en-IMO-2012-notes.jsonl b/IMO/segmented/en-IMO-2012-notes.jsonl
index bd674d6e67863138c4e3497447e9f1157be32373..f38d30b5fc7e50d1bc236bea637249a8e0e972ff 100644
--- a/IMO/segmented/en-IMO-2012-notes.jsonl
+++ b/IMO/segmented/en-IMO-2012-notes.jsonl
@@ -2,5 +2,5 @@
{"year": "2012", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "Let \\(a_{2}\\) , \\(a_{3}\\) , ..., \\(a_{n}\\) be positive reals with product 1, where \\(n \\geq 3\\) . Show that \n\n\\[(1 + a_{2})^{2}(1 + a_{3})^{3}\\dots (1 + a_{n})^{n} > n^{n}.\\]", "solution": "L \n\nTry the dumbest thing possible: by AM- GM, \n\n\\[(1 + a_{2})^{2}\\geq 2^{2}a_{2}\\] \\[(1 + a_{3})^{3} = \\left(\\frac{1}{2} +\\frac{1}{2} +a_{3}\\right)^{3}\\geq \\frac{3^{3}}{2^{2}}a_{3}\\] \\[(1 + a_{4})^{4} = \\left(\\frac{1}{3} +\\frac{1}{3} +\\frac{1}{3} +a_{4}\\right)^{4}\\geq \\frac{4^{4}}{3^{3}}a_{4}\\] \\[\\qquad \\vdots\\] \n\nand so on. Multiplying these all gives the result. The inequality is strict since it's not possible that \\(a_{2} = 1\\) , \\(a_{3} = \\frac{1}{2}\\) , et cetera.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2012-notes.jsonl", "problem_match": "2. ", "solution_match": "## \\(\\S 1.2\\) IMO 2012/2, proposed by Angelo di Pasquale (AUS) \n"}}
{"year": "2012", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "The liar's guessing game is a game played between two players \\(A\\) and \\(B\\) . The rules of the game depend on two fixed positive integers \\(k\\) and \\(n\\) which are known to both players. \n\nAt the start of the game \\(A\\) chooses integers \\(x\\) and \\(N\\) with \\(1 \\leq x \\leq N\\) . Player \\(A\\) keeps \\(x\\) secret, and truthfully tells \\(N\\) to player \\(B\\) . Player \\(B\\) now tries to obtain information about \\(x\\) by asking player \\(A\\) questions as follows: each question consists of \\(B\\) specifying an arbitrary set \\(S\\) of positive integers (possibly one specified in some previous question), and asking \\(A\\) whether \\(x\\) belongs to \\(S\\) . Player \\(B\\) may ask as many questions as he wishes. After each question, player \\(A\\) must immediately answer it with yes or no, but is allowed to lie as many times as she wants; the only restriction is that, among any \\(k + 1\\) consecutive answers, at least one answer must be truthful. \n\nAfter \\(B\\) has asked as many questions as he wants, he must specify a set \\(X\\) of at most \\(n\\) positive integers. If \\(x\\) belongs to \\(X\\) , then \\(B\\) wins; otherwise, he loses. Prove that: \n\n(a) If \\(n \\geq 2^{k}\\) , then \\(B\\) can guarantee a win. \n(b) For all sufficiently large \\(k\\) , there exists an integer \\(n \\geq (1.99)^{k}\\) such that \\(B\\) cannot guarantee a win.", "solution": "Call the players Alice and Bob. \n\nPart (a): We prove the following. \n\nClaim — If \\(N\\geq 2^{k} + 1\\) , then in \\(2k + 1\\) questions, Bob can rule out some number in \\(\\{1,\\ldots ,2^{k} + 1\\}\\) form being equal to \\(x\\) . \n\nProof. First, Bob asks the question \\(S_{0} = \\{2^{k} + 1\\}\\) until Alice answers \"yes\" or until Bob has asked \\(k + 1\\) questions. If Alice answers \"no\" to all of these then Bob rules out \\(2^{k} + 1\\) . So let's assume Alice just said \"yes\". \n\nNow let \\(T = \\{1,\\ldots ,2^{k}\\}\\) . Then, he asks \\(k\\) - follow up questions \\(S_{1}\\) , ..., \\(S_{k}\\) defined as follows: \n\n- \\(S_{1} = \\{1,3,5,7,\\ldots ,2^{k} - 1\\}\\) consists of all numbers in \\(T\\) whose least significant digit in binary is 1. \n- \\(S_{2} = \\{2,3,6,7,\\ldots ,2^{k} - 2,2^{k} - 1\\}\\) consists of all numbers in \\(T\\) whose second least significant digit in binary is 1. \n- More generally \\(S_{i}\\) consists of all numbers in \\(T\\) whose \\(i\\)th least significant digit in binary is 1. \n\nWLOG Alice answers these all as \"yes\" (the other cases are similar). Among the last \\(k + 1\\) answers at least one must be truthful, and the number \\(2^{k}\\) (having zeros in all relevant digits) does not appear in any of \\(S_{0}\\) , ..., \\(S_{k}\\) and is ruled out. \\(\\square\\)\n\n\n\nThus in this way Bob can repeatedly find non- possibilities for \\(x\\) (and then relabel the remaining candidates \\(1, \\ldots , N - 1\\) ) until he arrives at a set of at most \\(2^{k}\\) numbers. \n\nPart (b): It suffices to consider \\(n = \\lceil 1.99^{k}\\rceil\\) and \\(N = n + 1\\) for large \\(k\\) . At the \\(t\\) th step, Bob asks some question \\(S_{t}\\) ; we phrase each of Alice's answers in the form \" \\(x \\notin B_{t}\\) \", where \\(B_{t}\\) is either \\(S_{t}\\) or its complement. (You may think of these as \"bad sets\"; the idea is to show we can avoid having any number appear in \\(k + 1\\) consecutive bad sets, preventing Bob from ruling out any numbers.) \n\nMain idea: for every number \\(1 \\leq x \\leq N\\) , at time step \\(t\\) we define its weight to be \n\n\\[w(x) = 1.998^{e}\\] \n\nwhere \\(e\\) is the largest number such that \\(x \\in B_{t - 1} \\cap B_{t - 2} \\cap \\dots \\cap B_{t - e}\\) . \n\nClaim — Alice can ensure the total weight never exceeds \\(1.998^{k + 1}\\) for large \\(k\\) . \n\nProof. Let \\(W_{t}\\) denote the sum of weights after the \\(t\\) th question. We have \\(W_{0} = N < 1000n\\) . We will prove inductively that \\(W_{t} < 1000n\\) always. \n\nAt time \\(t\\) , Bob specifies a question \\(S_{t}\\) . We have Alice choose \\(B_{t}\\) as whichever of \\(S_{t}\\) or \\(\\overline{S_{t}}\\) has lesser total weight, hence at most \\(W_{t} / 2\\) . The weights of for \\(B_{t}\\) increase by a factor of 1.998, while the weights for \\(\\overline{B_{t}}\\) all reset to 1. So the new total weight after time \\(t\\) is \n\n\\[W_{t + 1} \\leq 1.998 \\cdot \\frac{W_{t}}{2} + \\# \\overline{B_{t}} \\leq 0.999W_{t} + n.\\] \n\nThus if \\(W_{t} < 1000n\\) then \\(W_{t + 1} < 1000n\\) . \n\nTo finish, note that \\(1000n < 1000 \\left(1.99^{k} + 1\\right) < 1.998^{k + 1}\\) for \\(k\\) large. \n\nIn particular, no individual number can have weight \\(1.998^{k + 1}\\) . Thus for every time step \\(t\\) we have \n\n\\[B_{t} \\cap B_{t + 1} \\cap \\dots \\cap B_{t + k} = \\emptyset .\\] \n\nThen once Bob stops, if he declares a set of \\(n\\) positive integers, and \\(x\\) is an integer Bob did not choose, then Alice's question history is consistent with \\(x\\) being Alice's number, as among any \\(k + 1\\) consecutive answers she claimed that \\(x \\in \\overline{B_{t}}\\) for some \\(t\\) in that range. \n\nRemark (Motivation). In our \\(B_{t}\\) setup, let's think backwards. The problem is equivalent to avoiding \\(e = k + 1\\) at any time step \\(t\\) , for any number \\(x\\) . That means \n\n- have at most two elements with \\(e = k\\) at time \\(t - 1\\) ,- thus have at most four elements with \\(e = k - 1\\) at time \\(t - 2\\) ,- thus have at most eight elements with \\(e = k - 2\\) at time \\(t - 3\\) ,- and so on. \n\nWe already exploited this in solving part (a). In any case it's now natural to try letting \\(w(x) = 2^{e}\\) , so that all the cases above sum to \"equally bad\" situations: since \\(8 \\cdot 2^{k - 2} = 4 \\cdot 2^{k - 1} = 2 \\cdot 2^{k}\\) , say. \n\nHowever, we then get \\(W_{t + 1} \\leq \\frac{1}{2} (2W_{t}) + n\\) , which can increase without bound due to contributions from numbers resetting to zero. The way to fix this is to change the weight to \\(w(x) = 1.998^{e}\\) , taking advantage of the little extra space we have due to having \\(n \\geq 1.99^{k}\\) rather than \\(n \\geq 2^{k}\\) .\n\n\n\n## \\(\\S 2\\) Solutions to Day 2", "metadata": {"resource_path": "IMO/segmented/en-IMO-2012-notes.jsonl", "problem_match": "3. ", "solution_match": "## \\(\\S 1.3\\) IMO 2012/3, proposed by David Arthur (CAN) \n"}}
{"year": "2012", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "Find all functions \\(f: \\mathbb{Z} \\to \\mathbb{Z}\\) such that, for all integers \\(a\\) , \\(b\\) , \\(c\\) that satisfy \\(a + b + c = 0\\) , the following equality holds: \n\n\\[f(a)^{2} + f(b)^{2} + f(c)^{2} = 2f(a)f(b) + 2f(b)f(c) + 2f(c)f(a).\\]", "solution": "arbitrary \\(k\\in \\mathbb{Z}\\) , we have \n\n(i) \\(f(x) = k x^{2}\\) \n\n(ii) \\(f(x) = 0\\) for even \\(x\\) , and \\(f(x) = k\\) for odd \\(x\\) , and \n\n(iii) \\(f(x) = 0\\) for \\(x\\equiv 0\\) (mod 4), \\(f(x) = k\\) for odd \\(x\\) , and \\(f(x) = 4k\\) for \\(x\\equiv 2\\) (mod 4). \n\nThese can be painfully seen to work. (It's more natural to think of these as \\(f(x) = x^{2}\\) , \\(f(x) = x^{2}\\) (mod 4), \\(f(x) = x^{2}\\) (mod 8), and multiples thereof.) \n\nSet \\(a = b = c = 0\\) to get \\(f(0) = 0\\) . Then set \\(c = 0\\) to get \\(f(a) = f(- a)\\) , so \\(f\\) is even. Now \n\n\\[f(a)^{2} + f(b)^{2} + f(a + b)^{2} = 2f(a + b)\\left(f(a) + f(b)\\right) + 2f(a)f(b)\\] \n\nor \n\n\\[(f(a + b) - (f(a) + f(b)))^{2} = 4f(a)f(b).\\] \n\nHence \\(f(a)f(b)\\) is a perfect square for all \\(a,b\\in \\mathbb{Z}\\) . So there exists a \\(\\lambda\\) such that \\(f(n) = \\lambda g(n)^{2}\\) , where \\(g(n)\\geq 0\\) . From here we recover \n\n\\[\\boxed{g(a + b) = \\pm g(a)\\pm g(b).}\\] \n\nAlso \\(g(0) = 0\\) \n\nLet \\(k = g(1)\\neq 0\\) . We now split into cases on \\(g(2)\\) \n\n- \\(g(2) = 0\\) . Put \\(b = 2\\) in original to get \\(g(a + 2) = \\pm g(a) = +g(a)\\) \n\n- \\(g(2) = 2k\\) . Cases on \\(g(4)\\) : \n\n- \\(g(4) = 0\\) , then we get \\((g(n))_{n\\geq 0} = (0,1,2,1,0,1,2,1,\\ldots)\\) . This works. \n\n- \\(g(4) = 4k\\) . This only happens when \\(g(1) = k\\) , \\(g(2) = 2k\\) , \\(g(3) = 3k\\) , \\(g(4) = 4k\\) . Then \n\n\\[*g(5) = \\pm 3k\\pm 2k = \\pm 4k\\pm k.\\] \n\n\\[*g(6) = \\pm 4k\\pm 2k = \\pm 5k\\pm k.\\] \n\n\\\\*... \n\nand so by induction \\(g(n) = nk\\)", "metadata": {"resource_path": "IMO/segmented/en-IMO-2012-notes.jsonl", "problem_match": "4. ", "solution_match": "## \\(\\S 2.1\\) IMO 2012/4, proposed by Liam Baker (SAF) \n"}}
-{"year": "2012", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "Let \\(A B C\\) be a triangle with \\(\\angle B C A = 90^{\\circ}\\) , and let \\(D\\) be the foot of the altitude from \\(C\\) . Let \\(X\\) be a point in the interior of the segment \\(C D\\) . Let \\(K\\) be the point on the segment \\(A X\\) such that \\(B K = B C\\) . Similarly, let \\(L\\) be the point on the segment \\(B X\\) such that \\(A L = A C\\) . Let \\(M = \\overline{A L} \\cap \\overline{B K}\\) . Prove that \\(M K = M L\\) .", "solution": "Let \\(\\omega_{A}\\) and \\(\\omega_{B}\\) be the circles through \\(C\\) centered at \\(A\\) and \\(B\\) ; extend rays \\(A K\\) and \\(B L\\) to hit \\(\\omega_{B}\\) and \\(\\omega_{A}\\) again at \\(K^{*}\\) , \\(L^{*}\\) . By radical center \\(X\\) , we have \\(K L K^{*}L^{*}\\) is cyclic, say with circumcircle \\(\\omega\\) . \n\n\n \n\nBy orthogonality of \\((A)\\) and \\((B)\\) we find that \\(\\overline{A L}\\) , \\(\\overline{A L^{*}}\\) , \\(\\overline{B K}\\) , \\(\\overline{B K^{*}}\\) are tangents to \\(\\omega\\) (in particular, \\(K L K^{*}L^{*}\\) is harmonic). In particular \\(\\overline{M K}\\) and \\(\\overline{M L}\\) are tangents to \\(\\omega\\) , so \\(M K = M L\\) .", "metadata": {"resource_path": "IMO/segmented/en-IMO-2012-notes.jsonl", "problem_match": "5. ", "solution_match": "## \\(\\S 2.2\\) IMO 2012/5, proposed by Josef Tkadlec (CZE) \n"}}
+{"year": "2012", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "Let \\(A B C\\) be a triangle with \\(\\angle B C A = 90^{\\circ}\\) , and let \\(D\\) be the foot of the altitude from \\(C\\) . Let \\(X\\) be a point in the interior of the segment \\(C D\\) . Let \\(K\\) be the point on the segment \\(A X\\) such that \\(B K = B C\\) . Similarly, let \\(L\\) be the point on the segment \\(B X\\) such that \\(A L = A C\\) . Let \\(M = \\overline{A L} \\cap \\overline{B K}\\) . Prove that \\(M K = M L\\) .", "solution": "Let \\(\\omega_{A}\\) and \\(\\omega_{B}\\) be the circles through \\(C\\) centered at \\(A\\) and \\(B\\) ; extend rays \\(A K\\) and \\(B L\\) to hit \\(\\omega_{B}\\) and \\(\\omega_{A}\\) again at \\(K^{*}\\) , \\(L^{*}\\) . By radical center \\(X\\) , we have \\(K L K^{*}L^{*}\\) is cyclic, say with circumcircle \\(\\omega\\) . \n\n\n \n\nBy orthogonality of \\((A)\\) and \\((B)\\) we find that \\(\\overline{A L}\\) , \\(\\overline{A L^{*}}\\) , \\(\\overline{B K}\\) , \\(\\overline{B K^{*}}\\) are tangents to \\(\\omega\\) (in particular, \\(K L K^{*}L^{*}\\) is harmonic). In particular \\(\\overline{M K}\\) and \\(\\overline{M L}\\) are tangents to \\(\\omega\\) , so \\(M K = M L\\) .", "metadata": {"resource_path": "IMO/segmented/en-IMO-2012-notes.jsonl", "problem_match": "5. ", "solution_match": "## \\(\\S 2.2\\) IMO 2012/5, proposed by Josef Tkadlec (CZE) \n"}}
{"year": "2012", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "Find all positive integers \\(n\\) for which there exist non-negative integers \\(a_{1}, a_{2}, \\ldots , a_{n}\\) such that \n\n\\[\\frac{1}{2^{a_{1}}} + \\frac{1}{2^{a_{2}}} + \\dots + \\frac{1}{2^{a_{n}}} = \\frac{1}{3^{a_{1}}} + \\frac{2}{3^{a_{2}}} + \\dots + \\frac{n}{3^{a_{n}}} = 1.\\]", "solution": "The answer is \\(n\\equiv 1,2\\) (mod 4). To see these are necessary, note that taking the latter equation modulo 2 gives \n\n\\[1 = \\frac{1}{3^{a_{1}}} +\\frac{2}{3^{a_{2}}} +\\dots +\\frac{n}{3^{a_{n}}}\\equiv 1 + 2 + \\ldots +n\\pmod {2}.\\] \n\nNow we prove these are sufficient. The following nice construction was posted on AOPs by the user cfheolipixin. \n\nClaim — If \\(n = 2k - 1\\) works then so does \\(n = 2k\\) \n\nProof. Replace \n\n\\[\\frac{k}{3^{r}} = \\frac{k}{3^{r + 1}} +\\frac{2k}{3^{r + 1}}. \\quad (*)\\] \n\nClaim — If \\(n = 4k + 2\\) works then so does \\(n = 4k + 13\\) \n\nProof. First use the identity \n\n\\[\\frac{k + 2}{3^{r}} = \\frac{k + 2}{3^{r + 2}} +\\frac{4k + 3}{3^{r + 3}} +\\frac{4k + 5}{3^{r + 3}} +\\frac{4k + 7}{3^{r + 3}} +\\frac{4k + 9}{3^{r + 3}} +\\frac{4k + 11}{3^{r + 3}} +\\frac{4k + 13}{3^{r + 3}}\\] \n\nto fill in the odd numbers. The even numbers can then be instantiated with \\((\\ast)\\) too. \n\nThus it suffices to construct base cases for \\(n = 1\\) , \\(n = 5\\) , \\(n = 9\\) . They are \n\n\\[1 = \\frac{1}{3^{0}}\\] \\[\\quad = \\frac{1}{3^{2}} +\\frac{2}{3^{2}} +\\frac{3}{3^{2}} +\\frac{4}{3^{3}} +\\frac{5}{3^{3}}\\] \\[\\quad = \\frac{1}{3^{2}} +\\frac{2}{3^{3}} +\\frac{3}{3^{3}} +\\frac{4}{3^{3}} +\\frac{5}{3^{3}} +\\frac{6}{3^{4}} +\\frac{7}{3^{4}} +\\frac{8}{3^{4}} +\\frac{9}{3^{4}}.\\]", "metadata": {"resource_path": "IMO/segmented/en-IMO-2012-notes.jsonl", "problem_match": "6. ", "solution_match": "## \\(\\S 2.3\\) IMO 2012/6, proposed by Dusan Djukic (SRB) \n"}}
diff --git a/IMO/segmented/en-IMO-2013-notes.jsonl b/IMO/segmented/en-IMO-2013-notes.jsonl
index 6f10ca8295f65594597cfe530200d1fe580d7fcd..53b6a334262c8df52f74e40a75178166f1d534d7 100644
--- a/IMO/segmented/en-IMO-2013-notes.jsonl
+++ b/IMO/segmented/en-IMO-2013-notes.jsonl
@@ -1,6 +1,6 @@
{"year": "2013", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "Let \\(k\\) and \\(n\\) be positive integers. Prove that there exist positive integers \\(m_{1}\\) , ..., \\(m_{k}\\) such that \n\n\\[1 + \\frac{2^{k} - 1}{n} = \\left(1 + \\frac{1}{m_{1}}\\right)\\left(1 + \\frac{1}{m_{2}}\\right)\\dots \\left(1 + \\frac{1}{m_{k}}\\right).\\]", "solution": "By induction on \\(k \\geq 1\\) . When \\(k = 1\\) there is nothing to prove. \n\nFor the inductive step, if \\(n\\) is even, write \n\n\\[\\frac{n + (2^{k} - 1)}{n} = \\left(1 + \\frac{1}{n + (2^{k} - 2)}\\right)\\cdot \\frac{\\frac{n}{2} + (2^{k - 1} - 1)}{\\frac{n}{2}}\\] \n\nand use inductive hypothesis on the second term. On the other hand if \\(n\\) is odd then write \n\n\\[\\frac{n + (2^{k} - 1)}{n} = \\left(1 + \\frac{1}{n}\\right)\\cdot \\frac{\\frac{n + 1}{2} + (2^{k - 1} - 1)}{\\frac{n + 1}{2}}\\] \n\nand use inductive hypothesis on the second term.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2013-notes.jsonl", "problem_match": "1. ", "solution_match": "## \\(\\S 1.1\\) IMO 2013/1, proposed by Japan \n"}}
{"year": "2013", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "A configuration of 4027 points in the plane is called Colombian if it consists of 2013 red points and 2014 blue points, and no three of the points of the configuration are collinear. By drawing some lines, the plane is divided into several regions. An arrangement of lines is good for a Colombian configuration if the following two conditions are satisfied: \n\n(i) No line passes through any point of the configuration. \n\n(ii) No region contains points of both colors. \n\nFind the least value of \\(k\\) such that for any Colombian configuration of 4027 points, there is a good arrangement of \\(k\\) lines.", "solution": "The answer is \\(k \\geq 2013\\) . \n\nTo see that \\(k = 2013\\) is necessary, consider a regular 4026- gon and alternatively color the points red and blue, then place the last blue point anywhere in general position (it doesn't matter). Each side of the 4026 is a red- blue line segment which needs to be cut by one of the \\(k\\) lines, and each line can cut at most two of the segments. \n\nNow, we prove that \\(k = 2013\\) lines is sufficient. Consider the convex hull of all the points. \n\n- If the convex hull has any red points, cut that red point off from everyone else by a single line. Then, for each of the remaining 2012 red points, break them into 1006 pairs arbitrarily, and for each pair \\(\\{A,B\\}\\) draw two lines parallel to \\(AB\\) and close to them.- If the convex hull has two consecutive blue points, cut those two blue points off from everyone else by a single line. Then repeat the above construction for the remaining 2012 blue points. \n\nThe end.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2013-notes.jsonl", "problem_match": "2. ", "solution_match": "## \\(\\S 1.2\\) IMO 2013/2, proposed by Ivan Guo (AUS) \n"}}
-{"year": "2013", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "Let the excircle of triangle \\(A B C\\) opposite the vertex \\(A\\) be tangent to the side \\(B C\\) at the point \\(A_{1}\\) . Define the points \\(B_{1}\\) on \\(C A\\) and \\(C_{1}\\) on \\(A B\\) analogously, using the excircles opposite \\(B\\) and \\(C\\) , respectively. Suppose that the circumcenter of triangle \\(A_{1}B_{1}C_{1}\\) lies on the circumcircle of triangle \\(A B C\\) . Prove that triangle \\(A B C\\) is right-angled.", "solution": "We ignore for now the given condition and prove the following important lemma. \n\n## Lemma \n\nLet \\((AB_{1}C_{1})\\) meet \\((ABC)\\) again at \\(X\\) . From \\(BC_{1} = B_{1}C\\) follows \\(XC_{1} = XB_{1}\\) , and \\(X\\) is the midpoint of major arc \\(\\widehat{BC}\\) . \n\nProof. This follows from the fact that we have a spiral similarity \\(\\triangle XBC_{1} \\sim \\triangle XCB_{1}\\) which must actually be a spiral congruence since \\(BC_{1} = B_{1}C\\) . \\(\\square\\) \n\nWe define the arc midpoints \\(Y\\) and \\(Z\\) similarly, which lie on the perpendicular bisectors of \\(\\overline{A_{1}C_{1}}\\) , \\(\\overline{A_{1}B_{1}}\\) . \n\n\n \n\nWe now turn to the problem condition which asserts the circumcenter \\(W\\) of \\(\\triangle A_{1}B_{1}C_{1}\\) lies on \\((ABC)\\) . \n\nClaim — We may assume WLOG that \\(W = X\\) . \n\nProof. This is just configuration analysis, since we already knew that the arc midpoints both lie on \\((ABC)\\) and the relevant perpendicular bisectors.\n\n\n\nPoint \\(W\\) lies on \\((ABC)\\) and hence outside \\(\\triangle ABC\\) , hence outside \\(\\triangle A_{1}B_{1}C_{1}\\) . Thus we may assume WLOG that \\(\\angle B_{1}A_{1}C_{1} > 90^{\\circ}\\) . Then \\(A\\) and \\(X\\) lie on the same side of line \\(\\overline{B_{1}C_{1}}\\) , and since \\(W\\) is supposed to lie both on \\((ABC)\\) and the perpendicular bisector of \\(\\overline{B_{1}C_{1}}\\) it follows \\(W = X\\) . \\(\\square\\) \n\nConsequently, \\(\\overline{XY}\\) and \\(\\overline{XZ}\\) are exactly the perpendicular bisectors of \\(\\overline{A_{1}C_{1}}\\) , \\(\\overline{A_{1}B_{1}}\\) . The rest is angle chase, the fastest one is \n\n\\[\\angle A = \\angle C_{1}X B_{1} = \\angle C_{1}X A_{1} + \\angle A_{1}X B_{1} = 2\\angle Y X A_{1} + 2\\angle A_{1}X Z\\] \\[\\qquad = 2\\angle Y X Z = 180^{\\circ} - \\angle A\\] \n\nwhich solves the problem. \n\nRemark. Angle chasing is also possible even without the points \\(Y\\) and \\(Z\\) , though it takes much longer. Introduce the Bevan point \\(V\\) and use the fact that \\(V A_{1}B_{1}C\\) is cyclic (with diameter \\(\\overline{VC}\\) ) and similarly \\(V A_{1}C_{1}B\\) is cyclic; a calculation then gives \\(\\angle C V B = 180^{\\circ} - \\frac{1}{2}\\angle A\\) . Thus \\(V\\) lies on the circle with diameter \\(\\overline{I_{b}I_{c}}\\) .\n\n\n\n## \\(\\S 2\\) Solutions to Day 2", "metadata": {"resource_path": "IMO/segmented/en-IMO-2013-notes.jsonl", "problem_match": "3. ", "solution_match": "## \\(\\S 1.3\\) IMO 2013/3, proposed by Alexander A. Polyansky (RUS) \n"}}
-{"year": "2013", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "Let \\(A B C\\) be an acute triangle with orthocenter \\(H\\) , and let \\(W\\) be a point on the side \\(\\overline{{B C}}\\) , between \\(B\\) and \\(C\\) . The points \\(M\\) and \\(N\\) are the feet of the altitudes drawn from \\(B\\) and \\(C\\) , respectively. Suppose \\(\\omega_{1}\\) is the circumcircle of triangle \\(B W N\\) and \\(X\\) is a point such that \\(\\overline{{W X}}\\) is a diameter of \\(\\omega_{1}\\) . Similarly, \\(\\omega_{2}\\) is the circumcircle of triangle \\(C W M\\) and \\(Y\\) is a point such that \\(\\overline{{W Y}}\\) is a diameter of \\(\\omega_{2}\\) . Show that the points \\(X\\) , \\(Y\\) , and \\(H\\) are collinear.", "solution": "## Problem stae present two solutions, an elementary one and then an advanced one by moving points. \n\n\\(\\P\\) First solution, classical. Let \\(P\\) be the second intersection of \\(\\omega_{1}\\) and \\(\\omega_{2}\\) ; this is the Miquel point, so \\(P\\) also lies on the circumcircle of \\(AMN\\) , which is the circle with diameter \\(\\overline{AH}\\) . \n\n\n \n\nWe now contend: \n\nClaim — Points \\(P\\) , \\(H\\) , \\(X\\) collinear. (Similarly, points \\(P\\) , \\(H\\) , \\(Y\\) are collinear.) \n\nProof using power of a point. By radical axis on \\(BNMC\\) , \\(\\omega_{1}\\) , \\(\\omega_{2}\\) , it follows that \\(A\\) , \\(P\\) , \\(W\\) are collinear. We know that \\(\\angle APH = 90^{\\circ}\\) , and also \\(\\angle XPW = 90^{\\circ}\\) by construction. Thus \\(P\\) , \\(H\\) , \\(X\\) are collinear. \\(\\square\\) \n\nProof using angle chasing. This is essentially Reim's theorem: \n\n\\[\\angle NPH = \\angle NAH = \\angle BAH = \\angle ABX = \\angle NBX = \\angle NPX\\] \n\nas desired. Alternatively, one may prove \\(A\\) , \\(P\\) , \\(W\\) are collinear by \\(\\angle NPA = \\angle NMA = \\angle NMC = \\angle NBC = \\angle NBW = \\angle NPW\\) . \\(\\square\\)\n\n\n\n\\(\\P\\) Second solution, by moving points. Fix \\(\\triangle ABC\\) and vary \\(W\\) . Let \\(\\infty\\) be the point at infinity perpendicular to \\(\\overline{BC}\\) for brevity. \n\nBy spiral similarity, the point \\(X\\) moves linearly on \\(\\overline{B\\infty}\\) as \\(W\\) varies linearly on \\(\\overline{BC}\\) . Similarly, so does \\(Y\\) . So in other words, the map \n\n\\[X\\mapsto W\\mapsto Y\\] \n\nis linear. However, the map \n\n\\[X\\mapsto Y^{\\prime}:= \\overline{XH}\\cap \\overline{C\\infty}\\] \n\nis linear too. \n\nTo show that these maps are the same, it suffices to check it thus at two points. \n\n- When \\(W = B\\) , the circle \\((BNW)\\) degenerates to the circle through \\(B\\) tangent to \\(\\overline{BC}\\) , and \\(X = \\overline{CN} \\cap \\overline{B\\infty}\\) . We have \\(Y = Y' = C\\) . \n\n- When \\(W = C\\) , the result is analogous. \n\n- Although we don't need to do so, it's also easy to check the result if \\(W\\) is the foot from \\(A\\) since then \\(XHWB\\) and \\(YHWC\\) are rectangles.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2013-notes.jsonl", "problem_match": "4. ", "solution_match": "## \\(\\S 2.1\\) IMO 2013/4, proposed by Warut Suksompong, Potcharapol Suteparuk (THA) \n"}}
+{"year": "2013", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "Let the excircle of triangle \\(A B C\\) opposite the vertex \\(A\\) be tangent to the side \\(B C\\) at the point \\(A_{1}\\) . Define the points \\(B_{1}\\) on \\(C A\\) and \\(C_{1}\\) on \\(A B\\) analogously, using the excircles opposite \\(B\\) and \\(C\\) , respectively. Suppose that the circumcenter of triangle \\(A_{1}B_{1}C_{1}\\) lies on the circumcircle of triangle \\(A B C\\) . Prove that triangle \\(A B C\\) is right-angled.", "solution": "We ignore for now the given condition and prove the following important lemma. \n\n## Lemma \n\nLet \\((AB_{1}C_{1})\\) meet \\((ABC)\\) again at \\(X\\) . From \\(BC_{1} = B_{1}C\\) follows \\(XC_{1} = XB_{1}\\) , and \\(X\\) is the midpoint of major arc \\(\\widehat{BC}\\) . \n\nProof. This follows from the fact that we have a spiral similarity \\(\\triangle XBC_{1} \\sim \\triangle XCB_{1}\\) which must actually be a spiral congruence since \\(BC_{1} = B_{1}C\\) . \\(\\square\\) \n\nWe define the arc midpoints \\(Y\\) and \\(Z\\) similarly, which lie on the perpendicular bisectors of \\(\\overline{A_{1}C_{1}}\\) , \\(\\overline{A_{1}B_{1}}\\) . \n\n\n \n\nWe now turn to the problem condition which asserts the circumcenter \\(W\\) of \\(\\triangle A_{1}B_{1}C_{1}\\) lies on \\((ABC)\\) . \n\nClaim — We may assume WLOG that \\(W = X\\) . \n\nProof. This is just configuration analysis, since we already knew that the arc midpoints both lie on \\((ABC)\\) and the relevant perpendicular bisectors.\n\n\n\nPoint \\(W\\) lies on \\((ABC)\\) and hence outside \\(\\triangle ABC\\) , hence outside \\(\\triangle A_{1}B_{1}C_{1}\\) . Thus we may assume WLOG that \\(\\angle B_{1}A_{1}C_{1} > 90^{\\circ}\\) . Then \\(A\\) and \\(X\\) lie on the same side of line \\(\\overline{B_{1}C_{1}}\\) , and since \\(W\\) is supposed to lie both on \\((ABC)\\) and the perpendicular bisector of \\(\\overline{B_{1}C_{1}}\\) it follows \\(W = X\\) . \\(\\square\\) \n\nConsequently, \\(\\overline{XY}\\) and \\(\\overline{XZ}\\) are exactly the perpendicular bisectors of \\(\\overline{A_{1}C_{1}}\\) , \\(\\overline{A_{1}B_{1}}\\) . The rest is angle chase, the fastest one is \n\n\\[\\angle A = \\angle C_{1}X B_{1} = \\angle C_{1}X A_{1} + \\angle A_{1}X B_{1} = 2\\angle Y X A_{1} + 2\\angle A_{1}X Z\\] \\[\\qquad = 2\\angle Y X Z = 180^{\\circ} - \\angle A\\] \n\nwhich solves the problem. \n\nRemark. Angle chasing is also possible even without the points \\(Y\\) and \\(Z\\) , though it takes much longer. Introduce the Bevan point \\(V\\) and use the fact that \\(V A_{1}B_{1}C\\) is cyclic (with diameter \\(\\overline{VC}\\) ) and similarly \\(V A_{1}C_{1}B\\) is cyclic; a calculation then gives \\(\\angle C V B = 180^{\\circ} - \\frac{1}{2}\\angle A\\) . Thus \\(V\\) lies on the circle with diameter \\(\\overline{I_{b}I_{c}}\\) .\n\n\n\n## \\(\\S 2\\) Solutions to Day 2", "metadata": {"resource_path": "IMO/segmented/en-IMO-2013-notes.jsonl", "problem_match": "3. ", "solution_match": "## \\(\\S 1.3\\) IMO 2013/3, proposed by Alexander A. Polyansky (RUS) \n"}}
+{"year": "2013", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "Let \\(A B C\\) be an acute triangle with orthocenter \\(H\\) , and let \\(W\\) be a point on the side \\(\\overline{{B C}}\\) , between \\(B\\) and \\(C\\) . The points \\(M\\) and \\(N\\) are the feet of the altitudes drawn from \\(B\\) and \\(C\\) , respectively. Suppose \\(\\omega_{1}\\) is the circumcircle of triangle \\(B W N\\) and \\(X\\) is a point such that \\(\\overline{{W X}}\\) is a diameter of \\(\\omega_{1}\\) . Similarly, \\(\\omega_{2}\\) is the circumcircle of triangle \\(C W M\\) and \\(Y\\) is a point such that \\(\\overline{{W Y}}\\) is a diameter of \\(\\omega_{2}\\) . Show that the points \\(X\\) , \\(Y\\) , and \\(H\\) are collinear.", "solution": "## Problem stae present two solutions, an elementary one and then an advanced one by moving points. \n\n\\(\\P\\) First solution, classical. Let \\(P\\) be the second intersection of \\(\\omega_{1}\\) and \\(\\omega_{2}\\) ; this is the Miquel point, so \\(P\\) also lies on the circumcircle of \\(AMN\\) , which is the circle with diameter \\(\\overline{AH}\\) . \n\n\n \n\nWe now contend: \n\nClaim — Points \\(P\\) , \\(H\\) , \\(X\\) collinear. (Similarly, points \\(P\\) , \\(H\\) , \\(Y\\) are collinear.) \n\nProof using power of a point. By radical axis on \\(BNMC\\) , \\(\\omega_{1}\\) , \\(\\omega_{2}\\) , it follows that \\(A\\) , \\(P\\) , \\(W\\) are collinear. We know that \\(\\angle APH = 90^{\\circ}\\) , and also \\(\\angle XPW = 90^{\\circ}\\) by construction. Thus \\(P\\) , \\(H\\) , \\(X\\) are collinear. \\(\\square\\) \n\nProof using angle chasing. This is essentially Reim's theorem: \n\n\\[\\angle NPH = \\angle NAH = \\angle BAH = \\angle ABX = \\angle NBX = \\angle NPX\\] \n\nas desired. Alternatively, one may prove \\(A\\) , \\(P\\) , \\(W\\) are collinear by \\(\\angle NPA = \\angle NMA = \\angle NMC = \\angle NBC = \\angle NBW = \\angle NPW\\) . \\(\\square\\)\n\n\n\n\\(\\P\\) Second solution, by moving points. Fix \\(\\triangle ABC\\) and vary \\(W\\) . Let \\(\\infty\\) be the point at infinity perpendicular to \\(\\overline{BC}\\) for brevity. \n\nBy spiral similarity, the point \\(X\\) moves linearly on \\(\\overline{B\\infty}\\) as \\(W\\) varies linearly on \\(\\overline{BC}\\) . Similarly, so does \\(Y\\) . So in other words, the map \n\n\\[X\\mapsto W\\mapsto Y\\] \n\nis linear. However, the map \n\n\\[X\\mapsto Y^{\\prime}:= \\overline{XH}\\cap \\overline{C\\infty}\\] \n\nis linear too. \n\nTo show that these maps are the same, it suffices to check it thus at two points. \n\n- When \\(W = B\\) , the circle \\((BNW)\\) degenerates to the circle through \\(B\\) tangent to \\(\\overline{BC}\\) , and \\(X = \\overline{CN} \\cap \\overline{B\\infty}\\) . We have \\(Y = Y' = C\\) . \n\n- When \\(W = C\\) , the result is analogous. \n\n- Although we don't need to do so, it's also easy to check the result if \\(W\\) is the foot from \\(A\\) since then \\(XHWB\\) and \\(YHWC\\) are rectangles.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2013-notes.jsonl", "problem_match": "4. ", "solution_match": "## \\(\\S 2.1\\) IMO 2013/4, proposed by Warut Suksompong, Potcharapol Suteparuk (THA) \n"}}
{"year": "2013", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "Suppose a function \\(f\\colon \\mathbb{Q}_{>0}\\to \\mathbb{R}\\) satisfies: \n\n(i) If \\(x,y\\in \\mathbb{Q}_{>0}\\) , then \\(f(x)f(y)\\geq f(xy)\\) \n\n(ii) If \\(x,y\\in \\mathbb{Q}_{>0}\\) , then \\(f(x + y)\\geq f(x) + f(y)\\) \n\n(iii) There exists a rational number \\(a > 1\\) with \\(f(a) = a\\) \n\nProve that \\(f(x) = x\\) for all positive rational numbers \\(x\\)", "solution": "First, we dispense of negative situations by proving: \n\nClaim — For any integer \\(n > 0\\) , we have \\(f(n) \\geq n\\) . \n\nProof. Note by induction on (ii) we have \\(f(nx) \\geq nf(x)\\) . Taking \\((x,y) = (a,1)\\) in (i) gives \\(f(1) \\geq 1\\) , and hence \\(f(n) \\geq n\\) . \\(\\square\\) \n\nClaim — The \\(f\\) takes only positive values, and hence by (ii) is strictly increasing. \n\nProof, suggested by Gopal Goel. Let \\(p,q > 0\\) be integers. Then \\(f(q)f(p / q)\\geq f(p)\\) , and since both \\(\\min (f(p),f(q)) > 0\\) it follows \\(f(p / q) > 0\\) . \\(\\square\\) \n\nClaim — For any \\(x > 1\\) we have \\(f(x) \\geq x\\) . \n\nProof. Note that \n\n\\[f(x)^{N}\\geq f(x^{N})\\geq f\\left(\\left\\lfloor x^{N}\\right\\rfloor\\right)\\geq \\left\\lfloor x^{N}\\right\\rfloor >x^{N} - 1\\] \n\nfor any integer \\(N\\) . Since \\(N\\) can be arbitrarily large, we conclude \\(f(x) \\geq x\\) for \\(x > 1\\) . \\(\\square\\) \n\nOn the other hand, \\(f\\) has arbitrarily large fixed points (namely powers of \\(a\\) ) so from (ii) we're essentially done. First, for \\(x > 1\\) pick a large \\(m\\) and note \n\n\\[a^{m} = f(a^{m})\\geq f(a^{m} - x) + f(x)\\geq (a^{m} - x) + x = a^{m}.\\] \n\nFinally, for \\(x \\leq 1\\) use \n\n\\[n f(x) = f(n)f(x)\\geq f(n x)\\geq n f(x)\\] \n\nfor large \\(n\\) . \n\nRemark. Note that \\(a > 1\\) is essential; if \\(b \\geq 1\\) then \\(f(x) = bx^{2}\\) works with unique fixed point \\(1 / b \\leq 1\\) .", "metadata": {"resource_path": "IMO/segmented/en-IMO-2013-notes.jsonl", "problem_match": "5. ", "solution_match": "## \\(\\S 2.2\\) IMO 2013/5, proposed by Nikolai Nikolov (BGR) \n"}}
-{"year": "2013", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "Let \\(n\\geq 3\\) be an integer, and consider a circle with \\(n + 1\\) equally spaced points marked on it. Consider all labellings of these points with the numbers \\(0,1,\\ldots ,n\\) such that each label is used exactly once; two such labellings are considered to be the same if one can be obtained from the other by a rotation of the circle. A labelling is called beautiful if, for any four labels \\(a< b< c< d\\) with \\(a + d = b + c\\) , the chord joining the points labelled \\(a\\) and \\(d\\) does not intersect the chord joining the points labelled \\(b\\) and \\(c\\) . Let \\(M\\) be the number of beautiful labellings, and let \\(N\\) be the number of ordered pairs \\((x,y)\\) of positive integers such that \\(x + y\\leq n\\) and \\(\\gcd (x,y) = 1\\) . Prove that \\(M = N + 1\\) .", "solution": ". \n\nFirst, here are half of the beautiful labellings up to reflection for \\(n = 6\\) , just for concreteness. \n\n\n \n\nAbbreviate \"beautiful labelling of points around a circle\" to ring. Moreover, throughout the solution we will allow degenerate chords that join a point to itself; this has no effect on the problem statement. \n\nThe idea is to proceed by induction in the following way. A ring of \\([0,n]\\) is called linear if it is an arithmetic progression modulo \\(n + 1\\) . For example, the first two rings in the diagram and the last one are linear for \\(n = 6\\) , while the other three are not. \n\nOf course we can move from any ring on \\([0,n]\\) to a ring on \\([0,n - 1]\\) by deleting \\(n\\) . We are going to prove that: \n\n- Each linear ring on \\([0,n - 1]\\) yields exactly two rings of \\([0,n]\\) , and\n\n\n\n- Each nonlinear ring on \\([0, n - 1]\\) yields exactly one rings of \\([0, n]\\) . \n\nIn light of the fact there are obviously \\(\\phi (n)\\) linear rings on \\([0, n]\\) , the conclusion will follow by induction. \n\nWe say a set of chords (possibly degenerate) is pseudo- parallel if for any three of them, one of them separates the two. (Pictorially, one can perturb the endpoints along the circle in order to make them parallel in Euclidean sense.) The main structure lemma is going to be: \n\n## Lemma \n\nIn any ring, the chords of sum \\(k\\) (even including degenerate ones) are pseudo- parallel. \n\nProof. By induction on \\(n\\) . By shifting, we may assume that one of the chords is \\(\\{0, k\\}\\) and discard all numbers exceeding \\(k\\) ; that is, assume \\(n = k\\) . Suppose the other two chords are \\(\\{a, n - a\\}\\) and \\(\\{b, n - b\\}\\) . \n\n\n \n\nWe consider the chord \\(\\{u, v\\}\\) directly above \\(\\{0, n\\}\\) , drawn in blue. There are now three cases. \n\n- If \\(u + v = n\\) , then delete 0 and \\(n\\) and decrease everything by 1. Then the chords \\(\\{a - 1, n - a - 1\\}\\) , \\(\\{b - 1, n - b - 1\\}\\) , \\(\\{u - 1, v - 1\\}\\) contradict the induction hypothesis.- If \\(u + v < n\\) , then search for the chord \\(\\{u + v, n - (u + v)\\}\\) . It lies on the other side of \\(\\{0, n\\}\\) in light of chord \\(\\{0, u + v\\}\\) . Now again delete 0 and \\(n\\) and decrease everything by 1. Then the chords \\(\\{a - 1, n - a - 1\\}\\) , \\(\\{b - 1, n - b - 1\\}\\) , \\(\\{u + v - 1, n - (u + v) - 1\\}\\) contradict the induction hypothesis.- If \\(u + v > n\\) , apply the map \\(t \\mapsto n - t\\) to the entire ring. This gives the previous case as now \\((n - u) + (n - v) < n\\) . \n\nNext, we give another characterization of linear rings. \n\n## Lemma \n\nA ring on \\([0, n - 1]\\) is linear if and only if the point 0 does not lie between two chords of sum \\(n\\) . \n\nProof. It's obviously true for linear rings. Conversely, assume the property holds for some ring. Note that the chords with sum \\(n - 1\\) are pseudo- parallel and encompass every point, so they are actually parallel. Similarly, the chords of sum \\(n\\) are actually parallel and encompass every point other than 0. So the map \n\n\\[t\\mapsto n - t\\mapsto (n - 1) - (n - t) = t - 1\\pmod {n}\\]\n\n\n\nis rotation as desired. \n\n## Lemma \n\nEvery nonlinear ring on \\([0, n - 1]\\) induces exactly one ring on \\([0, n]\\) . \n\nProof. Because the chords of sum \\(n\\) are pseudo- parallel, there is at most one possibility for the location \\(n\\) . \n\nConversely, we claim that this works. The chords of sum \\(n\\) (and less than \\(n\\) ) are OK by construction, so assume for contradiction that there exists \\(a, b, c \\in \\{1, \\ldots , n - 1\\}\\) such that \\(a + b = n + c\\) . Then, we can \"reflect\" them using the (pseudo- parallel) chords of length \\(n\\) to find that \\((n - a) + (n - b) = 0 + (n - c)\\) , and the chords joining 0 to \\(n - c\\) and \\(n - a\\) to \\(n - b\\) intersect, by definition. \n\n\n \n\nThis is a contradiction that the original numbers on \\([0, n - 1]\\) form a ring. \n\n## Lemma \n\nEvery linear ring on \\([0, n - 1]\\) induces exactly two rings on \\([0, n]\\) . \n\nProof. Because the chords of sum \\(n\\) are pseudo- parallel, the point \\(n\\) must lie either directly to the left or right of 0. For the same reason as in the previous proof, both of them work.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2013-notes.jsonl", "problem_match": "6. ", "solution_match": "## \\(\\S 2.3\\) IMO 2013/6, proposed by Alexander S. Golovanov and Mikhail A. Ivanov (RUS) \n"}}
+{"year": "2013", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "Let \\(n\\geq 3\\) be an integer, and consider a circle with \\(n + 1\\) equally spaced points marked on it. Consider all labellings of these points with the numbers \\(0,1,\\ldots ,n\\) such that each label is used exactly once; two such labellings are considered to be the same if one can be obtained from the other by a rotation of the circle. A labelling is called beautiful if, for any four labels \\(a< b< c< d\\) with \\(a + d = b + c\\) , the chord joining the points labelled \\(a\\) and \\(d\\) does not intersect the chord joining the points labelled \\(b\\) and \\(c\\) . Let \\(M\\) be the number of beautiful labellings, and let \\(N\\) be the number of ordered pairs \\((x,y)\\) of positive integers such that \\(x + y\\leq n\\) and \\(\\gcd (x,y) = 1\\) . Prove that \\(M = N + 1\\) .", "solution": ". \n\nFirst, here are half of the beautiful labellings up to reflection for \\(n = 6\\) , just for concreteness. \n\n\n \n\nAbbreviate \"beautiful labelling of points around a circle\" to ring. Moreover, throughout the solution we will allow degenerate chords that join a point to itself; this has no effect on the problem statement. \n\nThe idea is to proceed by induction in the following way. A ring of \\([0,n]\\) is called linear if it is an arithmetic progression modulo \\(n + 1\\) . For example, the first two rings in the diagram and the last one are linear for \\(n = 6\\) , while the other three are not. \n\nOf course we can move from any ring on \\([0,n]\\) to a ring on \\([0,n - 1]\\) by deleting \\(n\\) . We are going to prove that: \n\n- Each linear ring on \\([0,n - 1]\\) yields exactly two rings of \\([0,n]\\) , and\n\n\n\n- Each nonlinear ring on \\([0, n - 1]\\) yields exactly one rings of \\([0, n]\\) . \n\nIn light of the fact there are obviously \\(\\phi (n)\\) linear rings on \\([0, n]\\) , the conclusion will follow by induction. \n\nWe say a set of chords (possibly degenerate) is pseudo- parallel if for any three of them, one of them separates the two. (Pictorially, one can perturb the endpoints along the circle in order to make them parallel in Euclidean sense.) The main structure lemma is going to be: \n\n## Lemma \n\nIn any ring, the chords of sum \\(k\\) (even including degenerate ones) are pseudo- parallel. \n\nProof. By induction on \\(n\\) . By shifting, we may assume that one of the chords is \\(\\{0, k\\}\\) and discard all numbers exceeding \\(k\\) ; that is, assume \\(n = k\\) . Suppose the other two chords are \\(\\{a, n - a\\}\\) and \\(\\{b, n - b\\}\\) . \n\n\n \n\nWe consider the chord \\(\\{u, v\\}\\) directly above \\(\\{0, n\\}\\) , drawn in blue. There are now three cases. \n\n- If \\(u + v = n\\) , then delete 0 and \\(n\\) and decrease everything by 1. Then the chords \\(\\{a - 1, n - a - 1\\}\\) , \\(\\{b - 1, n - b - 1\\}\\) , \\(\\{u - 1, v - 1\\}\\) contradict the induction hypothesis.- If \\(u + v < n\\) , then search for the chord \\(\\{u + v, n - (u + v)\\}\\) . It lies on the other side of \\(\\{0, n\\}\\) in light of chord \\(\\{0, u + v\\}\\) . Now again delete 0 and \\(n\\) and decrease everything by 1. Then the chords \\(\\{a - 1, n - a - 1\\}\\) , \\(\\{b - 1, n - b - 1\\}\\) , \\(\\{u + v - 1, n - (u + v) - 1\\}\\) contradict the induction hypothesis.- If \\(u + v > n\\) , apply the map \\(t \\mapsto n - t\\) to the entire ring. This gives the previous case as now \\((n - u) + (n - v) < n\\) . \n\nNext, we give another characterization of linear rings. \n\n## Lemma \n\nA ring on \\([0, n - 1]\\) is linear if and only if the point 0 does not lie between two chords of sum \\(n\\) . \n\nProof. It's obviously true for linear rings. Conversely, assume the property holds for some ring. Note that the chords with sum \\(n - 1\\) are pseudo- parallel and encompass every point, so they are actually parallel. Similarly, the chords of sum \\(n\\) are actually parallel and encompass every point other than 0. So the map \n\n\\[t\\mapsto n - t\\mapsto (n - 1) - (n - t) = t - 1\\pmod {n}\\]\n\n\n\nis rotation as desired. \n\n## Lemma \n\nEvery nonlinear ring on \\([0, n - 1]\\) induces exactly one ring on \\([0, n]\\) . \n\nProof. Because the chords of sum \\(n\\) are pseudo- parallel, there is at most one possibility for the location \\(n\\) . \n\nConversely, we claim that this works. The chords of sum \\(n\\) (and less than \\(n\\) ) are OK by construction, so assume for contradiction that there exists \\(a, b, c \\in \\{1, \\ldots , n - 1\\}\\) such that \\(a + b = n + c\\) . Then, we can \"reflect\" them using the (pseudo- parallel) chords of length \\(n\\) to find that \\((n - a) + (n - b) = 0 + (n - c)\\) , and the chords joining 0 to \\(n - c\\) and \\(n - a\\) to \\(n - b\\) intersect, by definition. \n\n\n \n\nThis is a contradiction that the original numbers on \\([0, n - 1]\\) form a ring. \n\n## Lemma \n\nEvery linear ring on \\([0, n - 1]\\) induces exactly two rings on \\([0, n]\\) . \n\nProof. Because the chords of sum \\(n\\) are pseudo- parallel, the point \\(n\\) must lie either directly to the left or right of 0. For the same reason as in the previous proof, both of them work.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2013-notes.jsonl", "problem_match": "6. ", "solution_match": "## \\(\\S 2.3\\) IMO 2013/6, proposed by Alexander S. Golovanov and Mikhail A. Ivanov (RUS) \n"}}
diff --git a/IMO/segmented/en-IMO-2014-notes.jsonl b/IMO/segmented/en-IMO-2014-notes.jsonl
index e975f5584d4e0b9181d3c25cdd62e99e7b67d9a2..875c0e326b1b840b347bd6d071e40789993a5d9c 100644
--- a/IMO/segmented/en-IMO-2014-notes.jsonl
+++ b/IMO/segmented/en-IMO-2014-notes.jsonl
@@ -1,6 +1,6 @@
{"year": "2014", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "Let \\(a_{0}< a_{1}< a_{2}< \\dots\\) be an infinite sequence of positive integers. Prove that there exists a unique integer \\(n\\geq 1\\) such that \n\n\\[a_{n}< \\frac{a_{0} + a_{1} + a_{2} + \\cdots + a_{n}}{n}\\leq a_{n + 1}.\\]", "solution": "Fedor Petrov presents the following nice solution. Let us define the sequence \n\n\\[b_{n} = (a_{n} - a_{n - 1}) + \\cdot \\cdot \\cdot +(a_{n} - a_{1}).\\] \n\nSince \\((a_{n})_{n}\\) is increasing, we get \\((b_{n})_{n}\\) is strictly increasing, and moreover \\(b_{1} = 0\\) . The problem requires an \\(n\\) such that \n\n\\[b_{n}< a_{0}\\leq b_{n + 1}\\] \n\nwhich obviously exists and is unique.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2014-notes.jsonl", "problem_match": "1. ", "solution_match": "## \\(\\S 1.1\\) IMO 2014/1, proposed by Gerhard Woeginger (AUT) \n"}}
-{"year": "2014", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "Let \\(n\\geq 2\\) be an integer. Consider an \\(n\\times n\\) chessboard consisting of \\(n^{2}\\) unit squares. A configuration of \\(n\\) rooks on this board is peaceful if every row and every column contains exactly one rook. Find the greatest positive integer \\(k\\) such that, for each peaceful configuration of \\(n\\) rooks, there is a \\(k\\times k\\) square which does not contain a rook on any of its \\(k^{2}\\) unit squares.", "solution": "The answer is \\(k = \\lfloor \\sqrt{n - 1} \\rfloor\\) . \n\n\\(\\P\\) Proof that the property holds when \\(n\\geq k^{2} + 1\\) . First, assume \\(n > k^{2}\\) for some \\(k\\) . We will prove we can find an empty \\(k\\times k\\) square. Indeed, let \\(R\\) be a rook in the uppermost column, and draw \\(k\\) squares of size \\(k\\times k\\) directly below it, aligned. There are at most \\(k - 1\\) rooks among these squares, as desired. \n\n\n \n\n\\(\\P\\) Construction for all \\(n\\leq k^{2}\\) . We first give an example where for \\(n = k^{2}\\) showing there may be no empty \\(k\\times k\\) square. We draw the example for \\(k = 3\\) (with the generalization being obvious); \n\n\n \n\nTo show that this works, consider for each rook drawing an \\(k\\times k\\) square of \\(X\\) 's whose bottom- right hand corner is the rook (these may go off the board). These indicate\n\n\n\npositions where one cannot place the upper- left hand corner of any square. It’s easy to see that these cover the entire board, except parts of the last \\(k - 1\\) columns, which don’t matter anyways. \n\nIt remains to check that \\(n \\leq k^{2}\\) also all work (omitting this step is a common mistake). For this, we can delete rows and column to get an \\(n \\times n\\) board, and then fill in any gaps where we accidentally deleted a rook.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2014-notes.jsonl", "problem_match": "2. ", "solution_match": "## \\(\\S 1.2\\) IMO 2014/2, proposed by Tonči Kokan (HRV) \n"}}
-{"year": "2014", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "Convex quadrilateral \\(A B C D\\) has \\(\\angle A B C = \\angle C D A = 90^{\\circ}\\) . Point \\(H\\) is the foot of the perpendicular from \\(A\\) to \\(\\overline{{B D}}\\) . Points \\(S\\) and \\(T\\) lie on sides \\(A B\\) and \\(A D\\) , respectively, such that \\(H\\) lies inside triangle \\(S C T\\) and \n\n\\[\\angle C H S - \\angle C S B = 90^{\\circ},\\quad \\angle T H C - \\angle D T C = 90^{\\circ}.\\] \n\nProve that line \\(B D\\) is tangent to the circumcircle of triangle \\(T S H\\)", "solution": ") . \n\nFirst solution (mine). First we rewrite the angle condition in a suitable way. \n\nClaim — We have \\(\\angle A T H = \\angle T C H + 90^{\\circ}\\) . Thus the circumcenter of \\(\\triangle C T H\\) lies on \\(\\overline{{A D}}\\) . Similarly the circumcenter of \\(\\triangle C S H\\) lies on \\(\\overline{{A B}}\\) . \n\nProof. \n\n\\[\\angle A T H = \\angle D T H\\] \\[\\qquad = \\angle D T C + \\angle C T H\\] \\[\\qquad = \\angle D T C - \\angle T H C - \\angle H C T\\] \\[\\qquad = 90^{\\circ} - \\angle H C T = 90^{\\circ} + \\angle T C H.\\] \n\nwhich implies conclusion. \n\n\n \n\nLet the perpendicular bisector of \\(T H\\) meet \\(A H\\) at \\(P\\) now. It suffices to show that \\(\\frac{A P}{P H}\\) is symmetric in \\(b = A D\\) and \\(d = A B\\) , because then \\(P\\) will be the circumcenter of \\(\\triangle T S H\\) . To do this, set \\(A H = \\frac{b d}{2 R}\\) and \\(A C = 2 R\\) . \n\nLet \\(O\\) denote the circumcenter of \\(\\triangle C H T\\) . Use the Law of Cosines on \\(\\triangle A C O\\) and \\(\\triangle A H O\\) , using variables \\(x = A O\\) and \\(r = H O\\) . We get that \n\n\\[r^{2} = x^{2} + A H^{2} - 2x\\cdot A H\\cdot \\frac{d}{2R} = x^{2} + (2R)^{2} - 2b x.\\]\n\n\n\nBy the angle bisector theorem, \\(\\frac{AP}{PH} = \\frac{AO}{HO}\\) . \n\nThe rest is computation: notice that \n\n\\[r^{2} - x^{2} = h^{2} - 2xh\\cdot \\frac{d}{2R} = (2R)^{2} - 2bx\\] \n\nwhere \\(h = AH = \\frac{bd}{2R}\\) , whence \n\n\\[x = \\frac{(2R)^{2} - h^{2}}{2b - 2h\\cdot\\frac{d}{2R}}.\\] \n\nMoreover, \n\n\\[\\frac{1}{2}\\left(\\frac{r^{2}}{x^{2}} -1\\right) = \\frac{1}{x}\\left(\\frac{2}{x} R^{2} - b\\right).\\] \n\nNow, if we plug in the \\(x\\) in the right- hand side of the above, we obtain \n\n\\[\\frac{2b - 2h\\cdot\\frac{d}{2R}}{4R^{2} - h^{2}}\\left(\\frac{2b - 2h\\cdot\\frac{d}{2R}}{4R^{2}\\cdot h^{2}}\\cdot 2R^{2} - b\\right) = \\frac{2h}{(4R^{2} - h^{2})^{2}}\\left(b - h\\cdot \\frac{d}{2R}\\right)\\left(-2hdR + bh^{2}\\right).\\] \n\nPulling out a factor of \\(- 2Rh\\) from the rightmost term, we get something that is symmetric in \\(b\\) and \\(d\\) , as required. \n\n\\(\\P\\) Second solution (Victor Reis). Here is the fabled solution using inversion at \\(H\\) . First, we rephrase the angle conditions in the following ways: \n\n\\(\\overline{AD} \\perp (THC)\\) , which is equivalent to the claim from the first solution. \n\n\\(\\overline{AB} \\perp (SHC)\\) , by symmetry. \n\n\\(\\overline{AC} \\perp (ABCD)\\) , by definition. \n\nNow for concreteness we will use a negative inversion at \\(H\\) which swaps \\(B\\) and \\(D\\) and overlay it on the original diagram. As usual we denote inverses with stars. \n\nLet us describe the inverted problem. We let \\(M\\) and \\(N\\) denote the midpoints of \\(\\overline{A^{*}B^{*}}\\) and \\(\\overline{A^{*}D^{*}}\\) , which are the centers of \\((HA^{*}B^{*})\\) and \\((HA^{*}D^{*})\\) . From \\(\\overline{T^{*}C^{*}} \\perp (HA^{*}D^{*})\\) , we know have \\(C^{*}\\) , \\(M\\) , \\(T^{*}\\) collinear. Similarly, \\(C^{*}\\) , \\(N\\) , \\(S^{*}\\) are collinear. We have that \\((A^{*}HC^{*})\\) is orthogonal to \\((ABCD)\\) which remains fixed. We wish to show \\(\\overline{T^{*}S^{*}}\\) and \\(\\overline{MN}\\) are parallel. \n\n\n\n\n\n\nLot \\(\\omega\\) denote the circumcircle of \\(\\triangle A^{*}H C^{*}\\) , which is orthogonal to the original circle \\((A B C D)\\) . It would suffices to show \\((A^{*}H C^{*})\\) is an \\(H\\) - Apollonius circle with respect to \\(\\overline{{M N}}\\) , from which we would get \\(C^{*}M / H M = C^{*}N / H N\\) . \n\nHowever, \\(\\omega\\) through \\(H\\) and \\(A\\) , hence it center lies on line \\(M N\\) . Moreover \\(\\omega\\) is orthogonal to \\((A^{*}M N)\\) (since \\((A^{*}M N)\\) and \\((A^{*}B D)\\) are homothetic). This is enough (for example, if we let \\(O\\) denote the center of \\(\\omega\\) , we now have \\(\\mathrm{r}(\\omega)^{2} = O H^{2} = O M \\cdot O N\\) ). (Note in this proof that the fact that \\(C^{*}\\) lies on \\((A B C D)\\) is not relevant.)\n\n\n\n## \\(\\S 2\\) Solutions to Day 2", "metadata": {"resource_path": "IMO/segmented/en-IMO-2014-notes.jsonl", "problem_match": "3. ", "solution_match": "## \\(\\S 1.3\\) IMO 2014/3, proposed by ALi Zamani (IRN) \n"}}
-{"year": "2014", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "Let \\(P\\) and \\(Q\\) be on segment \\(B C\\) of an acute triangle \\(A B C\\) such that \\(\\angle P A B =\\) \\(\\angle B C A\\) and \\(\\angle C A Q = \\angle A B C\\) . Let \\(M\\) and \\(N\\) be points on \\(\\overline{{A P}}\\) and \\(\\overline{{A Q}}\\) , respectively, such that \\(P\\) is the midpoint of \\(\\overline{{A M}}\\) and \\(Q\\) is the midpoint of \\(\\overline{{A N}}\\) . Prove that \\(\\overline{{B M}}\\) and \\(\\overline{{C N}}\\) meet on the circumcircle of \\(\\triangle A B C\\)", "solution": ". \n\nWe give three solutions. \n\n\\(\\P\\) First solution by harmonic bundles. Let \\(\\overline{BM}\\) intersect the circumcircle again at \\(X\\) . \n\n\n \n\nThe angle conditions imply that the tangent to \\((ABC)\\) at \\(B\\) is parallel to \\(\\overline{AP}\\) . Let \\(\\infty\\) be the point at infinity along line \\(AP\\) . Then \n\n\\[-1 = (AM;P\\infty)\\stackrel {B}{=}(AX;BC).\\] \n\nSimilarly, if \\(\\overline{CN}\\) meets the circumcircle at \\(Y\\) then \\((AY;BC) = - 1\\) as well. Hence \\(X = Y\\) , which implies the problem. \n\n\\(\\P\\) Second solution by similar triangles. Once one observes \\(\\triangle CAQ \\sim \\triangle CBA\\) , one can construct \\(D\\) the reflection of \\(B\\) across \\(A\\) , so that \\(\\triangle CAN \\sim \\triangle CBD\\) . Similarly, letting \\(E\\) be the reflection of \\(C\\) across \\(A\\) , we get \\(\\triangle BAP \\sim \\triangle BCA \\Rightarrow \\triangle BAM \\sim \\triangle BCE\\) . Now to show \\(\\angle ABM + \\angle ACN = 180^\\circ\\) it suffices to show \\(\\angle EBC + \\angle BCD = 180^\\circ\\) , which follows since \\(BCDE\\) is a parallelogram.\n\n\n\nThird solution by barycentric coordinates. Since \\(PB = c^{2} / a\\) we have \n\n\\[P = (0:a^{2} - c^{2}:c^{2})\\] \n\nso the reflection \\(\\vec{M} = 2\\vec{P} - \\vec{A}\\) has coordinates \n\n\\[M = (-a^{2}:2(a^{2} - c^{2}):2c^{2}).\\] \n\nSimilarly \\(N = (-a^{2}:2b^{2}:2(b^{2} - a^{2}))\\) . Thus \n\n\\[\\overline{BM}\\cap \\overline{CN} = (-a^{2}:2b^{2}:2c^{2})\\] \n\nwhich clearly lies on the circumcircle, and is in fact the point identified in the first solution.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2014-notes.jsonl", "problem_match": "4. ", "solution_match": "## \\(\\S 2.1\\) IMO 2014/4, proposed by Giorgi Arabidze (GEO) \n"}}
+{"year": "2014", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "Let \\(n\\geq 2\\) be an integer. Consider an \\(n\\times n\\) chessboard consisting of \\(n^{2}\\) unit squares. A configuration of \\(n\\) rooks on this board is peaceful if every row and every column contains exactly one rook. Find the greatest positive integer \\(k\\) such that, for each peaceful configuration of \\(n\\) rooks, there is a \\(k\\times k\\) square which does not contain a rook on any of its \\(k^{2}\\) unit squares.", "solution": "The answer is \\(k = \\lfloor \\sqrt{n - 1} \\rfloor\\) . \n\n\\(\\P\\) Proof that the property holds when \\(n\\geq k^{2} + 1\\) . First, assume \\(n > k^{2}\\) for some \\(k\\) . We will prove we can find an empty \\(k\\times k\\) square. Indeed, let \\(R\\) be a rook in the uppermost column, and draw \\(k\\) squares of size \\(k\\times k\\) directly below it, aligned. There are at most \\(k - 1\\) rooks among these squares, as desired. \n\n\n \n\n\\(\\P\\) Construction for all \\(n\\leq k^{2}\\) . We first give an example where for \\(n = k^{2}\\) showing there may be no empty \\(k\\times k\\) square. We draw the example for \\(k = 3\\) (with the generalization being obvious); \n\n\n \n\nTo show that this works, consider for each rook drawing an \\(k\\times k\\) square of \\(X\\) 's whose bottom- right hand corner is the rook (these may go off the board). These indicate\n\n\n\npositions where one cannot place the upper- left hand corner of any square. It’s easy to see that these cover the entire board, except parts of the last \\(k - 1\\) columns, which don’t matter anyways. \n\nIt remains to check that \\(n \\leq k^{2}\\) also all work (omitting this step is a common mistake). For this, we can delete rows and column to get an \\(n \\times n\\) board, and then fill in any gaps where we accidentally deleted a rook.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2014-notes.jsonl", "problem_match": "2. ", "solution_match": "## \\(\\S 1.2\\) IMO 2014/2, proposed by Tonči Kokan (HRV) \n"}}
+{"year": "2014", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "Convex quadrilateral \\(A B C D\\) has \\(\\angle A B C = \\angle C D A = 90^{\\circ}\\) . Point \\(H\\) is the foot of the perpendicular from \\(A\\) to \\(\\overline{{B D}}\\) . Points \\(S\\) and \\(T\\) lie on sides \\(A B\\) and \\(A D\\) , respectively, such that \\(H\\) lies inside triangle \\(S C T\\) and \n\n\\[\\angle C H S - \\angle C S B = 90^{\\circ},\\quad \\angle T H C - \\angle D T C = 90^{\\circ}.\\] \n\nProve that line \\(B D\\) is tangent to the circumcircle of triangle \\(T S H\\)", "solution": ") . \n\nFirst solution (mine). First we rewrite the angle condition in a suitable way. \n\nClaim — We have \\(\\angle A T H = \\angle T C H + 90^{\\circ}\\) . Thus the circumcenter of \\(\\triangle C T H\\) lies on \\(\\overline{{A D}}\\) . Similarly the circumcenter of \\(\\triangle C S H\\) lies on \\(\\overline{{A B}}\\) . \n\nProof. \n\n\\[\\angle A T H = \\angle D T H\\] \\[\\qquad = \\angle D T C + \\angle C T H\\] \\[\\qquad = \\angle D T C - \\angle T H C - \\angle H C T\\] \\[\\qquad = 90^{\\circ} - \\angle H C T = 90^{\\circ} + \\angle T C H.\\] \n\nwhich implies conclusion. \n\n\n \n\nLet the perpendicular bisector of \\(T H\\) meet \\(A H\\) at \\(P\\) now. It suffices to show that \\(\\frac{A P}{P H}\\) is symmetric in \\(b = A D\\) and \\(d = A B\\) , because then \\(P\\) will be the circumcenter of \\(\\triangle T S H\\) . To do this, set \\(A H = \\frac{b d}{2 R}\\) and \\(A C = 2 R\\) . \n\nLet \\(O\\) denote the circumcenter of \\(\\triangle C H T\\) . Use the Law of Cosines on \\(\\triangle A C O\\) and \\(\\triangle A H O\\) , using variables \\(x = A O\\) and \\(r = H O\\) . We get that \n\n\\[r^{2} = x^{2} + A H^{2} - 2x\\cdot A H\\cdot \\frac{d}{2R} = x^{2} + (2R)^{2} - 2b x.\\]\n\n\n\nBy the angle bisector theorem, \\(\\frac{AP}{PH} = \\frac{AO}{HO}\\) . \n\nThe rest is computation: notice that \n\n\\[r^{2} - x^{2} = h^{2} - 2xh\\cdot \\frac{d}{2R} = (2R)^{2} - 2bx\\] \n\nwhere \\(h = AH = \\frac{bd}{2R}\\) , whence \n\n\\[x = \\frac{(2R)^{2} - h^{2}}{2b - 2h\\cdot\\frac{d}{2R}}.\\] \n\nMoreover, \n\n\\[\\frac{1}{2}\\left(\\frac{r^{2}}{x^{2}} -1\\right) = \\frac{1}{x}\\left(\\frac{2}{x} R^{2} - b\\right).\\] \n\nNow, if we plug in the \\(x\\) in the right- hand side of the above, we obtain \n\n\\[\\frac{2b - 2h\\cdot\\frac{d}{2R}}{4R^{2} - h^{2}}\\left(\\frac{2b - 2h\\cdot\\frac{d}{2R}}{4R^{2}\\cdot h^{2}}\\cdot 2R^{2} - b\\right) = \\frac{2h}{(4R^{2} - h^{2})^{2}}\\left(b - h\\cdot \\frac{d}{2R}\\right)\\left(-2hdR + bh^{2}\\right).\\] \n\nPulling out a factor of \\(- 2Rh\\) from the rightmost term, we get something that is symmetric in \\(b\\) and \\(d\\) , as required. \n\n\\(\\P\\) Second solution (Victor Reis). Here is the fabled solution using inversion at \\(H\\) . First, we rephrase the angle conditions in the following ways: \n\n\\(\\overline{AD} \\perp (THC)\\) , which is equivalent to the claim from the first solution. \n\n\\(\\overline{AB} \\perp (SHC)\\) , by symmetry. \n\n\\(\\overline{AC} \\perp (ABCD)\\) , by definition. \n\nNow for concreteness we will use a negative inversion at \\(H\\) which swaps \\(B\\) and \\(D\\) and overlay it on the original diagram. As usual we denote inverses with stars. \n\nLet us describe the inverted problem. We let \\(M\\) and \\(N\\) denote the midpoints of \\(\\overline{A^{*}B^{*}}\\) and \\(\\overline{A^{*}D^{*}}\\) , which are the centers of \\((HA^{*}B^{*})\\) and \\((HA^{*}D^{*})\\) . From \\(\\overline{T^{*}C^{*}} \\perp (HA^{*}D^{*})\\) , we know have \\(C^{*}\\) , \\(M\\) , \\(T^{*}\\) collinear. Similarly, \\(C^{*}\\) , \\(N\\) , \\(S^{*}\\) are collinear. We have that \\((A^{*}HC^{*})\\) is orthogonal to \\((ABCD)\\) which remains fixed. We wish to show \\(\\overline{T^{*}S^{*}}\\) and \\(\\overline{MN}\\) are parallel. \n\n\n\n\n\n\nLot \\(\\omega\\) denote the circumcircle of \\(\\triangle A^{*}H C^{*}\\) , which is orthogonal to the original circle \\((A B C D)\\) . It would suffices to show \\((A^{*}H C^{*})\\) is an \\(H\\) - Apollonius circle with respect to \\(\\overline{{M N}}\\) , from which we would get \\(C^{*}M / H M = C^{*}N / H N\\) . \n\nHowever, \\(\\omega\\) through \\(H\\) and \\(A\\) , hence it center lies on line \\(M N\\) . Moreover \\(\\omega\\) is orthogonal to \\((A^{*}M N)\\) (since \\((A^{*}M N)\\) and \\((A^{*}B D)\\) are homothetic). This is enough (for example, if we let \\(O\\) denote the center of \\(\\omega\\) , we now have \\(\\mathrm{r}(\\omega)^{2} = O H^{2} = O M \\cdot O N\\) ). (Note in this proof that the fact that \\(C^{*}\\) lies on \\((A B C D)\\) is not relevant.)\n\n\n\n## \\(\\S 2\\) Solutions to Day 2", "metadata": {"resource_path": "IMO/segmented/en-IMO-2014-notes.jsonl", "problem_match": "3. ", "solution_match": "## \\(\\S 1.3\\) IMO 2014/3, proposed by ALi Zamani (IRN) \n"}}
+{"year": "2014", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "Let \\(P\\) and \\(Q\\) be on segment \\(B C\\) of an acute triangle \\(A B C\\) such that \\(\\angle P A B =\\) \\(\\angle B C A\\) and \\(\\angle C A Q = \\angle A B C\\) . Let \\(M\\) and \\(N\\) be points on \\(\\overline{{A P}}\\) and \\(\\overline{{A Q}}\\) , respectively, such that \\(P\\) is the midpoint of \\(\\overline{{A M}}\\) and \\(Q\\) is the midpoint of \\(\\overline{{A N}}\\) . Prove that \\(\\overline{{B M}}\\) and \\(\\overline{{C N}}\\) meet on the circumcircle of \\(\\triangle A B C\\)", "solution": ". \n\nWe give three solutions. \n\n\\(\\P\\) First solution by harmonic bundles. Let \\(\\overline{BM}\\) intersect the circumcircle again at \\(X\\) . \n\n\n \n\nThe angle conditions imply that the tangent to \\((ABC)\\) at \\(B\\) is parallel to \\(\\overline{AP}\\) . Let \\(\\infty\\) be the point at infinity along line \\(AP\\) . Then \n\n\\[-1 = (AM;P\\infty)\\stackrel {B}{=}(AX;BC).\\] \n\nSimilarly, if \\(\\overline{CN}\\) meets the circumcircle at \\(Y\\) then \\((AY;BC) = - 1\\) as well. Hence \\(X = Y\\) , which implies the problem. \n\n\\(\\P\\) Second solution by similar triangles. Once one observes \\(\\triangle CAQ \\sim \\triangle CBA\\) , one can construct \\(D\\) the reflection of \\(B\\) across \\(A\\) , so that \\(\\triangle CAN \\sim \\triangle CBD\\) . Similarly, letting \\(E\\) be the reflection of \\(C\\) across \\(A\\) , we get \\(\\triangle BAP \\sim \\triangle BCA \\Rightarrow \\triangle BAM \\sim \\triangle BCE\\) . Now to show \\(\\angle ABM + \\angle ACN = 180^\\circ\\) it suffices to show \\(\\angle EBC + \\angle BCD = 180^\\circ\\) , which follows since \\(BCDE\\) is a parallelogram.\n\n\n\nThird solution by barycentric coordinates. Since \\(PB = c^{2} / a\\) we have \n\n\\[P = (0:a^{2} - c^{2}:c^{2})\\] \n\nso the reflection \\(\\vec{M} = 2\\vec{P} - \\vec{A}\\) has coordinates \n\n\\[M = (-a^{2}:2(a^{2} - c^{2}):2c^{2}).\\] \n\nSimilarly \\(N = (-a^{2}:2b^{2}:2(b^{2} - a^{2}))\\) . Thus \n\n\\[\\overline{BM}\\cap \\overline{CN} = (-a^{2}:2b^{2}:2c^{2})\\] \n\nwhich clearly lies on the circumcircle, and is in fact the point identified in the first solution.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2014-notes.jsonl", "problem_match": "4. ", "solution_match": "## \\(\\S 2.1\\) IMO 2014/4, proposed by Giorgi Arabidze (GEO) \n"}}
{"year": "2014", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "For every positive integer \\(n\\) , the Bank of Cape Town issues coins of denomination \\(\\frac{1}{n}\\) . Given a finite collection of such coins (of not necessarily different denominations) with total value at most \\(99 + \\frac{1}{2}\\) , prove that it is possible to split this collection into 100 or fewer groups, such that each group has total value at most 1.", "solution": "We'll prove the result for at most \\(k - \\frac{k}{2k + 1}\\) with \\(k\\) groups. First, perform the following optimizations. \n\nIf any coin of size \\(\\frac{1}{2m}\\) appears twice, then replace it with a single coin of size \\(\\frac{1}{m}\\) . If any coin of size \\(\\frac{1}{2m + 1}\\) appears \\(2m + 1\\) times, group it into a single group and induct downwards. \n\nApply this operation repeatedly until it cannot be done anymore. \n\nNow construct boxes \\(B_{0}\\) , \\(B_{1}\\) , ..., \\(B_{k - 1}\\) . In box \\(B_{0}\\) put any coins of size \\(\\frac{1}{2}\\) (clearly there is at most one). More generally for \\(m \\geq 0\\) , \\(B_{m}\\) , put coins of size \\(\\frac{1}{2m + 1}\\) and \\(\\frac{1}{2m + 2}\\) (there at most \\(2m\\) of the former and at most one of the latter). Note that \n\n\\[\\mathrm{total~weight~in~}B_{m}\\leq 2m\\cdot \\frac{1}{2m + 1} +\\frac{1}{2m + 2} < 1.\\] \n\nFinally, place the remaining \"light\" coins of size at most \\(\\frac{1}{2k + 1}\\) in a pile. \n\nRemark. One way to explain where this boxing comes from is to imagine what happens after applying the operation repeatedly until it can't be done any more. We expect the most troublesome situation would be if the leftover coins are as large as possible, which looks like \n\n\\[\\frac{1}{2},\\quad \\frac{1}{3}\\times 2,\\quad \\frac{1}{4},\\quad \\frac{1}{5}\\times 4,\\quad \\frac{1}{6},\\quad \\frac{1}{7}\\times 6,\\quad \\frac{1}{8},\\quad \\dots\\] \n\nSeeing this the boxing \\(B_{i}\\) is quite reasonable because it groups these into groups of almost 1 without too much effort: \n\n\\[B_{0}\\longrightarrow \\frac{1}{2\\] \\[B_{1}\\longrightarrow \\frac{1}{3}\\cdot 2 + \\frac{1}{4} = \\frac{11}{12\\] \\[B_{2}\\longrightarrow \\frac{1}{5}\\cdot 4 + \\frac{1}{6} = \\frac{29}{30\\] \\[B_{3}\\longrightarrow \\frac{1}{7}\\cdot 6 + \\frac{1}{8} = \\frac{55}{56\\] \\[\\vdots\\] \n\nThen just toss coins from the pile into the boxes arbitrarily, other than the proviso that no box should have its weight exceed 1. We claim this uses up all coins in the pile. Assume not, and that some coin remains in the pile when all the boxes are saturated. Then all the boxes must have at least \\(1 - \\frac{1}{2k + 1}\\) , meaning the total amount in the boxes is strictly greater than \n\n\\[k\\left(1 - \\frac{1}{2k + 1}\\right) > k - \\frac{1}{2}\\]\n\n\n\nwhich is a contradiction. \n\nRemark. This gets a stronger bound \\(k - \\frac{k}{2k + 1}\\) than the requested \\(k - \\frac{1}{2}\\) .", "metadata": {"resource_path": "IMO/segmented/en-IMO-2014-notes.jsonl", "problem_match": "5. ", "solution_match": "## \\(\\S 2.2\\) IMO 2014/5, proposed by Gerhard Woeginger (LUX) \n"}}
-{"year": "2014", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "A set of lines in the plane is in general position if no two are parallel and no three pass through the same point. A set of lines in general position cuts the plane into regions, some of which have finite area; we call these its finite regions. Prove that for all sufficiently large \\(n\\) , in any set of \\(n\\) lines in general position it is possible to colour at least \\(\\sqrt{n}\\) lines blue in such a way that none of its finite regions has a completely blue boundary.", "solution": "Suppose we have colored \\(k\\) of the lines blue, and that it is not possible to color any additional lines. That means any of the \\(n - k\\) non- blue lines is the side of some finite region with an otherwise entirely blue perimeter. For each such line \\(\\ell\\) , select one such region, and take the next counterclockwise vertex; this is the intersection of two blue lines \\(v\\) . We'll say \\(\\ell\\) is the eyelid of \\(v\\) . \n\n\n \n\nYou can prove without too much difficulty that every intersection of two blue lines has at most two eyelids. Since there are \\(\\binom{k}{2}\\) such intersections, we see that \n\n\\[n - k\\leq 2\\binom{k}{2} = k^{2} - k\\] \n\nso \\(n\\leq k^{2}\\) , as required. \n\nRemark. In fact, \\(k = \\sqrt{n}\\) is \"sharp for greedy algorithms\", as illustrated below for \\(k = 3\\) : \n\n", "metadata": {"resource_path": "IMO/segmented/en-IMO-2014-notes.jsonl", "problem_match": "6. ", "solution_match": "## \\(\\S 2.3\\) IMO 2014/6, proposed by Gerhard Woeginger (AUT) \n"}}
+{"year": "2014", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "A set of lines in the plane is in general position if no two are parallel and no three pass through the same point. A set of lines in general position cuts the plane into regions, some of which have finite area; we call these its finite regions. Prove that for all sufficiently large \\(n\\) , in any set of \\(n\\) lines in general position it is possible to colour at least \\(\\sqrt{n}\\) lines blue in such a way that none of its finite regions has a completely blue boundary.", "solution": "Suppose we have colored \\(k\\) of the lines blue, and that it is not possible to color any additional lines. That means any of the \\(n - k\\) non- blue lines is the side of some finite region with an otherwise entirely blue perimeter. For each such line \\(\\ell\\) , select one such region, and take the next counterclockwise vertex; this is the intersection of two blue lines \\(v\\) . We'll say \\(\\ell\\) is the eyelid of \\(v\\) . \n\n\n \n\nYou can prove without too much difficulty that every intersection of two blue lines has at most two eyelids. Since there are \\(\\binom{k}{2}\\) such intersections, we see that \n\n\\[n - k\\leq 2\\binom{k}{2} = k^{2} - k\\] \n\nso \\(n\\leq k^{2}\\) , as required. \n\nRemark. In fact, \\(k = \\sqrt{n}\\) is \"sharp for greedy algorithms\", as illustrated below for \\(k = 3\\) : \n\n", "metadata": {"resource_path": "IMO/segmented/en-IMO-2014-notes.jsonl", "problem_match": "6. ", "solution_match": "## \\(\\S 2.3\\) IMO 2014/6, proposed by Gerhard Woeginger (AUT) \n"}}
diff --git a/IMO/segmented/en-IMO-2015-notes.jsonl b/IMO/segmented/en-IMO-2015-notes.jsonl
index 66cbf5ecb88887d84cb82a5041931feb3fe57d9b..33568496ee834b796b08eef9fe6540a2704090f1 100644
--- a/IMO/segmented/en-IMO-2015-notes.jsonl
+++ b/IMO/segmented/en-IMO-2015-notes.jsonl
@@ -1,6 +1,6 @@
-{"year": "2015", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "We say that a finite set \\(S\\) of points in the plane is balanced if, for any two different points \\(A\\) and \\(B\\) in \\(S\\) , there is a point \\(C\\) in \\(S\\) such that \\(AC = BC\\) . We say that \\(S\\) is center-free if for any three different points \\(A\\) , \\(B\\) and \\(C\\) in \\(S\\) , there are no points \\(P\\) in \\(S\\) such that \\(PA = PB = PC\\) . \n\n(a) Show that for all integers \\(n \\geq 3\\) , there exists a balanced set consisting of \\(n\\) points. \n(b) Determine all integers \\(n \\geq 3\\) for which there exists a balanced center-free set consisting of \\(n\\) points.", "solution": "We \n\nFor part (a), take a circle centered at a point \\(O\\) , and add \\(n - 1\\) additional points by adding pairs of points separated by an arc of \\(60^{\\circ}\\) or similar triples. An example for \\(n = 6\\) is shown below. \n\n\n \n\nFor part (b), the answer is odd \\(n\\) , achieved by taking a regular \\(n\\) - gon. To show even \\(n\\) fail, note that some point is on the perpendicular bisector of \n\n\\[\\left\\lceil \\frac{1}{n} {\\binom{n}{2}}\\right\\rceil = \\frac{n}{2}\\] \n\npairs of points, which is enough. (This is a standard double- counting argument.) \n\nRemark. As an aside, there is a funny joke about this problem. There are two types of people in the world: \n\n- Those who solve (b) quickly and then take forever to solve (a), \n\n- those who solve (a) quickly and then can't solve (b) at all. \n\n(Empirically true when the Taiwan IMO 2014 team was working on it.)", "metadata": {"resource_path": "IMO/segmented/en-IMO-2015-notes.jsonl", "problem_match": "1. ", "solution_match": "## \\(\\S 1.1\\) IMO 2015/1, proposed by Merlijn Staps (NLD) \n"}}
+{"year": "2015", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "We say that a finite set \\(S\\) of points in the plane is balanced if, for any two different points \\(A\\) and \\(B\\) in \\(S\\) , there is a point \\(C\\) in \\(S\\) such that \\(AC = BC\\) . We say that \\(S\\) is center-free if for any three different points \\(A\\) , \\(B\\) and \\(C\\) in \\(S\\) , there are no points \\(P\\) in \\(S\\) such that \\(PA = PB = PC\\) . \n\n(a) Show that for all integers \\(n \\geq 3\\) , there exists a balanced set consisting of \\(n\\) points. \n(b) Determine all integers \\(n \\geq 3\\) for which there exists a balanced center-free set consisting of \\(n\\) points.", "solution": "We \n\nFor part (a), take a circle centered at a point \\(O\\) , and add \\(n - 1\\) additional points by adding pairs of points separated by an arc of \\(60^{\\circ}\\) or similar triples. An example for \\(n = 6\\) is shown below. \n\n\n \n\nFor part (b), the answer is odd \\(n\\) , achieved by taking a regular \\(n\\) - gon. To show even \\(n\\) fail, note that some point is on the perpendicular bisector of \n\n\\[\\left\\lceil \\frac{1}{n} {\\binom{n}{2}}\\right\\rceil = \\frac{n}{2}\\] \n\npairs of points, which is enough. (This is a standard double- counting argument.) \n\nRemark. As an aside, there is a funny joke about this problem. There are two types of people in the world: \n\n- Those who solve (b) quickly and then take forever to solve (a), \n\n- those who solve (a) quickly and then can't solve (b) at all. \n\n(Empirically true when the Taiwan IMO 2014 team was working on it.)", "metadata": {"resource_path": "IMO/segmented/en-IMO-2015-notes.jsonl", "problem_match": "1. ", "solution_match": "## \\(\\S 1.1\\) IMO 2015/1, proposed by Merlijn Staps (NLD) \n"}}
{"year": "2015", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "Find all positive integers \\(a\\) , \\(b\\) , \\(c\\) such that each of \\(ab - c\\) , \\(bc - a\\) , \\(ca - b\\) is a power of 2 (possibly including \\(2^0 = 1\\) ).", "solution": "Fi \n\nHere is the solution of Telv Cohl, which is the shortest solution I am aware of. We will prove the only solutions are \\((2,2,2)\\) , \\((2,2,3)\\) , \\((2,6,11)\\) and \\((3,5,7)\\) and permutations. \n\nWLOG assume \\(a \\geq b \\geq c > 1\\) , so \\(ab - c \\geq ca - b \\geq bc - a\\) . We consider the following cases: \n\n- If \\(a\\) is even, then \n\n\\[ca - b = \\gcd (ab - c,ca - b)\\leq \\gcd (ab - c,a(ca - b) + ab - c)\\] \\[\\qquad = \\gcd \\left(ab - c,c(a^{2} - 1)\\right).\\] \n\nAs \\(a^{2} - 1\\) is odd, we conclude \\(ca - b \\leq c\\) . This implies \\(a = b = c = 2\\) . \n\n- If \\(a\\) , \\(b\\) , \\(c\\) are all odd, then \\(a > b > c > 1\\) follows. Then as before \n\n\\[ca - b\\leq \\gcd (ab - c,c(a^{2} - 1))\\leq 2^{a_{2}(a^{2} - 1)}\\leq 2a + 2\\leq 3a - b\\] \n\nso \\(c = 3\\) and \\(a = b + 2\\) . As \\(3a - b = ca - b \\geq 2(bc - a) = 6b - 2a\\) we then conclude \\(a = 7\\) and \\(b = 5\\) . \n\n- If \\(a\\) is odd and \\(b\\) , \\(c\\) are even, then \\(bc - a = 1\\) and hence \\(bc^{2} - b - c = ca - b\\) . Then from the miraculous identity \n\n\\[c^{3} - b - c = (1 - c^{2})(ab - c) + a(\\underbrace{bc^{2} - b - c}_{= ca - b}) + (ca - b)\\] \n\nso we conclude \\(\\gcd (ab - c,ca - b) = \\gcd (ab - c,c^{3} - b - c)\\) , in other words \n\n\\[bc^{2} - b - c = ca - b = \\gcd (ab - c,ca - b) = \\gcd (ab - c,c^{3} - b - c).\\] \n\nWe thus consider two more cases: \n\n- If \\(c^{3} - b - c \\neq 0\\) then the above implies \\(|c^{3} - b - c| \\geq bc^{2} - b - c\\) . As \\(b \\geq c > 1\\) , we must actually have \\(b = c\\) , thus \\(a = c^{2} - 1\\) . Finally \\(ab - c = c(c^{2} - 2)\\) is a power of 2, hence \\(b = c = 2\\) , so \\(a = 3\\) . \n\n- In the second case, assume \\(c^{3} - b - c = 0\\) , hence \\(c^{3} - c\\) . From \\(bc - a = 1\\) we obtain \\(a = c^{4} - c^{2} - 1\\) , hence \n\n\\[ca - b = c^{5} - 2c^{3} = c^{3}(c^{2} - 2)\\] \n\nis a power of 2, hence again \\(c = 2\\) . Thus \\(a = 11\\) and \\(b = 6\\) . \n\nThis finishes all cases, so the proof is done.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2015-notes.jsonl", "problem_match": "2. ", "solution_match": "## \\(\\S 1.2\\) IMO 2015/2, proposed by Dušan Djukić (SRB) \n"}}
-{"year": "2015", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "Let \\(ABC\\) be an acute triangle with \\(AB > AC\\) . Let \\(\\Gamma\\) be its circumcircle, \\(H\\) its orthocenter, and \\(F\\) the foot of the altitude from \\(A\\) . Let \\(M\\) be the midpoint of \\(\\overline{BC}\\) . Let \\(Q\\) be the point on \\(\\Gamma\\) such that \\(\\angle HQA = 90^\\circ\\) and let \\(K\\) be the point on \\(\\Gamma\\) such that \\(\\angle HKQ = 90^\\circ\\) . Assume that the points \\(A\\) , \\(B\\) , \\(C\\) , \\(K\\) and \\(Q\\) are all different and lie on \\(\\Gamma\\) in this order. Prove that the circumcircles of triangles \\(KQH\\) and \\(FKM\\) are tangent to each other.", "solution": "Letr. \n\nLet \\(L\\) be on the nine- point circle with \\(\\angle HML = 90^{\\circ}\\) . The negative inversion at \\(H\\) swapping \\(\\Gamma\\) and nine- point circle maps \n\n\\[A\\longleftrightarrow F,\\quad Q\\longleftrightarrow M,\\quad K\\longleftrightarrow L.\\] \n\nIn the inverted statement, we want line \\(ML\\) to be tangent to \\((AQL)\\) . \n\n\n \n\nClaim — \\(\\overline{LM} \\parallel \\overline{AQ}\\) . \n\nProof. Both are perpendicular to \\(\\overline{MHQ}\\) . \n\nClaim — \\(LA = LQ\\) .\n\n\n\nProof. Let \\(N\\) and \\(T\\) be the midpoints of \\(\\overline{HQ}\\) and \\(\\overline{AH}\\) , and \\(O\\) the circumcenter. As \\(\\overline{MT}\\) is a diameter, we know \\(LTNM\\) is a rectangle, so \\(\\overline{LT}\\) passes through \\(O\\) . Since \\(\\overline{LOT} \\perp \\overline{AQ}\\) and \\(OA = OQ\\) , the proof is complete. \\(\\square\\) \n\nTogether these two claims solve the problem.\n\n\n\n## \\(\\S 2\\) Solutions to Day 2", "metadata": {"resource_path": "IMO/segmented/en-IMO-2015-notes.jsonl", "problem_match": "3. ", "solution_match": "## §1.3 IMO 2015/3, proposed by Danylo Khilko and Mykhailo Plotnikov (UKR) \n"}}
-{"year": "2015", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "Triangle \\(ABC\\) has circumcircle \\(\\Omega\\) and circumcenter \\(O\\) . A circle \\(\\Gamma\\) with center \\(A\\) intersects the segment \\(BC\\) at points \\(D\\) and \\(E\\) , such that \\(B\\) , \\(D\\) , \\(E\\) , and \\(C\\) are all different and lie on line \\(BC\\) in this order. Let \\(F\\) and \\(G\\) be the points of intersection of \\(\\Gamma\\) and \\(\\Omega\\) , such that \\(A\\) , \\(F\\) , \\(B\\) , \\(C\\) , and \\(G\\) lie on \\(\\Omega\\) in this order. Let \\(K = (BDF) \\cap \\overline{AB} \\neq B\\) and \\(L = (CGE) \\cap \\overline{AC} \\neq C\\) and assume these points do not lie on line \\(FG\\) . Define \\(X = \\overline{FK} \\cap \\overline{GL}\\) . Prove that \\(X\\) lies on the line \\(AO\\) .", "solution": "Since \\(\\overline{AO} \\perp \\overline{FG}\\) for obvious reasons, we will only need to show that \\(XF = XG\\) , or that \\(\\angle KFG = \\angle LGF\\) . \n\nLet line \\(FG\\) meet \\((BDF)\\) and \\((CGE)\\) again at \\(F_{2}\\) and \\(G_{2}\\) . \n\n\n \n\nClaim — Quadrilaterals \\(FBDF_{2}\\) and \\(G_{2}ECG\\) are similar, actually homothetic through \\(\\overline{FG} \\cap \\overline{BC}\\) . \n\nProof. This is essentially a repeated application of being \"anti- parallel\" through \\(\\angle (FG, BC)\\) . Note the four angle relations \n\n\\[\\angle (FD,FG) = \\angle (BC,GE) = \\angle (G_{2}C,FG)\\Rightarrow \\overline{FD}\\parallel \\overline{G_{2}C\\] \\[\\angle (F_{2}B,FG) = \\angle (BC,FD) = \\angle (GE,FG)\\Rightarrow \\overline{F_{2}B}\\parallel \\overline{GE\\] \\[\\angle (FB,FG) = \\angle (BC,GC) = \\angle (G_{2}E,FG)\\Rightarrow \\overline{FB}\\parallel \\overline{G_{2}E\\]\n\n\n\n\\[\\angle (F_{2}D,F G) = \\angle (B C,F B) = \\angle (G C,F G)\\Rightarrow \\overline{{F_{2}D}}\\parallel \\overline{{G C}}.\\] \n\nThis gives the desired homotheties. \n\nTo finish the angle chase, \n\n\\[\\angle G F K = \\angle F_{2}B K = \\angle F_{2}B F - \\angle A B F = \\angle F_{2}D F - \\angle A B F\\] \\[\\qquad = \\angle F_{2}D F - \\angle G C A = \\angle G C G_{2} - \\angle G C A\\] \\[\\qquad = \\angle L C G_{2} = \\angle L G F\\] \n\nas needed. (Here \\(\\angle A B F = \\angle G C A\\) since \\(A F = A G\\) .)", "metadata": {"resource_path": "IMO/segmented/en-IMO-2015-notes.jsonl", "problem_match": "4. ", "solution_match": "## \\(\\S 2.1\\) IMO 2015/4, proposed by Silouanos Brazitikos and Evangelos Psychas (HEL) \n"}}
+{"year": "2015", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "Let \\(ABC\\) be an acute triangle with \\(AB > AC\\) . Let \\(\\Gamma\\) be its circumcircle, \\(H\\) its orthocenter, and \\(F\\) the foot of the altitude from \\(A\\) . Let \\(M\\) be the midpoint of \\(\\overline{BC}\\) . Let \\(Q\\) be the point on \\(\\Gamma\\) such that \\(\\angle HQA = 90^\\circ\\) and let \\(K\\) be the point on \\(\\Gamma\\) such that \\(\\angle HKQ = 90^\\circ\\) . Assume that the points \\(A\\) , \\(B\\) , \\(C\\) , \\(K\\) and \\(Q\\) are all different and lie on \\(\\Gamma\\) in this order. Prove that the circumcircles of triangles \\(KQH\\) and \\(FKM\\) are tangent to each other.", "solution": "Letr. \n\nLet \\(L\\) be on the nine- point circle with \\(\\angle HML = 90^{\\circ}\\) . The negative inversion at \\(H\\) swapping \\(\\Gamma\\) and nine- point circle maps \n\n\\[A\\longleftrightarrow F,\\quad Q\\longleftrightarrow M,\\quad K\\longleftrightarrow L.\\] \n\nIn the inverted statement, we want line \\(ML\\) to be tangent to \\((AQL)\\) . \n\n\n \n\nClaim — \\(\\overline{LM} \\parallel \\overline{AQ}\\) . \n\nProof. Both are perpendicular to \\(\\overline{MHQ}\\) . \n\nClaim — \\(LA = LQ\\) .\n\n\n\nProof. Let \\(N\\) and \\(T\\) be the midpoints of \\(\\overline{HQ}\\) and \\(\\overline{AH}\\) , and \\(O\\) the circumcenter. As \\(\\overline{MT}\\) is a diameter, we know \\(LTNM\\) is a rectangle, so \\(\\overline{LT}\\) passes through \\(O\\) . Since \\(\\overline{LOT} \\perp \\overline{AQ}\\) and \\(OA = OQ\\) , the proof is complete. \\(\\square\\) \n\nTogether these two claims solve the problem.\n\n\n\n## \\(\\S 2\\) Solutions to Day 2", "metadata": {"resource_path": "IMO/segmented/en-IMO-2015-notes.jsonl", "problem_match": "3. ", "solution_match": "## §1.3 IMO 2015/3, proposed by Danylo Khilko and Mykhailo Plotnikov (UKR) \n"}}
+{"year": "2015", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "Triangle \\(ABC\\) has circumcircle \\(\\Omega\\) and circumcenter \\(O\\) . A circle \\(\\Gamma\\) with center \\(A\\) intersects the segment \\(BC\\) at points \\(D\\) and \\(E\\) , such that \\(B\\) , \\(D\\) , \\(E\\) , and \\(C\\) are all different and lie on line \\(BC\\) in this order. Let \\(F\\) and \\(G\\) be the points of intersection of \\(\\Gamma\\) and \\(\\Omega\\) , such that \\(A\\) , \\(F\\) , \\(B\\) , \\(C\\) , and \\(G\\) lie on \\(\\Omega\\) in this order. Let \\(K = (BDF) \\cap \\overline{AB} \\neq B\\) and \\(L = (CGE) \\cap \\overline{AC} \\neq C\\) and assume these points do not lie on line \\(FG\\) . Define \\(X = \\overline{FK} \\cap \\overline{GL}\\) . Prove that \\(X\\) lies on the line \\(AO\\) .", "solution": "Since \\(\\overline{AO} \\perp \\overline{FG}\\) for obvious reasons, we will only need to show that \\(XF = XG\\) , or that \\(\\angle KFG = \\angle LGF\\) . \n\nLet line \\(FG\\) meet \\((BDF)\\) and \\((CGE)\\) again at \\(F_{2}\\) and \\(G_{2}\\) . \n\n\n \n\nClaim — Quadrilaterals \\(FBDF_{2}\\) and \\(G_{2}ECG\\) are similar, actually homothetic through \\(\\overline{FG} \\cap \\overline{BC}\\) . \n\nProof. This is essentially a repeated application of being \"anti- parallel\" through \\(\\angle (FG, BC)\\) . Note the four angle relations \n\n\\[\\angle (FD,FG) = \\angle (BC,GE) = \\angle (G_{2}C,FG)\\Rightarrow \\overline{FD}\\parallel \\overline{G_{2}C\\] \\[\\angle (F_{2}B,FG) = \\angle (BC,FD) = \\angle (GE,FG)\\Rightarrow \\overline{F_{2}B}\\parallel \\overline{GE\\] \\[\\angle (FB,FG) = \\angle (BC,GC) = \\angle (G_{2}E,FG)\\Rightarrow \\overline{FB}\\parallel \\overline{G_{2}E\\]\n\n\n\n\\[\\angle (F_{2}D,F G) = \\angle (B C,F B) = \\angle (G C,F G)\\Rightarrow \\overline{{F_{2}D}}\\parallel \\overline{{G C}}.\\] \n\nThis gives the desired homotheties. \n\nTo finish the angle chase, \n\n\\[\\angle G F K = \\angle F_{2}B K = \\angle F_{2}B F - \\angle A B F = \\angle F_{2}D F - \\angle A B F\\] \\[\\qquad = \\angle F_{2}D F - \\angle G C A = \\angle G C G_{2} - \\angle G C A\\] \\[\\qquad = \\angle L C G_{2} = \\angle L G F\\] \n\nas needed. (Here \\(\\angle A B F = \\angle G C A\\) since \\(A F = A G\\) .)", "metadata": {"resource_path": "IMO/segmented/en-IMO-2015-notes.jsonl", "problem_match": "4. ", "solution_match": "## \\(\\S 2.1\\) IMO 2015/4, proposed by Silouanos Brazitikos and Evangelos Psychas (HEL) \n"}}
{"year": "2015", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "Solve the functional equation \n\n\\[f(x + f(x + y)) + f(xy) = x + f(x + y) + yf(x)\\] \n\nfor \\(f: \\mathbb{R} \\to \\mathbb{R}\\) .", "solution": "\\) \n\nThe answers are \\(f(x)\\equiv x\\) and \\(f(x)\\equiv 2 - x\\) . Obviously, both of them work. \n\nLet \\(P(x,y)\\) be the given assertion. We also will let \\(S = \\{t\\mid f(t) = t\\}\\) be the set of fixed points of \\(f\\) \n\nFrom \\(P(0,0)\\) we get \\(f(f(0)) = 0\\) \n\nFrom \\(P(0,f(0))\\) we get \\(2f(0) = f(0)^{2}\\) and hence \\(f(0)\\in \\{0,2\\}\\) \n\nFrom \\(P(x,1)\\) we find that \\(x + f(x + 1)\\in S\\) for all \\(x\\) \n\nWe now solve the case \\(f(0) = 2\\) \n\nClaim If \\(f(0) = 2\\) then \\(f(x)\\equiv 2 - x\\) \n\nProof. Let \\(t\\in S\\) be any fixed point. Then \\(P(0,t)\\) gives \\(2 = 2t\\) or \\(t = 1\\) ; so \\(S = \\{1\\}\\) . But we also saw \\(x + f(x + 1)\\in S\\) , which implies \\(f(x)\\equiv 2 - x\\) . \\(\\square\\) \n\nHenceforth, assume \\(f(0) = 0\\) \n\nClaim If \\(f(0) = 0\\) then \\(f\\) is odd. \n\nProof. Note that \\(P(1, - 1)\\Rightarrow f(1) + f(- 1) = 1 - f(1)\\) and \\(P(- 1,1)\\Rightarrow f(- 1)+\\) \\(f(- 1) = - 1 + f(1)\\) , together giving \\(f(1) = 1\\) and \\(f(- 1) = - 1\\) . To prove \\(f\\) odd we now obtain more fixed points: \n\nFrom \\(P(x,0)\\) we find that \\(x + f(x)\\in S\\) for all \\(x\\in \\mathbb{R}\\) \n\nFrom \\(P(x - 1,1)\\) we find that \\(x - 1 + f(x)\\in S\\) for all \\(x\\in \\mathbb{R}\\) \n\nFrom \\(P(1,f(x) + x - 1)\\) we find \\(x + 1 + f(x)\\in S\\) for all \\(x\\in \\mathbb{R}\\) \n\nFinally \\(P(x, - 1)\\) gives \\(f\\) odd. \n\nTo finish from \\(f\\) odd, notice that \n\n\\[P(x, - x)\\Rightarrow f(x) + f(- x^{2}) = x - xf(x)\\] \\[P(- x,x)\\Rightarrow f(- x) + f(- x^{2}) = - x + xf(- x)\\] \n\nwhich upon subtracting gives \\(f(x)\\equiv x\\)", "metadata": {"resource_path": "IMO/segmented/en-IMO-2015-notes.jsonl", "problem_match": "5. ", "solution_match": "## \\(\\S 2.2\\) IMO 2015/5, proposed by Dorlir Ahmeti (ALB) \n"}}
{"year": "2015", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "The sequence \\(a_1, a_2, \\ldots\\) of integers satisfies the conditions: \n\n(i) \\(1 \\leq a_j \\leq 2015\\) for all \\(j \\geq 1\\) , \n\n(ii) \\(k + a_k \\neq \\ell + a_\\ell\\) for all \\(1 \\leq k < \\ell\\) . \n\nProve that there exist two positive integers \\(b\\) and \\(N\\) for which \n\n\\[\\left|\\sum_{j = m + 1}^{n}(a_{j} - b)\\right| \\leq 1007^{2}\\] \n\nfor all integers \\(m\\) and \\(n\\) such that \\(n > m \\geq N\\) .", "solution": "give two equivalent solutions with different presentations, one with \"arrows\" and the other by \"juggling\". \n\n\\(\\P\\) First solution (arrows). Consider the map \n\n\\[f\\colon k\\mapsto k + a_{k}.\\] \n\nThis map is injective, so if we draw all arrows of the form \\(k\\mapsto f(k)\\) we get a partition of \\(\\mathbb{N}\\) into one or more ascending chains (which skip by at most 2015). \n\nThere are at most 2015 such chains, since among any 2015 consecutive points in \\(\\mathbb{N}\\) every chain must have an element. \n\nWe claim we may take \\(b\\) to be the number of such chains, and \\(N\\) to be the largest of the start- points of all the chains. \n\nConsider an interval \\(I = [m + 1,n]\\) . We have that \n\n\\[\\sum_{m< j\\leq n}a_{j} = \\sum_{\\mathrm{chain~}c}[\\min \\{x > n,x\\in c\\} -\\min \\{x > m,x\\in c\\} ].\\] \n\nThus the upper bound is proved by the calculation \n\n\\[\\sum_{m< j\\leq n}(a_{j} - b) = \\sum_{\\mathrm{chain~}c}[(\\min \\{x > n,x\\in c\\} -n) - (\\min \\{x > m,x\\in c\\} -m)]\\] \\[\\qquad = \\sum_{\\mathrm{chain~}c}[(\\min \\{x > n,x\\in c\\} -n)] - \\sum_{\\mathrm{chain~}c}[\\min \\{x > m,x\\in c\\} -m]\\] \\[\\qquad \\leq (1 + 2015 + 2014 + \\dots +(2015 - (b - 2))) - (1 + 2 + \\dots +b)\\] \\[\\qquad = (b - 1)(2015 - b)\\] \n\nfrom above (noting that \\(n + 1\\) has to belong to some chain). The lower bound is similar. \n\n\\(\\P\\) Second solution (juggling). This solution is essentially the same, but phrased as a juggling problem. Here is a solution in this interpretation: we will consider several balls thrown in the air, which may be at heights 0, 1, 2, ..., 2014. The process is as follows:\n\n\n\n- Initially, at time \\(t = 0\\) , there are no balls in the air. \n\n- Then at each integer time \\(t\\) thereafter, if there is a ball at height 0, it is caught; otherwise a ball is added to the juggler's hand. This ball (either caught or added) is then thrown to a height of \\(a_{t}\\) . \n\n- Immediately afterwards, all balls have their height decreased by one. \n\nThe condition \\(a_{k} + k \\neq \\ell + a_{\\ell}\\) thus ensures that no two balls are ever at the same height. In particular, there will never be more than 2016 balls, since there are only 2015 possible heights. \n\nWe claim we may set. \n\n\\(b =\\) number of balls in entire process \n\n\\(N =\\) last moment in time at which a ball is added. \n\nIndeed, the key fact is that if we let \\(S_{t}\\) denote the sum of the height of all the balls just after time \\(t + \\frac{1}{2}\\) , then \n\n\\[S_{t + 1} - S_{t} = a_{t + 1} - b\\] \n\nAfter all, at each time step \\(t\\) , the caught ball is thrown to height \\(a_{t}\\) , and then all balls have their height decreased by 1, from which the conclusion follows. Hence the quantity in the problem is exactly equal to \n\n\\[\\left|\\sum_{j = m + 1}^{n}(a_{j} - b)\\right| = |S_{m} - S_{n}|.\\] \n\nFor a fixed \\(b\\) , we easily have the inequalities \\(0 + 1 + \\dots +(b - 1) \\leq S_{t} \\leq 2014 + 2013 + \\dots +(2015 - b)\\) . Hence \\(|S_{m} - S_{n}| \\leq (b - 1)(2015 - b) \\leq 1007^{2}\\) as desired.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2015-notes.jsonl", "problem_match": "6. ", "solution_match": "## \\(\\S 2.3\\) IMO 2015/6, proposed by Ross Atkins and Ivan Guo (AUS) \n"}}
diff --git a/IMO/segmented/en-IMO-2016-notes.jsonl b/IMO/segmented/en-IMO-2016-notes.jsonl
index c924f58f4c401030b9f85af4abe63ee902e648ea..759db1e67ad86efc136eff3f5c6e491b35427e3b 100644
--- a/IMO/segmented/en-IMO-2016-notes.jsonl
+++ b/IMO/segmented/en-IMO-2016-notes.jsonl
@@ -1,6 +1,6 @@
-{"year": "2016", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "In convex pentagon \\(A B C D E\\) with \\(\\angle B > 90^{\\circ}\\) , let \\(F\\) be a point on \\(\\overline{{A C}}\\) such that \\(\\angle F B C = 90^{\\circ}\\) . It is given that \\(F A = F B\\) \\(D A = D C\\) \\(E A = E D\\) , and rays \\(\\overline{{A C}}\\) and \\(\\overline{{A D}}\\) trisect \\(\\angle B A E\\) . Let \\(M\\) be the midpoint of \\(\\overline{{C F}}\\) . Let \\(X\\) be the point such that \\(A M X E\\) is a parallelogram. Show that \\(\\overline{{F X}}\\) \\(\\overline{{E M}}\\) \\(\\overline{{B D}}\\) are concurrent.", "solution": "Here is a \"long\" solution which I think shows where the \"power\" in the configuration comes from (it should be possible to come up with shorter solutions by cutting more directly to the desired conclusion). Throughout the proof, we let \n\n\\[\\theta = \\angle FAB = \\angle FBA = \\angle DAC = \\angle DCA = \\angle EAD = \\angle EDA.\\] \n\nWe begin by focusing just on \\(ABCD\\) with point \\(F\\) , ignoring for now the points \\(E\\) and \\(X\\) (and to some extent even point \\(M\\) ). It turns out this is a very familiar configuration. \n\n## Lemma (Central lemma) \n\nThe points \\(F\\) and \\(C\\) are the incenter and \\(A\\) - excenter of \\(\\triangle DAB\\) . Moreover, \\(\\triangle DAB\\) is isosceles with \\(DA = DB\\) . \n\nProof. The proof uses three observations: \n\n- We already know that \\(\\overline{FAC}\\) is the angle bisector of \\(\\angle ABD\\) . \n\n- We were given \\(\\angle FBC = 90^{\\circ}\\) . \n\n- Next, note that \\(\\triangle AFB \\sim \\triangle ADC\\) (they are similar isosceles triangles). From this it follows that \\(AF \\cdot AC = AB \\cdot AD\\) . \n\nThese three facts, together with \\(F\\) lying inside \\(\\triangle ABD\\) , are enough to imply the result. \\(\\square\\) \n\n\n\n\n\n\n## Corollary \n\nThe point \\(M\\) is the midpoint of arc \\(\\overline{BD}\\) of \\((DAB)\\) , and the center of cyclic quadrilateral \\(FDCB\\) . \n\nProof. Fact 5. \n\nUsing these observations as the anchor for everything that follows, we now prove several claims about \\(X\\) and \\(E\\) in succession. \n\n\n \n\nClaim — Point \\(E\\) is the midpoint of arc \\(\\overline{AD}\\) in \\((ABMD)\\) , and hence lies on ray \\(BF\\) . \n\nProof. This follows from \\(\\angle EDA = \\theta = \\angle EBA\\) . \n\nClaim — Points \\(X\\) is the second intersection of ray \\(\\overline{ED}\\) with \\((BFDC)\\) . \n\nProof. First, \\(\\overline{ED} \\parallel \\overline{AC}\\) already since \\(\\angle AED = 180^\\circ - 2\\theta\\) and \\(\\angle CAE = 2\\theta\\) . Now since \\(DB = DA\\) , we get \\(MB = MD = ED = EA\\) . Thus, \\(MX = AE = MB\\) , so \\(X\\) also lies on the circle \\((BFDC)\\) centered at \\(M\\) . \n\nClaim — The quadrilateral \\(EXMF\\) is an isosceles trapezoid. \n\nProof. We already know \\(\\overline{EX} \\parallel \\overline{FM}\\) . Since \\(\\angle EFA = 180^\\circ - \\angle AFB = 2\\theta = \\angle FAE\\) , we have \\(EF = EA\\) as well (and \\(F \\neq A\\) ). As \\(EXMA\\) was a parallelogram, it follows \\(EXMF\\) is an isosceles trapezoid. \n\nThe problem then follows by radical axis theorem on the three circles \\((AEDMB)\\) , \\((BFDXC)\\) and \\((EXMF)\\) .", "metadata": {"resource_path": "IMO/segmented/en-IMO-2016-notes.jsonl", "problem_match": "1. ", "solution_match": "## \\(\\S 1.1\\) IMO 2016/1, proposed by Art Waeterschoot (BEL) \n"}}
+{"year": "2016", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "In convex pentagon \\(A B C D E\\) with \\(\\angle B > 90^{\\circ}\\) , let \\(F\\) be a point on \\(\\overline{{A C}}\\) such that \\(\\angle F B C = 90^{\\circ}\\) . It is given that \\(F A = F B\\) \\(D A = D C\\) \\(E A = E D\\) , and rays \\(\\overline{{A C}}\\) and \\(\\overline{{A D}}\\) trisect \\(\\angle B A E\\) . Let \\(M\\) be the midpoint of \\(\\overline{{C F}}\\) . Let \\(X\\) be the point such that \\(A M X E\\) is a parallelogram. Show that \\(\\overline{{F X}}\\) \\(\\overline{{E M}}\\) \\(\\overline{{B D}}\\) are concurrent.", "solution": "Here is a \"long\" solution which I think shows where the \"power\" in the configuration comes from (it should be possible to come up with shorter solutions by cutting more directly to the desired conclusion). Throughout the proof, we let \n\n\\[\\theta = \\angle FAB = \\angle FBA = \\angle DAC = \\angle DCA = \\angle EAD = \\angle EDA.\\] \n\nWe begin by focusing just on \\(ABCD\\) with point \\(F\\) , ignoring for now the points \\(E\\) and \\(X\\) (and to some extent even point \\(M\\) ). It turns out this is a very familiar configuration. \n\n## Lemma (Central lemma) \n\nThe points \\(F\\) and \\(C\\) are the incenter and \\(A\\) - excenter of \\(\\triangle DAB\\) . Moreover, \\(\\triangle DAB\\) is isosceles with \\(DA = DB\\) . \n\nProof. The proof uses three observations: \n\n- We already know that \\(\\overline{FAC}\\) is the angle bisector of \\(\\angle ABD\\) . \n\n- We were given \\(\\angle FBC = 90^{\\circ}\\) . \n\n- Next, note that \\(\\triangle AFB \\sim \\triangle ADC\\) (they are similar isosceles triangles). From this it follows that \\(AF \\cdot AC = AB \\cdot AD\\) . \n\nThese three facts, together with \\(F\\) lying inside \\(\\triangle ABD\\) , are enough to imply the result. \\(\\square\\) \n\n\n\n\n\n\n## Corollary \n\nThe point \\(M\\) is the midpoint of arc \\(\\overline{BD}\\) of \\((DAB)\\) , and the center of cyclic quadrilateral \\(FDCB\\) . \n\nProof. Fact 5. \n\nUsing these observations as the anchor for everything that follows, we now prove several claims about \\(X\\) and \\(E\\) in succession. \n\n\n \n\nClaim — Point \\(E\\) is the midpoint of arc \\(\\overline{AD}\\) in \\((ABMD)\\) , and hence lies on ray \\(BF\\) . \n\nProof. This follows from \\(\\angle EDA = \\theta = \\angle EBA\\) . \n\nClaim — Points \\(X\\) is the second intersection of ray \\(\\overline{ED}\\) with \\((BFDC)\\) . \n\nProof. First, \\(\\overline{ED} \\parallel \\overline{AC}\\) already since \\(\\angle AED = 180^\\circ - 2\\theta\\) and \\(\\angle CAE = 2\\theta\\) . Now since \\(DB = DA\\) , we get \\(MB = MD = ED = EA\\) . Thus, \\(MX = AE = MB\\) , so \\(X\\) also lies on the circle \\((BFDC)\\) centered at \\(M\\) . \n\nClaim — The quadrilateral \\(EXMF\\) is an isosceles trapezoid. \n\nProof. We already know \\(\\overline{EX} \\parallel \\overline{FM}\\) . Since \\(\\angle EFA = 180^\\circ - \\angle AFB = 2\\theta = \\angle FAE\\) , we have \\(EF = EA\\) as well (and \\(F \\neq A\\) ). As \\(EXMA\\) was a parallelogram, it follows \\(EXMF\\) is an isosceles trapezoid. \n\nThe problem then follows by radical axis theorem on the three circles \\((AEDMB)\\) , \\((BFDXC)\\) and \\((EXMF)\\) .", "metadata": {"resource_path": "IMO/segmented/en-IMO-2016-notes.jsonl", "problem_match": "1. ", "solution_match": "## \\(\\S 1.1\\) IMO 2016/1, proposed by Art Waeterschoot (BEL) \n"}}
{"year": "2016", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "Find all integers \\(n\\) for which each cell of \\(n\\times n\\) table can be filled with one of the letters \\(I\\) \\(M\\) and \\(o\\) in such a way that: \n\nIn each row and column, one third of the entries are \\(I\\) , one third are \\(M\\) and one third are \\(o\\) ; and in any diagonal, if the number of entries on the diagonal is a multiple of three, then one third of the entries are \\(I\\) , one third are \\(M\\) and one third are \\(o\\) . \n\nNote that an \\(n\\times n\\) table has \\(4n - 2\\) diagonals.", "solution": "Fin \n\nThe answer is \\(n\\) divisible by 9. \n\nFirst we construct \\(n = 9\\) and by extension every multiple of 9. \n\n| I | I | I | M | M | M | O | O | O |
| M | M | M | O | O | O | I | I | I |
| O | O | O | I | I | I | M | M | M |
| I | I | I | M | M | M | O | O | O |
| M | M | M | O | O | O | I | I | I |
| O | O | O | I | I | I | M | M | M |
| I | I | I | M | M | M | O | O | O |
| M | M | M | O | O | O | I | I | I |
| O | O | O | I | I | I | M | M | M |
\n\nWe now prove \\(9\\mid n\\) is necessary. \n\nLet \\(n = 3k\\) , which divides the given grid into \\(k^{2}\\) sub- boxes (of size \\(3\\times 3\\) each). We say a multiset of squares \\(S\\) is clean if the letters distribute equally among them; note that unions of clean multisets are clean. \n\nConsider the following clean sets (given to us by problem statement): \n\nAll columns indexed 2 (mod 3), All rows indexed 2 (mod 3), and All \\(4k - 2\\) diagonals mentioned in the problem. \n\nTake their union. This covers the center of each box four times, and every other cell exactly once. We conclude the set of \\(k^{2}\\) center squares are clean, hence \\(3\\mid k^{2}\\) and so \\(9\\mid n\\) , as desired.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2016-notes.jsonl", "problem_match": "2. ", "solution_match": "## \\(\\S 1.2\\) IMO 2016/2, proposed by Trevor Tao (AUS) \n"}}
{"year": "2016", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "Let \\(P = A_{1}A_{2}\\ldots A_{k}\\) be a convex polygon in the plane. The vertices \\(A_{1}\\) \\(A_{2}\\) ,..., \\(A_{k}\\) have integral coordinates and lie on a circle. Let \\(S\\) be the area of \\(P\\) . An odd positive integer \\(n\\) is given such that the squares of the side lengths of \\(P\\) are integers divisible by \\(n\\) . Prove that \\(2S\\) is an integer divisible by \\(n\\) .", "solution": "L . \n\nSolution by Jeck Lim: We will prove the result just for \\(n = p^{e}\\) where \\(p\\) is an odd prime and \\(e \\geq 1\\) . The case \\(k = 3\\) is resolved by Heron's formula directly: we have \\(S = \\frac{1}{4} \\sqrt{2(a^{2}b^{2} + b^{2}c^{2} + c^{2}a^{2}) - a^{4} - b^{4} - c^{4}}\\) , so if \\(p^{e} \\mid \\gcd (a^{2}, b^{2}, c^{2})\\) then \\(p^{2e} \\mid S^{2}\\) . \n\nNow we show we can pick a diagonal and induct down on \\(k\\) by using inversion. \n\nLet the polygon be \\(A_{1}A_{2}\\ldots A_{k + 1}\\) and suppose for contradiction that all sides are divisible by \\(p^{e}\\) but no diagonals are. Let \\(O = A_{k + 1}\\) for notational convenience. By applying inversion around \\(O\\) with radius 1, we get the \"generalized Ptolemy theorem\" \n\n\\[\\frac{A_{1}A_{2}}{OA_{1}\\cdot OA_{2}} +\\frac{A_{2}A_{3}}{OA_{2}\\cdot OA_{3}} +\\dots +\\frac{A_{k - 1}A_{k}}{OA_{k - 1}\\cdot OA_{k}} = \\frac{A_{1}A_{k}}{OA_{1}\\cdot OA_{k}}\\] \n\nor, making use of square roots, \n\n\\[\\sqrt{\\frac{A_{1}A_{2}^{2}}{OA_{1}^{2}\\cdot OA_{2}^{2}}} +\\sqrt{\\frac{A_{2}A_{3}^{2}}{OA_{2}^{2}\\cdot OA_{3}^{2}}} +\\dots +\\sqrt{\\frac{A_{k - 1}A_{k}^{2}}{OA_{k - 1}^{2}\\cdot OA_{k}^{2}}} = \\sqrt{\\frac{A_{1}A_{k}^{2}}{OA_{1}^{2}\\cdot OA_{k}^{2}}}\\] \n\nSuppose \\(\\nu_{p}\\) of all diagonals is strictly less than \\(e\\) . Then the relation becomes \n\n\\[\\sqrt{q_{1}} +\\dots +\\sqrt{q_{k - 1}} = \\sqrt{q}\\] \n\nwhere \\(q_{i}\\) are positive rational numbers. Since there are no nontrivial relations between square roots (see this link) there is a positive rational number \\(b\\) such that \\(r_{i} = \\sqrt{q_{i} / b}\\) and \\(r = \\sqrt{q / b}\\) are all rational numbers. Then \n\n\\[\\sum r_{i} = r.\\] \n\nHowever, the condition implies that \\(\\nu_{p}(q_{i}^{2}) > \\nu_{p}(q^{2})\\) for all \\(i\\) (check this for \\(i = 1\\) , \\(i = k - 1\\) and \\(2 \\leq i \\leq k - 2\\) ), and hence \\(\\nu_{p}(r_{i}) > \\nu_{p}(r)\\) . This is absurd. \n\nRemark. I think you basically have to use some Ptolemy- like geometric property, and also all correct solutions I know of for \\(n = p^{e}\\) depend on finding a diagonal and inducting down. (Actually, the case \\(k = 4\\) is pretty motivating; Ptolemy implies one can cut in two.)\n\n\n\n## \\(\\S 2\\) Solutions to Day 2", "metadata": {"resource_path": "IMO/segmented/en-IMO-2016-notes.jsonl", "problem_match": "3. ", "solution_match": "## \\(\\S 1.3\\) IMO 2016/3, proposed by Aleksandr Gaifullin (RUS) \n"}}
{"year": "2016", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "A set of positive integers is called fragrant if it contains at least two elements and each of its elements has a prime factor in common with at least one of the other elements. Let \\(P(n) = n^{2} + n + 1\\) . What is the smallest possible positive integer value of \\(b\\) such that there exists a non-negative integer \\(a\\) for which the set \n\n\\[\\{P(a + 1),P(a + 2),\\ldots ,P(a + b)\\}\\] \n\nis fragrant?", "solution": "A \n\nThe answer is \\(b = 6\\) \n\nFirst, we prove \\(b\\geq 6\\) must hold. It is not hard to prove the following divisibilities by Euclid: \n\n\\[\\gcd (P(n),P(n + 1))\\mid 1\\] \\[\\gcd (P(n),P(n + 2))\\mid 7\\] \\[\\gcd (P(n),P(n + 3))\\mid 3\\] \\[\\gcd (P(n),P(n + 4))\\mid 19.\\] \n\nNow assume for contradiction \\(b\\leq 5\\) . Then any GCD's among \\(P(a + 1)\\) , ..., \\(P(a + b)\\) must be among \\(\\{3,7,19\\}\\) . Consider a multi- graph on \\(\\{a + 1,\\ldots ,a + b\\}\\) where we join two elements with nontrivial GCD and label the edge with the corresponding prime. Then we readily see there is at most one edge each of \\(\\{3,7,19\\}\\) : id est at most one edge of gap 2, 3, 4 (and no edges of gap 1). (By the gap of an edge \\(e = \\{u,v\\}\\) we mean \\(|u - v|\\) .) But one can see that it's now impossible for every vertex to have nonzero degree, contradiction. \n\nTo construct \\(b = 6\\) we use the Chinese remainder theorem: select \\(a\\) with \n\n\\[a + 1\\equiv 7\\pmod{19\\] \\[a + 5\\equiv 11\\pmod{19\\] \\[a + 2\\equiv 2\\pmod{7\\] \\[a + 4\\equiv 4\\pmod{7\\] \\[a + 3\\equiv 1\\pmod{3\\] \\[a + 6\\equiv 1\\pmod{3\\]", "metadata": {"resource_path": "IMO/segmented/en-IMO-2016-notes.jsonl", "problem_match": "4. ", "solution_match": "## \\(\\S 2.1\\) IMO 2016/4, proposed by Gerhard Woeginger (LUX) \n"}}
{"year": "2016", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "The equation \n\n\\[(x - 1)(x - 2)\\ldots (x - 2016) = (x - 1)(x - 2)\\ldots (x - 2016)\\] \n\nis written on the board, with 2016 linear factors on each side. What is the least possible value of \\(k\\) for which it is possible to erase exactly \\(k\\) of these 4032 linear factors so that at least one factor remains on each side and the resulting equation has no real solutions?", "solution": "The answer is 2016. Obviously this is necessary in order to delete duplicated factors. We now prove it suffices to deleted 2 (mod 4) and 3 (mod 4) guys from the left- hand side, and 0 (mod 4), 1 (mod 4) from the right- hand side. \n\nConsider the 1008 inequalities \n\n\\[(x - 1)(x - 4)< (x - 2)(x - 3)\\] \\[(x - 5)(x - 8)< (x - 6)(x - 7)\\] \\[(x - 9)(x - 12)< (x - 10)(x - 11)\\] \\[\\vdots\\] \\[(x - 2013)(x - 2016)< (x - 2014)(x - 2015).\\] \n\nNotice that in all these inequalities, at most one of them has non- positive numbers in it, and we never have both zero. If there is exactly one negative term among the \\(1008\\cdot 2 = 2016\\) sides, it is on the left and we can multiply all together. Thus the only case that remains is if \\(x\\in (4m - 2,4m - 1)\\) for some \\(m\\) , say the \\(m\\) th inequality. In that case, the two sides of that inequality differ by a factor of at least 9. \n\nClaim — We have \n\n\\[\\prod_{k\\geq 0}\\frac{(4k + 2)(4k + 3)}{(4k + 1)(4k + 4)} < e.\\] \n\nProof of claim using logarithms. It's equivalent to prove \n\n\\[\\sum_{k\\geq 0}\\log \\left(1 + \\frac{2}{(4k + 1)(4k + 4)}\\right)< 1.\\] \n\nTo this end, we use the deep fact that \\(\\log (1 + t)\\leq t\\) , and thus it follows from \\(\\begin{array}{r}{\\sum_{k\\geq 0}\\frac{1}{(4k + 1)(4k + 4)} < \\frac{1}{2}} \\end{array}\\) , which one can obtain for example by noticing it's less than \\(\\frac{1}{4}\\frac{\\pi^{2}}{6}\\) . \\(\\square\\) \n\nRemark (Elementary proof of claim, given by Espen Slettnes). To avoid calculus as above,\n\n\n\nfor each \\(N \\geq 0\\) , note the partial product is bounded by \n\n\\[\\prod_{k = 0}^{N}\\frac{(4k + 2)(4k + 3)}{(4k + 1)(4k + 4)} = \\frac{2}{1}\\cdot \\left(\\frac{3}{4}\\cdot \\frac{6}{5}\\right)\\cdot \\left(\\frac{7}{8}\\cdot \\frac{10}{9}\\right)\\cdot \\cdot \\cdot \\cdot \\cdot \\cdot \\cdot \\cdot \\cdot \\cdot \\cdot\\] \\[\\qquad < 2\\cdot 1\\cdot 1\\cdot \\cdot \\cdot \\cdot \\cdot \\cdot \\cdot \\cdot \\cdot \\cdot \\cdot \\frac{4N + 3}{4N + 4} < 2< e.\\] \n\nThis solves the problem, because then the factors being multiplied on by the positive inequalities before the \\(m\\) th one are both less than \\(e\\) , and \\(e^{2} < 9\\) . In symbols, for \\(4m - 2 < x < 4m - 1\\) we should have \n\n\\[\\frac{(x - (4m - 6))(x - (4m - 5))}{(x - (4m - 7))(x - (4m - 4))} \\times \\dots \\times \\frac{(x - 2)(x - 3)}{(x - 1)(x - 4)} < e\\] \n\nand \n\n\\[\\frac{(x - (4m + 2))(x - (4m + 3))}{(x - (4m + 1))(x - (4m + 4))} \\times \\dots \\times \\frac{(x - 2014)(x - 2015)}{(x - 2013)(x - 2016)} < e\\] \n\nbecause the \\((k + 1)\\) st term of each left- hand side is at most \\(\\frac{(4k + 2)(4k + 3)}{(4k + 1)(4k + 4)}\\) , for \\(k \\geq 0\\) . As \\(e^{2} < 9\\) , we're okay.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2016-notes.jsonl", "problem_match": "5. ", "solution_match": "## \\(\\S 2.2\\) IMO 2016/5, proposed by Russia \n"}}
-{"year": "2016", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "There are \\(n\\geq 2\\) line segments in the plane such that every two segments cross and no three segments meet at a point. Geoff has to choose an endpoint of each segment and place a frog on it facing the other endpoint. Then he will clap his hands \\(n - 1\\) times. Every time he claps, each frog will immediately jump forward to the next intersection point on its segment. Frogs never change the direction of their jumps. Geoff wishes to place the frogs in such a way that no two of them will ever occupy the same intersection point at the same time. \n\n(a) Prove that Geoff can always fulfill his wish if \\(n\\) is odd. \n\n(b) Prove that Geoff can never fulfill his wish if \\(n\\) is even.", "solution": "The following solution was communicated to me by Yang Liu. \n\nImagine taking a larger circle \\(\\omega\\) encasing all \\(\\binom{n}{2}\\) intersection points. Denote by \\(P_{1}\\) , \\(P_{2}\\) , ..., \\(P_{2n}\\) the order of the points on \\(\\omega\\) in clockwise order; we imagine placing the frogs on \\(P_{i}\\) instead. Observe that, in order for every pair of segments to meet, each line segment must be of the form \\(P_{i}P_{i + n}\\) . \n\n\n \n\nThen: \n\n(a) Place the frogs on \\(P_{1}\\) , \\(P_{3}\\) , ..., \\(P_{2n - 1}\\) . A simple parity arguments shows this works. \n\n(b) Observe that we cannot place frogs on consecutive \\(P_{i}\\) , so the \\(n\\) frogs must be placed on alternating points. But since we also are supposed to not place frogs on diametrically opposite points, for even \\(n\\) we immediately get a contradiction.\n\n\n\nRemark. Yang says: this is easy to guess if you just do a few small cases and notice that the pairs of \"violating points\" just forms a large cycle around.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2016-notes.jsonl", "problem_match": "6. ", "solution_match": "## \\(\\S 2.3\\) IMO 2016/6, proposed by Josef Tkadlec (CZE) \n"}}
+{"year": "2016", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "There are \\(n\\geq 2\\) line segments in the plane such that every two segments cross and no three segments meet at a point. Geoff has to choose an endpoint of each segment and place a frog on it facing the other endpoint. Then he will clap his hands \\(n - 1\\) times. Every time he claps, each frog will immediately jump forward to the next intersection point on its segment. Frogs never change the direction of their jumps. Geoff wishes to place the frogs in such a way that no two of them will ever occupy the same intersection point at the same time. \n\n(a) Prove that Geoff can always fulfill his wish if \\(n\\) is odd. \n\n(b) Prove that Geoff can never fulfill his wish if \\(n\\) is even.", "solution": "The following solution was communicated to me by Yang Liu. \n\nImagine taking a larger circle \\(\\omega\\) encasing all \\(\\binom{n}{2}\\) intersection points. Denote by \\(P_{1}\\) , \\(P_{2}\\) , ..., \\(P_{2n}\\) the order of the points on \\(\\omega\\) in clockwise order; we imagine placing the frogs on \\(P_{i}\\) instead. Observe that, in order for every pair of segments to meet, each line segment must be of the form \\(P_{i}P_{i + n}\\) . \n\n\n \n\nThen: \n\n(a) Place the frogs on \\(P_{1}\\) , \\(P_{3}\\) , ..., \\(P_{2n - 1}\\) . A simple parity arguments shows this works. \n\n(b) Observe that we cannot place frogs on consecutive \\(P_{i}\\) , so the \\(n\\) frogs must be placed on alternating points. But since we also are supposed to not place frogs on diametrically opposite points, for even \\(n\\) we immediately get a contradiction.\n\n\n\nRemark. Yang says: this is easy to guess if you just do a few small cases and notice that the pairs of \"violating points\" just forms a large cycle around.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2016-notes.jsonl", "problem_match": "6. ", "solution_match": "## \\(\\S 2.3\\) IMO 2016/6, proposed by Josef Tkadlec (CZE) \n"}}
diff --git a/IMO/segmented/en-IMO-2017-notes.jsonl b/IMO/segmented/en-IMO-2017-notes.jsonl
index 9e65ad86e372bc1e47f33c8a478a33b46ace808a..ebbee536a696b37f20d98e960ec19408e2e80910 100644
--- a/IMO/segmented/en-IMO-2017-notes.jsonl
+++ b/IMO/segmented/en-IMO-2017-notes.jsonl
@@ -1,6 +1,6 @@
{"year": "2017", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "For each integer \\(a_{0} > 1\\) , define the sequence \\(a_{0}\\) , \\(a_{1}\\) , \\(a_{2}\\) , ..., by \n\n\\[a_{n + 1} = \\left\\{ \\begin{array}{ll}\\sqrt{a_{n}} & \\mathrm{if~}\\sqrt{a_{n}} \\mathrm{is~an~integer,}\\\\ a_{n} + 3 & \\mathrm{otherwise} \\end{array} \\right.\\] \n\nfor each \\(n \\geq 0\\) . Determine all values of \\(a_{0}\\) for which there is a number \\(A\\) such that \\(a_{n} = A\\) for infinitely many values of \\(n\\) .", "solution": "F \n\nThe answer is \\(a_{0} \\equiv 0\\) (mod 3) only. \n\n\\(\\P\\) First solution. We first compute the minimal term of any sequence, periodic or not. \n\n## Lemma \n\nLet \\(c\\) be the smallest term in \\(a_{n}\\) . Then either \\(c \\equiv 2\\) (mod 3) or \\(c = 3\\) . \n\nProof. Clearly \\(c \\neq 1,4\\) . Assume \\(c \\neq 2\\) (mod 3). As \\(c\\) is not itself a square, the next perfect square after \\(c\\) in the sequence is one of \\((\\lfloor \\sqrt{c} \\rfloor +1)^{2}\\) , \\((\\lfloor \\sqrt{c} \\rfloor +2)^{2}\\) , or \\((\\lfloor \\sqrt{c} \\rfloor +3)^{2}\\) . So by minimality we require \n\n\\[c \\leq \\lfloor \\sqrt{c} \\rfloor +3 \\leq \\sqrt{c} +3\\] \n\nwhich requires \\(c \\leq 5\\) . Since \\(c \\neq 1,2,4,5\\) we conclude \\(c = 3\\) . \n\nNow we split the problem into two cases: \n\n- If \\(a_{0} \\equiv 0\\) (mod 3), then all terms of the sequence are 0 (mod 3). The smallest term of the sequence is thus 3 by the lemma and we have \n\n\\[3 \\rightarrow 6 \\rightarrow 9 \\rightarrow 3\\] \n\nso \\(A = 3\\) works fine. \n\n- If \\(a_{0} \\neq 0\\) (mod 3), then no term of the sequence is 0 (mod 3), and so in particular 3 does not appear in the sequence. So the smallest term of the sequence is 2 (mod 3) by lemma. But since no squares are 2 (mod 3), the sequence \\(a_{k}\\) grows without bound forever after, so no such \\(A\\) can exist. \n\nHence the answer is \\(a_{0} \\equiv 0\\) (mod 3) only.\n\n\n\nSecond solution. We clean up the argument by proving the following lemma. \n\n## Lemma \n\nIf \\(a_{n}\\) is constant modulo 3 and not 2 (mod 3), then \\(a_{n}\\) must eventually cycle in the form \\((m, m + 3, m + 6, \\ldots , m^{2})\\) , with no squares inside the cycle except \\(m^{2}\\) . \n\nProof. Observe that \\(a_{n}\\) must eventually hit a square, say \\(a_{k} = c^{2}\\) ; the next term is \\(a_{k + 1} = c\\) . Then it is forever impossible to exceed \\(c^{2}\\) again, by what is essentially discrete intermediate value theorem. Indeed, suppose \\(a_{\\ell} > c^{2}\\) and take \\(\\ell > k\\) minimal (in particular \\(a_{\\ell} \\neq \\sqrt{a_{\\ell - 1}}\\) ). Thus \\(a_{\\ell - 1} \\in \\{c^{2} - 2, c^{2} - 1, c^{2}\\}\\) and thus for modulo 3 reasons we have \\(a_{\\ell - 1} = c^{2}\\) . But that should imply \\(a_{\\ell} = c < c^{2}\\) , contradiction. \n\nWe therefore conclude \\(\\sup \\{a_{n}, a_{n + 1}, \\ldots \\}\\) is a decreasing integer sequence in \\(n\\) . It must eventually stabilize, say at \\(m^{2}\\) . Now we can't hit a square between \\(m\\) and \\(m^{2}\\) , and so we are done. \\(\\square\\) \n\nNow, we contend that all \\(a_{0} \\equiv 0\\) (mod 3) work. Indeed, for such \\(a_{0}\\) we have \\(a_{n} \\equiv 0\\) (mod 3) for all \\(n\\) , so the lemma implies that the problem statement is valid. \n\nNext, we observe that if \\(a_{i} \\equiv 2\\) (mod 3), then the sequence grows without bound afterwards since no squares are 2 (mod 3). In particular, if \\(a_{0} \\equiv 2\\) (mod 3) the answer is no. \n\nFinally, we claim that if \\(a_{0} \\equiv 1\\) (mod 3), then eventually some term is 2 (mod 3). Assume for contradiction this is not so; then \\(a_{n} \\equiv 1\\) (mod 3) must hold forever, and the lemma applies to give us a cycle of the form \\((m, m + 3, \\ldots , m^{2})\\) where \\(m \\equiv 1\\) (mod 3). In particular \\(m \\geq 4\\) and \n\n\\[m \\leq (m - 2)^{2} < m^{2}\\] \n\nbut \\((m - 2)^{2} \\equiv 1\\) (mod 3) which is a contradiction.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2017-notes.jsonl", "problem_match": "1. ", "solution_match": "## \\(\\S 1.1\\) IMO 2017/1, proposed by Stephan Wagner (SAF) \n"}}
{"year": "2017", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "Solve over \\(\\mathbb{R}\\) the functional equation \n\n\\[f(f(x)f(y)) + f(x + y) = f(xy).\\]", "solution": "The only solutions are \\(f(x) = 0\\) , \\(f(x) = x - 1\\) and \\(f(x) = 1 - x\\) , which clearly work. Note that \n\n- If \\(f\\) is a solution, so is \\(-f\\) . \n\n- Moreover, if \\(f(0) = 0\\) then setting \\(y = 0\\) gives \\(f \\equiv 0\\) . So henceforth we assume \\(f(0) > 0\\) . \n\nClaim — We have \\(f(z) = 0 \\iff z = 1\\) . Also, \\(f(0) = 1\\) and \\(f(1) = 0\\) . \n\nProof. For the forwards direction, if \\(f(z) = 0\\) and \\(z \\neq 1\\) one may put \\((x, y) = (z, z(z - 1)^{- 1})\\) (so that \\(x + y = xy\\) ) we deduce \\(f(0) = 0\\) which is a contradiction. \n\nFor the reverse, \\(f(f(0)^2) = 0\\) by setting \\(x = y = 0\\) , and use the previous part. We also conclude \\(f(1) = 0\\) , \\(f(0) = 1\\) . \\(\\square\\) \n\nClaim — If \\(f\\) is injective, we are done. \n\nProof. Setting \\(y = 0\\) in the original equation gives \\(f(f(x)) = 1 - f(x)\\) . We apply this three times on the expression \\(f^3 (x)\\) : \n\n\\[f(1 - f(x)) = f(f(f(x))) = 1 - f(f(x)) = f(x).\\] \n\nHence \\(1 - f(x) = x\\) or \\(f(x) = 1 - x\\) . \\(\\square\\) \n\nRemark. The result \\(f(f(x)) + f(x) = 1\\) also implies that surjectivity would solve the problem. \n\nClaim — \\(f\\) is injective. \n\nProof. Setting \\(y = 1\\) in the original equation gives \\(f(x + 1) = f(x) - 1\\) , and by induction \n\n\\[f(x + n) = f(x) - n. \\quad (1)\\] \n\nAssume now \\(f(a) = f(b)\\) . By using (1) we may shift \\(a\\) and \\(b\\) to be large enough that we may find \\(x\\) and \\(y\\) obeying \\(x + y = a + 1\\) , \\(xy = b\\) . Setting these gives \n\n\\[f(f(x)f(y)) = f(xy) - f(x + y) = f(b) - f(a + 1)\\] \\[= f(b) + 1 - f(a) = 1\\]\n\n\n\nfrom which we conclude \n\n\\[f(f(x)f(y) + 1) = 0.\\] \n\nHence by the first claim we have \\(f(x)f(y) + 1 = 1\\) , so \\(f(x)f(y) = 0\\) . Applying the first claim again gives \\(1 \\in \\{x, y\\}\\) . But that implies \\(a = b\\) . \\(\\square\\) \n\nRemark. Jessica Wan points out that for any \\(a \\neq b\\) , at least one of \\(a^2 > 4(b - 1)\\) and \\(b^2 > 4(a - 1)\\) is true. So shifting via (1) is actually unnecessary for this proof. \n\nRemark. One can solve the problem over \\(\\mathbb{Q}\\) using only (1) and the easy parts. Indeed, that already implies \\(f(n) = 1 - n\\) for all \\(n\\) . Now we induct to show \\(f(p / q) = 1 - p / q\\) for all \\(0 < p < q\\) (on \\(q\\) ). By choosing \\(x = 1 + p / q\\) , \\(y = 1 + q / p\\) , we cause \\(xy = x + y\\) , and hence \\(0 = f(f(1 + p / q)f(1 + q / p))\\) or \\(1 = f(1 + p / q)f(1 + q / p)\\) . \n\nBy induction we compute \\(f(1 + q / p)\\) and this gives \\(f(p / q + 1) = f(p / q) - 1\\) .", "metadata": {"resource_path": "IMO/segmented/en-IMO-2017-notes.jsonl", "problem_match": "2. ", "solution_match": "## \\(\\S 1.2\\) IMO 2017/2, proposed by Dorlir Ahmeti (ALB) \n"}}
-{"year": "2017", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "A hunter and an invisible rabbit play a game in the plane. The rabbit and hunter start at points \\(A_{0} = B_{0}\\) . In the \\(n\\) th round of the game ( \\(n \\geq 1\\) ), three things occur in order: \n\n(i) The rabbit moves invisibly from \\(A_{n - 1}\\) to a point \\(A_{n}\\) such that \\(A_{n - 1}A_{n} = 1\\) . \n\n(ii) The hunter has a tracking device (e.g. dog) which reports an approximate location \\(P_{n}\\) of the rabbit, such that \\(P_{n}A_{n} \\leq 1\\) . \n\n(iii) The hunter moves visibly from \\(B_{n - 1}\\) to a point \\(B_{n}\\) such that \\(B_{n - 1}B_{n} = 1\\) . \n\nLet \\(N = 10^{9}\\) . Can the hunter guarantee that \\(A_{N}B_{N} < 100\\) ?", "solution": "No, the hunter cannot. We will show how to increase the distance in the following way: \n\nClaim — Suppose the rabbit is at a distance \\(d \\geq 1\\) from the hunter at some point in time. Then it can increase its distance to at least \\(\\sqrt{d^{2} + \\frac{1}{2}}\\) in \\(4d\\) steps regardless of what the hunter already knows about the rabbit. \n\nProof. Consider a positive integer \\(n > d\\) , to be chosen later. Let the hunter start at \\(B\\) and the rabbit at \\(A\\) , as shown. Let \\(\\ell\\) denote line \\(AB\\) . \n\nNow, we may assume the rabbit reveals its location \\(A\\) , so that all previous information becomes irrelevant. \n\nThe rabbit chooses two points \\(X\\) and \\(Y\\) symmetric about \\(\\ell\\) such that \\(XY = 2\\) and \\(AX = AY = n\\) , as shown. The rabbit can then hop to either \\(X\\) or \\(Y\\) , pinging the point \\(P_{n}\\) on the \\(\\ell\\) each time. This takes \\(n\\) hops. \n\n\n \n\nNow among all points \\(H\\) the hunter can go to, \\(\\min \\max \\{HX, HY\\}\\) is clearly minimized with \\(H \\in \\ell\\) by symmetry. So the hunter moves to a point \\(H\\) such that \\(BH = n\\) as well. In that case the new distance is \\(HX = HY\\) . \n\nWe now compute \n\n\\[H X^{2} = 1 + H M^{2} = 1 + \\left(\\sqrt{A X^{2} - 1} -A H\\right)^{2}\\] \\[\\qquad = 1 + \\left(\\sqrt{n^{2} - 1} - (n - d)\\right)^{2}\\] \\[\\qquad \\geq 1 + \\left(\\left(n - \\frac{1}{n}\\right) - (n - d)\\right)^{2}\\] \\[\\qquad = 1 + (d - 1 / n)^{2}\\] \n\nwhich exceeds \\(d^{2} + \\frac{1}{2}\\) whenever \\(n \\geq 4d\\) .\n\n\n\nIn particular we can always take \\(n = 400\\) even very crudely; applying the lemma \\(2 \\cdot 100^{2}\\) times, this gives a bound of \\(400 \\cdot 2 \\cdot 100^{2} < 10^{9}\\) , as desired. \n\nRemark. The step of revealing the location of the rabbit seems critical because as far as I am aware it is basically impossible to keep track of ping locations in the problem. \n\nRemark. Reasons to believe the answer is \"no\": the \\(10^{9}\\) constant, and also that \"follow the last ping\" is losing for the hunter. \n\nRemark. I think there are roughly two ways you can approach the problem once you recognize the answer. \n\n(i) Try and control the location of the pings \n\n(ii) Abandon the notion of controlling possible locations, and try to increase the distance by a little bit, say from \\(d\\) to \\(\\sqrt{d^{2} + \\epsilon}\\) . This involves revealing the location of the rabbit before each iteration of several jumps. \n\nI think it's clear that the difficulty of my approach is realizing that (ii) is possible; once you do, the two- point approach is more or less the only one possible. \n\nMy opinion is that (ii) is not that magical; as I said it was the first idea I had. But I am biased, because when I test- solved the problem at the IMO it was called \"C5\" and not \"IMO3\"; this effectively told me it was unlikely that the official solution was along the lines of (i), because otherwise it would have been placed much later in the shortlist.\n\n\n\n## \\(\\S 2\\) Solutions to Day 2", "metadata": {"resource_path": "IMO/segmented/en-IMO-2017-notes.jsonl", "problem_match": "3. ", "solution_match": "## \\(\\S 1.3\\) IMO 2017/3, proposed by Gerhard Woeginger (AUT) \n"}}
-{"year": "2017", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "Let \\(R\\) and \\(S\\) be different points on a circle \\(\\Omega\\) such that \\(\\overline{RS}\\) is not a diameter. Let \\(\\ell\\) be the tangent line to \\(\\Omega\\) at \\(R\\) . Point \\(T\\) is such that \\(S\\) is the midpoint of \\(\\overline{RT}\\) . Point \\(J\\) is chosen on minor arc \\(RS\\) of \\(\\Omega\\) so that the circumcircle \\(\\Gamma\\) of triangle \\(JST\\) intersects \\(\\ell\\) at two distinct points. Let \\(A\\) be the common point of \\(\\Gamma\\) and \\(\\ell\\) closer to \\(R\\) . Line \\(AJ\\) meets \\(\\Omega\\) again at \\(K\\) . Prove that line \\(KT\\) is tangent to \\(\\Gamma\\) .", "solution": "First solution (elementary). First, note \n\n\\[\\angle RKA = \\angle RKJ = \\angle RSJ = \\angle TSJ = \\angle TAJ = \\angle TAK\\] \n\nso \\(\\overline{RK} \\parallel \\overline{AT}\\) . Now, \n\n\\(\\overline{RA}\\) is tangent at \\(R\\) iff \\(\\triangle KRS \\sim \\triangle RTA\\) (oppositely), because both equate to \\(- \\angle RKS = \\angle SKR = \\angle SRA = \\angle TRA\\) . \n\nSimilarly, \\(\\overline{TK}\\) is tangent at \\(T\\) iff \\(\\triangle ATS \\sim \\triangle TRK\\) . \n\nThe two similarities are equivalent because \\(RS = ST\\) the SAS gives \\(KR \\cdot TA = RS \\cdot RT = TS \\cdot TR\\) . \n\n\n \n\nRemark. The problem is actually symmetric with respect to two circles; \\(\\overline{RA}\\) is tangent at \\(R\\) if and only if \\(\\overline{TK}\\) at \\(T\\) . \n\nSecond solution (inversion). Consider an inversion at \\(R\\) fixing the circumcircle \\(\\Gamma\\) of \\(TSJA\\) . Then: \n\n- \\(T\\) and \\(S\\) swap,\n\n\n\n- \\(A\\) and \\(B\\) swap, where \\(B\\) is the second intersection of \\(\\ell\\) with \\(\\Gamma\\) . \n\n- Circle \\(\\Omega\\) inverts to the line through \\(T\\) parallel to \\(\\overline{RAB}\\) , call it \\(\\ell\\) . \n\n- \\(J^{*}\\) is the second intersection of \\(\\ell\\) with \\(\\Gamma\\) . \n\n- \\(K^{*}\\) is the intersection of \\(\\ell\\) with the circumcircle of \\(RBJ^{*}\\) ; this implies \\(RK^{*}J^{*}B\\) is an isosceles trapezoid. In particular, one reads \\(\\overline{RK^{*}} \\parallel \\overline{AT}\\) from this, hence \\(RK^{*}TA\\) is a parallelogram. \n\nThus we wish to show the circumcircle of \\(RSK^{*}\\) is tangent to \\(\\Gamma\\) . But that follows from the final parallelogram observed: \\(S\\) is the center of the parallelogram since it is the midpoint of the diagonal. \n\nRemark. This also implies \\(RKTB\\) is cyclic, from \\(\\overline{K^{*}SA}\\) collinear. Moreover, quadrilateral \\(KK^{*}TS\\) is cyclic (by power of a point); this leads to the second official solution to the problem.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2017-notes.jsonl", "problem_match": "4. ", "solution_match": "## \\(\\S 2.1\\) IMO 2017/4, proposed by Charles Leytem (LUX) \n"}}
-{"year": "2017", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "Fix \\(N \\geq 1\\) . A collection of \\(N(N + 1)\\) soccer players of distinct heights stand in a row. Sir Alex Song wishes to remove \\(N(N - 1)\\) players from this row to obtain a new row of \\(2N\\) players in which the following \\(N\\) conditions hold: no one stands between the two tallest players, no one stands between the third and fourth tallest players, ..., no one stands between the two shortest players. Prove that this is possible.", "solution": "Some opening remarks: location and height are symmetric to each other, if one thinks about this problem as permutation pattern avoidance. So while officially there are multiple solutions, they are basically isomorphic to one another, and I am not aware of any solution otherwise. \n\n\n \n\nTake a partition of \\(N\\) groups in order by height: \\(G_{1} = \\{1,\\ldots ,N + 1\\}\\) , \\(G_{2} = \\{N+\\) \\(2,\\ldots ,2N + 2\\}\\) , and so on. We will pick two people from each group \\(G_{k}\\) \n\nScan from the left until we find two people in the same group \\(G_{k}\\) . Delete all people scanned and also everyone in \\(G_{k}\\) . All the groups still have at least \\(N\\) people left, so we can induct down with the non- deleted people; the chosen pair is to the far left anyways. \n\nRemark. The important bit is to scan by position but group by height, and moreover not change the groups as we scan. Dually, one can have a solution which scans by height but groups by position.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2017-notes.jsonl", "problem_match": "5. ", "solution_match": "## \\(\\S 2.2\\) IMO 2017/5, proposed by Grigory Chelnokov (RUS) \n"}}
+{"year": "2017", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "A hunter and an invisible rabbit play a game in the plane. The rabbit and hunter start at points \\(A_{0} = B_{0}\\) . In the \\(n\\) th round of the game ( \\(n \\geq 1\\) ), three things occur in order: \n\n(i) The rabbit moves invisibly from \\(A_{n - 1}\\) to a point \\(A_{n}\\) such that \\(A_{n - 1}A_{n} = 1\\) . \n\n(ii) The hunter has a tracking device (e.g. dog) which reports an approximate location \\(P_{n}\\) of the rabbit, such that \\(P_{n}A_{n} \\leq 1\\) . \n\n(iii) The hunter moves visibly from \\(B_{n - 1}\\) to a point \\(B_{n}\\) such that \\(B_{n - 1}B_{n} = 1\\) . \n\nLet \\(N = 10^{9}\\) . Can the hunter guarantee that \\(A_{N}B_{N} < 100\\) ?", "solution": "No, the hunter cannot. We will show how to increase the distance in the following way: \n\nClaim — Suppose the rabbit is at a distance \\(d \\geq 1\\) from the hunter at some point in time. Then it can increase its distance to at least \\(\\sqrt{d^{2} + \\frac{1}{2}}\\) in \\(4d\\) steps regardless of what the hunter already knows about the rabbit. \n\nProof. Consider a positive integer \\(n > d\\) , to be chosen later. Let the hunter start at \\(B\\) and the rabbit at \\(A\\) , as shown. Let \\(\\ell\\) denote line \\(AB\\) . \n\nNow, we may assume the rabbit reveals its location \\(A\\) , so that all previous information becomes irrelevant. \n\nThe rabbit chooses two points \\(X\\) and \\(Y\\) symmetric about \\(\\ell\\) such that \\(XY = 2\\) and \\(AX = AY = n\\) , as shown. The rabbit can then hop to either \\(X\\) or \\(Y\\) , pinging the point \\(P_{n}\\) on the \\(\\ell\\) each time. This takes \\(n\\) hops. \n\n\n \n\nNow among all points \\(H\\) the hunter can go to, \\(\\min \\max \\{HX, HY\\}\\) is clearly minimized with \\(H \\in \\ell\\) by symmetry. So the hunter moves to a point \\(H\\) such that \\(BH = n\\) as well. In that case the new distance is \\(HX = HY\\) . \n\nWe now compute \n\n\\[H X^{2} = 1 + H M^{2} = 1 + \\left(\\sqrt{A X^{2} - 1} -A H\\right)^{2}\\] \\[\\qquad = 1 + \\left(\\sqrt{n^{2} - 1} - (n - d)\\right)^{2}\\] \\[\\qquad \\geq 1 + \\left(\\left(n - \\frac{1}{n}\\right) - (n - d)\\right)^{2}\\] \\[\\qquad = 1 + (d - 1 / n)^{2}\\] \n\nwhich exceeds \\(d^{2} + \\frac{1}{2}\\) whenever \\(n \\geq 4d\\) .\n\n\n\nIn particular we can always take \\(n = 400\\) even very crudely; applying the lemma \\(2 \\cdot 100^{2}\\) times, this gives a bound of \\(400 \\cdot 2 \\cdot 100^{2} < 10^{9}\\) , as desired. \n\nRemark. The step of revealing the location of the rabbit seems critical because as far as I am aware it is basically impossible to keep track of ping locations in the problem. \n\nRemark. Reasons to believe the answer is \"no\": the \\(10^{9}\\) constant, and also that \"follow the last ping\" is losing for the hunter. \n\nRemark. I think there are roughly two ways you can approach the problem once you recognize the answer. \n\n(i) Try and control the location of the pings \n\n(ii) Abandon the notion of controlling possible locations, and try to increase the distance by a little bit, say from \\(d\\) to \\(\\sqrt{d^{2} + \\epsilon}\\) . This involves revealing the location of the rabbit before each iteration of several jumps. \n\nI think it's clear that the difficulty of my approach is realizing that (ii) is possible; once you do, the two- point approach is more or less the only one possible. \n\nMy opinion is that (ii) is not that magical; as I said it was the first idea I had. But I am biased, because when I test- solved the problem at the IMO it was called \"C5\" and not \"IMO3\"; this effectively told me it was unlikely that the official solution was along the lines of (i), because otherwise it would have been placed much later in the shortlist.\n\n\n\n## \\(\\S 2\\) Solutions to Day 2", "metadata": {"resource_path": "IMO/segmented/en-IMO-2017-notes.jsonl", "problem_match": "3. ", "solution_match": "## \\(\\S 1.3\\) IMO 2017/3, proposed by Gerhard Woeginger (AUT) \n"}}
+{"year": "2017", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "Let \\(R\\) and \\(S\\) be different points on a circle \\(\\Omega\\) such that \\(\\overline{RS}\\) is not a diameter. Let \\(\\ell\\) be the tangent line to \\(\\Omega\\) at \\(R\\) . Point \\(T\\) is such that \\(S\\) is the midpoint of \\(\\overline{RT}\\) . Point \\(J\\) is chosen on minor arc \\(RS\\) of \\(\\Omega\\) so that the circumcircle \\(\\Gamma\\) of triangle \\(JST\\) intersects \\(\\ell\\) at two distinct points. Let \\(A\\) be the common point of \\(\\Gamma\\) and \\(\\ell\\) closer to \\(R\\) . Line \\(AJ\\) meets \\(\\Omega\\) again at \\(K\\) . Prove that line \\(KT\\) is tangent to \\(\\Gamma\\) .", "solution": "First solution (elementary). First, note \n\n\\[\\angle RKA = \\angle RKJ = \\angle RSJ = \\angle TSJ = \\angle TAJ = \\angle TAK\\] \n\nso \\(\\overline{RK} \\parallel \\overline{AT}\\) . Now, \n\n\\(\\overline{RA}\\) is tangent at \\(R\\) iff \\(\\triangle KRS \\sim \\triangle RTA\\) (oppositely), because both equate to \\(- \\angle RKS = \\angle SKR = \\angle SRA = \\angle TRA\\) . \n\nSimilarly, \\(\\overline{TK}\\) is tangent at \\(T\\) iff \\(\\triangle ATS \\sim \\triangle TRK\\) . \n\nThe two similarities are equivalent because \\(RS = ST\\) the SAS gives \\(KR \\cdot TA = RS \\cdot RT = TS \\cdot TR\\) . \n\n\n \n\nRemark. The problem is actually symmetric with respect to two circles; \\(\\overline{RA}\\) is tangent at \\(R\\) if and only if \\(\\overline{TK}\\) at \\(T\\) . \n\nSecond solution (inversion). Consider an inversion at \\(R\\) fixing the circumcircle \\(\\Gamma\\) of \\(TSJA\\) . Then: \n\n- \\(T\\) and \\(S\\) swap,\n\n\n\n- \\(A\\) and \\(B\\) swap, where \\(B\\) is the second intersection of \\(\\ell\\) with \\(\\Gamma\\) . \n\n- Circle \\(\\Omega\\) inverts to the line through \\(T\\) parallel to \\(\\overline{RAB}\\) , call it \\(\\ell\\) . \n\n- \\(J^{*}\\) is the second intersection of \\(\\ell\\) with \\(\\Gamma\\) . \n\n- \\(K^{*}\\) is the intersection of \\(\\ell\\) with the circumcircle of \\(RBJ^{*}\\) ; this implies \\(RK^{*}J^{*}B\\) is an isosceles trapezoid. In particular, one reads \\(\\overline{RK^{*}} \\parallel \\overline{AT}\\) from this, hence \\(RK^{*}TA\\) is a parallelogram. \n\nThus we wish to show the circumcircle of \\(RSK^{*}\\) is tangent to \\(\\Gamma\\) . But that follows from the final parallelogram observed: \\(S\\) is the center of the parallelogram since it is the midpoint of the diagonal. \n\nRemark. This also implies \\(RKTB\\) is cyclic, from \\(\\overline{K^{*}SA}\\) collinear. Moreover, quadrilateral \\(KK^{*}TS\\) is cyclic (by power of a point); this leads to the second official solution to the problem.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2017-notes.jsonl", "problem_match": "4. ", "solution_match": "## \\(\\S 2.1\\) IMO 2017/4, proposed by Charles Leytem (LUX) \n"}}
+{"year": "2017", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "Fix \\(N \\geq 1\\) . A collection of \\(N(N + 1)\\) soccer players of distinct heights stand in a row. Sir Alex Song wishes to remove \\(N(N - 1)\\) players from this row to obtain a new row of \\(2N\\) players in which the following \\(N\\) conditions hold: no one stands between the two tallest players, no one stands between the third and fourth tallest players, ..., no one stands between the two shortest players. Prove that this is possible.", "solution": "Some opening remarks: location and height are symmetric to each other, if one thinks about this problem as permutation pattern avoidance. So while officially there are multiple solutions, they are basically isomorphic to one another, and I am not aware of any solution otherwise. \n\n\n \n\nTake a partition of \\(N\\) groups in order by height: \\(G_{1} = \\{1,\\ldots ,N + 1\\}\\) , \\(G_{2} = \\{N+\\) \\(2,\\ldots ,2N + 2\\}\\) , and so on. We will pick two people from each group \\(G_{k}\\) \n\nScan from the left until we find two people in the same group \\(G_{k}\\) . Delete all people scanned and also everyone in \\(G_{k}\\) . All the groups still have at least \\(N\\) people left, so we can induct down with the non- deleted people; the chosen pair is to the far left anyways. \n\nRemark. The important bit is to scan by position but group by height, and moreover not change the groups as we scan. Dually, one can have a solution which scans by height but groups by position.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2017-notes.jsonl", "problem_match": "5. ", "solution_match": "## \\(\\S 2.2\\) IMO 2017/5, proposed by Grigory Chelnokov (RUS) \n"}}
{"year": "2017", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "An irreducible lattice point is an ordered pair of integers \\((x, y)\\) satisfying \\(\\gcd (x, y) = 1\\) . Prove that if \\(S\\) is a finite set of irreducible lattice points then there exists a nonconstant homogeneous polynomial \\(f(x, y)\\) with integer coefficients such that \\(f(x, y) = 1\\) for each \\((x, y) \\in S\\) .", "solution": "## Problem state \n\nWe present two solutions. \n\n\\(\\P\\) First solution (Dan Carmon, Israel). We prove the result by induction on \\(|S|\\) , with the base case being Bezout's Lemma ( \\(n = 1\\) ). For the inductive step, suppose we want to add a given pair \\((a_{m + 1},b_{m + 1})\\) to \\(\\{(a_{1},\\ldots ,a_{m}),(b_{1},\\ldots ,b_{m})\\}\\) . \n\nClaim (Standard) — By a suitable linear transformation we may assume \n\n\\[(a_{m + 1},b_{m + 1}) = (1,0).\\] \n\nOutline of proof. It would be sufficient to show there exists a \\(2\\times 2\\) matrix \\(T = \\left[ \\begin{array}{ll}u & v\\\\ s & t \\end{array} \\right]\\) with integer entries such that \\(\\operatorname *{det}T = 1\\) and \\(T\\cdot \\left[ \\begin{array}{l}a_{m + 1}\\\\ b_{m + 1} \\end{array} \\right] = \\left[ \\begin{array}{l}1\\\\ 0 \\end{array} \\right]\\) . Then we could apply \\(T\\) to all the ordered pairs in \\(S\\) (viewed as column vectors); if \\(f\\) was a polynomial that works on the transformed ordered pairs, then \\(f(ux + vy,sx + ty)\\) works for the original ordered pairs. (Here, the condition that \\(\\operatorname *{det}T = 1\\) ensures that \\(T^{- 1}\\) has integer entries, and hence that \\(T\\) maps irreducible lattice points to irreducible lattice points.) \n\nTo generate such a matrix \\(T\\) , choose \\(T = \\left[ \\begin{array}{ll}u & v\\\\ - b_{m + 1} & a_{m + 1} \\end{array} \\right]\\) where \\(u\\) and \\(v\\) are chosen via Bezout lemma so that \\(ua_{m + 1} + vb_{m + 1} = 1\\) . This matrix \\(T\\) is rigged so that \\(\\operatorname *{det}T = 1\\) and the leftmost column of \\(T^{- 1}\\) is \\(\\left[ \\begin{array}{l}a_{m + 1}\\\\ b_{m + 1} \\end{array} \\right]\\) . \\(\\square\\) \n\nRemark. This transformation claim is not necessary to proceed; the solution below can be rewritten to avoid it with only cosmetic edits. However, it makes things a bit easier to read. \n\nLet \\(g(x,y)\\) be a polynomial which works on the latter set. We claim we can choose the new polynomial \\(f\\) of the form \n\n\\[f(x,y) = g(x,y)^{M} - Cx^{\\deg g\\cdot M - m}\\prod_{i = 1}^{m}(b_{i}x - a_{i}y).\\] \n\nwhere \\(C\\) and \\(M\\) are integer parameters we may adjust. \n\nSince \\(f(a_{i},b_{i}) = 1\\) by construction we just need \n\n\\[1 = f(1,0) = g(1,0)^{M} - C\\prod b_{i}.\\] \n\nIf \\(\\prod b_{i} = 0\\) we are done, since \\(b_{i} = 0 \\implies a_{i} = \\pm 1\\) in that case and so \\(g(1,0) = \\pm 1\\) , thus take \\(M = 2\\) . So it suffices to prove:\n\n\n\nClaim — We have \\(\\gcd (g(1,0),b_{i}) = 1\\) when \\(b_{i}\\neq 0\\) . \n\nProof. Fix \\(i\\) . If \\(b_{i} = 0\\) then \\(a_{i} = \\pm 1\\) and \\(g(\\pm 1,0) = \\pm 1\\) . Otherwise know \n\n\\[1 = g(a_{i},b_{i})\\equiv g(a_{i},0)\\pmod {b_{i}}\\] \n\nand since the polynomial is homogeneous with \\(\\gcd (a_{i},b_{i}) = 1\\) it follows \\(g(1,0)\\neq 0\\) (mod \\(b_{i}\\) ) as well. \\(\\square\\) \n\nThen take \\(M\\) a large multiple of \\(\\phi (\\prod |b_{i}|)\\) and we're done. \n\nSecond solution (Lagrange). The main claim is that: \n\nClaim — For every positive integer \\(N\\) , there is a homogeneous polynomial \\(P(x,y)\\) such that \\(P(x,y)\\equiv 1\\) (mod \\(N\\) ) whenever \\(\\gcd (x,y) = 1\\) . \n\n(This claim is actually implied by the problem.) \n\nProof. For \\(N = p^{e}\\) a prime take \\((x^{p - 1} + y^{p - 1})^{\\phi (N)}\\) when \\(p\\) is odd, and \\((x^{2} + xy + y^{2})^{\\phi (N)}\\) for \\(p = 2\\) . \n\nNow, if \\(N\\) is a product of primes, we can collate coefficient by coefficient using the Chinese remainder theorem. \\(\\square\\) \n\nLet \\(S = \\{(a_{i},b_{i}) \\mid i = 1,\\ldots ,m\\}\\) . We have the natural homogeneous \"Lagrange polynomials\" \n\n\\[L_{k}(x,y):= \\prod_{i\\neq k}(b_{i}x - a_{i}y)\\] \n\nNow let \n\n\\[N:= \\prod_{k}L_{k}(x_{k},y_{k})\\] \n\nand take \\(P\\) as in the claim. Then we can take a large power of \\(P\\) , and for each \\(i\\) subtract an appropriate multiple of \\(L_{i}(x,y)\\) ; that is, choose \n\n\\[f(x,y) = P(x,y)^{C} - \\sum_{i}L_{i}(x,y)\\cdot Q_{i}(x,y)\\] \n\nwhere \\(C\\) is large enough that \\(C\\deg P > \\max_{i}\\deg L_{i}\\) , and \\(Q_{i}(x,y)\\) is any homogeneous polynomial of degree \\(C\\deg P - \\deg L_{i}\\) such that \\(L_{k}(x_{k},y_{k})Q_{k}(x_{k},y_{k}) = \\frac{P(x_{k},y_{k})^{C - 1}}{N}\\) . \\(L_{k}(x_{k},y_{k})\\) (which is an integer).", "metadata": {"resource_path": "IMO/segmented/en-IMO-2017-notes.jsonl", "problem_match": "6. ", "solution_match": "## \\(\\S 2.3\\) IMO 2017/6, proposed by John Berman (USA) \n"}}
diff --git a/IMO/segmented/en-IMO-2018-notes.jsonl b/IMO/segmented/en-IMO-2018-notes.jsonl
index 6c9c759a14c7592183295793ba47b4006809495e..6f602aea60bc2c90176b2638b10a996a85c7b545 100644
--- a/IMO/segmented/en-IMO-2018-notes.jsonl
+++ b/IMO/segmented/en-IMO-2018-notes.jsonl
@@ -1,6 +1,6 @@
-{"year": "2018", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "Let \\(\\Gamma\\) be the circumcircle of acute triangle \\(A B C\\) . Points \\(D\\) and \\(E\\) lie on segments \\(A B\\) and \\(A C\\) , respectively, such that \\(A D = A E\\) . The perpendicular bisectors of \\(\\overline{{B D}}\\) and \\(\\overline{{C E}}\\) intersect the minor arcs \\(A B\\) and \\(A C\\) of \\(\\Gamma\\) at points \\(F\\) and \\(G\\) , respectively. Prove that the lines \\(D E\\) and \\(F G\\) are parallel.", "solution": "## Problem ste present a synthetic solution from the IMO shortlist as well as a complex numbers approach. We also outline a trig solution (the one I found at IMO), and a fourth solution from Derek Liu. \n\n\\(\\P\\) Synthetic solution (from Shortlist). Construct parallelograms \\(AXFD\\) and \\(AEGY\\) , noting that \\(X\\) and \\(Y\\) lie on \\(\\Gamma\\) . As \\(\\overline{XF} \\parallel \\overline{AB}\\) we can let \\(M\\) denote the midpoint of minor arcs \\(\\overline{XF}\\) and \\(\\overline{AB}\\) (which coincide). Define \\(N\\) similarly. \n\n\n \n\nObserve that \\(XF = AD = AE = YG\\) , so arcs \\(\\widehat{XF}\\) and \\(\\widehat{YG}\\) have equal measure; hence arcs \\(\\widehat{MF}\\) and \\(\\widehat{NG}\\) have equal measure; therefore \\(\\widehat{MN} \\parallel \\widehat{FG}\\) . \n\nSince \\(\\widehat{MN}\\) and \\(\\widehat{DE}\\) are both perpendicular to the \\(\\angle A\\) bisector, we're done. \n\n\\(\\P\\) Complex numbers solution. Let \\(b\\) , \\(c\\) , \\(f\\) , \\(g\\) , \\(a\\) be as usual. Note that \n\n\\[d - a = \\left(2\\cdot \\frac{f + a + b - ab\\overline{f}}{2} -b\\right) - a = f - \\frac{ab}{f}\\] \\[e - a = g - \\frac{ac}{g}\\]\n\n\n\nWe are given \\(AD = AE\\) from which one deduces \n\n\\[\\left(\\frac{e - a}{d - a}\\right)^{2} = \\frac{c}{b}\\Longrightarrow \\frac{(g^{2} - a c)^{2}}{(f^{2} - a b)^{2}} = \\frac{g^{2}c}{f^{2}b}\\] \\[\\Longrightarrow b c(b g^{2} - c f^{2})a^{2} = g^{2}f^{4}c - f^{2}g^{4}b = f^{2}g^{2}(f^{2}c - g^{2}b)\\] \\[\\Longrightarrow b c\\cdot a^{2} = (f g)^{2}\\Longrightarrow \\left(-\\frac{f g}{a}\\right)^{2} = b c.\\] \n\nSince \\(\\frac{- f g}{a}\\) is the point \\(X\\) on the circle with \\(\\overline{{A X}}\\perp \\overline{{F G}}\\) , we conclude \\(\\overline{{F G}}\\) is either parallel or perpendicular to the \\(\\angle A\\) - bisector; it must the latter since the \\(\\angle A\\) - bisector separates the two minor arcs. \n\n\\(\\P\\) Trig solution (outline). Let \\(\\ell\\) denote the \\(\\angle A\\) bisector. Fix \\(D\\) and \\(F\\) . We define the phantom point \\(G^{\\prime}\\) such that \\(\\overline{{F G^{\\prime}}}\\perp \\ell\\) and \\(E^{\\prime}\\) on side \\(\\overline{{A C}}\\) such that \\(G E^{\\prime} = G C\\) \n\nClaim (Converse of the IMO problem) — We have \\(AD = AE^{\\prime}\\) , so that \\(E = E^{\\prime}\\) . \n\nProof. Since \\(\\overline{{F G^{\\prime}}}\\perp \\ell\\) , one can deduce \\(\\angle F B D = \\frac{1}{2} C + x\\) and \\(\\angle G C A = \\frac{1}{2} B + x\\) for some \\(x\\) . (One fast way to see this is to note that \\(\\overline{{F G}}\\parallel \\overline{{M N}}\\) where \\(M\\) and \\(N\\) are in the first solution.) Then \\(\\angle F A B = \\frac{1}{2} C - x\\) and \\(\\angle G A C = \\frac{1}{2} B - x\\) . \n\nLet \\(R\\) be the circumradius. Now, by the law of sines, \n\n\\[B F = 2R\\sin \\left(\\frac{1}{2} C - x\\right).\\] \n\nFrom there we get \n\n\\[B D = 2\\cdot B F\\cos \\left(\\frac{1}{2} C + x\\right) = 4R\\cos \\left(\\frac{1}{2} C + x\\right)\\sin \\left(\\frac{1}{2} C - x\\right)\\] \\[D A = A B - B D = 2R\\sin C - 4R\\cos \\left(\\frac{1}{2} C + x\\right)\\sin \\left(\\frac{1}{2} C - x\\right)\\] \\[\\qquad = 2R\\left[\\sin C - 2\\cos \\left(\\frac{1}{2} C + x\\right)\\sin \\left(\\frac{1}{2} C - x\\right)\\right]\\] \\[\\qquad = 2R\\left[\\sin C - (\\sin C - \\sin 2x)\\right] = 2R\\sin 2x.\\] \n\nA similar calculation gives \\(A E^{\\prime} = 2R\\sin 2x\\) as needed. \n\nThus, \\(\\overline{{F G^{\\prime}}}\\parallel \\overline{{D E}}\\) , so \\(G = G^{\\prime}\\) as well. This concludes the proof. \n\n\\(\\P\\) Synthetic solution from Derek Liu. Let lines \\(F D\\) and \\(G E\\) intersect \\(\\Gamma\\) again at \\(J\\) and \\(K\\) , respectively.\n\n\n\n \n\nNotice that \\(\\triangle BFD \\sim \\triangle JAD\\) ; as \\(FB = FD\\) , it follows that \\(AJ = AD\\) . Likewise, \\(\\triangle CGE \\sim \\triangle KAE\\) and \\(GC = GE\\) , so \\(AK = AE\\) . Hence, \n\n\\[AK = AE = AD = AJ,\\] \n\nso \\(DEJK\\) is cyclic with center \\(A\\) . \n\nIt follows that \n\n\\[\\angle KED = \\angle KJD = \\angle KJF = \\angle KGF,\\] \n\nso we're done. \n\nRemark. Note that \\(K\\) and \\(J\\) must be distinct for this solution to work. Since \\(G\\) and \\(K\\) lie on opposite sides of \\(AC\\) , \\(K\\) is on major arc \\(ABC\\) . As \\(AK = AD = AE \\leq \\min (AB, AC)\\) , \\(K\\) lies on minor arc \\(AB\\) . Similarly, \\(J\\) lies on minor arc \\(AC\\) , so \\(K \\neq J\\) .", "metadata": {"resource_path": "IMO/segmented/en-IMO-2018-notes.jsonl", "problem_match": "1. ", "solution_match": "## \\(\\S 1.1\\) IMO 2018/1, proposed by Silouanos Brazitikos, Vangelis Psyxas, Michael Sarantis (HEL) \n"}}
+{"year": "2018", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "Let \\(\\Gamma\\) be the circumcircle of acute triangle \\(A B C\\) . Points \\(D\\) and \\(E\\) lie on segments \\(A B\\) and \\(A C\\) , respectively, such that \\(A D = A E\\) . The perpendicular bisectors of \\(\\overline{{B D}}\\) and \\(\\overline{{C E}}\\) intersect the minor arcs \\(A B\\) and \\(A C\\) of \\(\\Gamma\\) at points \\(F\\) and \\(G\\) , respectively. Prove that the lines \\(D E\\) and \\(F G\\) are parallel.", "solution": "## Problem ste present a synthetic solution from the IMO shortlist as well as a complex numbers approach. We also outline a trig solution (the one I found at IMO), and a fourth solution from Derek Liu. \n\n\\(\\P\\) Synthetic solution (from Shortlist). Construct parallelograms \\(AXFD\\) and \\(AEGY\\) , noting that \\(X\\) and \\(Y\\) lie on \\(\\Gamma\\) . As \\(\\overline{XF} \\parallel \\overline{AB}\\) we can let \\(M\\) denote the midpoint of minor arcs \\(\\overline{XF}\\) and \\(\\overline{AB}\\) (which coincide). Define \\(N\\) similarly. \n\n\n \n\nObserve that \\(XF = AD = AE = YG\\) , so arcs \\(\\widehat{XF}\\) and \\(\\widehat{YG}\\) have equal measure; hence arcs \\(\\widehat{MF}\\) and \\(\\widehat{NG}\\) have equal measure; therefore \\(\\widehat{MN} \\parallel \\widehat{FG}\\) . \n\nSince \\(\\widehat{MN}\\) and \\(\\widehat{DE}\\) are both perpendicular to the \\(\\angle A\\) bisector, we're done. \n\n\\(\\P\\) Complex numbers solution. Let \\(b\\) , \\(c\\) , \\(f\\) , \\(g\\) , \\(a\\) be as usual. Note that \n\n\\[d - a = \\left(2\\cdot \\frac{f + a + b - ab\\overline{f}}{2} -b\\right) - a = f - \\frac{ab}{f}\\] \\[e - a = g - \\frac{ac}{g}\\]\n\n\n\nWe are given \\(AD = AE\\) from which one deduces \n\n\\[\\left(\\frac{e - a}{d - a}\\right)^{2} = \\frac{c}{b}\\Longrightarrow \\frac{(g^{2} - a c)^{2}}{(f^{2} - a b)^{2}} = \\frac{g^{2}c}{f^{2}b}\\] \\[\\Longrightarrow b c(b g^{2} - c f^{2})a^{2} = g^{2}f^{4}c - f^{2}g^{4}b = f^{2}g^{2}(f^{2}c - g^{2}b)\\] \\[\\Longrightarrow b c\\cdot a^{2} = (f g)^{2}\\Longrightarrow \\left(-\\frac{f g}{a}\\right)^{2} = b c.\\] \n\nSince \\(\\frac{- f g}{a}\\) is the point \\(X\\) on the circle with \\(\\overline{{A X}}\\perp \\overline{{F G}}\\) , we conclude \\(\\overline{{F G}}\\) is either parallel or perpendicular to the \\(\\angle A\\) - bisector; it must the latter since the \\(\\angle A\\) - bisector separates the two minor arcs. \n\n\\(\\P\\) Trig solution (outline). Let \\(\\ell\\) denote the \\(\\angle A\\) bisector. Fix \\(D\\) and \\(F\\) . We define the phantom point \\(G^{\\prime}\\) such that \\(\\overline{{F G^{\\prime}}}\\perp \\ell\\) and \\(E^{\\prime}\\) on side \\(\\overline{{A C}}\\) such that \\(G E^{\\prime} = G C\\) \n\nClaim (Converse of the IMO problem) — We have \\(AD = AE^{\\prime}\\) , so that \\(E = E^{\\prime}\\) . \n\nProof. Since \\(\\overline{{F G^{\\prime}}}\\perp \\ell\\) , one can deduce \\(\\angle F B D = \\frac{1}{2} C + x\\) and \\(\\angle G C A = \\frac{1}{2} B + x\\) for some \\(x\\) . (One fast way to see this is to note that \\(\\overline{{F G}}\\parallel \\overline{{M N}}\\) where \\(M\\) and \\(N\\) are in the first solution.) Then \\(\\angle F A B = \\frac{1}{2} C - x\\) and \\(\\angle G A C = \\frac{1}{2} B - x\\) . \n\nLet \\(R\\) be the circumradius. Now, by the law of sines, \n\n\\[B F = 2R\\sin \\left(\\frac{1}{2} C - x\\right).\\] \n\nFrom there we get \n\n\\[B D = 2\\cdot B F\\cos \\left(\\frac{1}{2} C + x\\right) = 4R\\cos \\left(\\frac{1}{2} C + x\\right)\\sin \\left(\\frac{1}{2} C - x\\right)\\] \\[D A = A B - B D = 2R\\sin C - 4R\\cos \\left(\\frac{1}{2} C + x\\right)\\sin \\left(\\frac{1}{2} C - x\\right)\\] \\[\\qquad = 2R\\left[\\sin C - 2\\cos \\left(\\frac{1}{2} C + x\\right)\\sin \\left(\\frac{1}{2} C - x\\right)\\right]\\] \\[\\qquad = 2R\\left[\\sin C - (\\sin C - \\sin 2x)\\right] = 2R\\sin 2x.\\] \n\nA similar calculation gives \\(A E^{\\prime} = 2R\\sin 2x\\) as needed. \n\nThus, \\(\\overline{{F G^{\\prime}}}\\parallel \\overline{{D E}}\\) , so \\(G = G^{\\prime}\\) as well. This concludes the proof. \n\n\\(\\P\\) Synthetic solution from Derek Liu. Let lines \\(F D\\) and \\(G E\\) intersect \\(\\Gamma\\) again at \\(J\\) and \\(K\\) , respectively.\n\n\n\n \n\nNotice that \\(\\triangle BFD \\sim \\triangle JAD\\) ; as \\(FB = FD\\) , it follows that \\(AJ = AD\\) . Likewise, \\(\\triangle CGE \\sim \\triangle KAE\\) and \\(GC = GE\\) , so \\(AK = AE\\) . Hence, \n\n\\[AK = AE = AD = AJ,\\] \n\nso \\(DEJK\\) is cyclic with center \\(A\\) . \n\nIt follows that \n\n\\[\\angle KED = \\angle KJD = \\angle KJF = \\angle KGF,\\] \n\nso we're done. \n\nRemark. Note that \\(K\\) and \\(J\\) must be distinct for this solution to work. Since \\(G\\) and \\(K\\) lie on opposite sides of \\(AC\\) , \\(K\\) is on major arc \\(ABC\\) . As \\(AK = AD = AE \\leq \\min (AB, AC)\\) , \\(K\\) lies on minor arc \\(AB\\) . Similarly, \\(J\\) lies on minor arc \\(AC\\) , so \\(K \\neq J\\) .", "metadata": {"resource_path": "IMO/segmented/en-IMO-2018-notes.jsonl", "problem_match": "1. ", "solution_match": "## \\(\\S 1.1\\) IMO 2018/1, proposed by Silouanos Brazitikos, Vangelis Psyxas, Michael Sarantis (HEL) \n"}}
{"year": "2018", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "Find all integers \\(n\\geq 3\\) for which there exist real numbers \\(a_{1},a_{2},\\ldots ,a_{n}\\) satisfying \n\n\\[a_{i}a_{i + 1} + 1 = a_{i + 2}\\] \n\nfor \\(i = 1,2,\\ldots ,n\\) , where indices are taken modulo \\(n\\)", "solution": "The answer is \\(3 \\mid n\\) , achieved by \\((- 1, - 1,2, - 1, - 1,2,\\ldots)\\) . We present two solutions. \n\n\\(\\P\\) First solution by inequalities. We compute \\(a_{i}a_{i + 1}a_{i + 2}\\) in two ways: \n\n\\[a_{i}a_{i + 1}a_{i + 2} = [a_{i + 2} - 1]a_{i + 2} = a_{i + 2}^{2} - a_{i + 2}\\] \\[\\qquad = a_{i}[a_{i + 3} - 1] = a_{i}a_{i + 3} - a_{i}.\\] \n\nCyclically summing \\(a_{i + 2}^{2} - a_{i + 2} = a_{i}a_{i + 3} - a_{i}\\) then gives \n\n\\[\\sum_{i}a_{i + 2}^{2} = \\sum_{i}a_{i}a_{i + 3}\\iff \\sum_{\\mathrm{cyc}}(a_{i} - a_{i + 3})^{2} = 0.\\] \n\nThis means for inequality reasons the sequence is 3- periodic. Since the sequence is clearly not 1- periodic, as \\(x^{2} + 1 = x\\) has no real solutions. Thus \\(3 \\mid n\\) . \n\n\\(\\P\\) Second solution by sign counting. Extend \\(a_{n}\\) to be a periodic sequence. The idea is to look at the signs, and show the sequence of the signs must be \\(- - +\\) repeated. This takes several steps: \n\n- The pattern \\(- - -\\) is impossible. Obvious, since the third term should be \\(>1\\) . \n\n- The pattern \\(+ +\\) is impossible. Then the sequence becomes strictly increasing, hence may not be periodic. \n\n- Zeros are impossible. If \\(a_{1} = 0\\) , then \\(a_{2} = 0\\) , \\(a_{3} > 0\\) , \\(a_{4} > 0\\) , which gives the impossible \\(+ +\\) . \n\n- The pattern \\(- - + - +\\) is impossible. Compute the terms: \n\n\\[a_{1} = -x< 0\\] \\[a_{2} = -y< 0\\] \\[a_{3} = 1 + xy > 1\\] \\[a_{4} = 1 - y(1 + xy)< 0\\] \\[a_{5} = 1 + (1 + xy)(1 - y(1 + xy))< 1.\\] \n\nBut now \n\n\\[a_{6} - a_{5} = (1 + a_{5}a_{4}) - (1 + a_{3}a_{4}) = a_{4}(a_{5} - a_{3}) > 0\\] \n\nsince \\(a_{5} > 1 > a_{3}\\) . This means we have the impossible \\(+ +\\) pattern.\n\n\n\n- The infinite alternating pattern \\(- + - + - + - + \\ldots\\) is impossible. Note that \n\n\\[a_{1}a_{2} + 1 = a_{3}< 0< a_{4} = 1 + a_{2}a_{3}\\Rightarrow a_{1}< a_{3}\\] \n\nsince \\(a_{2} > 0\\) ; extending this we get \\(a_{1}< a_{3}< a_{5}< \\ldots\\) which contradicts the periodicity. \n\nWe finally collate the logic of sign patterns. Since the pattern is not alternating, there must be - - somewhere. Afterwards must be \\(^+\\) , and then after that must be two minus signs (since one minus sign is impossible by impossibility of \\(- - + - +\\) and \\(- - -\\) is also forbidden); thus we get the periodic \\(- - +\\) as desired.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2018-notes.jsonl", "problem_match": "2. ", "solution_match": "## \\(\\S 1.2\\) IMO 2018/2, proposed by Patrik Bak (SVK) \n"}}
-{"year": "2018", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "An anti-Pascal triangle is an equilateral triangular array of numbers such that, except for the numbers in the bottom row, each number is the absolute value of the difference of the two numbers immediately below it. For example, the following array is an anti-Pascal triangle with four rows which contains every integer from 1 to 10. \n\n\\[4\\] \\[2 6\\] \\[5 7 1\\] \\[8 3 10 9\\] \n\nDoes there exist an anti- Pascal triangle with 2018 rows which contains every integer from 1 to \\(1 + 2 + \\dots +2018\\) ?", "solution": "The answer is no, there is no anti- Pascal triangle with the required properties. \n\nLet \\(n = 2018\\) and \\(N = 1 + 2 + \\dots +n\\) . For every number \\(d\\) not in the bottom row, draw an arrow from \\(d\\) to the larger of the two numbers below it (i.e. if \\(d = a - b\\) , draw \\(d\\rightarrow a\\) ). This creates an oriented forest (which looks like lightning strikes). \n\nConsider the directed path starting from the top vertex \\(A\\) . Starting from the first number, it increments by at least \\(1 + 2 + \\dots +n\\) , since the increments at each step in the path are distinct; therefore equality must hold and thus the path from the top ends at \\(N = 1 + 2 + \\dots +n\\) with all the numbers \\(\\{1,2,\\ldots ,n\\}\\) being close by. Let \\(B\\) be that position. \n\n\n \n\nConsider the two left/right neighbors \\(X\\) and \\(Y\\) of the endpoint \\(B\\) . Assume that \\(B\\) is to the right of the midpoint of the bottom side, and complete the equilateral triangle\n\n\n\nas shown to an apex \\(C\\) . Consider the lightning strike from \\(C\\) hitting the bottom at \\(D\\) . It travels at least \\(\\lfloor n / 2 - 1 \\rfloor\\) steps, by construction. But the increases must be at least \\(n + 1\\) , \\(n + 2\\) , ...since \\(1, 2, \\ldots , n\\) are close to the \\(A \\to B\\) lightning path. Then the number at \\(D\\) is at least \n\n\\[(n + 1) + (n + 2) + \\dots +(n + (\\lfloor n / 2 - 1\\rfloor)) > 1 + 2 + \\dots +n\\] \n\nfor \\(n \\geq 2018\\) , contradiction.\n\n\n\n## \\(\\S 2\\) Solutions to Day 2", "metadata": {"resource_path": "IMO/segmented/en-IMO-2018-notes.jsonl", "problem_match": "3. ", "solution_match": "## \\(\\S 1.3\\) IMO 2018/3, proposed by Morteza Saghafian (IRN) \n"}}
+{"year": "2018", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "An anti-Pascal triangle is an equilateral triangular array of numbers such that, except for the numbers in the bottom row, each number is the absolute value of the difference of the two numbers immediately below it. For example, the following array is an anti-Pascal triangle with four rows which contains every integer from 1 to 10. \n\n\\[4\\] \\[2 6\\] \\[5 7 1\\] \\[8 3 10 9\\] \n\nDoes there exist an anti- Pascal triangle with 2018 rows which contains every integer from 1 to \\(1 + 2 + \\dots +2018\\) ?", "solution": "The answer is no, there is no anti- Pascal triangle with the required properties. \n\nLet \\(n = 2018\\) and \\(N = 1 + 2 + \\dots +n\\) . For every number \\(d\\) not in the bottom row, draw an arrow from \\(d\\) to the larger of the two numbers below it (i.e. if \\(d = a - b\\) , draw \\(d\\rightarrow a\\) ). This creates an oriented forest (which looks like lightning strikes). \n\nConsider the directed path starting from the top vertex \\(A\\) . Starting from the first number, it increments by at least \\(1 + 2 + \\dots +n\\) , since the increments at each step in the path are distinct; therefore equality must hold and thus the path from the top ends at \\(N = 1 + 2 + \\dots +n\\) with all the numbers \\(\\{1,2,\\ldots ,n\\}\\) being close by. Let \\(B\\) be that position. \n\n\n \n\nConsider the two left/right neighbors \\(X\\) and \\(Y\\) of the endpoint \\(B\\) . Assume that \\(B\\) is to the right of the midpoint of the bottom side, and complete the equilateral triangle\n\n\n\nas shown to an apex \\(C\\) . Consider the lightning strike from \\(C\\) hitting the bottom at \\(D\\) . It travels at least \\(\\lfloor n / 2 - 1 \\rfloor\\) steps, by construction. But the increases must be at least \\(n + 1\\) , \\(n + 2\\) , ...since \\(1, 2, \\ldots , n\\) are close to the \\(A \\to B\\) lightning path. Then the number at \\(D\\) is at least \n\n\\[(n + 1) + (n + 2) + \\dots +(n + (\\lfloor n / 2 - 1\\rfloor)) > 1 + 2 + \\dots +n\\] \n\nfor \\(n \\geq 2018\\) , contradiction.\n\n\n\n## \\(\\S 2\\) Solutions to Day 2", "metadata": {"resource_path": "IMO/segmented/en-IMO-2018-notes.jsonl", "problem_match": "3. ", "solution_match": "## \\(\\S 1.3\\) IMO 2018/3, proposed by Morteza Saghafian (IRN) \n"}}
{"year": "2018", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "A site is any point \\((x,y)\\) in the plane for which \\(x,y\\in \\{1,\\ldots ,20\\}\\) . Initially, each of the 400 sites is unoccupied. Any and Ben take turns placing stones on unoccupied sites, with Amy going first; Amy has the additional restriction that no two of her stones may be at a distance equal to \\(\\sqrt{5}\\) . They stop once either player cannot move. Find the greatest \\(K\\) such that Amy can ensure that she places at least \\(K\\) stones.", "solution": "The answer is \\(K = 100\\) \n\nFirst, we show Amy can always place at least 100 stones. Indeed, treat the problem as a grid with checkerboard coloring. Then Amy can choose to always play on one of the 200 black squares. In this way, she can guarantee half the black squares, i.e. she can get \\(\\frac{1}{2}\\cdot 200 = 100\\) stones. \n\nSecond, we show Ben can prevent Amy from placing more than 100 stones. Divide into several \\(4\\times 4\\) squares and then further partition each \\(4\\times 4\\) squares as shown in the grid below. \n\n \n\nThe squares with each label form 4- cycles by knight jumps. For each such cycle, whenever Amy plays in the cycle, Ben plays in the opposite point of the cycle, preventing Amy from playing any more stones in that original cycle. Hence Amy can play at most in \\(1 / 4\\) of the stones, as desired.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2018-notes.jsonl", "problem_match": "4. ", "solution_match": "## \\(\\S 2.1\\) IMO 2018/4, proposed by Gurgen Asatryan (ARM) \n"}}
-{"year": "2018", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "Let \\(a_{1}\\) , \\(a_{2}\\) , ... be an infinite sequence of positive integers, and \\(N\\) a positive integer. Suppose that for all integers \\(n\\geq N\\) , the expression \n\n\\[\\frac{a_{1}}{a_{2}} +\\frac{a_{2}}{a_{3}} +\\dots +\\frac{a_{n - 1}}{a_{n}} +\\frac{a_{n}}{a_{1}}\\] \n\nis an integer. Prove that \\((a_{n})\\) is eventually constant.", "solution": "L \n\nThe condition implies that the difference \n\n\\[S(n) = \\frac{a_{n + 1} - a_{n}}{a_{1}} +\\frac{a_{n}}{a_{n + 1}}\\] \n\nis an integer for all \\(n > N\\) . We proceed by \\(p\\) - adic valuation only henceforth; fix a prime \\(p\\) . Then analyzing the \\(\\nu_{p}\\) , we immediately get that for \\(n > N\\) : \n\n- If \\(\\nu_{p}(a_{n}) < \\nu_{p}(a_{n + 1})\\) , then \\(\\nu_{p}(a_{n + 1}) = \\nu_{p}(a_{1})\\) . \n\n- If \\(\\nu_{p}(a_{n}) = \\nu_{p}(a_{n + 1})\\) , no conclusion. \n\n- If \\(\\nu_{p}(a_{n}) > \\nu_{p}(a_{n + 1})\\) , then \\(\\nu_{p}(a_{n + 1}) \\geq \\nu_{p}(a_{1})\\) . \n\nIn other words: \n\nClaim — Let \\(p\\) be a prime. Consider the sequence \\(\\nu_{p}(a_{N + 1})\\) , \\(\\nu_{p}(a_{N + 2})\\) , .... Then either: \n\n- We have \\(\\nu_{p}(a_{N + 1}) \\geq \\nu_{p}(a_{N + 2}) \\geq \\ldots\\) and so on, i.e. the sequence is weakly decreasing immediately; or \n\n- For some index \\(K > N\\) we have \\(\\nu_{p}(a_{K}) < \\nu_{p}(a_{K + 1}) = \\nu_{p}(a_{K + 2}) = \\dots = \\nu_{p}(a_{1})\\) , i.e. the sequence \"jumps\" to \\(\\nu_{p}(a_{1})\\) at some point and then stays there forever after. Note this requires \\(\\nu_{p}(a_{1}) > 0\\) . \n\nA cartoon of the situation is drawn below. \n\n\n\n\n\n\nAs only finitely many primes \\(p\\) divide \\(a_{1}\\) , after some time \\(\\nu_{p}(a_{n})\\) is fixed for all such \\(p \\mid a_{1}\\) . Afterwards, the sequence satisfies \\(a_{n + 1} \\mid a_{n}\\) for each \\(n\\) , and thus must be eventually constant. \n\nRemark. This solution is almost completely \\(p\\) - adic, in the sense that I think a similar result holds if one replaces \\(a_{n} \\in \\mathbb{Z}\\) by \\(a_{n} \\in \\mathbb{Z}_{p}\\) for any particular prime \\(p\\) . In other words, the primes almost do not talk to each other. \n\nThere is one caveat: if \\(x_{n}\\) is an integer sequence such that \\(\\nu_{p}(x_{n})\\) is eventually constant for each prime then \\(x_{n}\\) may not be constant. For example, take \\(x_{n}\\) to be the \\(n\\) th prime! That's why in the first claim (applied to co- finitely many of the primes), we need the stronger non- decreasing condition, rather than just eventually constant. \n\nRemark. An alternative approach is to show that, when the fractions \\(a_{n} / a_{1}\\) is written in simplest form for \\(n = N + 1, N + 2, \\ldots\\) , the numerator and denominator are both weakly decreasing. Hence it must eventually be constant; in which case it equals \\(\\frac{1}{1}\\) .", "metadata": {"resource_path": "IMO/segmented/en-IMO-2018-notes.jsonl", "problem_match": "5. ", "solution_match": "## \\(\\S 2.2\\) IMO 2018/5, proposed by Bayarmagnai Gombodorj (MNG) \n"}}
-{"year": "2018", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "A convex quadrilateral \\(A B C D\\) satisfies \\(A B\\cdot C D = B C\\cdot D A\\) . Point \\(X\\) lies inside \\(A B C D\\) so that \n\n\\[\\angle X A B = \\angle X C D\\quad \\mathrm{and}\\quad \\angle X B C = \\angle X D A.\\] \n\nProve that \\(\\angle B X A + \\angle D X C = 180^{\\circ}\\)", "solution": "solutions by inversion. The first is the official one. The second is a solution via inversion, completed by USA5 Michael Ren. \n\nOfficial solution by inversion. In what follows a convex quadrilateral is called quasi- harmonic if \\(AB\\cdot CD = BC\\cdot DA\\) . \n\nClaim — A quasi- harmonic quadrilateral is determined up to similarity by its angles. \n\nProof. Do some inequalities. \n\nRemark. This could be expected by degrees of freedom; a quadrilateral has four degrees of freedom up to similarity; the pseudo- harmonic condition is one while the measures of angles \\(\\angle A\\) , \\(\\angle B\\) , \\(\\angle C\\) , \\(\\angle D\\) (summing to \\(360^{\\circ}\\) ) provide three degrees of freedom. (Note that the point \\(X\\) plays no role in this comment.) \n\nPerforming an inversion at \\(X\\) , one obtains a second quasi- harmonic quadrilateral \\(A^{*}B^{*}C^{*}D^{*}\\) which has the same angles as the original one, \\(\\angle D^{*} = \\angle A\\) , \\(\\angle A^{*} = \\angle B\\) , and so on. Thus by the claim we obtain similarity \n\n\\[D^{*}A^{*}B^{*}C^{*}\\sim ABCD.\\] \n\nIf one then maps \\(D^{*}A^{*}B^{*}C^{*}\\) , onto \\(ABCD\\) , the image of \\(X^{*}\\) becomes a point isogonally conjugate to \\(X\\) . In other words, \\(X\\) has an isogonal conjugate in \\(ABCD\\) . \n\nIt is well- known that this is equivalent to \\(\\angle BXA + \\angle DXC = 180^{\\circ}\\) , for example by inscribing an ellipse with foci \\(X\\) and \\(X^{*}\\) . \n\nSecond solution: \"rhombus inversion\", by Michael Ren. Since \n\n\\[\\frac{AB}{AD} = \\frac{CB}{CD}\\] \n\nand \n\n\\[\\frac{BA}{BC} = \\frac{DA}{DC}\\] \n\nit follows that \\(B\\) and \\(D\\) lie on an Apollonian circle \\(\\omega_{AC}\\) through \\(A\\) and \\(C\\) , while \\(A\\) and \\(C\\) lie on an Apollonian circle \\(\\omega_{BD}\\) through \\(B\\) and \\(D\\) . We let these two circles intersect at a point \\(P\\) inside \\(ABCD\\) . \n\nThe main idea is then to perform an inversion about \\(P\\) with radius 1. We obtain:\n\n\n\n## Lemma \n\nThe image of \\(ABCD\\) is a rhombus. \n\nProof. By the inversion distance formula, we have \n\n\\[\\frac{1}{A^{\\prime}B^{\\prime}} = \\frac{PA}{AB}\\cdot PB = \\frac{PC}{BC}\\cdot PB = \\frac{1}{B^{\\prime}C^{\\prime}}\\] \n\nand so \\(A^{\\prime}B^{\\prime} = B^{\\prime}C^{\\prime}\\) . In a similar way, we derive \\(B^{\\prime}C^{\\prime} = C^{\\prime}D^{\\prime} = D^{\\prime}A^{\\prime}\\) , so the image is a rhombus as claimed. \\(\\square\\) \n\nLet us now translate the angle conditions. We were given that \\(\\angle XAB = \\angle XCD\\) , but \n\n\\[\\angle XAB = \\angle XAP + \\angle PAB = \\angle PX'A' + \\angle A'B'P\\] \n\n\\[\\angle XCD = \\angle XCP + \\angle PCD = \\angle PX'C' + \\angle C'D'P\\] \n\nso subtracting these gives \n\n\\[\\begin{array}{r l} & {\\angle A^{\\prime}X^{\\prime}C^{\\prime} = \\angle A^{\\prime}B^{\\prime}P + \\angle P D^{\\prime}C^{\\prime} = \\angle (A^{\\prime}B^{\\prime},B^{\\prime}P) + \\angle (P D^{\\prime},C^{\\prime}D^{\\prime})}\\\\ & {\\qquad = \\angle (A^{\\prime}B^{\\prime},B^{\\prime}P) + \\angle (P D^{\\prime},A^{\\prime}B^{\\prime}) = \\angle D^{\\prime}P B^{\\prime}.} \\end{array} \\quad (1)\\] \n\nsince \\(\\overline{{A^{\\prime}B^{\\prime}}}\\parallel \\overline{{C^{\\prime}D^{\\prime}}}\\) . Similarly, we obtain \n\n\\[\\angle B^{\\prime}X^{\\prime}D^{\\prime} = \\angle A^{\\prime}PC^{\\prime}. \\quad (2)\\] \n\nWe now translate the desired condition. Since \n\n\\[\\angle A X B = \\angle A X P + \\angle P X B = \\angle P A^{\\prime}X^{\\prime} + \\angle X^{\\prime}B^{\\prime}P\\] \\[\\angle C X D = \\angle C X P + \\angle P X D = \\angle P C^{\\prime}X^{\\prime} + \\angle X^{\\prime}D P^{\\prime}\\] \n\nwe compute \n\n\\[\\angle A X B + \\angle C X D = (\\angle P A^{\\prime}X^{\\prime} + \\angle X^{\\prime}B^{\\prime}P) + (\\angle P C^{\\prime}X^{\\prime} + \\angle X^{\\prime}D^{\\prime}P)\\] \\[\\qquad = - \\left[(\\angle A^{\\prime}X^{\\prime}P + \\angle X^{\\prime}P A^{\\prime}) + (\\angle P X^{\\prime}B^{\\prime} + \\angle B^{\\prime}P X^{\\prime})\\right]\\] \\[\\qquad - \\left[(\\angle C^{\\prime}X^{\\prime}P + \\angle X^{\\prime}P C^{\\prime}) + (\\angle P X^{\\prime}D^{\\prime} + \\angle D^{\\prime}P X^{\\prime})\\right]\\] \\[\\qquad = \\left[\\angle P X^{\\prime}A^{\\prime} + \\angle B X^{\\prime}P + \\angle P X^{\\prime}C^{\\prime} + \\angle D^{\\prime}X^{\\prime}P\\right]\\] \\[\\qquad +\\left[\\angle A^{\\prime}P X^{\\prime} + \\angle X^{\\prime}P B^{\\prime} + \\angle C^{\\prime}P X^{\\prime} + \\angle X^{\\prime}P D^{\\prime}\\right]\\] \\[\\qquad = \\angle A^{\\prime}P B^{\\prime} + \\angle C^{\\prime}P D^{\\prime} + \\angle B^{\\prime}X^{\\prime}C + \\angle D^{\\prime}X^{\\prime}A\\] \n\nand we wish to show this is equal to zero, i.e. the desired becomes \n\n\\[\\angle A^{\\prime}P B^{\\prime} + \\angle C^{\\prime}P D^{\\prime} + \\angle B^{\\prime}X^{\\prime}C + \\angle D^{\\prime}X^{\\prime}A = 0. \\quad (3)\\] \n\nIn other words, the problem is to show (1) and (2) implies (3). \n\nHenceforth drop apostrophes. Here is the inverted diagram (with apostrophes dropped).\n\n\n\n \n\nLet \\(Q\\) denote the reflection of \\(P\\) and let \\(Y\\) denote the second intersection of \\((BQC)\\) and \\((AQD)\\) . Then \n\n\\[-\\angle AXC = -\\angle DPB = \\angle BQD = \\angle BQY + \\angle YQD = \\angle BCY + \\angle YAD\\] \\[= \\angle (BC,CY) + \\angle (YA,AD) = \\angle YCA = -\\angle AYC.\\] \n\nHence \\(XACY\\) is concyclic; similarly \\(XBDY\\) is concyclic. \n\nClaim — \\(X \\neq Y\\) . \n\nProof. To see this: Work pre- inversion assuming \\(AB < AC\\) . Then \\(Q\\) was the center of \\(\\omega_{BD}\\) . If \\(T\\) was the second intersection of \\(BA\\) with \\((QBC)\\) , then \\(QB = QD = QT = \\sqrt{QA \\cdot QC}\\) , by shooting lemma. Since \\(\\angle BAD < 180^\\circ\\) , it follows \\((QBCY)\\) encloses \\(ABCD\\) (pre- inversion). (This part is where the hypothesis that \\(ABCD\\) is convex with \\(X\\) inside is used.) \\(\\square\\) \n\nFinally, we do an angle chase to finish: \n\n\\[\\begin{array}{r l} & {\\angle D X A = \\angle D X Y + \\angle Y X A = \\angle D B Y + \\angle Y C A}\\\\ & {\\qquad = \\angle (D B,Y B) + \\angle (C Y,C A) = \\angle C Y B + 90^{\\circ}}\\\\ & {\\qquad = \\angle C Q B + 90^{\\circ} = -\\angle A P B + 90^{\\circ}.} \\end{array} \\quad (4)\\] \n\nSimilarly, \n\n\\[\\angle BXC = \\angle DPC + 90^\\circ . \\quad (5)\\] \n\nSumming (4) and (5) gives (3). \n\nRemark. A difficult part of the problem in many solutions is that the conclusion is false in the directed sense, if the point \\(X\\) is allowed to lie outside the quadrilateral. We are saved in the first solution because the equivalence of the isogonal conjugation requires \\(X\\) inside the quadrilateral. On the other hand, in the second solution, the issue appears in the presence\n\n\n\nof the second point \\(Y\\) .", "metadata": {"resource_path": "IMO/segmented/en-IMO-2018-notes.jsonl", "problem_match": "6. ", "solution_match": "## \\(\\S 2.3\\) IMO 2018/6, proposed by Tomasz Ciesla (POL) \n"}}
+{"year": "2018", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "Let \\(a_{1}\\) , \\(a_{2}\\) , ... be an infinite sequence of positive integers, and \\(N\\) a positive integer. Suppose that for all integers \\(n\\geq N\\) , the expression \n\n\\[\\frac{a_{1}}{a_{2}} +\\frac{a_{2}}{a_{3}} +\\dots +\\frac{a_{n - 1}}{a_{n}} +\\frac{a_{n}}{a_{1}}\\] \n\nis an integer. Prove that \\((a_{n})\\) is eventually constant.", "solution": "L \n\nThe condition implies that the difference \n\n\\[S(n) = \\frac{a_{n + 1} - a_{n}}{a_{1}} +\\frac{a_{n}}{a_{n + 1}}\\] \n\nis an integer for all \\(n > N\\) . We proceed by \\(p\\) - adic valuation only henceforth; fix a prime \\(p\\) . Then analyzing the \\(\\nu_{p}\\) , we immediately get that for \\(n > N\\) : \n\n- If \\(\\nu_{p}(a_{n}) < \\nu_{p}(a_{n + 1})\\) , then \\(\\nu_{p}(a_{n + 1}) = \\nu_{p}(a_{1})\\) . \n\n- If \\(\\nu_{p}(a_{n}) = \\nu_{p}(a_{n + 1})\\) , no conclusion. \n\n- If \\(\\nu_{p}(a_{n}) > \\nu_{p}(a_{n + 1})\\) , then \\(\\nu_{p}(a_{n + 1}) \\geq \\nu_{p}(a_{1})\\) . \n\nIn other words: \n\nClaim — Let \\(p\\) be a prime. Consider the sequence \\(\\nu_{p}(a_{N + 1})\\) , \\(\\nu_{p}(a_{N + 2})\\) , .... Then either: \n\n- We have \\(\\nu_{p}(a_{N + 1}) \\geq \\nu_{p}(a_{N + 2}) \\geq \\ldots\\) and so on, i.e. the sequence is weakly decreasing immediately; or \n\n- For some index \\(K > N\\) we have \\(\\nu_{p}(a_{K}) < \\nu_{p}(a_{K + 1}) = \\nu_{p}(a_{K + 2}) = \\dots = \\nu_{p}(a_{1})\\) , i.e. the sequence \"jumps\" to \\(\\nu_{p}(a_{1})\\) at some point and then stays there forever after. Note this requires \\(\\nu_{p}(a_{1}) > 0\\) . \n\nA cartoon of the situation is drawn below. \n\n\n\n\n\n\nAs only finitely many primes \\(p\\) divide \\(a_{1}\\) , after some time \\(\\nu_{p}(a_{n})\\) is fixed for all such \\(p \\mid a_{1}\\) . Afterwards, the sequence satisfies \\(a_{n + 1} \\mid a_{n}\\) for each \\(n\\) , and thus must be eventually constant. \n\nRemark. This solution is almost completely \\(p\\) - adic, in the sense that I think a similar result holds if one replaces \\(a_{n} \\in \\mathbb{Z}\\) by \\(a_{n} \\in \\mathbb{Z}_{p}\\) for any particular prime \\(p\\) . In other words, the primes almost do not talk to each other. \n\nThere is one caveat: if \\(x_{n}\\) is an integer sequence such that \\(\\nu_{p}(x_{n})\\) is eventually constant for each prime then \\(x_{n}\\) may not be constant. For example, take \\(x_{n}\\) to be the \\(n\\) th prime! That's why in the first claim (applied to co- finitely many of the primes), we need the stronger non- decreasing condition, rather than just eventually constant. \n\nRemark. An alternative approach is to show that, when the fractions \\(a_{n} / a_{1}\\) is written in simplest form for \\(n = N + 1, N + 2, \\ldots\\) , the numerator and denominator are both weakly decreasing. Hence it must eventually be constant; in which case it equals \\(\\frac{1}{1}\\) .", "metadata": {"resource_path": "IMO/segmented/en-IMO-2018-notes.jsonl", "problem_match": "5. ", "solution_match": "## \\(\\S 2.2\\) IMO 2018/5, proposed by Bayarmagnai Gombodorj (MNG) \n"}}
+{"year": "2018", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "A convex quadrilateral \\(A B C D\\) satisfies \\(A B\\cdot C D = B C\\cdot D A\\) . Point \\(X\\) lies inside \\(A B C D\\) so that \n\n\\[\\angle X A B = \\angle X C D\\quad \\mathrm{and}\\quad \\angle X B C = \\angle X D A.\\] \n\nProve that \\(\\angle B X A + \\angle D X C = 180^{\\circ}\\)", "solution": "solutions by inversion. The first is the official one. The second is a solution via inversion, completed by USA5 Michael Ren. \n\nOfficial solution by inversion. In what follows a convex quadrilateral is called quasi- harmonic if \\(AB\\cdot CD = BC\\cdot DA\\) . \n\nClaim — A quasi- harmonic quadrilateral is determined up to similarity by its angles. \n\nProof. Do some inequalities. \n\nRemark. This could be expected by degrees of freedom; a quadrilateral has four degrees of freedom up to similarity; the pseudo- harmonic condition is one while the measures of angles \\(\\angle A\\) , \\(\\angle B\\) , \\(\\angle C\\) , \\(\\angle D\\) (summing to \\(360^{\\circ}\\) ) provide three degrees of freedom. (Note that the point \\(X\\) plays no role in this comment.) \n\nPerforming an inversion at \\(X\\) , one obtains a second quasi- harmonic quadrilateral \\(A^{*}B^{*}C^{*}D^{*}\\) which has the same angles as the original one, \\(\\angle D^{*} = \\angle A\\) , \\(\\angle A^{*} = \\angle B\\) , and so on. Thus by the claim we obtain similarity \n\n\\[D^{*}A^{*}B^{*}C^{*}\\sim ABCD.\\] \n\nIf one then maps \\(D^{*}A^{*}B^{*}C^{*}\\) , onto \\(ABCD\\) , the image of \\(X^{*}\\) becomes a point isogonally conjugate to \\(X\\) . In other words, \\(X\\) has an isogonal conjugate in \\(ABCD\\) . \n\nIt is well- known that this is equivalent to \\(\\angle BXA + \\angle DXC = 180^{\\circ}\\) , for example by inscribing an ellipse with foci \\(X\\) and \\(X^{*}\\) . \n\nSecond solution: \"rhombus inversion\", by Michael Ren. Since \n\n\\[\\frac{AB}{AD} = \\frac{CB}{CD}\\] \n\nand \n\n\\[\\frac{BA}{BC} = \\frac{DA}{DC}\\] \n\nit follows that \\(B\\) and \\(D\\) lie on an Apollonian circle \\(\\omega_{AC}\\) through \\(A\\) and \\(C\\) , while \\(A\\) and \\(C\\) lie on an Apollonian circle \\(\\omega_{BD}\\) through \\(B\\) and \\(D\\) . We let these two circles intersect at a point \\(P\\) inside \\(ABCD\\) . \n\nThe main idea is then to perform an inversion about \\(P\\) with radius 1. We obtain:\n\n\n\n## Lemma \n\nThe image of \\(ABCD\\) is a rhombus. \n\nProof. By the inversion distance formula, we have \n\n\\[\\frac{1}{A^{\\prime}B^{\\prime}} = \\frac{PA}{AB}\\cdot PB = \\frac{PC}{BC}\\cdot PB = \\frac{1}{B^{\\prime}C^{\\prime}}\\] \n\nand so \\(A^{\\prime}B^{\\prime} = B^{\\prime}C^{\\prime}\\) . In a similar way, we derive \\(B^{\\prime}C^{\\prime} = C^{\\prime}D^{\\prime} = D^{\\prime}A^{\\prime}\\) , so the image is a rhombus as claimed. \\(\\square\\) \n\nLet us now translate the angle conditions. We were given that \\(\\angle XAB = \\angle XCD\\) , but \n\n\\[\\angle XAB = \\angle XAP + \\angle PAB = \\angle PX'A' + \\angle A'B'P\\] \n\n\\[\\angle XCD = \\angle XCP + \\angle PCD = \\angle PX'C' + \\angle C'D'P\\] \n\nso subtracting these gives \n\n\\[\\begin{array}{r l} & {\\angle A^{\\prime}X^{\\prime}C^{\\prime} = \\angle A^{\\prime}B^{\\prime}P + \\angle P D^{\\prime}C^{\\prime} = \\angle (A^{\\prime}B^{\\prime},B^{\\prime}P) + \\angle (P D^{\\prime},C^{\\prime}D^{\\prime})}\\\\ & {\\qquad = \\angle (A^{\\prime}B^{\\prime},B^{\\prime}P) + \\angle (P D^{\\prime},A^{\\prime}B^{\\prime}) = \\angle D^{\\prime}P B^{\\prime}.} \\end{array} \\quad (1)\\] \n\nsince \\(\\overline{{A^{\\prime}B^{\\prime}}}\\parallel \\overline{{C^{\\prime}D^{\\prime}}}\\) . Similarly, we obtain \n\n\\[\\angle B^{\\prime}X^{\\prime}D^{\\prime} = \\angle A^{\\prime}PC^{\\prime}. \\quad (2)\\] \n\nWe now translate the desired condition. Since \n\n\\[\\angle A X B = \\angle A X P + \\angle P X B = \\angle P A^{\\prime}X^{\\prime} + \\angle X^{\\prime}B^{\\prime}P\\] \\[\\angle C X D = \\angle C X P + \\angle P X D = \\angle P C^{\\prime}X^{\\prime} + \\angle X^{\\prime}D P^{\\prime}\\] \n\nwe compute \n\n\\[\\angle A X B + \\angle C X D = (\\angle P A^{\\prime}X^{\\prime} + \\angle X^{\\prime}B^{\\prime}P) + (\\angle P C^{\\prime}X^{\\prime} + \\angle X^{\\prime}D^{\\prime}P)\\] \\[\\qquad = - \\left[(\\angle A^{\\prime}X^{\\prime}P + \\angle X^{\\prime}P A^{\\prime}) + (\\angle P X^{\\prime}B^{\\prime} + \\angle B^{\\prime}P X^{\\prime})\\right]\\] \\[\\qquad - \\left[(\\angle C^{\\prime}X^{\\prime}P + \\angle X^{\\prime}P C^{\\prime}) + (\\angle P X^{\\prime}D^{\\prime} + \\angle D^{\\prime}P X^{\\prime})\\right]\\] \\[\\qquad = \\left[\\angle P X^{\\prime}A^{\\prime} + \\angle B X^{\\prime}P + \\angle P X^{\\prime}C^{\\prime} + \\angle D^{\\prime}X^{\\prime}P\\right]\\] \\[\\qquad +\\left[\\angle A^{\\prime}P X^{\\prime} + \\angle X^{\\prime}P B^{\\prime} + \\angle C^{\\prime}P X^{\\prime} + \\angle X^{\\prime}P D^{\\prime}\\right]\\] \\[\\qquad = \\angle A^{\\prime}P B^{\\prime} + \\angle C^{\\prime}P D^{\\prime} + \\angle B^{\\prime}X^{\\prime}C + \\angle D^{\\prime}X^{\\prime}A\\] \n\nand we wish to show this is equal to zero, i.e. the desired becomes \n\n\\[\\angle A^{\\prime}P B^{\\prime} + \\angle C^{\\prime}P D^{\\prime} + \\angle B^{\\prime}X^{\\prime}C + \\angle D^{\\prime}X^{\\prime}A = 0. \\quad (3)\\] \n\nIn other words, the problem is to show (1) and (2) implies (3). \n\nHenceforth drop apostrophes. Here is the inverted diagram (with apostrophes dropped).\n\n\n\n \n\nLet \\(Q\\) denote the reflection of \\(P\\) and let \\(Y\\) denote the second intersection of \\((BQC)\\) and \\((AQD)\\) . Then \n\n\\[-\\angle AXC = -\\angle DPB = \\angle BQD = \\angle BQY + \\angle YQD = \\angle BCY + \\angle YAD\\] \\[= \\angle (BC,CY) + \\angle (YA,AD) = \\angle YCA = -\\angle AYC.\\] \n\nHence \\(XACY\\) is concyclic; similarly \\(XBDY\\) is concyclic. \n\nClaim — \\(X \\neq Y\\) . \n\nProof. To see this: Work pre- inversion assuming \\(AB < AC\\) . Then \\(Q\\) was the center of \\(\\omega_{BD}\\) . If \\(T\\) was the second intersection of \\(BA\\) with \\((QBC)\\) , then \\(QB = QD = QT = \\sqrt{QA \\cdot QC}\\) , by shooting lemma. Since \\(\\angle BAD < 180^\\circ\\) , it follows \\((QBCY)\\) encloses \\(ABCD\\) (pre- inversion). (This part is where the hypothesis that \\(ABCD\\) is convex with \\(X\\) inside is used.) \\(\\square\\) \n\nFinally, we do an angle chase to finish: \n\n\\[\\begin{array}{r l} & {\\angle D X A = \\angle D X Y + \\angle Y X A = \\angle D B Y + \\angle Y C A}\\\\ & {\\qquad = \\angle (D B,Y B) + \\angle (C Y,C A) = \\angle C Y B + 90^{\\circ}}\\\\ & {\\qquad = \\angle C Q B + 90^{\\circ} = -\\angle A P B + 90^{\\circ}.} \\end{array} \\quad (4)\\] \n\nSimilarly, \n\n\\[\\angle BXC = \\angle DPC + 90^\\circ . \\quad (5)\\] \n\nSumming (4) and (5) gives (3). \n\nRemark. A difficult part of the problem in many solutions is that the conclusion is false in the directed sense, if the point \\(X\\) is allowed to lie outside the quadrilateral. We are saved in the first solution because the equivalence of the isogonal conjugation requires \\(X\\) inside the quadrilateral. On the other hand, in the second solution, the issue appears in the presence\n\n\n\nof the second point \\(Y\\) .", "metadata": {"resource_path": "IMO/segmented/en-IMO-2018-notes.jsonl", "problem_match": "6. ", "solution_match": "## \\(\\S 2.3\\) IMO 2018/6, proposed by Tomasz Ciesla (POL) \n"}}
diff --git a/IMO/segmented/en-IMO-2019-notes.jsonl b/IMO/segmented/en-IMO-2019-notes.jsonl
index e52b139c90c4facc7eb87d1724e29747f76d8f9c..737115520ac85b98629b64bbfd388b90a62b2bbb 100644
--- a/IMO/segmented/en-IMO-2019-notes.jsonl
+++ b/IMO/segmented/en-IMO-2019-notes.jsonl
@@ -1,6 +1,6 @@
{"year": "2019", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "Solve over \\(\\mathbb{Z}\\) the functional equation \\(f(2a) + 2f(b) = f(f(a + b))\\) .", "solution": "Notice that \\(f(x)\\equiv 0\\) or \\(f(x)\\equiv 2x + k\\) work and are clearly the only linear solutions. We now prove all solutions are linear. \n\nLet \\(P(a,b)\\) be the assertion. \n\nClaim — For each \\(x\\in \\mathbb{Z}\\) we have \\(f(2x) = 2f(x) - f(0)\\) \n\nProof. Compare \\(P(0,x)\\) and \\(P(x,0)\\) \n\nNow, \\(P(a,b)\\) and \\(P(0,a + b)\\) give \n\n\\[f(f(a + b)) = f(2a) + 2f(b) = f(0) + 2f(a + b)\\] \\[\\Rightarrow [2f(a) - f(0)] + 2f(b) = f(0) + 2f(a + b)\\] \\[(f(a) - f(0)) + (f(b) - f(0)) = (f(a + b) - f(0)).\\] \n\nThus the map \\(x\\mapsto f(x) - f(0)\\) is additive, therefore linear. \n\nRemark. The same proof works on the functional equation \n\n\\[f(2a) + 2f(b) = g(a + b)\\] \n\nwhere \\(g\\) is an arbitrary function (it implies that \\(f\\) is linear).", "metadata": {"resource_path": "IMO/segmented/en-IMO-2019-notes.jsonl", "problem_match": "1. ", "solution_match": "## \\(\\S 1.1\\) IMO 2019/1, proposed by Liam Baker (SAF) \n"}}
-{"year": "2019", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "In triangle \\(A B C\\) point \\(A_{1}\\) lies on side \\(B C\\) and point \\(B_{1}\\) lies on side \\(A C\\) . Let \\(P\\) and \\(Q\\) be points on segments \\(A A_{1}\\) and \\(B B_{1}\\) , respectively, such that \\(\\overline{{P Q}}\\parallel \\overline{{A B}}\\) . Point \\(P_{1}\\) is chosen on ray \\(P B_{1}\\) beyond \\(B_{1}\\) such that \\(\\angle P P_{1}C = \\angle B A C\\) . Point \\(Q_{1}\\) is chosen on ray \\(Q A_{1}\\) beyond \\(A_{1}\\) such that \\(\\angle C Q_{1}Q = \\angle C B A\\) . Prove that points \\(P_{1}\\) , \\(Q_{1}\\) , \\(P\\) , \\(Q\\) are cyclic.", "solution": "We present two solutions. \n\nFirst solution by bary (Evan Chen). Let \\(P B_{1}\\) and \\(Q A_{1}\\) meet line \\(A B\\) at \\(X\\) and \\(Y\\) . Since \\(\\overline{{X Y}}\\parallel \\overline{{P Q}}\\) it is equivalent to show \\(P_{1}X Y Q_{1}\\) is cyclic (Reim's theorem). \n\nNote the angle condition implies \\(P_{1}C X A\\) and \\(Q_{1}C Y B\\) are cyclic. \n\nLetting \\(T = \\overline{{P X}}\\cap \\overline{{Q Y}}\\) (possibly at infinity), it suffices to show that the radical axis of \\(\\triangle C X A\\) and \\(\\triangle C Y B\\) passes through \\(T\\) , because that would imply \\(P_{1}X Y Q_{1}\\) is cyclic (by power of a point when \\(T\\) is Euclidean, and because it is an isosceles trapezoid if \\(T\\) is at infinity). \n\n\n \n\nTo this end we use barycentric coordinates on \\(\\triangle A B C\\) . We begin by writing \n\n\\[P = (u + t:s:r),\\quad Q = (t:u + s:r)\\] \n\nfrom which it follows that \\(A_{1} = (0:s:r)\\) and \\(B_{1} = (t:0:r)\\) . \n\nNext, compute \\(X = \\left(\\operatorname *{det}\\left[\\begin{array}{l}{u+t}\\\\ {t}\\end{array}\\right]:\\operatorname *{det}\\left[\\begin{array}{l}{s}\\\\ {0}\\end{array}\\right]:0\\right) = (u:s:0)\\) . Similarly, \\(Y = (t:u:0)\\) . So we have computed all points. \n\nClaim — Line \\(B_{1}X\\) has equation \\(- r s\\cdot x + r u\\cdot y + s t\\cdot z = 0\\) , while line \\(C_{1}Y\\) has equation \\(r u\\cdot x - r t\\cdot y + s t\\cdot z = 0\\) . \n\nProof. Line \\(B_{1}X\\) is \\(0 = \\operatorname *{det}(B_{1},X, - ) = \\operatorname *{det}\\left[\\begin{array}{l l}{t} & {0}\\\\ {u} & {s}\\\\ {x} & {y}\\end{array}\\right]\\) . Line \\(C_{1}Y\\) is analogous.\n\n\n\nClaim — The radical axis \\((u + t)y - (u + s)x = 0\\) . \n\nProof. Circle \\((A X C)\\) is given by \\(- a^{2}y z - b^{2}z x - c^{2}x y + (x + y + z)\\cdot \\frac{c^{2}\\cdot u}{u + s}y = 0\\) . Similarly, circle \\((B Y C)\\) has equation \\(- a^{2}y z - b^{2}z x - c^{2}x y + (x + y + z)\\cdot \\frac{c^{2}\\cdot u}{u + t}x = 0\\) . Subtracting gives the radical axis. \\(\\square\\) \n\nFinally, to see these three lines are concurrent, we now compute \n\n\\[\\operatorname *{det}\\left[ \\begin{array}{c c c}{-r s r u s t\\] \\[r u -r t s t\\] \\[-(u + s) u + t 0} \\end{array} \\right] = r s t\\left[\\left[u(u + t) - t(u + s)\\right] + \\left[s(u + t) - u(u + s)\\right]\\right]\\] \\[= r s t\\left[\\left(u^{2} - s t\\right) + (s t - u^{2})\\right] = 0.\\] \n\nThis completes the proof. \n\n\\(\\P\\) Second official solution by tricky angle chasing. Let lines \\(A A_{1}\\) and \\(B B_{1}\\) meet at the circumcircle of \\(\\triangle A B C\\) again at points \\(A_{2}\\) and \\(B_{2}\\) . By Reim's theorem, \\(P Q A_{2}B_{2}\\) are cyclic. \n\n\n \n\nClaim — The points \\(P\\) , \\(Q\\) , \\(A_{2}\\) , \\(Q_{1}\\) are cyclic. Similarly the points \\(P\\) , \\(Q\\) , \\(B_{2}\\) , \\(P_{1}\\) are cyclic. \n\nProof. Note that \\(C A_{1}A_{2}Q_{1}\\) is cyclic since \\(\\angle C Q_{1}A_{1} = \\angle C Q_{1}Q = \\angle C B A = \\angle C A_{2}A =\\) \\(\\angle C A_{2}A_{1}\\) . Then \\(\\angle Q Q_{1}A_{2} = \\angle A_{1}Q_{1}A_{2} = \\angle A_{1}C A_{2} = \\angle B C A_{2} = \\angle B A A_{2} = \\angle Q P A_{2}\\) . \\(\\square\\) \n\nThis claim obviously solves the problem.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2019-notes.jsonl", "problem_match": "2. ", "solution_match": "## \\(\\S 1.2\\) IMO 2019/2, proposed by Anton Trygub (UKR) \n"}}
+{"year": "2019", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "In triangle \\(A B C\\) point \\(A_{1}\\) lies on side \\(B C\\) and point \\(B_{1}\\) lies on side \\(A C\\) . Let \\(P\\) and \\(Q\\) be points on segments \\(A A_{1}\\) and \\(B B_{1}\\) , respectively, such that \\(\\overline{{P Q}}\\parallel \\overline{{A B}}\\) . Point \\(P_{1}\\) is chosen on ray \\(P B_{1}\\) beyond \\(B_{1}\\) such that \\(\\angle P P_{1}C = \\angle B A C\\) . Point \\(Q_{1}\\) is chosen on ray \\(Q A_{1}\\) beyond \\(A_{1}\\) such that \\(\\angle C Q_{1}Q = \\angle C B A\\) . Prove that points \\(P_{1}\\) , \\(Q_{1}\\) , \\(P\\) , \\(Q\\) are cyclic.", "solution": "We present two solutions. \n\nFirst solution by bary (Evan Chen). Let \\(P B_{1}\\) and \\(Q A_{1}\\) meet line \\(A B\\) at \\(X\\) and \\(Y\\) . Since \\(\\overline{{X Y}}\\parallel \\overline{{P Q}}\\) it is equivalent to show \\(P_{1}X Y Q_{1}\\) is cyclic (Reim's theorem). \n\nNote the angle condition implies \\(P_{1}C X A\\) and \\(Q_{1}C Y B\\) are cyclic. \n\nLetting \\(T = \\overline{{P X}}\\cap \\overline{{Q Y}}\\) (possibly at infinity), it suffices to show that the radical axis of \\(\\triangle C X A\\) and \\(\\triangle C Y B\\) passes through \\(T\\) , because that would imply \\(P_{1}X Y Q_{1}\\) is cyclic (by power of a point when \\(T\\) is Euclidean, and because it is an isosceles trapezoid if \\(T\\) is at infinity). \n\n\n \n\nTo this end we use barycentric coordinates on \\(\\triangle A B C\\) . We begin by writing \n\n\\[P = (u + t:s:r),\\quad Q = (t:u + s:r)\\] \n\nfrom which it follows that \\(A_{1} = (0:s:r)\\) and \\(B_{1} = (t:0:r)\\) . \n\nNext, compute \\(X = \\left(\\operatorname *{det}\\left[\\begin{array}{l}{u+t}\\\\ {t}\\end{array}\\right]:\\operatorname *{det}\\left[\\begin{array}{l}{s}\\\\ {0}\\end{array}\\right]:0\\right) = (u:s:0)\\) . Similarly, \\(Y = (t:u:0)\\) . So we have computed all points. \n\nClaim — Line \\(B_{1}X\\) has equation \\(- r s\\cdot x + r u\\cdot y + s t\\cdot z = 0\\) , while line \\(C_{1}Y\\) has equation \\(r u\\cdot x - r t\\cdot y + s t\\cdot z = 0\\) . \n\nProof. Line \\(B_{1}X\\) is \\(0 = \\operatorname *{det}(B_{1},X, - ) = \\operatorname *{det}\\left[\\begin{array}{l l}{t} & {0}\\\\ {u} & {s}\\\\ {x} & {y}\\end{array}\\right]\\) . Line \\(C_{1}Y\\) is analogous.\n\n\n\nClaim — The radical axis \\((u + t)y - (u + s)x = 0\\) . \n\nProof. Circle \\((A X C)\\) is given by \\(- a^{2}y z - b^{2}z x - c^{2}x y + (x + y + z)\\cdot \\frac{c^{2}\\cdot u}{u + s}y = 0\\) . Similarly, circle \\((B Y C)\\) has equation \\(- a^{2}y z - b^{2}z x - c^{2}x y + (x + y + z)\\cdot \\frac{c^{2}\\cdot u}{u + t}x = 0\\) . Subtracting gives the radical axis. \\(\\square\\) \n\nFinally, to see these three lines are concurrent, we now compute \n\n\\[\\operatorname *{det}\\left[ \\begin{array}{c c c}{-r s r u s t\\] \\[r u -r t s t\\] \\[-(u + s) u + t 0} \\end{array} \\right] = r s t\\left[\\left[u(u + t) - t(u + s)\\right] + \\left[s(u + t) - u(u + s)\\right]\\right]\\] \\[= r s t\\left[\\left(u^{2} - s t\\right) + (s t - u^{2})\\right] = 0.\\] \n\nThis completes the proof. \n\n\\(\\P\\) Second official solution by tricky angle chasing. Let lines \\(A A_{1}\\) and \\(B B_{1}\\) meet at the circumcircle of \\(\\triangle A B C\\) again at points \\(A_{2}\\) and \\(B_{2}\\) . By Reim's theorem, \\(P Q A_{2}B_{2}\\) are cyclic. \n\n\n \n\nClaim — The points \\(P\\) , \\(Q\\) , \\(A_{2}\\) , \\(Q_{1}\\) are cyclic. Similarly the points \\(P\\) , \\(Q\\) , \\(B_{2}\\) , \\(P_{1}\\) are cyclic. \n\nProof. Note that \\(C A_{1}A_{2}Q_{1}\\) is cyclic since \\(\\angle C Q_{1}A_{1} = \\angle C Q_{1}Q = \\angle C B A = \\angle C A_{2}A =\\) \\(\\angle C A_{2}A_{1}\\) . Then \\(\\angle Q Q_{1}A_{2} = \\angle A_{1}Q_{1}A_{2} = \\angle A_{1}C A_{2} = \\angle B C A_{2} = \\angle B A A_{2} = \\angle Q P A_{2}\\) . \\(\\square\\) \n\nThis claim obviously solves the problem.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2019-notes.jsonl", "problem_match": "2. ", "solution_match": "## \\(\\S 1.2\\) IMO 2019/2, proposed by Anton Trygub (UKR) \n"}}
{"year": "2019", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "A social network has 2019 users, some pairs of which are friends (friendship is symmetric). If \\(A\\) , \\(B\\) , \\(C\\) are three users such that \\(A B\\) are friends and \\(A C\\) are friends but \\(B C\\) is not, then the administrator may perform the following operation: change the friendships such that \\(B C\\) are friends, but \\(A B\\) and \\(A C\\) are no longer friends. \n\nInitially, 1009 users have 1010 friends and 1010 users have 1009 friends. Prove that the administrator can make a sequence of operations such that all users have at most 1 friend.", "solution": "## Problem stateme \n\nWe take the obvious graph formulation and call the move a toggle. \n\nClaim — Let \\(G\\) be a connected graph. Then one can toggle \\(G\\) without disconnecting the graph, unless \\(G\\) is a clique, a cycle, or a tree. \n\nProof. Assume \\(G\\) is connected and not a tree, so it has a cycle. Take the smallest cycle \\(C\\) ; by hypothesis \\(C \\neq G\\) . \n\nIf \\(C\\) is not a triangle (equivalently, \\(G\\) is triangle- free), then let \\(b \\notin C\\) be a vertex adjacent to \\(C\\) , say at \\(a\\) . Take a vertex \\(c\\) of the cycle adjacent to \\(a\\) (hence not to \\(b\\) ). Then we can toggle \\(abc\\) . \n\nNow assume there exists a triangle; let \\(K\\) be the maximal clique. By hypothesis, \\(K \\neq G\\) . We take an edge \\(e = ab\\) dangling off the clique, with \\(a \\in K\\) and \\(b \\notin K\\) . Note some vertex \\(c\\) of \\(K\\) is not adjacent to \\(b\\) ; now toggle \\(abc\\) . \\(\\square\\) \n\nBack to the original problem; let \\(G_{\\mathrm{imo}}\\) be the given graph. The point is that we can apply toggles (by the claim) repeatedly, without disconnecting the graph, until we get a tree. This is because \n\n- \\(G_{\\mathrm{imo}}\\) is connected, since any two vertices which are not adjacent have a common neighbor by pigeonhole \\((1009 + 1009 + 2 > 2019)\\) .- \\(G_{\\mathrm{imo}}\\) cannot become a cycle, because it initially has an odd-degree vertex, and toggles preserve parity of degree!- \\(G_{\\mathrm{imo}}\\) is obviously not a clique initially (and hence not afterwards). \n\nSo, we can eventually get \\(G_{\\mathrm{imo}}\\) to be a tree. \n\nOnce \\(G_{\\mathrm{imo}}\\) is a tree the problem follows by repeatedly applying toggles arbitrarily until no more are possible; the graph (although now disconnected) remains acyclic (in particular having no triangles) and therefore can only terminate in the desired situation. \n\nRemark. The above proof in fact shows the following better result: \n\nThe task is possible if and only if \\(G_{\\mathrm{imo}}\\) is a connected graph which is not a clique and has any vertex of odd degree. \n\nThe \"only if\" follows from the observation that toggles preserve parity of degree.\n\n\n\nThus the given condition about the degrees of vertices being 1009 and 1010 is largely a red herring; it's a somewhat strange way of masking the correct and more natural both- sufficient- and- necessary condition.\n\n\n\n## \\(\\S 2\\) Solutions to Day 2", "metadata": {"resource_path": "IMO/segmented/en-IMO-2019-notes.jsonl", "problem_match": "3. ", "solution_match": "## \\(\\S 1.3\\) IMO 2019/3, proposed by Adrian Beker (HRV) \n"}}
{"year": "2019", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "Solve over positive integers the equation \n\n\\[k! = \\prod_{i = 0}^{n - 1}(2^{n} - 2^{i}) = (2^{n} - 1)(2^{n} - 2)(2^{n} - 4)\\dots (2^{n} - 2^{n - 1}).\\]", "solution": "] \n\nThe answer is \\((n,k) = (1,1)\\) and \\((n,k) = (2,3)\\) which work. \n\nLet \\(A = \\prod_{i}(2^{n} - 2^{k})\\) , and assume \\(A = k!\\) for some \\(k\\geq 3\\) . Recall by exponent lifting that \n\n\\[\\nu_{3}(2^{t} - 1) = \\left\\{ \\begin{array}{ll}0 & t\\mathrm{odd}\\\\ 1 + \\nu_{3}(t / 2) & t\\mathrm{even}. \\end{array} \\right.\\] \n\nThus, we can compute \n\n\\[k > \\nu_{2}(k!) = \\nu_{2}(A) = 1 + 2 + \\cdot \\cdot \\cdot +(n - 1) = \\frac{n(n - 1)}{2}\\] \\[\\left\\lfloor \\frac{k}{3}\\right\\rfloor \\leq \\nu_{3}(k!) = \\nu_{3}(A) = \\left\\lfloor \\frac{n}{2}\\right\\rfloor +\\left\\lfloor \\frac{n}{6}\\right\\rfloor +\\cdot \\cdot \\cdot < \\frac{3}{4} n.\\] \n\nwhere the first inequality follows by Legendre's formula \\(\\nu_{2}(k!) = k - s_{2}(k)\\) \n\nIn this way, we get \n\n\\[\\frac{9}{4} n + 3 > k > \\frac{n(n - 1)}{2}\\] \n\nwhich means \\(n\\leq 6\\) ; a manual check then shows the solutions we claimed earlier are the only ones. \n\nRemark. An amusing corollary of the problem pointed out in the shortlist is that the symmetric group \\(S_{k}\\) cannot be isomorphic to the group \\(\\mathrm{GL}_{n}(\\mathbb{F}_{2})\\) unless \\((n,k) = (1,1)\\) or \\((n,k) = (2,3)\\) , which indeed produce isomorphisms.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2019-notes.jsonl", "problem_match": "4. ", "solution_match": "## \\(\\S 2.1\\) IMO 2019/4, proposed by Gabriel Chicas Reyes (SLV) \n"}}
-{"year": "2019", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "Let \\(n\\) be a positive integer. Harry has \\(n\\) coins lined up on his desk, which can show either heads or tails. He does the following operation: if there are \\(k\\) coins which show heads and \\(k > 0\\) , then he flips the \\(k\\) th coin over; otherwise he stops the process. (For example, the process starting with \\(T H T\\) would be \\(T H T\\rightarrow H H T\\rightarrow H T T\\rightarrow T T T\\) , which takes three steps.) \n\nProve the process will always terminate, and determine the average number of steps this takes over all \\(2^{n}\\) configurations.", "solution": "The answer is \n\n\\[E_{n} = \\frac{1}{2} (1 + \\dots +n) = \\frac{1}{4} n(n + 1)\\] \n\nwhich is finite. \n\nWe'll represent the operation by a directed graph \\(G_{n}\\) on vertices \\(\\{0,1\\}^{n}\\) (each string points to its successor) with 1 corresponding to heads and 0 corresponding to tails. For \\(b\\in \\{0,1\\}\\) we let \\(\\overline{{b}} = 1 - b\\) , and denote binary strings as a sequence of \\(n\\) symbols. \n\nThe main claim is that \\(G_{n}\\) can be described explicitly in terms of \\(G_{n - 1}\\) : \n\n- We take two copies \\(X\\) and \\(Y\\) of \\(G_{n - 1}\\) . \n\n- In \\(X\\) , we take each string of length \\(n - 1\\) and just append a 0 to it. In symbols, we replace \\(s_{1}\\dots s_{n - 1}\\mapsto s_{1}\\dots s_{n - 1}0\\) . \n\n- In \\(Y\\) , we toggle every bit, then reverse the order, and then append a 1 to it. In symbols, we replace \\(s_{1}\\dots s_{n - 1}\\mapsto \\overline{s}_{n - 1}\\overline{s}_{n - 2}\\dots \\overline{s}_{1}1\\) . \n\n- Finally, we add one new edge from \\(Y\\) to \\(X\\) by \\(11\\dots 1\\mapsto 11\\dots 110\\) . \n\nAn illustration of \\(G_{4}\\) is given below. \n\n\n \n\nTo prove this claim, we need only show the arrows of this directed graph remain valid. The graph \\(X\\) is correct as a subgraph of \\(G_{n}\\) , since the extra 0 makes no difference. As for \\(Y\\) , note that if \\(s = s_{1}\\dots s_{n - 1}\\) had \\(k\\) ones, then the modified string has \\((n - 1 - k) + 1 = n - k\\)\n\n\n\nones, ergo \\(\\overline{s}_{n - 1}\\ldots \\overline{s}_{1}1\\mapsto \\overline{s}_{n - 1}\\ldots \\overline{s}_{k + 1}s_{k}\\overline{s}_{k - 1}\\ldots \\overline{s}_{1}1\\) which is what we wanted. Finally, the one edge from \\(Y\\) to \\(X\\) is obviously correct. \n\nTo finish, let \\(E_{n}\\) denote the desired expected value. Since \\(1\\ldots 1\\) takes \\(n\\) steps to finish we have \n\n\\[E_{n} = \\frac{1}{2} [E_{n - 1} + (E_{n - 1} + n)]\\] \n\nbased on cases on whether the chosen string is in \\(X\\) or \\(Y\\) or not. By induction, we have \\(E_{n} = \\frac{1}{2} (1 + \\dots +n) = \\frac{1}{4} n(n + 1)\\) , as desired. \n\nRemark. Actually, the following is true: if the indices of the 1's are \\(1\\leq i_{1}< \\dots < i_{\\ell}\\leq n\\) then the number of operations required is \n\n\\[2(i_{1} + \\dots +i_{\\ell}) - \\ell^{2}.\\] \n\nThis problem also has an interpretation as a Turing machine: the head starts at a position on the tape (the binary string). If it sees a 1, it changes the cell to a 0 and moves left; if it sees a 0, it changes the cell to a 1 and moves right.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2019-notes.jsonl", "problem_match": "5. ", "solution_match": "## \\(\\S 2.2\\) IMO 2019/5, proposed by David Altizio (USA) \n"}}
-{"year": "2019", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "Let \\(A B C\\) be a triangle with incenter \\(I\\) and incircle \\(\\omega\\) . Let \\(D\\) , \\(E\\) , \\(F\\) denote the tangency points of \\(\\omega\\) with \\(\\overline{{B C}}\\) , \\(\\overline{{C A}}\\) , \\(\\overline{{A B}}\\) . The line through \\(D\\) perpendicular to \\(\\overline{{E F}}\\) meets \\(\\omega\\) again at \\(R\\) (other than \\(D\\) ), and line \\(A R\\) meets \\(\\omega\\) again at \\(P\\) (other than \\(R\\) ). Suppose the circumcircles of \\(\\triangle P C E\\) and \\(\\triangle P B F\\) meet again at \\(Q\\) (other than \\(P\\) ). Prove that lines \\(D I\\) and \\(P Q\\) meet on the external \\(\\angle A\\) -bisector.", "solution": "L \n\nWe present five solutions. \n\n\\(\\P\\) First solution by complex numbers (Evan Chen, with Yang Liu). We use complex numbers with \\(D = x\\) , \\(E = y\\) , \\(F = z\\) . \n\n\n \n\nThen \\(A = \\frac{2y z}{y + z}\\) , \\(R = \\frac{- y z}{x}\\) and so \n\n\\[P = \\frac{A - R}{1 - R\\overline{{A}}} = \\frac{\\frac{2y z}{y + z} + \\frac{y z}{x}}{1 + \\frac{y z}{x}\\cdot\\frac{2}{y + z}} = \\frac{y z(2x + y + z)}{2y z + x(y + z)}.\\] \n\nWe now compute \n\n\\[O_{B} = \\operatorname *{det}\\left[ \\begin{array}{l l l}{P P P 1\\] \\[F F F 1\\] \\[B B B 1} \\end{array} \\right]\\div \\operatorname *{det}\\left[ \\begin{array}{l l l}{P P 1\\] \\[F F 1\\] \\[B B 1} \\end{array} \\right] = \\operatorname *{det}\\left[ \\begin{array}{l l l}{P 1 1\\] \\[z 1 1\\] \\[\\frac{2x z}{x + z}\\frac{4x z}{(x + z)^{2}} 1} \\end{array} \\right]\\div \\operatorname *{det}\\left[ \\begin{array}{l l l}{P 1 / P 1\\] \\[z 1 / z 1\\] \\[\\frac{2x z}{x + z}\\frac{2}{x + z} 1} \\end{array} \\right]\\] \\[\\quad = \\frac{1}{x + z}\\operatorname *{det}\\left[ \\begin{array}{l l l}{P 0 1\\] \\[z 0 1\\] \\[2x z(x + z) - (x - z)^{2} (x + z)^{2}} \\end{array} \\right]\\div \\operatorname *{det}\\left[ \\begin{array}{l l l}{P 1 / P 1\\] \\[z 1 / z 1\\] \\[\\frac{2x z}{x + z}\\frac{2}{x + z} 1} \\end{array} \\right]\\] \\[\\quad = \\frac{(x - z)^{2}}{x + z}\\cdot \\frac{P - z}{(x + z)(P / z - z / P) + 2z - 2x + \\frac{2x z}{P} - 2P\\]\n\n\n\n\\[= \\frac{(x - z)^{2}}{x + z}\\cdot \\frac{P - z}{(\\frac{x}{z} - 1)P - 2(x - z) + (xz - z^{2})\\frac{1}{P}}\\] \\[= \\frac{x - z}{x + z}\\cdot \\frac{P - z}{P / z + z / P - 2} = \\frac{x - z}{x + z}\\cdot \\frac{P - z}{\\frac{(P - z)^{2}}{P z}} = \\frac{x - z}{x + z}\\cdot \\frac{1}{\\frac{1}{z} - \\frac{1}{P}}\\] \\[= \\frac{x - z}{x + z}\\cdot \\frac{y(2x + y + z)}{y(2x + y + z) - (2yz + xy + xz)} = \\frac{x - z}{x + z}\\cdot \\frac{yz(2x + y + z)}{xy + y^{2} - yz - xz}\\] \\[= \\frac{x - z}{x + z}\\cdot \\frac{yz(2x + y + z)}{(y - z)(x + y)}.\\] \n\nSimilarly \n\n\\[O_{C} = \\frac{x - y}{x + y}\\cdot \\frac{yz(2x + y + z)}{(z - y)(x + z)}.\\] \n\nTherefore, subtraction gives \n\n\\[O_{B} - O_{C} = \\frac{yz(2x + y + z)}{(x + y)(x + z)(y - z)} [(x - z) + (x - y)] = \\frac{yz(2x + y + z)(2x - y - z)}{(x + y)(x + z)(z - y)}.\\] \n\nIt remains to compute \\(T\\) . Since \\(T \\in \\overline{ID}\\) we have \\(t / x \\in \\mathbb{R}\\) so \\(\\overline{t} = t / x^{2}\\) . Also, \n\n\\[t - \\frac{2yz}{y + z}\\in i\\mathbb{R}\\Rightarrow 0 = \\frac{t - \\frac{2yz}{y + z} + \\frac{t}{x^{2}} - \\frac{2}{y + z}}{\\frac{1}{y} + \\frac{1}{z}}\\] \\[\\qquad = \\frac{1 + \\frac{yz}{x^{2}}}{y + z} t - \\frac{2yz}{(y + z)^{2}} - \\frac{2yz}{(y + z)^{2}}\\] \\[\\qquad \\Rightarrow t = \\frac{x^{2}}{x^{2} + yz}\\cdot \\frac{4yz}{y + z}\\] \n\nThus \n\n\\[P - T = \\frac{yz(2x + y + z)}{2yz + x(y + z)} -\\frac{4x^{2}yz}{(x^{2} + yz)(y + z)}\\] \\[\\qquad = yz\\cdot \\frac{(2x + y + z)(x^{2} + yz)(y + z) - 4x^{2}(2yz + xy + xz)}{(y + z)(x^{2} + yz)(2yz + xy + xz)}\\] \\[\\qquad = -yz\\cdot \\frac{(2x - y - z)(x^{2}y + x^{2}z + 4xyz + y^{2}z + yz^{2})}{(y + z)(x^{2} + yz)(2yz + xy + xz)}.\\] \n\nThis gives \\(\\overline{PT} \\perp \\overline{O_{B}O_{C}}\\) as needed. \n\n\\(\\P\\) Second solution by tethered moving points, with optimization (Evan Chen). Fix \\(\\triangle DEF\\) and \\(\\omega\\) , with \\(B = \\overline{DD} \\cap \\overline{FF}\\) and \\(C = \\overline{DD} \\cap \\overline{EE}\\) . We consider a variable point \\(M\\) on \\(\\omega\\) and let \\(X\\) , \\(Y\\) be on \\(\\overline{EF}\\) with \\(\\overline{CY} \\cap \\| \\overline{ME}\\) , \\(\\overline{BX} \\cap \\| \\overline{MF}\\) . We define \\(W = \\overline{CY} \\cap \\overline{BX}\\) . Also, let line \\(MW\\) meet \\(\\omega\\) again at \\(V\\) .\n\n\n\n \n\nClaim (Angle chasing) — Pentagons \\(C V W X E\\) and \\(B V W Y F\\) are cyclic. \n\nProof. By \\(\\angle E V W = \\angle E V M = \\angle E F M = \\angle C E M = \\angle E C W\\) and \\(\\angle E X W = \\angle E F M =\\) \\(\\angle C E M = \\angle E C W\\) . \\(\\square\\) \n\nLet \\(N = \\overline{D M} \\cap \\overline{E F}\\) and \\(R'\\) be the \\(D\\) - antipode on \\(\\omega\\) . \n\nClaim (Black magic) — The points \\(V\\) , \\(N\\) , \\(R'\\) are collinear. \n\nProof. We use tethered moving points with \\(\\triangle D E F\\) fixed. \n\nObviously the map \\(\\omega \\mapsto \\overline{E F} \\mapsto \\omega\\) by \\(M \\mapsto N \\mapsto \\overline{R'N} \\cap \\omega\\) is projective. Also, the map \\(\\omega \\mapsto \\overline{E F} \\mapsto \\omega\\) by \\(M \\mapsto X \\mapsto V\\) is also projective (the first by projection to the line at infinity at back; the second say by inversion at \\(E\\) ). \n\nSo it suffices to check for three points. When \\(M = E\\) we get \\(N = E\\) so \\(\\overline{R'N} \\cap \\omega = E\\) , while \\(W = E\\) and thus \\(V = E\\) . The case \\(M = F\\) is similar. Finally, if \\(M = R'\\) , then \\(W\\) is the center of \\(\\omega\\) and so \\(V = \\overline{R'N} \\cap \\overline{E F} = D\\) . \\(\\square\\) \n\nWe now address the original problem by specializing \\(M\\) : choose it so that \\(N\\) is the midpoint of \\(\\overline{E F}\\) . Let \\(M' = \\overline{D A} \\cap (D E F)\\) . \n\nClaim — After this specialization, \\(V = P\\) and \\(W = Q\\) . \n\nProof. Thus \\(\\overline{R R'}\\) and \\(\\overline{M M'}\\) are parallel to \\(\\overline{E F}\\) . From \\((E F; P R) = - 1 = (E F; N \\infty) \\frac{R'}{2}\\) \\((E F; N V)\\) , we derive that \\(P = V\\) and \\(Q = R\\) , proving (i). \\(\\square\\) \n\nFinally, the concurrence requested follows by Pascal theorem on \\(M' M D R' P R\\) . \n\nThird solution by power of a point linearity (Luke Robitaille). Let us define \n\n\\[f(\\bullet) = \\mathrm{Pow}(\\bullet ,(C P E)) - \\mathrm{Pow}(\\bullet ,(B P F))\\] \n\nwhich is a linear function from the plane to \\(\\mathbb{R}\\) .\n\n\n\nDefine \\(W = \\overline{BA} \\cap \\overline{PE}\\) , \\(V = \\overline{AC} \\cap \\overline{PF}\\) . Also, let \\(W_{1} = \\overline{ER} \\cap \\overline{AB}\\) , \\(V_{1} = \\overline{FR} \\cap \\overline{AC}\\) . Note that \n\n\\[-1 = (PR;EF) \\stackrel{E}{=} (WA;W_{1}F)\\] \n\nand similarly \\((VA;V_{1}E) = - 1\\) . \n\nClaim — We have \n\n\\[f(F) = \\frac{|EF|\\cdot(s - c)\\sin C / 2}{\\sin B / 2}\\] \\[f(E) = -\\frac{|EF|\\cdot(s - b)\\sin B / 2}{\\sin C / 2}.\\] \n\nProof. We have \n\n\\[f(W) = WF^{2} - WB\\cdot WF = WF\\cdot BF\\] \n\nwhere lengths are directed. Next, \n\n\\[f(F) = \\frac{AF\\cdot f(W) + FW\\cdot f(A)}{AW}\\] \\[= \\frac{AF\\cdot WF\\cdot BF + FW\\cdot (AE\\cdot AC - AF\\cdot AB)}{AW}\\] \\[= \\frac{WF(AF\\cdot BF + AF\\cdot AB) + FW\\cdot AE\\cdot AC}{AW}\\] \\[= \\frac{WF\\cdot AF^{2} - WF\\cdot AE\\cdot AC}{AW} = \\frac{WF}{AW}\\cdot (AE^{2} - AE\\cdot AC)\\] \\[= \\frac{WF}{AW}\\cdot AE\\cdot CE = -\\frac{W_{1}F}{AW_{1}}\\cdot AE\\cdot CE.\\] \n\nSince \\(\\triangle DEF\\) is acute, the point \\(R\\) lies inside \\(\\triangle AEF\\) . Thus \\(W_{1}\\) lies inside segment \\(\\overline{AF}\\) and the ratio \\(\\frac{W_{1}F}{AW_{1}}\\) is positive. We now determine its value: by the ratio lemma \n\n\\[\\frac{|W_{1}F|}{|AW_{1}|} = \\frac{|EF|\\sin\\angle W_{1}EF}{|AE|\\sin\\angle AEW_{1}}\\] \\[= \\frac{|EF|\\sin\\angle REF}{|AE|\\sin\\angle AER}\\] \\[= \\frac{|EF|\\sin\\angle RDF}{|AE|\\sin\\angle EDR}\\] \\[= \\frac{|EF|\\sin C / 2}{|AE|\\sin B / 2}.\\] \n\nAlso, we have \\(AE \\cdot CE < 0\\) since \\(E\\) lies inside \\(\\overline{AC}\\) . Hence \n\n\\[f(F) = -\\frac{|EF|\\sin C / 2}{|AE|\\sin B / 2}\\cdot AE\\cdot CE = |EF|\\cdot \\frac{|CE|\\sin B / 2}{\\sin C / 2} = |EF|\\cdot \\frac{(s - c)\\sin B / 2}{\\sin C / 2}.\\] \n\nThe calculation for \\(f(E)\\) is similar, (noting the sign flips since \\(f\\) is anti- symmetric in terms of \\(B\\) and \\(C\\) ). \\(\\square\\) \n\nLet \\(Z \\in \\overline{DI}\\) with \\(\\angle ZAI = 90^{\\circ}\\) be the point requested in the problem now. Our goal is to show \\(f(Z) = 0\\) . We assume WLOG that \\(AB < AC\\) , so \\(\\frac{ZA}{EF} > 0\\) . Then \n\n\\[|ZA| = |AI|\\cdot \\tan \\angle AIZ\\]\n\n\n\n\\[= |A I|\\cdot \\tan \\angle (\\overline{{A I}},\\overline{{D I}})\\] \\[= \\frac{s - a}{\\cos A / 2}\\cdot \\tan (\\overline{{B C}},\\overline{{E F}})\\] \\[= \\frac{s - a}{\\cos A / 2}\\tan (B / 2 - C / 2).\\] \n\nTo this end we compute \n\n\\[\\begin{array}{r l} & {f(Z) = f(A) + [f(Z) - f(A)] = f(A) + \\frac{Z A}{E F} [f(E) - f(F)]}\\\\ & {\\quad = f(A) - \\frac{Z A}{E F}\\left[\\frac{|E F|\\cdot(s - b)\\sin B / 2}{\\sin C / 2} +\\frac{|E F|\\cdot(s - c)\\sin C / 2}{\\sin B / 2}\\right]}\\\\ & {\\quad = f(A) - |Z A|\\left[\\frac{(s - b)\\sin B / 2}{\\sin C / 2} +\\frac{(s - c)\\sin C / 2}{\\sin B / 2}\\right]}\\\\ & {\\quad = [b(s - a) - c(s - a)] - |Z A|\\left[\\frac{(s - b)\\sin B / 2}{\\sin C / 2} +\\frac{(s - c)\\sin C / 2}{\\sin B / 2}\\right]}\\\\ & {\\quad = (b - c)(s - a) - \\frac{s - a}{\\cos A / 2}\\tan (B / 2 - C / 2)\\left[\\frac{(s - b)\\sin B / 2}{\\sin C / 2} +\\frac{(s - c)\\sin C / 2}{\\sin B / 2}\\right].} \\end{array} \\quad (1)\\] \n\nDividing out, \n\n\\[\\frac{f(Z)}{s - a} = (b - c) - \\frac{1}{\\cos A / 2}\\tan (B / 2 - C / 2)\\left[\\frac{r\\cos B / 2}{\\sin C / 2} +\\frac{r\\cos C / 2}{\\sin B / 2}\\right]\\] \\[\\qquad = (b - c) - \\frac{r\\tan(B / 2 - C / 2)}{\\cos A / 2}\\cdot \\frac{\\cos B / 2\\sin B / 2 + \\cos C / 2\\sin C / 2}{\\sin C / 2\\sin B / 2}\\] \\[\\qquad = (b - c) - \\frac{r\\tan(B / 2 - C / 2)}{\\cos A / 2}\\cdot \\frac{\\sin B + \\sin C}{2\\sin C / 2\\sin B / 2}\\] \\[\\qquad = (b - c) - \\frac{r\\tan(B / 2 - C / 2)}{\\cos A / 2}\\cdot \\frac{\\sin(B / 2 + C / 2)\\cos(B / 2 - C / 2)}{\\sin C / 2\\sin B / 2}\\] \\[\\qquad = (b - c) - r\\frac{\\sin(B / 2 - C / 2)}{\\sin B / 2\\sin C / 2}\\] \\[\\qquad = (b - c) - r(\\cot C / 2 - \\cot B / 2) = (b - c) - ((s - c) - (s - b)) = 0.\\] \n\n\\(\\P\\) Fourth solution by incircle inversion (USA IMO live stream, led by Andrew Gu). Let \\(T\\) be the intersection of line \\(D I\\) and the external \\(\\angle A\\) - bisector. Also, let \\(G\\) be the antipode of \\(D\\) on \\(\\omega\\) . \n\nWe perform inversion around \\(\\omega\\) , using \\(\\bullet^{*}\\) for the inverse. Then \\(\\triangle A^{*}B^{*}C^{*}\\) is the medial triangle of \\(\\triangle D E F\\) , and \\(T^{*}\\) is the foot from \\(A^{*}\\) on to \\(\\overline{{D I}}\\) . If we denote \\(Q^{*}\\) as the second intersection of \\((P C^{*}E)\\) and \\((P B^{*}F)\\) , then the goal is to show that \\(Q^{*}\\) lies on \\((P I T^{*})\\) .\n\n\n\n \n\nClaim — Points \\(Q^{*}\\) , \\(B^{*}\\) , \\(C^{*}\\) are collinear. \n\nProof. \\(\\angle P Q^{*}C^{*} = \\angle P E C^{*} = \\angle P E D = \\angle P F D = \\angle P F B^{*} = \\angle P Q^{*}B^{*}\\) . \n\nClaim (cf Brazil 2011/5) — Points \\(P\\) , \\(A^{*}\\) , \\(G\\) are collinear. \n\nProof. Project harmonic quadrilateral \\(PERF\\) through \\(G\\) , noting \\(\\overline{GR} \\parallel \\overline{EF}\\) . \n\nDenote by \\(M\\) the center of parallelogram \\(DC^{*}A^{*}B^{*}\\) . Note that it is the center of the circle with diameter \\(\\overline{DA^{*}}\\) , which passes through \\(P\\) and \\(T^{*}\\) . Also, \\(\\overline{MI} \\parallel \\overline{PA^{*}G}\\) . \n\nClaim — Points \\(P\\) , \\(M\\) , \\(I\\) , \\(T^{*}\\) are cyclic. \n\nProof. \\(\\angle IT^{*}P = \\angle DT^{*}P = \\angle DA^{*}P = \\angle MA^{*}P = \\angle A^{*}PM = \\angle IMP\\) . \n\nClaim — Points \\(P\\) , \\(M\\) , \\(I\\) , \\(Q^{*}\\) are cyclic. \n\nProof. \\(\\angle MQ^{*}P = \\angle C^{*}Q^{*}P = \\angle C^{*}EP = \\angle DEP = \\angle DGP = \\angle GPI = \\angle MIP\\) . \n\nFifth solution by double inversion (Brandon Wang, Luke Robitaille, Michael Ren, Evan Chen). We outline one final approach. After inverting about \\(\\omega\\) as in the previous approach, we then apply another inversion around \\(P\\) . Dropping the apostrophes/stars/etc now one can check that the problem we arrive at becomes the following. \n\n## Proposition (Doubly inverted problem) \n\nIn \\(\\triangle PEF\\) , the \\(P\\) - symmedian meets \\(\\overline{EF}\\) and \\((PEF)\\) at \\(K\\) , \\(L\\) . Let \\(D \\in \\overline{EF}\\) with \\(\\angle DPK = 90^{\\circ}\\) , and let \\(T\\) be the foot from \\(K\\) to \\(\\overline{DL}\\) . Denote by \\(I\\) the reflection of \\(P\\) about \\(\\overline{EF}\\) . Finally, let \\(PDNE\\) and \\(PDMF\\) be cyclic harmonic quadrilaterals. Then lines \\(EN\\) , \\(MF\\) , \\(TI\\) , are concurrent.\n\n\n\nThe proof proceeds in three steps. Suppose the line through \\(L\\) perpendicular to \\(\\overline{EF}\\) meets \\(\\overline{EF}\\) at \\(W\\) and \\((PEF)\\) at \\(Z\\) . \n\n\n \n\n1. Since \\(\\angle ZEP = \\angle WLP = \\angle WDP\\) , it follows \\(\\overline{ZE}\\) is tangent to \\((PDNE)\\) . Similarly, \\(\\overline{ZF}\\) is tangent to \\((PDMF)\\) . \n\n2. \\(\\Delta WTP\\) is the orthic triangle of \\(\\triangle DKL\\) , so \\(\\overline{WD}\\) bisects \\(\\angle PWT\\) and \\(\\overline{WTI}\\) collinear. \n\n3. \\(-1 = E(PN; DZ) = F(PM; DZ) = W(PI; DZ)\\) , so \\(\\overline{EN}\\) , \\(\\overline{FM}\\) , \\(\\overline{WI}\\) meet on \\(\\overline{PZ}\\) .", "metadata": {"resource_path": "IMO/segmented/en-IMO-2019-notes.jsonl", "problem_match": "6. ", "solution_match": "## \\(\\S 2.3\\) IMO 2019/6, proposed by Anant Mudgal (IND) \n"}}
+{"year": "2019", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "Let \\(n\\) be a positive integer. Harry has \\(n\\) coins lined up on his desk, which can show either heads or tails. He does the following operation: if there are \\(k\\) coins which show heads and \\(k > 0\\) , then he flips the \\(k\\) th coin over; otherwise he stops the process. (For example, the process starting with \\(T H T\\) would be \\(T H T\\rightarrow H H T\\rightarrow H T T\\rightarrow T T T\\) , which takes three steps.) \n\nProve the process will always terminate, and determine the average number of steps this takes over all \\(2^{n}\\) configurations.", "solution": "The answer is \n\n\\[E_{n} = \\frac{1}{2} (1 + \\dots +n) = \\frac{1}{4} n(n + 1)\\] \n\nwhich is finite. \n\nWe'll represent the operation by a directed graph \\(G_{n}\\) on vertices \\(\\{0,1\\}^{n}\\) (each string points to its successor) with 1 corresponding to heads and 0 corresponding to tails. For \\(b\\in \\{0,1\\}\\) we let \\(\\overline{{b}} = 1 - b\\) , and denote binary strings as a sequence of \\(n\\) symbols. \n\nThe main claim is that \\(G_{n}\\) can be described explicitly in terms of \\(G_{n - 1}\\) : \n\n- We take two copies \\(X\\) and \\(Y\\) of \\(G_{n - 1}\\) . \n\n- In \\(X\\) , we take each string of length \\(n - 1\\) and just append a 0 to it. In symbols, we replace \\(s_{1}\\dots s_{n - 1}\\mapsto s_{1}\\dots s_{n - 1}0\\) . \n\n- In \\(Y\\) , we toggle every bit, then reverse the order, and then append a 1 to it. In symbols, we replace \\(s_{1}\\dots s_{n - 1}\\mapsto \\overline{s}_{n - 1}\\overline{s}_{n - 2}\\dots \\overline{s}_{1}1\\) . \n\n- Finally, we add one new edge from \\(Y\\) to \\(X\\) by \\(11\\dots 1\\mapsto 11\\dots 110\\) . \n\nAn illustration of \\(G_{4}\\) is given below. \n\n\n \n\nTo prove this claim, we need only show the arrows of this directed graph remain valid. The graph \\(X\\) is correct as a subgraph of \\(G_{n}\\) , since the extra 0 makes no difference. As for \\(Y\\) , note that if \\(s = s_{1}\\dots s_{n - 1}\\) had \\(k\\) ones, then the modified string has \\((n - 1 - k) + 1 = n - k\\)\n\n\n\nones, ergo \\(\\overline{s}_{n - 1}\\ldots \\overline{s}_{1}1\\mapsto \\overline{s}_{n - 1}\\ldots \\overline{s}_{k + 1}s_{k}\\overline{s}_{k - 1}\\ldots \\overline{s}_{1}1\\) which is what we wanted. Finally, the one edge from \\(Y\\) to \\(X\\) is obviously correct. \n\nTo finish, let \\(E_{n}\\) denote the desired expected value. Since \\(1\\ldots 1\\) takes \\(n\\) steps to finish we have \n\n\\[E_{n} = \\frac{1}{2} [E_{n - 1} + (E_{n - 1} + n)]\\] \n\nbased on cases on whether the chosen string is in \\(X\\) or \\(Y\\) or not. By induction, we have \\(E_{n} = \\frac{1}{2} (1 + \\dots +n) = \\frac{1}{4} n(n + 1)\\) , as desired. \n\nRemark. Actually, the following is true: if the indices of the 1's are \\(1\\leq i_{1}< \\dots < i_{\\ell}\\leq n\\) then the number of operations required is \n\n\\[2(i_{1} + \\dots +i_{\\ell}) - \\ell^{2}.\\] \n\nThis problem also has an interpretation as a Turing machine: the head starts at a position on the tape (the binary string). If it sees a 1, it changes the cell to a 0 and moves left; if it sees a 0, it changes the cell to a 1 and moves right.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2019-notes.jsonl", "problem_match": "5. ", "solution_match": "## \\(\\S 2.2\\) IMO 2019/5, proposed by David Altizio (USA) \n"}}
+{"year": "2019", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "Let \\(A B C\\) be a triangle with incenter \\(I\\) and incircle \\(\\omega\\) . Let \\(D\\) , \\(E\\) , \\(F\\) denote the tangency points of \\(\\omega\\) with \\(\\overline{{B C}}\\) , \\(\\overline{{C A}}\\) , \\(\\overline{{A B}}\\) . The line through \\(D\\) perpendicular to \\(\\overline{{E F}}\\) meets \\(\\omega\\) again at \\(R\\) (other than \\(D\\) ), and line \\(A R\\) meets \\(\\omega\\) again at \\(P\\) (other than \\(R\\) ). Suppose the circumcircles of \\(\\triangle P C E\\) and \\(\\triangle P B F\\) meet again at \\(Q\\) (other than \\(P\\) ). Prove that lines \\(D I\\) and \\(P Q\\) meet on the external \\(\\angle A\\) -bisector.", "solution": "L \n\nWe present five solutions. \n\n\\(\\P\\) First solution by complex numbers (Evan Chen, with Yang Liu). We use complex numbers with \\(D = x\\) , \\(E = y\\) , \\(F = z\\) . \n\n\n \n\nThen \\(A = \\frac{2y z}{y + z}\\) , \\(R = \\frac{- y z}{x}\\) and so \n\n\\[P = \\frac{A - R}{1 - R\\overline{{A}}} = \\frac{\\frac{2y z}{y + z} + \\frac{y z}{x}}{1 + \\frac{y z}{x}\\cdot\\frac{2}{y + z}} = \\frac{y z(2x + y + z)}{2y z + x(y + z)}.\\] \n\nWe now compute \n\n\\[O_{B} = \\operatorname *{det}\\left[ \\begin{array}{l l l}{P P P 1\\] \\[F F F 1\\] \\[B B B 1} \\end{array} \\right]\\div \\operatorname *{det}\\left[ \\begin{array}{l l l}{P P 1\\] \\[F F 1\\] \\[B B 1} \\end{array} \\right] = \\operatorname *{det}\\left[ \\begin{array}{l l l}{P 1 1\\] \\[z 1 1\\] \\[\\frac{2x z}{x + z}\\frac{4x z}{(x + z)^{2}} 1} \\end{array} \\right]\\div \\operatorname *{det}\\left[ \\begin{array}{l l l}{P 1 / P 1\\] \\[z 1 / z 1\\] \\[\\frac{2x z}{x + z}\\frac{2}{x + z} 1} \\end{array} \\right]\\] \\[\\quad = \\frac{1}{x + z}\\operatorname *{det}\\left[ \\begin{array}{l l l}{P 0 1\\] \\[z 0 1\\] \\[2x z(x + z) - (x - z)^{2} (x + z)^{2}} \\end{array} \\right]\\div \\operatorname *{det}\\left[ \\begin{array}{l l l}{P 1 / P 1\\] \\[z 1 / z 1\\] \\[\\frac{2x z}{x + z}\\frac{2}{x + z} 1} \\end{array} \\right]\\] \\[\\quad = \\frac{(x - z)^{2}}{x + z}\\cdot \\frac{P - z}{(x + z)(P / z - z / P) + 2z - 2x + \\frac{2x z}{P} - 2P\\]\n\n\n\n\\[= \\frac{(x - z)^{2}}{x + z}\\cdot \\frac{P - z}{(\\frac{x}{z} - 1)P - 2(x - z) + (xz - z^{2})\\frac{1}{P}}\\] \\[= \\frac{x - z}{x + z}\\cdot \\frac{P - z}{P / z + z / P - 2} = \\frac{x - z}{x + z}\\cdot \\frac{P - z}{\\frac{(P - z)^{2}}{P z}} = \\frac{x - z}{x + z}\\cdot \\frac{1}{\\frac{1}{z} - \\frac{1}{P}}\\] \\[= \\frac{x - z}{x + z}\\cdot \\frac{y(2x + y + z)}{y(2x + y + z) - (2yz + xy + xz)} = \\frac{x - z}{x + z}\\cdot \\frac{yz(2x + y + z)}{xy + y^{2} - yz - xz}\\] \\[= \\frac{x - z}{x + z}\\cdot \\frac{yz(2x + y + z)}{(y - z)(x + y)}.\\] \n\nSimilarly \n\n\\[O_{C} = \\frac{x - y}{x + y}\\cdot \\frac{yz(2x + y + z)}{(z - y)(x + z)}.\\] \n\nTherefore, subtraction gives \n\n\\[O_{B} - O_{C} = \\frac{yz(2x + y + z)}{(x + y)(x + z)(y - z)} [(x - z) + (x - y)] = \\frac{yz(2x + y + z)(2x - y - z)}{(x + y)(x + z)(z - y)}.\\] \n\nIt remains to compute \\(T\\) . Since \\(T \\in \\overline{ID}\\) we have \\(t / x \\in \\mathbb{R}\\) so \\(\\overline{t} = t / x^{2}\\) . Also, \n\n\\[t - \\frac{2yz}{y + z}\\in i\\mathbb{R}\\Rightarrow 0 = \\frac{t - \\frac{2yz}{y + z} + \\frac{t}{x^{2}} - \\frac{2}{y + z}}{\\frac{1}{y} + \\frac{1}{z}}\\] \\[\\qquad = \\frac{1 + \\frac{yz}{x^{2}}}{y + z} t - \\frac{2yz}{(y + z)^{2}} - \\frac{2yz}{(y + z)^{2}}\\] \\[\\qquad \\Rightarrow t = \\frac{x^{2}}{x^{2} + yz}\\cdot \\frac{4yz}{y + z}\\] \n\nThus \n\n\\[P - T = \\frac{yz(2x + y + z)}{2yz + x(y + z)} -\\frac{4x^{2}yz}{(x^{2} + yz)(y + z)}\\] \\[\\qquad = yz\\cdot \\frac{(2x + y + z)(x^{2} + yz)(y + z) - 4x^{2}(2yz + xy + xz)}{(y + z)(x^{2} + yz)(2yz + xy + xz)}\\] \\[\\qquad = -yz\\cdot \\frac{(2x - y - z)(x^{2}y + x^{2}z + 4xyz + y^{2}z + yz^{2})}{(y + z)(x^{2} + yz)(2yz + xy + xz)}.\\] \n\nThis gives \\(\\overline{PT} \\perp \\overline{O_{B}O_{C}}\\) as needed. \n\n\\(\\P\\) Second solution by tethered moving points, with optimization (Evan Chen). Fix \\(\\triangle DEF\\) and \\(\\omega\\) , with \\(B = \\overline{DD} \\cap \\overline{FF}\\) and \\(C = \\overline{DD} \\cap \\overline{EE}\\) . We consider a variable point \\(M\\) on \\(\\omega\\) and let \\(X\\) , \\(Y\\) be on \\(\\overline{EF}\\) with \\(\\overline{CY} \\cap \\| \\overline{ME}\\) , \\(\\overline{BX} \\cap \\| \\overline{MF}\\) . We define \\(W = \\overline{CY} \\cap \\overline{BX}\\) . Also, let line \\(MW\\) meet \\(\\omega\\) again at \\(V\\) .\n\n\n\n \n\nClaim (Angle chasing) — Pentagons \\(C V W X E\\) and \\(B V W Y F\\) are cyclic. \n\nProof. By \\(\\angle E V W = \\angle E V M = \\angle E F M = \\angle C E M = \\angle E C W\\) and \\(\\angle E X W = \\angle E F M =\\) \\(\\angle C E M = \\angle E C W\\) . \\(\\square\\) \n\nLet \\(N = \\overline{D M} \\cap \\overline{E F}\\) and \\(R'\\) be the \\(D\\) - antipode on \\(\\omega\\) . \n\nClaim (Black magic) — The points \\(V\\) , \\(N\\) , \\(R'\\) are collinear. \n\nProof. We use tethered moving points with \\(\\triangle D E F\\) fixed. \n\nObviously the map \\(\\omega \\mapsto \\overline{E F} \\mapsto \\omega\\) by \\(M \\mapsto N \\mapsto \\overline{R'N} \\cap \\omega\\) is projective. Also, the map \\(\\omega \\mapsto \\overline{E F} \\mapsto \\omega\\) by \\(M \\mapsto X \\mapsto V\\) is also projective (the first by projection to the line at infinity at back; the second say by inversion at \\(E\\) ). \n\nSo it suffices to check for three points. When \\(M = E\\) we get \\(N = E\\) so \\(\\overline{R'N} \\cap \\omega = E\\) , while \\(W = E\\) and thus \\(V = E\\) . The case \\(M = F\\) is similar. Finally, if \\(M = R'\\) , then \\(W\\) is the center of \\(\\omega\\) and so \\(V = \\overline{R'N} \\cap \\overline{E F} = D\\) . \\(\\square\\) \n\nWe now address the original problem by specializing \\(M\\) : choose it so that \\(N\\) is the midpoint of \\(\\overline{E F}\\) . Let \\(M' = \\overline{D A} \\cap (D E F)\\) . \n\nClaim — After this specialization, \\(V = P\\) and \\(W = Q\\) . \n\nProof. Thus \\(\\overline{R R'}\\) and \\(\\overline{M M'}\\) are parallel to \\(\\overline{E F}\\) . From \\((E F; P R) = - 1 = (E F; N \\infty) \\frac{R'}{2}\\) \\((E F; N V)\\) , we derive that \\(P = V\\) and \\(Q = R\\) , proving (i). \\(\\square\\) \n\nFinally, the concurrence requested follows by Pascal theorem on \\(M' M D R' P R\\) . \n\nThird solution by power of a point linearity (Luke Robitaille). Let us define \n\n\\[f(\\bullet) = \\mathrm{Pow}(\\bullet ,(C P E)) - \\mathrm{Pow}(\\bullet ,(B P F))\\] \n\nwhich is a linear function from the plane to \\(\\mathbb{R}\\) .\n\n\n\nDefine \\(W = \\overline{BA} \\cap \\overline{PE}\\) , \\(V = \\overline{AC} \\cap \\overline{PF}\\) . Also, let \\(W_{1} = \\overline{ER} \\cap \\overline{AB}\\) , \\(V_{1} = \\overline{FR} \\cap \\overline{AC}\\) . Note that \n\n\\[-1 = (PR;EF) \\stackrel{E}{=} (WA;W_{1}F)\\] \n\nand similarly \\((VA;V_{1}E) = - 1\\) . \n\nClaim — We have \n\n\\[f(F) = \\frac{|EF|\\cdot(s - c)\\sin C / 2}{\\sin B / 2}\\] \\[f(E) = -\\frac{|EF|\\cdot(s - b)\\sin B / 2}{\\sin C / 2}.\\] \n\nProof. We have \n\n\\[f(W) = WF^{2} - WB\\cdot WF = WF\\cdot BF\\] \n\nwhere lengths are directed. Next, \n\n\\[f(F) = \\frac{AF\\cdot f(W) + FW\\cdot f(A)}{AW}\\] \\[= \\frac{AF\\cdot WF\\cdot BF + FW\\cdot (AE\\cdot AC - AF\\cdot AB)}{AW}\\] \\[= \\frac{WF(AF\\cdot BF + AF\\cdot AB) + FW\\cdot AE\\cdot AC}{AW}\\] \\[= \\frac{WF\\cdot AF^{2} - WF\\cdot AE\\cdot AC}{AW} = \\frac{WF}{AW}\\cdot (AE^{2} - AE\\cdot AC)\\] \\[= \\frac{WF}{AW}\\cdot AE\\cdot CE = -\\frac{W_{1}F}{AW_{1}}\\cdot AE\\cdot CE.\\] \n\nSince \\(\\triangle DEF\\) is acute, the point \\(R\\) lies inside \\(\\triangle AEF\\) . Thus \\(W_{1}\\) lies inside segment \\(\\overline{AF}\\) and the ratio \\(\\frac{W_{1}F}{AW_{1}}\\) is positive. We now determine its value: by the ratio lemma \n\n\\[\\frac{|W_{1}F|}{|AW_{1}|} = \\frac{|EF|\\sin\\angle W_{1}EF}{|AE|\\sin\\angle AEW_{1}}\\] \\[= \\frac{|EF|\\sin\\angle REF}{|AE|\\sin\\angle AER}\\] \\[= \\frac{|EF|\\sin\\angle RDF}{|AE|\\sin\\angle EDR}\\] \\[= \\frac{|EF|\\sin C / 2}{|AE|\\sin B / 2}.\\] \n\nAlso, we have \\(AE \\cdot CE < 0\\) since \\(E\\) lies inside \\(\\overline{AC}\\) . Hence \n\n\\[f(F) = -\\frac{|EF|\\sin C / 2}{|AE|\\sin B / 2}\\cdot AE\\cdot CE = |EF|\\cdot \\frac{|CE|\\sin B / 2}{\\sin C / 2} = |EF|\\cdot \\frac{(s - c)\\sin B / 2}{\\sin C / 2}.\\] \n\nThe calculation for \\(f(E)\\) is similar, (noting the sign flips since \\(f\\) is anti- symmetric in terms of \\(B\\) and \\(C\\) ). \\(\\square\\) \n\nLet \\(Z \\in \\overline{DI}\\) with \\(\\angle ZAI = 90^{\\circ}\\) be the point requested in the problem now. Our goal is to show \\(f(Z) = 0\\) . We assume WLOG that \\(AB < AC\\) , so \\(\\frac{ZA}{EF} > 0\\) . Then \n\n\\[|ZA| = |AI|\\cdot \\tan \\angle AIZ\\]\n\n\n\n\\[= |A I|\\cdot \\tan \\angle (\\overline{{A I}},\\overline{{D I}})\\] \\[= \\frac{s - a}{\\cos A / 2}\\cdot \\tan (\\overline{{B C}},\\overline{{E F}})\\] \\[= \\frac{s - a}{\\cos A / 2}\\tan (B / 2 - C / 2).\\] \n\nTo this end we compute \n\n\\[\\begin{array}{r l} & {f(Z) = f(A) + [f(Z) - f(A)] = f(A) + \\frac{Z A}{E F} [f(E) - f(F)]}\\\\ & {\\quad = f(A) - \\frac{Z A}{E F}\\left[\\frac{|E F|\\cdot(s - b)\\sin B / 2}{\\sin C / 2} +\\frac{|E F|\\cdot(s - c)\\sin C / 2}{\\sin B / 2}\\right]}\\\\ & {\\quad = f(A) - |Z A|\\left[\\frac{(s - b)\\sin B / 2}{\\sin C / 2} +\\frac{(s - c)\\sin C / 2}{\\sin B / 2}\\right]}\\\\ & {\\quad = [b(s - a) - c(s - a)] - |Z A|\\left[\\frac{(s - b)\\sin B / 2}{\\sin C / 2} +\\frac{(s - c)\\sin C / 2}{\\sin B / 2}\\right]}\\\\ & {\\quad = (b - c)(s - a) - \\frac{s - a}{\\cos A / 2}\\tan (B / 2 - C / 2)\\left[\\frac{(s - b)\\sin B / 2}{\\sin C / 2} +\\frac{(s - c)\\sin C / 2}{\\sin B / 2}\\right].} \\end{array} \\quad (1)\\] \n\nDividing out, \n\n\\[\\frac{f(Z)}{s - a} = (b - c) - \\frac{1}{\\cos A / 2}\\tan (B / 2 - C / 2)\\left[\\frac{r\\cos B / 2}{\\sin C / 2} +\\frac{r\\cos C / 2}{\\sin B / 2}\\right]\\] \\[\\qquad = (b - c) - \\frac{r\\tan(B / 2 - C / 2)}{\\cos A / 2}\\cdot \\frac{\\cos B / 2\\sin B / 2 + \\cos C / 2\\sin C / 2}{\\sin C / 2\\sin B / 2}\\] \\[\\qquad = (b - c) - \\frac{r\\tan(B / 2 - C / 2)}{\\cos A / 2}\\cdot \\frac{\\sin B + \\sin C}{2\\sin C / 2\\sin B / 2}\\] \\[\\qquad = (b - c) - \\frac{r\\tan(B / 2 - C / 2)}{\\cos A / 2}\\cdot \\frac{\\sin(B / 2 + C / 2)\\cos(B / 2 - C / 2)}{\\sin C / 2\\sin B / 2}\\] \\[\\qquad = (b - c) - r\\frac{\\sin(B / 2 - C / 2)}{\\sin B / 2\\sin C / 2}\\] \\[\\qquad = (b - c) - r(\\cot C / 2 - \\cot B / 2) = (b - c) - ((s - c) - (s - b)) = 0.\\] \n\n\\(\\P\\) Fourth solution by incircle inversion (USA IMO live stream, led by Andrew Gu). Let \\(T\\) be the intersection of line \\(D I\\) and the external \\(\\angle A\\) - bisector. Also, let \\(G\\) be the antipode of \\(D\\) on \\(\\omega\\) . \n\nWe perform inversion around \\(\\omega\\) , using \\(\\bullet^{*}\\) for the inverse. Then \\(\\triangle A^{*}B^{*}C^{*}\\) is the medial triangle of \\(\\triangle D E F\\) , and \\(T^{*}\\) is the foot from \\(A^{*}\\) on to \\(\\overline{{D I}}\\) . If we denote \\(Q^{*}\\) as the second intersection of \\((P C^{*}E)\\) and \\((P B^{*}F)\\) , then the goal is to show that \\(Q^{*}\\) lies on \\((P I T^{*})\\) .\n\n\n\n \n\nClaim — Points \\(Q^{*}\\) , \\(B^{*}\\) , \\(C^{*}\\) are collinear. \n\nProof. \\(\\angle P Q^{*}C^{*} = \\angle P E C^{*} = \\angle P E D = \\angle P F D = \\angle P F B^{*} = \\angle P Q^{*}B^{*}\\) . \n\nClaim (cf Brazil 2011/5) — Points \\(P\\) , \\(A^{*}\\) , \\(G\\) are collinear. \n\nProof. Project harmonic quadrilateral \\(PERF\\) through \\(G\\) , noting \\(\\overline{GR} \\parallel \\overline{EF}\\) . \n\nDenote by \\(M\\) the center of parallelogram \\(DC^{*}A^{*}B^{*}\\) . Note that it is the center of the circle with diameter \\(\\overline{DA^{*}}\\) , which passes through \\(P\\) and \\(T^{*}\\) . Also, \\(\\overline{MI} \\parallel \\overline{PA^{*}G}\\) . \n\nClaim — Points \\(P\\) , \\(M\\) , \\(I\\) , \\(T^{*}\\) are cyclic. \n\nProof. \\(\\angle IT^{*}P = \\angle DT^{*}P = \\angle DA^{*}P = \\angle MA^{*}P = \\angle A^{*}PM = \\angle IMP\\) . \n\nClaim — Points \\(P\\) , \\(M\\) , \\(I\\) , \\(Q^{*}\\) are cyclic. \n\nProof. \\(\\angle MQ^{*}P = \\angle C^{*}Q^{*}P = \\angle C^{*}EP = \\angle DEP = \\angle DGP = \\angle GPI = \\angle MIP\\) . \n\nFifth solution by double inversion (Brandon Wang, Luke Robitaille, Michael Ren, Evan Chen). We outline one final approach. After inverting about \\(\\omega\\) as in the previous approach, we then apply another inversion around \\(P\\) . Dropping the apostrophes/stars/etc now one can check that the problem we arrive at becomes the following. \n\n## Proposition (Doubly inverted problem) \n\nIn \\(\\triangle PEF\\) , the \\(P\\) - symmedian meets \\(\\overline{EF}\\) and \\((PEF)\\) at \\(K\\) , \\(L\\) . Let \\(D \\in \\overline{EF}\\) with \\(\\angle DPK = 90^{\\circ}\\) , and let \\(T\\) be the foot from \\(K\\) to \\(\\overline{DL}\\) . Denote by \\(I\\) the reflection of \\(P\\) about \\(\\overline{EF}\\) . Finally, let \\(PDNE\\) and \\(PDMF\\) be cyclic harmonic quadrilaterals. Then lines \\(EN\\) , \\(MF\\) , \\(TI\\) , are concurrent.\n\n\n\nThe proof proceeds in three steps. Suppose the line through \\(L\\) perpendicular to \\(\\overline{EF}\\) meets \\(\\overline{EF}\\) at \\(W\\) and \\((PEF)\\) at \\(Z\\) . \n\n\n \n\n1. Since \\(\\angle ZEP = \\angle WLP = \\angle WDP\\) , it follows \\(\\overline{ZE}\\) is tangent to \\((PDNE)\\) . Similarly, \\(\\overline{ZF}\\) is tangent to \\((PDMF)\\) . \n\n2. \\(\\Delta WTP\\) is the orthic triangle of \\(\\triangle DKL\\) , so \\(\\overline{WD}\\) bisects \\(\\angle PWT\\) and \\(\\overline{WTI}\\) collinear. \n\n3. \\(-1 = E(PN; DZ) = F(PM; DZ) = W(PI; DZ)\\) , so \\(\\overline{EN}\\) , \\(\\overline{FM}\\) , \\(\\overline{WI}\\) meet on \\(\\overline{PZ}\\) .", "metadata": {"resource_path": "IMO/segmented/en-IMO-2019-notes.jsonl", "problem_match": "6. ", "solution_match": "## \\(\\S 2.3\\) IMO 2019/6, proposed by Anant Mudgal (IND) \n"}}
diff --git a/IMO/segmented/en-IMO-2020-notes.jsonl b/IMO/segmented/en-IMO-2020-notes.jsonl
index 107905eb58e0c7a89d65a34b65fc81a4780da9da..c36d6e406fa20fedfe8dd60c031a34306d387539 100644
--- a/IMO/segmented/en-IMO-2020-notes.jsonl
+++ b/IMO/segmented/en-IMO-2020-notes.jsonl
@@ -1,6 +1,6 @@
-{"year": "2020", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "Consider the convex quadrilateral \\(ABCD\\) . The point \\(P\\) is in the interior of \\(ABCD\\) . The following ratio equalities hold: \n\n\\[\\angle PAD:\\angle PBA:\\angle DPA = 1:2:3 = \\angle CBP:\\angle BAP:\\angle BPC.\\] \n\nProve that the following three lines meet in a point: the internal bisectors of angles \\(\\angle ADP\\) and \\(\\angle PCB\\) and the perpendicular bisector of segment \\(AB\\) .", "solution": "Let \\(O\\) denote the circumcenter of \\(\\triangle PAB\\) . We claim it is the desired concurrency point. \n\n\n \n\nIndeed, \\(O\\) obviously lies on the perpendicular bisector of \\(AB\\) . Now \n\n\\[\\angle BCP = \\angle CBP + \\angle BPC\\] \\[\\qquad = 2\\angle BAP = \\angle BOP\\] \n\nit follows \\(BOPC\\) are cyclic. And since \\(OP = OB\\) , it follows that \\(O\\) is on the bisector of \\(\\angle PCB\\) , as needed. \n\nRemark. The angle equality is only used insomuch \\(\\angle BAP\\) is the average of \\(\\angle CBP\\) and \\(\\angle BPC\\) , i.e. only \\(\\frac{1 + 3}{2} = 2\\) matters.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2020-notes.jsonl", "problem_match": "1. ", "solution_match": "## \\(\\S 1.1\\) IMO 2020/1, proposed by Dominik Burek (POL) \n"}}
+{"year": "2020", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "Consider the convex quadrilateral \\(ABCD\\) . The point \\(P\\) is in the interior of \\(ABCD\\) . The following ratio equalities hold: \n\n\\[\\angle PAD:\\angle PBA:\\angle DPA = 1:2:3 = \\angle CBP:\\angle BAP:\\angle BPC.\\] \n\nProve that the following three lines meet in a point: the internal bisectors of angles \\(\\angle ADP\\) and \\(\\angle PCB\\) and the perpendicular bisector of segment \\(AB\\) .", "solution": "Let \\(O\\) denote the circumcenter of \\(\\triangle PAB\\) . We claim it is the desired concurrency point. \n\n\n \n\nIndeed, \\(O\\) obviously lies on the perpendicular bisector of \\(AB\\) . Now \n\n\\[\\angle BCP = \\angle CBP + \\angle BPC\\] \\[\\qquad = 2\\angle BAP = \\angle BOP\\] \n\nit follows \\(BOPC\\) are cyclic. And since \\(OP = OB\\) , it follows that \\(O\\) is on the bisector of \\(\\angle PCB\\) , as needed. \n\nRemark. The angle equality is only used insomuch \\(\\angle BAP\\) is the average of \\(\\angle CBP\\) and \\(\\angle BPC\\) , i.e. only \\(\\frac{1 + 3}{2} = 2\\) matters.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2020-notes.jsonl", "problem_match": "1. ", "solution_match": "## \\(\\S 1.1\\) IMO 2020/1, proposed by Dominik Burek (POL) \n"}}
{"year": "2020", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "Let \\(a \\geq b \\geq c \\geq d > 0\\) be real numbers satisfying \\(a + b + c + d = 1\\) . Prove that \n\n\\[(a + 2b + 3c + 4d)a^{a}b^{b}c^{c}d^{d}< 1.\\]", "solution": "By weighted AM- GM we have \n\n\\[a^{a}b^{b}c^{c}d^{d}\\leq \\sum_{\\mathrm{cyc}}\\frac{a}{a + b + c + d}\\cdot a = a^{2} + b^{2} + c^{2} + d^{2}.\\] \n\nSo, it is enough to prove that \n\n\\[(a^{2} + b^{2} + c^{2} + d^{2})(a + 2b + 3c + 4d)\\leq 1 = (a + b + c + d)^{3}.\\] \n\nExpand both sides to get \n\n\\[+a^{3} + b^{2}a +c^{2}a +d^{2}a +a^{3} + 3b^{2}a +3c^{2}a +3d^{2}a\\] \\[+2a^{2}b +2b^{3} + 2bc^{2} + 2d^{2}b +3a^{2}b +b^{3} + 3bc^{2} + 3d^{2}b\\] \\[+3a^{2}c +3b^{2}c +3c^{3} + 3d^{2}c +3a^{2}c +3b^{2}c +c^{3} + 3d^{2}c\\] \\[+4a^{2}d +4b^{2}d +4c^{2}d +4d^{3} + 6abc +6bcd +6cda +6dab.\\] \n\nIn other words, we need to prove that \n\n\\[+b^{3} + 2b^{2}a +2c^{2}a +2d^{2}a\\] \\[+a^{2}b +b c^{2} + d^{2}b\\] \\[+a^{2}d +b^{2}d +c^{2}d +3d^{3} + 6abc +6bcd +6cda +6dab\\] \n\nThis follows since \n\n\\[2b^{2}a\\geq b^{3} + b^{2}d\\] \\[2c^{2}a\\geq 2c^{3}\\] \\[2d^{2}a\\geq 2d^{3}\\] \\[a^{2}b\\geq a^{2}d\\] \\[bc^{2}\\geq c^{2}d\\] \\[d^{2}b\\geq d^{3}\\] \n\nand \\(6(abc + bcd + cda + dab) > 0\\) \n\nRemark. Some students complained this problem was \"unfair\" and they couldn't solve it because they didn't think an IMO problem would be solved only by expansion. I don't agree with this. Fedor Petrov provides the following motivational comments for why the existence of this solution should not be that surprising: \n\nBetter to think about mathematics. You have to bound from above a product \\((a + 2b + 3c + 4d)(a^{2} + b^{2} + c^{2} + d^{2})\\) , the coefficients \\(1,2,3,4\\) are increasing and so play on your side, so plausibly \\((a + b + c + d)^{3}\\) should majorize this term- wise, you check it and this appears to be true.\n\n\n\nHe also gave the following advice: \n\nThe general advice is to study mathematics, not olympiad problems in past years. If the IMO problems set the students and their teachers on this path, I am more than satisfied.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2020-notes.jsonl", "problem_match": "2. ", "solution_match": "## \\(\\S 1.2\\) IMO 2020/2, proposed by Belarus \n"}}
-{"year": "2020", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "There are \\(4n\\) pebbles of weights \\(1,2,3,\\ldots ,4n\\) . Each pebble is coloured in one of \\(n\\) colours and there are four pebbles of each colour. Show that we can arrange the pebbles into two piles so the total weights of both piles are the same, and each pile contains two pebbles of each colour.", "solution": "The first key idea is the deep fact that \n\n\\[1 + 4n = 2 + (4n - 1) = 3 + (4n - 2) = \\dots .\\] \n\nSo, place all four pebbles of the same colour in a box (hence \\(n\\) boxes). For each \\(k = 1,2,\\ldots ,2n\\) we tape a piece of string between pebble \\(k\\) and \\(4n + 1 - k\\) . To solve the problem, it suffices to paint each string either blue or green such that each box has two blue strings and two green strings (where a string between two pebbles in the same box counts double). \n\n\n \n\nWe can therefore rephrase the problem as follows, if we view boxes as vertices and strings as edges:\n\n\n\nClaim — Given a 4- regular multigraph on \\(n\\) vertices (where self- loops are allowed and have degree 2), one can color the edges blue and green such that each vertex has two blue and two green edges. \n\nProof. Each connected component of the graph can be decomposed into an Eulerian circuit, since 4 is even. A connected component with \\(k\\) vertices has \\(2k\\) edges in its Eulerian circuit, so we may color the edges in this circuit alternating green and blue. This may be checked to work. \\(\\square\\)\n\n\n\n## \\(\\S 2\\) Solutions to Day 2", "metadata": {"resource_path": "IMO/segmented/en-IMO-2020-notes.jsonl", "problem_match": "3. ", "solution_match": "## \\(\\S 1.3\\) IMO 2020/3, proposed by Milan Haiman (HUN), Carl Schildkraut (USA) \n"}}
-{"year": "2020", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "There is an integer \\(n > 1\\) . There are \\(n^2\\) stations on a slope of a mountain, all at different altitudes. Each of two cable car companies, \\(A\\) and \\(B\\) , operates \\(k\\) cable cars; each cable car provides a transfer from one of the stations to a higher one (with no intermediate stops). The \\(k\\) cable cars of \\(A\\) have \\(k\\) different starting points and \\(k\\) different finishing points, and a cable car which starts higher also finishes higher. The same conditions hold for \\(B\\) . We say that two stations are linked by a company if one can start from the lower station and reach the higher one by using one or more cars of that company (no other movements between stations are allowed). Determine the smallest positive integer \\(k\\) for which one can guarantee that there are two stations that are linked by both companies.", "solution": "Answer: \\(k = n^2 - n + 1\\) \n\nWhen \\(k = n^2 - n\\) , the construction for \\(n = 4\\) is shown below which generalizes readily. (We draw \\(A\\) in red and \\(B\\) in blue.) \n\n\n \n\nTo see this is sharp, view \\(A\\) and \\(B\\) as graphs whose connected components are paths (possibly with 0 edges; the direction of these edges is irrelevant). Now, if \\(k = n^2 - n + 1\\) it follows that \\(A\\) and \\(B\\) each have exactly \\(n - 1\\) connected components. \n\nBut in particular some component of \\(A\\) has at least \\(n + 1\\) vertices. This component has two vertices in the same component of \\(B\\) , as desired. \n\nRemark. The main foothold for this problem is the hypothesis that the number of stations should be \\(n^2\\) rather than, say, \\(n\\) . This gives a big hint towards finding the construction which in turn shows how the bound can be computed. \n\nOn the other hand, the hypothesis that \"a cable car which starts higher also finishes\n\n\n\nhigher\" appears to be superfluous.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2020-notes.jsonl", "problem_match": "4. ", "solution_match": "## \\(\\S 2.1\\) IMO 2020/4, proposed by Tejaswi Navilarekallu (IND) \n"}}
+{"year": "2020", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "There are \\(4n\\) pebbles of weights \\(1,2,3,\\ldots ,4n\\) . Each pebble is coloured in one of \\(n\\) colours and there are four pebbles of each colour. Show that we can arrange the pebbles into two piles so the total weights of both piles are the same, and each pile contains two pebbles of each colour.", "solution": "The first key idea is the deep fact that \n\n\\[1 + 4n = 2 + (4n - 1) = 3 + (4n - 2) = \\dots .\\] \n\nSo, place all four pebbles of the same colour in a box (hence \\(n\\) boxes). For each \\(k = 1,2,\\ldots ,2n\\) we tape a piece of string between pebble \\(k\\) and \\(4n + 1 - k\\) . To solve the problem, it suffices to paint each string either blue or green such that each box has two blue strings and two green strings (where a string between two pebbles in the same box counts double). \n\n\n \n\nWe can therefore rephrase the problem as follows, if we view boxes as vertices and strings as edges:\n\n\n\nClaim — Given a 4- regular multigraph on \\(n\\) vertices (where self- loops are allowed and have degree 2), one can color the edges blue and green such that each vertex has two blue and two green edges. \n\nProof. Each connected component of the graph can be decomposed into an Eulerian circuit, since 4 is even. A connected component with \\(k\\) vertices has \\(2k\\) edges in its Eulerian circuit, so we may color the edges in this circuit alternating green and blue. This may be checked to work. \\(\\square\\)\n\n\n\n## \\(\\S 2\\) Solutions to Day 2", "metadata": {"resource_path": "IMO/segmented/en-IMO-2020-notes.jsonl", "problem_match": "3. ", "solution_match": "## \\(\\S 1.3\\) IMO 2020/3, proposed by Milan Haiman (HUN), Carl Schildkraut (USA) \n"}}
+{"year": "2020", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "There is an integer \\(n > 1\\) . There are \\(n^2\\) stations on a slope of a mountain, all at different altitudes. Each of two cable car companies, \\(A\\) and \\(B\\) , operates \\(k\\) cable cars; each cable car provides a transfer from one of the stations to a higher one (with no intermediate stops). The \\(k\\) cable cars of \\(A\\) have \\(k\\) different starting points and \\(k\\) different finishing points, and a cable car which starts higher also finishes higher. The same conditions hold for \\(B\\) . We say that two stations are linked by a company if one can start from the lower station and reach the higher one by using one or more cars of that company (no other movements between stations are allowed). Determine the smallest positive integer \\(k\\) for which one can guarantee that there are two stations that are linked by both companies.", "solution": "Answer: \\(k = n^2 - n + 1\\) \n\nWhen \\(k = n^2 - n\\) , the construction for \\(n = 4\\) is shown below which generalizes readily. (We draw \\(A\\) in red and \\(B\\) in blue.) \n\n\n \n\nTo see this is sharp, view \\(A\\) and \\(B\\) as graphs whose connected components are paths (possibly with 0 edges; the direction of these edges is irrelevant). Now, if \\(k = n^2 - n + 1\\) it follows that \\(A\\) and \\(B\\) each have exactly \\(n - 1\\) connected components. \n\nBut in particular some component of \\(A\\) has at least \\(n + 1\\) vertices. This component has two vertices in the same component of \\(B\\) , as desired. \n\nRemark. The main foothold for this problem is the hypothesis that the number of stations should be \\(n^2\\) rather than, say, \\(n\\) . This gives a big hint towards finding the construction which in turn shows how the bound can be computed. \n\nOn the other hand, the hypothesis that \"a cable car which starts higher also finishes\n\n\n\nhigher\" appears to be superfluous.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2020-notes.jsonl", "problem_match": "4. ", "solution_match": "## \\(\\S 2.1\\) IMO 2020/4, proposed by Tejaswi Navilarekallu (IND) \n"}}
{"year": "2020", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "A deck of \\(n > 1\\) cards is given. A positive integer is written on each card. The deck has the property that the arithmetic mean of the numbers on each pair of cards is also the geometric mean of the numbers on some collection of one or more cards. For which \\(n\\) does it follow that the numbers on the cards are all equal?", "solution": "The assertion is true for all \\(n\\) . \n\nSetup (boilerplate). Suppose that \\(a_{1},\\ldots ,a_{n}\\) satisfy the required properties but are not all equal. Let \\(d = \\gcd (a_{1},\\ldots ,a_{n}) > 1\\) then replace \\(a_{1},\\ldots ,a_{n}\\) by \\(\\frac{a_{1}}{d},\\ldots ,\\frac{a_{n}}{d}\\) . Hence without loss of generality we may assume \n\n\\[\\gcd (a_{1},a_{2},\\ldots ,a_{n}) = 1.\\] \n\nWLOG we also assume \n\n\\[a_{1}\\geq a_{2}\\geq \\dots \\geq a_{n}.\\] \n\nMain proof. As \\(a_{1}\\geq 2\\) , let \\(p\\) be a prime divisor of \\(a_{1}\\) . Let \\(k\\) be smallest index such that \\(p\\nmid a_{k}\\) (which must exist). In particular, note that \\(a_{1}\\neq a_{k}\\) . \n\nConsider the mean \\(x = \\frac{a_{1} + a_{k}}{2}\\) ; by assumption, it equals some geometric mean, hence \n\n\\[\\frac{n}{\\sqrt[n]{a_{i_{1}}\\cdot\\cdot\\cdot a_{i_{m}}}} = \\frac{a_{1} + a_{k}}{2} >a_{k}.\\] \n\nSince the arithmetic mean is an integer not divisible by \\(p\\) , all the indices \\(i_{1},i_{2},\\ldots ,i_{m}\\) must be at least \\(k\\) . But then the GM is at most \\(a_{k}\\) , contradiction. \n\nRemark. A similar approach could be attempted by using the smallest numbers rather than the largest ones, but one must then handle the edge case \\(a_{n} = 1\\) separately since no prime divides 1. \n\nRemark. Since \\(\\frac{27 + 9}{2} = 18 = \\sqrt[3]{27\\cdot 27\\cdot 8}\\) , it is not true that in general the AM of two largest different cards is not the GM of other numbers in the sequence (say the cards are \\(27,27,9,8,\\ldots)\\) .", "metadata": {"resource_path": "IMO/segmented/en-IMO-2020-notes.jsonl", "problem_match": "5. ", "solution_match": "## \\(\\S 2.2\\) IMO 2020/5, proposed by Oleg Kosik (EST) \n"}}
-{"year": "2020", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "Consider an integer \\(n > 1\\) , and a set \\(S\\) of \\(n\\) points in the plane such that the distance between any two different points in \\(S\\) is at least 1. Prove there is a line \\(\\ell\\) separating \\(S\\) such that the distance from any point of \\(S\\) to \\(\\ell\\) is at least \\(\\Omega (n^{-1 / 3})\\) . \n\n(A line \\(\\ell\\) separates a set of points \\(S\\) if some segment joining two points in \\(S\\) crosses \\(\\ell\\) .)", "solution": "C \n\nWe present the official solution given by the Problem Selection Committee. \n\nLet's suppose that among all projections of points in \\(S\\) onto some line \\(m\\) , the maximum possible distance between two consecutive projections is \\(\\delta\\) . We will prove that \\(\\delta \\geq \\Omega (n^{- 1 / 3})\\) , solving the problem. \n\nWe make the following the definitions: \n\n- Define \\(A\\) and \\(B\\) as the two points farthest apart in \\(S\\) . This means that all points lie in the intersections of the circles centered at \\(A\\) and \\(B\\) with radius \\(R = AB \\geq 1\\) .- We pick chord \\(\\overline{XY}\\) of \\(\\odot (B)\\) such that \\(\\overline{XY} \\perp \\overline{AB}\\) and the distance from \\(A\\) to \\(\\overline{XY}\\) is exactly \\(\\frac{1}{2}\\) .- We denote by \\(\\mathcal{T}\\) the smaller region bound by \\(\\odot (B)\\) and chord \\(\\overline{XY}\\) . \n\nThe figure is shown below with \\(\\mathcal{T}\\) drawn in yellow, and points of \\(S\\) drawn in blue. \n\n\n\n\n\n\nClaim (Length of \\(AB + \\text{Pythagorean theorem}\\) ) — We have \\(XY < 2\\sqrt{nd}\\) . \n\nProof. First, note that we have \\(R = AB < (n - 1) \\cdot \\delta\\) , since the \\(n\\) projections of points onto \\(AB\\) are spaced at most \\(\\delta\\) apart. The Pythagorean theorem gives \n\n\\[XY = 2\\sqrt{R^{2} - \\left(R - \\frac{1}{2}\\right)^{2}} = 2\\sqrt{R - \\frac{1}{4}} < 2\\sqrt{nd}.\\] \n\nClaim ( \\(|T|\\) lower bound + narrowness) — We have \\(XY > \\frac{\\sqrt{3}}{2} \\left(\\frac{1}{2}\\delta^{- 1} - 1\\right)\\) . \n\nProof. Because \\(\\mathcal{T}\\) is so narrow (has width \\(\\frac{1}{2}\\) only), the projections of points in \\(\\mathcal{T}\\) onto line \\(XY\\) are spaced at least \\(\\frac{\\sqrt{3}}{2}\\) apart (more than just \\(\\delta\\) ). This means \n\n\\[XY > \\frac{\\sqrt{3}}{2} (|T| - 1).\\] \n\nBut projections of points in \\(\\mathcal{T}\\) onto the segment of length \\(\\frac{1}{2}\\) are spaced at most \\(\\delta\\) apart, so apparently \n\n\\[|T| > \\frac{1}{2} \\cdot \\delta^{-1}.\\] \n\nThis implies the result. \n\nCombining these two this implies \\(\\delta \\geq \\Omega (n^{- 1 / 3})\\) as needed. \n\nRemark. The constant \\(1 / 3\\) in the problem is actually optimal and cannot be improved; the constructions give an example showing \\(\\Theta (n^{- 1 / 3} \\log n)\\) .", "metadata": {"resource_path": "IMO/segmented/en-IMO-2020-notes.jsonl", "problem_match": "6. ", "solution_match": "## \\(\\S 2.3\\) IMO 2020/6, proposed by Ting-Feng Lin, Hung-Hsun Hans Yu (TWN) \n"}}
+{"year": "2020", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "Consider an integer \\(n > 1\\) , and a set \\(S\\) of \\(n\\) points in the plane such that the distance between any two different points in \\(S\\) is at least 1. Prove there is a line \\(\\ell\\) separating \\(S\\) such that the distance from any point of \\(S\\) to \\(\\ell\\) is at least \\(\\Omega (n^{-1 / 3})\\) . \n\n(A line \\(\\ell\\) separates a set of points \\(S\\) if some segment joining two points in \\(S\\) crosses \\(\\ell\\) .)", "solution": "C \n\nWe present the official solution given by the Problem Selection Committee. \n\nLet's suppose that among all projections of points in \\(S\\) onto some line \\(m\\) , the maximum possible distance between two consecutive projections is \\(\\delta\\) . We will prove that \\(\\delta \\geq \\Omega (n^{- 1 / 3})\\) , solving the problem. \n\nWe make the following the definitions: \n\n- Define \\(A\\) and \\(B\\) as the two points farthest apart in \\(S\\) . This means that all points lie in the intersections of the circles centered at \\(A\\) and \\(B\\) with radius \\(R = AB \\geq 1\\) .- We pick chord \\(\\overline{XY}\\) of \\(\\odot (B)\\) such that \\(\\overline{XY} \\perp \\overline{AB}\\) and the distance from \\(A\\) to \\(\\overline{XY}\\) is exactly \\(\\frac{1}{2}\\) .- We denote by \\(\\mathcal{T}\\) the smaller region bound by \\(\\odot (B)\\) and chord \\(\\overline{XY}\\) . \n\nThe figure is shown below with \\(\\mathcal{T}\\) drawn in yellow, and points of \\(S\\) drawn in blue. \n\n\n\n\n\n\nClaim (Length of \\(AB + \\text{Pythagorean theorem}\\) ) — We have \\(XY < 2\\sqrt{nd}\\) . \n\nProof. First, note that we have \\(R = AB < (n - 1) \\cdot \\delta\\) , since the \\(n\\) projections of points onto \\(AB\\) are spaced at most \\(\\delta\\) apart. The Pythagorean theorem gives \n\n\\[XY = 2\\sqrt{R^{2} - \\left(R - \\frac{1}{2}\\right)^{2}} = 2\\sqrt{R - \\frac{1}{4}} < 2\\sqrt{nd}.\\] \n\nClaim ( \\(|T|\\) lower bound + narrowness) — We have \\(XY > \\frac{\\sqrt{3}}{2} \\left(\\frac{1}{2}\\delta^{- 1} - 1\\right)\\) . \n\nProof. Because \\(\\mathcal{T}\\) is so narrow (has width \\(\\frac{1}{2}\\) only), the projections of points in \\(\\mathcal{T}\\) onto line \\(XY\\) are spaced at least \\(\\frac{\\sqrt{3}}{2}\\) apart (more than just \\(\\delta\\) ). This means \n\n\\[XY > \\frac{\\sqrt{3}}{2} (|T| - 1).\\] \n\nBut projections of points in \\(\\mathcal{T}\\) onto the segment of length \\(\\frac{1}{2}\\) are spaced at most \\(\\delta\\) apart, so apparently \n\n\\[|T| > \\frac{1}{2} \\cdot \\delta^{-1}.\\] \n\nThis implies the result. \n\nCombining these two this implies \\(\\delta \\geq \\Omega (n^{- 1 / 3})\\) as needed. \n\nRemark. The constant \\(1 / 3\\) in the problem is actually optimal and cannot be improved; the constructions give an example showing \\(\\Theta (n^{- 1 / 3} \\log n)\\) .", "metadata": {"resource_path": "IMO/segmented/en-IMO-2020-notes.jsonl", "problem_match": "6. ", "solution_match": "## \\(\\S 2.3\\) IMO 2020/6, proposed by Ting-Feng Lin, Hung-Hsun Hans Yu (TWN) \n"}}
diff --git a/IMO/segmented/en-IMO-2021-notes.jsonl b/IMO/segmented/en-IMO-2021-notes.jsonl
index fae40e253e69ce98fbd2816099cc4dfcaf70e7f5..61ac4e632c296a229ec5a746029b429c365f6ccb 100644
--- a/IMO/segmented/en-IMO-2021-notes.jsonl
+++ b/IMO/segmented/en-IMO-2021-notes.jsonl
@@ -1,6 +1,6 @@
{"year": "2021", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "Let \\(n \\geq 100\\) be an integer. Ivan writes the numbers \\(n, n + 1, \\ldots , 2n\\) each on different cards. He then shuffles these \\(n + 1\\) cards, and divides them into two piles. Prove that at least one of the piles contains two cards such that the sum of their numbers is a perfect square.", "solution": "We will find three cards \\(a< b< c\\) such that \n\n\\[b + c = (2k + 1)^2\\] \\[c + a = (2k)^2\\] \\[a + b = (2k - 1)^2\\] \n\nfor some integer \\(k\\) . Solving for \\(a\\) , \\(b\\) , \\(c\\) gives \n\n\\[a = \\frac{(2k)^2 + (2k - 1)^2 - (2k + 1)^2}{2} = 2k^2 -4k\\] \\[b = \\frac{(2k + 1)^2 + (2k - 1)^2 - (2k)^2}{2} = 2k^2 +1\\] \\[c = \\frac{(2k + 1)^2 + (2k)^2 - (2k - 1)^2}{2} = 2k^2 +4k\\] \n\nWe need to show that when \\(n\\geq 100\\) , one can find a suitable \\(k\\) . \n\nLet \n\n\\[I_{k}:= \\{n\\in \\mathbb{Z}\\mid n\\leq a< b< c\\leq 2n\\}\\] \\[\\qquad = \\{n\\in \\mathbb{Z}\\mid k^{2} + 2k\\leq n\\leq 2k^{2} - 4k\\}\\] \n\nbe the interval such that when \\(n\\in I_{k}\\) , the problem dies for that choice of \\(k\\) . It would be sufficient to show these intervals \\(I_{k}\\) cover all the integers \\(\\geq 100\\) . Starting from \\(I_{9} = \\{99\\leq n\\leq 126\\}\\) , we have \n\n\\[k\\geq 9\\Longrightarrow 2k^{2} - 4k\\geq (k + 1)^{2} + 2(k + 1)\\] \n\nwhich means the right endpoint of \\(I_{k}\\) exceeds the left endpoint of \\(I_{k + 1}\\) . Hence for \\(n\\geq 99\\) in fact the problem is true. \n\nRemark. The problem turns out to be false for \\(n = 98\\) , surprisingly. The counterexample is for one pile to be \n\n\\[\\{98,100,102,\\ldots ,126\\} \\cup \\{129,131,135,\\ldots ,161\\} \\cup \\{162,164,\\ldots ,196\\} .\\]", "metadata": {"resource_path": "IMO/segmented/en-IMO-2021-notes.jsonl", "problem_match": "1. ", "solution_match": "## \\(\\S 1.1\\) IMO 2021/1, proposed by Australia \n"}}
{"year": "2021", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "Show that the inequality \n\n\\[\\sum_{i = 1}^{n}\\sum_{j = 1}^{n}\\sqrt{|x_{i} - x_{j}|}\\leq \\sum_{i = 1}^{n}\\sum_{j = 1}^{n}\\sqrt{|x_{i} + x_{j}|}\\] \n\nholds for all real numbers \\(x_{1}, x_{2}, \\ldots , x_{n}\\) .", "solution": "Sho) . \n\nThe proof is by induction on \\(n \\geq 1\\) with the base cases \\(n = 1\\) and \\(n = 2\\) being easy to verify by hand. \n\nIn the general situation, consider replacing the tuple \\((x_{i})_{i}\\) with \\((x_{i} + t)_{i}\\) for some parameter \\(t \\in \\mathbb{R}\\) . The inequality becomes \n\n\\[\\sum_{i = 1}^{n}\\sum_{j = 1}^{n}\\sqrt{|x_{i} - x_{j}|}\\leq \\sum_{i = 1}^{n}\\sum_{j = 1}^{n}\\sqrt{|x_{i} + x_{j} + 2t|}.\\] \n\nThe left- hand side is independent of \\(t\\) . \n\nClaim — The right- hand side, viewed as a function \\(F(t)\\) of \\(t\\) , is minimized when \\(2t = -(x_{i} + x_{j})\\) for some \\(i\\) and \\(j\\) . \n\nProof. Since \\(F(t)\\) is the sum of piecewise concave functions, it is hence itself piecewise concave. Moreover \\(F\\) increases without bound if \\(|t| \\to \\infty\\) . \n\nOn each of the finitely many intervals on which \\(F(t)\\) is concave, the function is minimized at its endpoints. Hence the minimum value must occur at one of the endpoints. \n\nIf \\(t = - x_{i}\\) for some \\(i\\) , this is the same as shifting all the variables so that \\(x_{i} = 0\\) . In that case, we may apply induction on \\(n - 1\\) variables, deleting the variable \\(x_{i}\\) . \n\nIf \\(t = - \\frac{x_{i} + x_{j}}{2}\\) , then notice \n\n\\[x_{i} + t = -(x_{j} + t)\\] \n\nso it's the same as shifting all the variables such that \\(x_{i} = - x_{j}\\) . In that case, we may apply induction on \\(n - 2\\) variables, after deleting \\(x_{i}\\) and \\(x_{j}\\) .", "metadata": {"resource_path": "IMO/segmented/en-IMO-2021-notes.jsonl", "problem_match": "2. ", "solution_match": "## \\(\\S 1.2\\) IMO 2021/2, proposed by Calvin Deng \n"}}
-{"year": "2021", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "Let \\(D\\) be an interior point of the acute triangle \\(ABC\\) with \\(AB > AC\\) so that \\(\\angle DAB = \\angle CAD\\) . The point \\(E\\) on the segment \\(AC\\) satisfies \\(\\angle ADE = \\angle BCD\\) , the point \\(F\\) on the segment \\(AB\\) satisfies \\(\\angle FDA = \\angle DBC\\) , and the point \\(X\\) on the line \\(AC\\) satisfies \\(CX = BX\\) . Let \\(O_{1}\\) and \\(O_{2}\\) be the circumcenters of the triangles \\(ADC\\) and \\(EXD\\) , respectively. Prove that the lines \\(BC\\) , \\(EF\\) , and \\(O_{1}O_{2}\\) are concurrent.", "solution": "This solution was contributed by Abdullahi Kafi. \n\nClaim — Quadrilateral \\(BCEF\\) is cyclic. \n\nProof. Let \\(D'\\) be the isogonal conjugate of the point \\(D\\) . The angle condition implies quadrilateral \\(CEDD'\\) and \\(BFDD'\\) are cyclic. By power of point we have \n\n\\[AE\\cdot AC = AD\\cdot AD' = AF\\cdot AB\\] \n\nSo \\(BCEF\\) is cyclic. \n\nClaim — Line \\(ZD\\) is tangent to the circles \\((BCD)\\) and \\((DEF)\\) where \\(Z = EF\\cap BC\\) . \n\nProof. Let \\(\\angle CAD = \\angle BAD = \\alpha\\) , \\(\\angle BCD = \\beta\\) , \\(\\angle DBC = \\gamma\\) , \\(\\angle ACD = \\phi\\) , \\(\\angle ABD = \\epsilon\\) . From \\(\\triangle ABC\\) we have \\(2\\alpha + \\beta + \\gamma + \\phi + \\epsilon = 180^{\\circ}\\) . Let \\(\\ell\\) be a line tangent to \\((BCD)\\) and \\(K\\) be a point on it in the same side of \\(AD\\) as \\(C\\) and \\(L = AD\\cap BC\\) . From our labeling we have, \n\n\\[\\angle AFE = \\beta +\\phi \\qquad \\angle BFD = \\alpha +\\gamma \\qquad \\angle DFE = \\alpha +\\phi \\qquad \\angle CDL = \\alpha +\\phi\\] \n\nNow \\(\\angle CDJ = 180^{\\circ} - \\gamma - \\beta - (\\alpha +\\phi) = \\alpha +\\epsilon\\) . So \\(\\angle DFE = \\angle EDK = \\alpha +\\epsilon\\) , which means \\(\\ell\\) is also tangent to \\((DEF)\\) . Now by the radical center theorem we have \\(\\ell\\) passes through \\(Z\\) . \\(\\square\\) \n\nLet \\(M\\) be the Miquel point of the cyclic quadrilateral \\(BCEF\\) . From the Miquel configuration we have \\(A\\) , \\(M\\) , \\(Z\\) are collinear and \\((AFEM)\\) , \\((ZCEM)\\) are cyclic. \n\nClaim — Points \\(B\\) , \\(X\\) , \\(M\\) , \\(E\\) are cyclic. \n\nProof. Notice that \\(\\angle EMB = 180^{\\circ} - \\angle AMB - \\angle EMZ = 180^{\\circ} - 2\\angle ACB = \\angle EXB\\) . \\(\\square\\) \n\nLet \\(N\\) be the other intersection of circles \\((ACD)\\) and \\((DEX)\\) and let \\(R\\) be the intersection of \\(AC\\) and \\(BM\\) .\n\n\n\n \n\nClaim — Points \\(B\\) , \\(D\\) , \\(M\\) , \\(N\\) are cyclic. \n\nProof. By power of point we have \n\n\\[\\operatorname {Pow}(R,(ACD)) = RC\\cdot RA = RM\\cdot RB = RE\\cdot RX = \\operatorname {Pow}(R,(DEX)).\\] \n\nHence \\(R\\) lies on the radical axis of \\((ACD)\\) and \\((DEX)\\) , so \\(N\\) , \\(R\\) , \\(D\\) are collinear. Also \n\n\\[R N\\cdot R D = R A\\cdot R C = R M\\cdot R B\\] \n\nSo \\(BDMN\\) is cyclic. \n\nNotice that \\((ACD)\\) , \\((BDMN)\\) , \\((DEX)\\) are coaxial so their centers are collinear. Now we just need to prove the centers of \\((ACD)\\) , \\((BDMN)\\) and \\(Z\\) are collinear. To prove this, take a circle \\(\\omega\\) with radius \\(ZD\\) centered at \\(Z\\) . Notice that by power of point \n\n\\[ZC\\cdot ZB = ZD^{2} = ZE\\cdot ZF = ZM\\cdot ZA\\] \n\nwhich means inversion circle \\(\\omega\\) swaps \\((ACD)\\) and \\((BDMN)\\) . So the centers of \\((ACD)\\) and \\((BDMN)\\) must have to be collinear with the center of inversion circle, as desired.\n\n\n\n## \\(\\S 2\\) Solutions to Day 2", "metadata": {"resource_path": "IMO/segmented/en-IMO-2021-notes.jsonl", "problem_match": "3. ", "solution_match": "## \\(\\S 1.3\\) IMO 2021/3, proposed by Mykhailo Shtandenko (UKR) \n"}}
-{"year": "2021", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "Let \\(\\Gamma\\) be a circle with center \\(I\\) , and \\(ABCD\\) a convex quadrilateral such that each of the segments \\(AB\\) , \\(BC\\) , \\(CD\\) and \\(DA\\) is tangent to \\(\\Gamma\\) . Let \\(\\Omega\\) be the circumcircle of the triangle \\(AIC\\) . The extension of \\(BA\\) beyond \\(A\\) meets \\(\\Omega\\) at \\(X\\) , and the extension of \\(BC\\) beyond \\(C\\) meets \\(\\Omega\\) at \\(Z\\) . The extensions of \\(AD\\) and \\(CD\\) beyond \\(D\\) meet \\(\\Omega\\) at \\(Y\\) and \\(T\\) , respectively. Prove that \n\n\\[AD + DT + TX + XA = CD + DY + YZ + ZC.\\]", "solution": "Let \\(PQR S\\) be the contact points of \\(\\Gamma\\) an \\(\\overline{AB}\\) , \\(\\overline{BC}\\) , \\(\\overline{CD}\\) , \\(\\overline{DA}\\) . \n\n\n \n\nClaim — We have \\(\\triangle IQZ \\cong \\triangle IRT\\) . Similarly, \\(\\triangle IPX \\cong \\triangle ISY\\) . \n\nProof. By considering \\((CQIR)\\) and \\((CITZ)\\) , there is a spiral similarity similarity mapping \\(\\triangle IQZ\\) to \\(\\triangle IRT\\) . Since \\(IQ = IR\\) , it is in fact a congruence. \\(\\square\\) \n\nThis congruence essentially solves the problem. First, it implies: \n\nClaim — \\(TX = YZ\\) . \n\nProof. Because we saw \\(IX = IY\\) and \\(IT = IZ\\) . \\(\\square\\)\n\n\n\nThen, we can compute \n\n\\[AD + DT + XA = AD + (RT - RD) + (XP - AP)\\] \\[\\qquad = (AD - RD - AP) + RT + XP = RT + XP\\] \n\nand \n\n\\[CD + DY + ZC = CD + (SY - SD) + (ZQ - QC)\\] \\[\\qquad = (CD - SD - QC) + SY + ZQ = SY + ZQ\\] \n\nbut \\(ZQ = RT\\) and \\(XP = SY\\) , as needed.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2021-notes.jsonl", "problem_match": "4. ", "solution_match": "## \\(\\S 2.1\\) IMO 2021/4, proposed by Dominik Burek (POL) and Tomasz Ciesla (POL) \n"}}
-{"year": "2021", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "Two squirrels, Bushy and Jumpy, have collected 2021 walnuts for the winter. Jumpy numbers the walnuts from 1 through 2021, and digs 2021 little holes in a circular pattern in the ground around their favourite tree. The next morning Jumpy notices that Bushy had placed one walnut into each hole, but had paid no attention to the numbering. Unhappy, Jumpy decides to reorder the walnuts by performing a sequence of 2021 moves. In the \\(k\\) th move, Jumpy swaps the positions of the two walnuts adjacent to walnut \\(k\\) . \n\nProve that there exists a value of \\(k\\) such that, on the \\(k\\) th move, Jumpy swaps some walnuts \\(a\\) and \\(b\\) such that \\(a < k < b\\) .", "solution": "Assume for contradiction no such \\(k\\) exists. We will use a so- called \"threshold trick\". \n\nThis process takes exactly 2021 steps. Right after the \\(k\\) th move, we consider a situation where we color walnut \\(k\\) red as well, so at the \\(k\\) th step there are \\(k\\) ones. For brevity, a non- red walnut is called black. An example is illustrated below with 2021 replaced by 6. \n\n\n \n\nClaim — At each step, the walnut that becomes red is between two non- red or two red walnuts. \n\nProof. By definition. \n\nOn the other hand, if there are 2021 walnuts, one obtains a parity obstruction to this simplified process: \n\nClaim — After the first step, there is always a consecutive block of black walnuts positive even length. \n\nProof. After the first step, there is a block of 2020 black walnuts.\n\n\n\nThereafter, note that a length 2 block of black walnuts can never be changed. Meanwhile for even lengths at least 4, if one places a red walnut inside it, the even length block splits into an odd length block and an even length block. \\(\\square\\) \n\nRemark. The statement is true with 2021 replaced by any odd number, and false for any even number. \n\nThe motivation comes from the following rephrasing of the problem: \n\nStart with all 0's and at each step change a 0 between two matching numbers from a 0 to a 1. \n\nAlthough the coloring (or 0/1) argument may appear to lose information at first, I think it should be equivalent to the original process; the \"extra\" information comes down to the choice of which walnut to color red at each step.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2021-notes.jsonl", "problem_match": "5. ", "solution_match": "## \\(\\S 2.2\\) IMO 2021/5, proposed by Spain \n"}}
+{"year": "2021", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "Let \\(D\\) be an interior point of the acute triangle \\(ABC\\) with \\(AB > AC\\) so that \\(\\angle DAB = \\angle CAD\\) . The point \\(E\\) on the segment \\(AC\\) satisfies \\(\\angle ADE = \\angle BCD\\) , the point \\(F\\) on the segment \\(AB\\) satisfies \\(\\angle FDA = \\angle DBC\\) , and the point \\(X\\) on the line \\(AC\\) satisfies \\(CX = BX\\) . Let \\(O_{1}\\) and \\(O_{2}\\) be the circumcenters of the triangles \\(ADC\\) and \\(EXD\\) , respectively. Prove that the lines \\(BC\\) , \\(EF\\) , and \\(O_{1}O_{2}\\) are concurrent.", "solution": "This solution was contributed by Abdullahi Kafi. \n\nClaim — Quadrilateral \\(BCEF\\) is cyclic. \n\nProof. Let \\(D'\\) be the isogonal conjugate of the point \\(D\\) . The angle condition implies quadrilateral \\(CEDD'\\) and \\(BFDD'\\) are cyclic. By power of point we have \n\n\\[AE\\cdot AC = AD\\cdot AD' = AF\\cdot AB\\] \n\nSo \\(BCEF\\) is cyclic. \n\nClaim — Line \\(ZD\\) is tangent to the circles \\((BCD)\\) and \\((DEF)\\) where \\(Z = EF\\cap BC\\) . \n\nProof. Let \\(\\angle CAD = \\angle BAD = \\alpha\\) , \\(\\angle BCD = \\beta\\) , \\(\\angle DBC = \\gamma\\) , \\(\\angle ACD = \\phi\\) , \\(\\angle ABD = \\epsilon\\) . From \\(\\triangle ABC\\) we have \\(2\\alpha + \\beta + \\gamma + \\phi + \\epsilon = 180^{\\circ}\\) . Let \\(\\ell\\) be a line tangent to \\((BCD)\\) and \\(K\\) be a point on it in the same side of \\(AD\\) as \\(C\\) and \\(L = AD\\cap BC\\) . From our labeling we have, \n\n\\[\\angle AFE = \\beta +\\phi \\qquad \\angle BFD = \\alpha +\\gamma \\qquad \\angle DFE = \\alpha +\\phi \\qquad \\angle CDL = \\alpha +\\phi\\] \n\nNow \\(\\angle CDJ = 180^{\\circ} - \\gamma - \\beta - (\\alpha +\\phi) = \\alpha +\\epsilon\\) . So \\(\\angle DFE = \\angle EDK = \\alpha +\\epsilon\\) , which means \\(\\ell\\) is also tangent to \\((DEF)\\) . Now by the radical center theorem we have \\(\\ell\\) passes through \\(Z\\) . \\(\\square\\) \n\nLet \\(M\\) be the Miquel point of the cyclic quadrilateral \\(BCEF\\) . From the Miquel configuration we have \\(A\\) , \\(M\\) , \\(Z\\) are collinear and \\((AFEM)\\) , \\((ZCEM)\\) are cyclic. \n\nClaim — Points \\(B\\) , \\(X\\) , \\(M\\) , \\(E\\) are cyclic. \n\nProof. Notice that \\(\\angle EMB = 180^{\\circ} - \\angle AMB - \\angle EMZ = 180^{\\circ} - 2\\angle ACB = \\angle EXB\\) . \\(\\square\\) \n\nLet \\(N\\) be the other intersection of circles \\((ACD)\\) and \\((DEX)\\) and let \\(R\\) be the intersection of \\(AC\\) and \\(BM\\) .\n\n\n\n \n\nClaim — Points \\(B\\) , \\(D\\) , \\(M\\) , \\(N\\) are cyclic. \n\nProof. By power of point we have \n\n\\[\\operatorname {Pow}(R,(ACD)) = RC\\cdot RA = RM\\cdot RB = RE\\cdot RX = \\operatorname {Pow}(R,(DEX)).\\] \n\nHence \\(R\\) lies on the radical axis of \\((ACD)\\) and \\((DEX)\\) , so \\(N\\) , \\(R\\) , \\(D\\) are collinear. Also \n\n\\[R N\\cdot R D = R A\\cdot R C = R M\\cdot R B\\] \n\nSo \\(BDMN\\) is cyclic. \n\nNotice that \\((ACD)\\) , \\((BDMN)\\) , \\((DEX)\\) are coaxial so their centers are collinear. Now we just need to prove the centers of \\((ACD)\\) , \\((BDMN)\\) and \\(Z\\) are collinear. To prove this, take a circle \\(\\omega\\) with radius \\(ZD\\) centered at \\(Z\\) . Notice that by power of point \n\n\\[ZC\\cdot ZB = ZD^{2} = ZE\\cdot ZF = ZM\\cdot ZA\\] \n\nwhich means inversion circle \\(\\omega\\) swaps \\((ACD)\\) and \\((BDMN)\\) . So the centers of \\((ACD)\\) and \\((BDMN)\\) must have to be collinear with the center of inversion circle, as desired.\n\n\n\n## \\(\\S 2\\) Solutions to Day 2", "metadata": {"resource_path": "IMO/segmented/en-IMO-2021-notes.jsonl", "problem_match": "3. ", "solution_match": "## \\(\\S 1.3\\) IMO 2021/3, proposed by Mykhailo Shtandenko (UKR) \n"}}
+{"year": "2021", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "Let \\(\\Gamma\\) be a circle with center \\(I\\) , and \\(ABCD\\) a convex quadrilateral such that each of the segments \\(AB\\) , \\(BC\\) , \\(CD\\) and \\(DA\\) is tangent to \\(\\Gamma\\) . Let \\(\\Omega\\) be the circumcircle of the triangle \\(AIC\\) . The extension of \\(BA\\) beyond \\(A\\) meets \\(\\Omega\\) at \\(X\\) , and the extension of \\(BC\\) beyond \\(C\\) meets \\(\\Omega\\) at \\(Z\\) . The extensions of \\(AD\\) and \\(CD\\) beyond \\(D\\) meet \\(\\Omega\\) at \\(Y\\) and \\(T\\) , respectively. Prove that \n\n\\[AD + DT + TX + XA = CD + DY + YZ + ZC.\\]", "solution": "Let \\(PQR S\\) be the contact points of \\(\\Gamma\\) an \\(\\overline{AB}\\) , \\(\\overline{BC}\\) , \\(\\overline{CD}\\) , \\(\\overline{DA}\\) . \n\n\n \n\nClaim — We have \\(\\triangle IQZ \\cong \\triangle IRT\\) . Similarly, \\(\\triangle IPX \\cong \\triangle ISY\\) . \n\nProof. By considering \\((CQIR)\\) and \\((CITZ)\\) , there is a spiral similarity similarity mapping \\(\\triangle IQZ\\) to \\(\\triangle IRT\\) . Since \\(IQ = IR\\) , it is in fact a congruence. \\(\\square\\) \n\nThis congruence essentially solves the problem. First, it implies: \n\nClaim — \\(TX = YZ\\) . \n\nProof. Because we saw \\(IX = IY\\) and \\(IT = IZ\\) . \\(\\square\\)\n\n\n\nThen, we can compute \n\n\\[AD + DT + XA = AD + (RT - RD) + (XP - AP)\\] \\[\\qquad = (AD - RD - AP) + RT + XP = RT + XP\\] \n\nand \n\n\\[CD + DY + ZC = CD + (SY - SD) + (ZQ - QC)\\] \\[\\qquad = (CD - SD - QC) + SY + ZQ = SY + ZQ\\] \n\nbut \\(ZQ = RT\\) and \\(XP = SY\\) , as needed.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2021-notes.jsonl", "problem_match": "4. ", "solution_match": "## \\(\\S 2.1\\) IMO 2021/4, proposed by Dominik Burek (POL) and Tomasz Ciesla (POL) \n"}}
+{"year": "2021", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "Two squirrels, Bushy and Jumpy, have collected 2021 walnuts for the winter. Jumpy numbers the walnuts from 1 through 2021, and digs 2021 little holes in a circular pattern in the ground around their favourite tree. The next morning Jumpy notices that Bushy had placed one walnut into each hole, but had paid no attention to the numbering. Unhappy, Jumpy decides to reorder the walnuts by performing a sequence of 2021 moves. In the \\(k\\) th move, Jumpy swaps the positions of the two walnuts adjacent to walnut \\(k\\) . \n\nProve that there exists a value of \\(k\\) such that, on the \\(k\\) th move, Jumpy swaps some walnuts \\(a\\) and \\(b\\) such that \\(a < k < b\\) .", "solution": "Assume for contradiction no such \\(k\\) exists. We will use a so- called \"threshold trick\". \n\nThis process takes exactly 2021 steps. Right after the \\(k\\) th move, we consider a situation where we color walnut \\(k\\) red as well, so at the \\(k\\) th step there are \\(k\\) ones. For brevity, a non- red walnut is called black. An example is illustrated below with 2021 replaced by 6. \n\n\n \n\nClaim — At each step, the walnut that becomes red is between two non- red or two red walnuts. \n\nProof. By definition. \n\nOn the other hand, if there are 2021 walnuts, one obtains a parity obstruction to this simplified process: \n\nClaim — After the first step, there is always a consecutive block of black walnuts positive even length. \n\nProof. After the first step, there is a block of 2020 black walnuts.\n\n\n\nThereafter, note that a length 2 block of black walnuts can never be changed. Meanwhile for even lengths at least 4, if one places a red walnut inside it, the even length block splits into an odd length block and an even length block. \\(\\square\\) \n\nRemark. The statement is true with 2021 replaced by any odd number, and false for any even number. \n\nThe motivation comes from the following rephrasing of the problem: \n\nStart with all 0's and at each step change a 0 between two matching numbers from a 0 to a 1. \n\nAlthough the coloring (or 0/1) argument may appear to lose information at first, I think it should be equivalent to the original process; the \"extra\" information comes down to the choice of which walnut to color red at each step.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2021-notes.jsonl", "problem_match": "5. ", "solution_match": "## \\(\\S 2.2\\) IMO 2021/5, proposed by Spain \n"}}
{"year": "2021", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "Let \\(m \\geq 2\\) be an integer, \\(A\\) a finite set of integers (not necessarily positive) and \\(B_{1}, B_{2}, \\ldots , B_{m}\\) subsets of \\(A\\) . Suppose that, for every \\(k = 1, 2, \\ldots , m\\) , the sum of the elements of \\(B_{k}\\) is \\(m^{k}\\) . Prove that \\(A\\) contains at least \\(\\frac{m}{2}\\) elements.", "solution": "L. \n\nIf \\(0\\leq X< m^{m + 1}\\) is a multiple of \\(m\\) , then write it in base \\(m\\) as \n\n\\[X = \\sum_{i = 1}^{m}c_{i}m^{i}\\qquad c_{i}\\in \\{0,1,2,\\ldots ,m - 1\\}\\] \n\nThen swapping the summation to over \\(A\\) through the \\(B_{i}\\) 's gives \n\n\\[X = \\sum_{i = 1}^{n}\\left(\\sum_{b\\in B_{i}}b\\right)c_{i} = \\sum_{a\\in A}f_{a}(X)a\\quad \\mathrm{where}\\quad f_{a}(X):= \\sum_{i:a\\in B_{i}}c_{i}.\\] \n\nEvidently, \\(0\\leq f_{a}(X)\\leq n(m - 1)\\) for any \\(a\\) and \\(X\\) . So, setting \\(|A| = n\\) , the right- hand side of the display takes on at most \\((n(m - 1) + 1)^{n}\\) distinct values. This means \n\n\\[m^{m}\\leq (n(m - 1))^{n}\\] \n\nwhich implies \\(n\\geq m / 2\\) \n\nRemark (Motivation comments from USJL). In linear algebra terms, we have some \\(n\\) - dimensional \\(0 / 1\\) vectors \\(\\vec{v_{1}}\\) , ..., \\(\\vec{v_{m}}\\) and an \\(n\\) - dimensional vector \\(\\vec{a}\\) such that \\(\\vec{v_{i}}\\cdot \\vec{a} = m^{i}\\) for \\(i = 1,\\ldots ,m\\) . The intuition is that if \\(n\\) is too small, then there should be lots of linear dependences between \\(\\vec{v_{i}}\\) . \n\nIn fact, Siegel's lemma is a result that says, if there are many more vectors than the dimension of the ambient space, there exist linear dependences whose coefficients are not- too- big integers. On the other hand, any linear dependence between \\(m\\) , \\(m^{2}\\) , ..., \\(m^{m}\\) is going to have coefficients that are pretty big; at least one of them needs to exceed \\(m\\) . \n\nApplying Siegel's lemma turns out to solve the problem (and is roughly equivalent to the solution above). \n\nRemark. In https://aops.com/community/p23185192, dgrozev shows the stronger bound \\(n\\geq \\left(\\frac{2}{3} +\\frac{c}{\\log m}\\right)m\\) elements, for some absolute constant \\(c > 0\\) .", "metadata": {"resource_path": "IMO/segmented/en-IMO-2021-notes.jsonl", "problem_match": "6. ", "solution_match": "## \\(\\S 2.3\\) IMO 2021/6, proposed by Austria \n"}}
diff --git a/IMO/segmented/en-IMO-2022-notes.jsonl b/IMO/segmented/en-IMO-2022-notes.jsonl
index b784693b5b03942326d5def9d6381a0d366c728d..f9986db0f7cee872109e5917eefa5f1fea2986bd 100644
--- a/IMO/segmented/en-IMO-2022-notes.jsonl
+++ b/IMO/segmented/en-IMO-2022-notes.jsonl
@@ -1,6 +1,6 @@
{"year": "2022", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "The Bank of Oslo issues two types of coin: aluminum (denoted \\(A\\) ) and bronze (denoted \\(B\\) ). Marianne has \\(n\\) aluminum coins and \\(n\\) bronze coins arranged in a row in some arbitrary initial order. A chain is any subsequence of consecutive coins of the same type. Given a fixed positive integer \\(k \\leq 2n\\) , Gilberty repeatedly performs the following operation: he identifies the longest chain containing the \\(k^{\\mathrm{th}}\\) coin from the left and moves all coins in that chain to the left end of the row. For example, if \\(n = 4\\) and \\(k = 4\\) , the process starting from the ordering \\(AABBBA\\) would be \\(AABBBA\\) \\(A\\) \\(BBBA\\) \\(A\\) \\(AAABBB\\) \\(A\\) \\(BBBBAA\\) \\(A\\) \\(BBBBAA\\) \\(A\\) \\(\\dots\\) \n\nFind all pairs \\((n,k)\\) with \\(1\\leq k\\leq 2n\\) such that for every initial ordering, at some moment during the process, the leftmost \\(n\\) coins will all be of the same type.", "solution": "Tsame type. \n\nAnswer: \\(n\\leq k\\leq \\left\\lceil \\frac{3}{2} n\\right\\rceil\\) \n\nCall a maximal chain a block. Then the line can be described as a sequence of blocks: it's one of: \n\n\\[\\underbrace{A\\ldots A}_{e_1}\\underbrace{B\\ldots B}_{e_2}\\underbrace{A\\ldots A}_{e_3}\\ldots \\underbrace{A\\ldots A}_{e_m}\\mathrm{~for~odd~}m\\] \\[\\underbrace{A\\ldots A}_{e_1}\\underbrace{B\\ldots B}_{e_2}\\underbrace{A\\ldots A}_{e_3}\\ldots \\underbrace{B\\ldots B}_{e_m}\\mathrm{~for~even~}m\\] \n\nor the same thing with the roles of \\(A\\) and \\(B\\) flipped. \n\nThe main claim is the following: \n\nClaim — The number \\(m\\) of blocks will never increase after an operation. Moreover, it stays the same if and only if \n\n- \\(k \\leq e_1\\) ; or \n\n- \\(m\\) is even and \\(e_m \\geq 2n + 1 - k\\) . \n\nProof. This is obvious, just run the operation and see! \n\nThe problem asks for which values of \\(k\\) we always reach \\(m = 2\\) eventually; we already know that it's non- increasing. We consider a few cases: \n\n- If \\(k < n\\) , then any configuration with \\(e_1 = n - 1\\) will never change. \n\n- If \\(k > \\lceil 3n / 2 \\rceil\\) , then take \\(m = 4\\) and \\(e_1 = e_2 = \\lfloor n / 2 \\rfloor\\) and \\(e_3 = e_4 = \\lceil n / 2 \\rceil\\) . This configuration retains \\(m = 4\\) always: the blocks simply rotate. \n\n- Conversely, suppose \\(k \\geq n\\) has the property that \\(m > 2\\) stays fixed. If after the first three operations \\(m\\) hasn't changed, then we must have \\(m \\geq 4\\) even, and \\(e_m, e_{m-1}, e_{m-2} \\geq 2n + 1 - k\\) . Now, \n\n\\[n \\geq e_m + e_{m-2} \\geq 2(2n + 1 - k) \\implies k \\geq \\frac{3}{2} n + 1\\]\n\n\n\nso this completes the proof.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2022-notes.jsonl", "problem_match": "1. ", "solution_match": "## \\(\\S 1.1\\) IMO 2022/1, proposed by Baptiste Serraille (FRA) \n"}}
{"year": "2022", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "Find all functions \\(f\\colon \\mathbb{R}^{+}\\to \\mathbb{R}^{+}\\) such that for each \\(x\\in \\mathbb{R}^{+}\\) , there is exactly one \\(y\\in \\mathbb{R}^{+}\\) satisfying \n\n\\[x f(y) + y f(x)\\leq 2.\\]", "solution": "The answer is \\(f(x)\\equiv 1 / x\\) which obviously works (here \\(y = x\\) ). \n\nFor the converse, assume we have \\(f\\) such that each \\(x\\in \\mathbb{R}^{+}\\) has a friend \\(y\\) with \\(x f(y) + y f(x)\\leq 2\\) . By symmetry \\(y\\) is also the friend of \\(x\\) . \n\nClaim — In fact every number is its own friend. \n\nProof. Assume for contradiction \\(a\\neq b\\) are friends. Then we know that \\(a f(a) + a f(a)>\\) \\(2\\Longrightarrow f(a) > \\frac{1}{a}\\) . Analogously, \\(f(b) > \\frac{1}{b}\\) . However, we then get \n\n\\[2\\geq a f(b) + b f(a) > \\frac{a}{b} +\\frac{b}{a}\\stackrel {\\mathrm{AMGM}}{\\geq}2\\] \n\nwhich is impossible. \n\nThe problem condition now simplifies to saying \n\n\\[f(x)\\leq \\frac{1}{x}\\mathrm{~for~all~}x,\\qquad x f(y) + y f(x) > 2\\mathrm{~for~all~}x\\neq y.\\] \n\nIn particular, for any \\(x > 0\\) and \\(\\epsilon >0\\) we have \n\n\\[2< x f(x + \\epsilon) + (x + \\epsilon)f(x)\\leq \\frac{x}{x + \\epsilon} +(x + \\epsilon)f(x)\\] \\[\\Longrightarrow f(x) > \\frac{x + 2\\epsilon}{(x + \\epsilon)^{2}} = \\frac{1}{x + \\frac{\\epsilon^{2}}{x + 2\\epsilon}}.\\] \n\nSince this holds for all \\(\\epsilon >0\\) this forces \\(f(x)\\geq \\frac{1}{x}\\) as well. We're done. \n\nRemark. Alternatively, instead of using \\(x + \\epsilon\\) , it also works to consider \\(y = \\frac{1}{f(x)}\\) . For such a \\(y\\) , we would have \n\n\\[x f\\left(\\frac{1}{f(x)}\\right) + \\frac{1}{f(x)}\\cdot f(x) = x f\\left(\\frac{1}{f(x)}\\right) + 1\\leq x\\cdot f(x) + 1\\leq 1 + 1 = 2\\] \n\nwhich gives a similar contradiction.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2022-notes.jsonl", "problem_match": "2. ", "solution_match": "## \\(\\S 1.2\\) IMO 2022/2, proposed by Merlijn Staps (NLD) \n"}}
{"year": "2022", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "Let \\(k\\) be a positive integer and let \\(S\\) be a finite set of odd prime numbers. Prove that there is at most one way (up to rotation and reflection) to place the elements of \\(S\\) around the circle such that the product of any two neighbors is of the form \\(x^{2} + x + k\\) for some positive integer \\(x\\) .", "solution": "We replace \"positive integer \\(x\\) \" with \"nonnegative integer \\(x\\) \", and say numbers of the form \\(x^{2} + x + k\\) are good. We could also replace \"nonnegative integer \\(x\\) \" with \"integer \\(x\\) \" owing to the obvious map \\(x \\mapsto 1 - x\\) . \n\nClaim — If \\(p\\) is an odd prime, there are at most two odd primes \\(q\\) and \\(r\\) less than \\(p\\) for which \\(pq = x^{2} + x + k\\) and \\(pr = y^{2} + y + k\\) are good. \n\nMoreover, if the above occurs and \\(x, y \\geq 0\\) , then \\(x + y + 1 = p\\) and \\(xy \\equiv k \\pmod{p}\\) . \n\nProof. The equation \\(T^{2} + T + k \\equiv 0 \\pmod{p}\\) has at most two solutions modulo \\(p\\) , i.e. at most two solutions in the interval \\([0, p - 1]\\) . Because \\(0 \\leq x, y < p\\) from \\(p > \\max(q, r)\\) and \\(k > 0\\) , the first half follows. \n\nFor the second half, Vieta also gives \\(x + y \\equiv - 1 \\pmod{p}\\) and \\(xy \\equiv k \\pmod{p}\\) , and we know \\(0 < x + y < 2p\\) . \\(\\square\\) \n\nClaim — If two such primes do exist as above, then \\(qr\\) is also good (!). \n\nProof. Let \\(pq = x^{2} + x + k\\) and \\(pr = y^{2} + y + k\\) for \\(x, y \\geq 0\\) as before. Fix \\(\\alpha \\in \\mathbb{C}\\) such that \\(\\alpha^{2} + \\alpha + k = 0\\) ; then for any \\(n \\in \\mathbb{Z}\\) , we have \n\n\\[n^{2} + n + k = \\mathrm{Norm}(n - \\alpha).\\] \n\nHence \n\n\\[pq \\cdot pr = \\mathrm{Norm} \\left((x - \\alpha)(y - \\alpha)\\right) = \\mathrm{Norm} \\left((xy - k) - (x + y + 1)\\alpha\\right)\\] \n\nBut \\(\\mathrm{Norm}(p) = p^{2}\\) , so combining with the second half of the previous claim gives \n\n\\[qr = \\mathrm{Norm} \\left(\\frac{1}{p} (xy - k) - \\alpha\\right)\\] \n\nas needed. \n\nThese two claims imply the claim directly by induction on \\(|S|\\) , since one can now delete the largest element of \\(S\\) . \n\nRemark. To show that the condition is not vacuous, the author gave a ring of 385 primes for \\(k = 41\\) ; see https://aops.com/community/p26068963.\n\n\n\n## \\(\\S 2\\) Solutions to Day 2", "metadata": {"resource_path": "IMO/segmented/en-IMO-2022-notes.jsonl", "problem_match": "3. ", "solution_match": "## \\(\\S 1.3\\) IMO 2022/3, proposed by Ankan Bhattacharya (USA) \n"}}
-{"year": "2022", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "Let \\(A B C D E\\) be a convex pentagon such that \\(B C = D E\\) . Assume that there is a point \\(T\\) inside \\(A B C D E\\) with \\(T B = T D\\) \\(T C = T E\\) and \\(\\angle A B T = \\angle T E A\\) . Let line \\(A B\\) intersect lines \\(C D\\) and \\(C T\\) at points \\(P\\) and \\(Q\\) , respectively. Assume that the points \\(P\\) , \\(B\\) , \\(A\\) , \\(Q\\) occur on their line in that order. Let line \\(A E\\) intersect \\(C D\\) and \\(D T\\) at points \\(R\\) and \\(S\\) , respectively. Assume that the points \\(R\\) , \\(E\\) , \\(A\\) , \\(S\\) occur on their line in that order. Prove that the points \\(P\\) , \\(S\\) , \\(Q\\) , \\(R\\) lie on a circle.", "solution": "## Problem statemens imply \n\n\\[\\triangle BTC\\cong \\triangle DTE,\\qquad \\mathrm{and}\\qquad \\triangle BTY\\cong \\triangle ETX.\\] \n\nDefine \\(K = \\overline{CT} \\cap \\overline{AE}\\) , \\(L = \\overline{DT} \\cap \\overline{AB}\\) , \\(X = \\overline{BT} \\cap \\overline{AE}\\) , \\(Y = \\overline{ET} \\cap \\overline{BY}\\) . \n\n\n \n\nClaim (Main claim) — We have \n\n\\[\\triangle BTQ \\sim \\triangle ETS, \\qquad \\text{and} \\qquad BY: YL: LQ = EX: XK: KS.\\]\n\n\n\nIn other words, \\(TBYLQ \\sim TEXKS\\) . \n\nProof. We know \\(\\triangle BTY \\sim \\triangle ETX\\) . Also, \\(\\triangle BTL = \\triangle BTD = \\triangle CTE = \\triangle KTE\\) and \\(\\triangle BTQ = \\triangle BTC = \\triangle DTE = \\triangle STE\\) . \\(\\square\\) \n\nIt follows from the claim that: \n\n\\(\\cdot \\mathit{T L} / \\mathit{T Q} = \\mathit{T K} / \\mathit{T S}\\) , ergo \\(T L\\cdot T S = T K\\cdot T Q\\) , so \\(K L S Q\\) is cyclic; and \n\n\\(\\cdot \\mathit{T C} / \\mathit{T K} = \\mathit{T E} / \\mathit{T K} = \\mathit{T B} / \\mathit{T L} = \\mathit{T D} / \\mathit{T L}\\) , so \\(\\overline{{K L}}\\parallel \\overline{{P C D R}}\\) \n\nWith these two bullets, we're done by Reim theorem.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2022-notes.jsonl", "problem_match": "4. ", "solution_match": "## \\(\\S 2.1\\) IMO 2022/4, proposed by Patrik Bak (SVK) \n"}}
+{"year": "2022", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "Let \\(A B C D E\\) be a convex pentagon such that \\(B C = D E\\) . Assume that there is a point \\(T\\) inside \\(A B C D E\\) with \\(T B = T D\\) \\(T C = T E\\) and \\(\\angle A B T = \\angle T E A\\) . Let line \\(A B\\) intersect lines \\(C D\\) and \\(C T\\) at points \\(P\\) and \\(Q\\) , respectively. Assume that the points \\(P\\) , \\(B\\) , \\(A\\) , \\(Q\\) occur on their line in that order. Let line \\(A E\\) intersect \\(C D\\) and \\(D T\\) at points \\(R\\) and \\(S\\) , respectively. Assume that the points \\(R\\) , \\(E\\) , \\(A\\) , \\(S\\) occur on their line in that order. Prove that the points \\(P\\) , \\(S\\) , \\(Q\\) , \\(R\\) lie on a circle.", "solution": "## Problem statemens imply \n\n\\[\\triangle BTC\\cong \\triangle DTE,\\qquad \\mathrm{and}\\qquad \\triangle BTY\\cong \\triangle ETX.\\] \n\nDefine \\(K = \\overline{CT} \\cap \\overline{AE}\\) , \\(L = \\overline{DT} \\cap \\overline{AB}\\) , \\(X = \\overline{BT} \\cap \\overline{AE}\\) , \\(Y = \\overline{ET} \\cap \\overline{BY}\\) . \n\n\n \n\nClaim (Main claim) — We have \n\n\\[\\triangle BTQ \\sim \\triangle ETS, \\qquad \\text{and} \\qquad BY: YL: LQ = EX: XK: KS.\\]\n\n\n\nIn other words, \\(TBYLQ \\sim TEXKS\\) . \n\nProof. We know \\(\\triangle BTY \\sim \\triangle ETX\\) . Also, \\(\\triangle BTL = \\triangle BTD = \\triangle CTE = \\triangle KTE\\) and \\(\\triangle BTQ = \\triangle BTC = \\triangle DTE = \\triangle STE\\) . \\(\\square\\) \n\nIt follows from the claim that: \n\n\\(\\cdot \\mathit{T L} / \\mathit{T Q} = \\mathit{T K} / \\mathit{T S}\\) , ergo \\(T L\\cdot T S = T K\\cdot T Q\\) , so \\(K L S Q\\) is cyclic; and \n\n\\(\\cdot \\mathit{T C} / \\mathit{T K} = \\mathit{T E} / \\mathit{T K} = \\mathit{T B} / \\mathit{T L} = \\mathit{T D} / \\mathit{T L}\\) , so \\(\\overline{{K L}}\\parallel \\overline{{P C D R}}\\) \n\nWith these two bullets, we're done by Reim theorem.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2022-notes.jsonl", "problem_match": "4. ", "solution_match": "## \\(\\S 2.1\\) IMO 2022/4, proposed by Patrik Bak (SVK) \n"}}
{"year": "2022", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "Find all triples \\((a,b,p)\\) of positive integers with \\(p\\) prime and \n\n\\[a^{p} = b! + p.\\]", "solution": "The answer is \\((2,2,2)\\) and \\((3,4,3)\\) only, which work. \n\nIn what follows we assume \\(a\\geq 2\\) \n\nClaim — We have \\(b\\leq 2p - 2\\) , and hence \\(a< p^{2}\\) \n\nProof. For the first half, assume first for contradiction that \\(b\\geq 2p\\) . Then \\(b! + p\\equiv p\\) (mod \\(p^{2}\\) ), so \\(\\nu_{p}(b! + p) = 1\\) , but \\(\\nu_{p}(a^{p}) = 1\\) never occurs. \n\nWe can also rule out \\(b = 2p - 1\\) since that would give \n\n\\[(2p - 1)! + p = p[(p - 1)!(p + 1)(p + 2)\\dots (2p - 1) + 1]\\] \n\nBy Wilson theorem the inner bracket is \\((- 1)^{2} + 1\\equiv 2\\) (mod \\(p\\) ) exactly, contradiction for \\(p > 2\\) . And when \\(p = 2\\) , \\(3! + 2 = 8\\) is not a perfect square. \n\nThe second half follows as \\(a^{p}\\leq (2p - 2)! + p< p^{2p}\\) . (Here we used the crude estimate \\((2p - 2)! = \\prod_{k = 1}^{p - 1}k\\cdot (2p - 1 - k)< (p(p - 1))^{p - 1})\\) . \\(\\square\\) \n\nClaim — We have \\(a\\geq p\\) , and hence \\(b\\geq p\\) \n\nProof. For the first half, assume for contradiction that \\(p > a\\) . Then \n\n\\[b! + p = a^{p}\\geq a^{p - 1} + p\\geq a^{a} + p > a! + p\\Rightarrow b > a.\\] \n\nThen taking modulo \\(a\\) now gives \\(0\\equiv 0 + p\\) (mod \\(a\\) ), which is obviously impossible. \n\nThe second half follows from \\(b! = a^{p} - p\\geq p! - p > (p - 1)!\\) . \\(\\square\\) \n\nClaim — We have \\(a = p\\) exactly. \n\nProof. We know \\(p\\geq b\\) hence \\(p\\mid b! + p\\) , so let \\(a = pk\\) for \\(k< p\\) . Then \\(k\\mid b!\\) yet \\(k\\nmid a^{p} - p\\) contradiction. \\(\\square\\) \n\nLet's get the small \\(p\\) out of the way: \n\n- For \\(p = 2\\) , checking \\(2\\leq b\\leq 3\\) gives \\((a,b) = (2,2)\\) only. \n\n- For \\(p = 3\\) , checking \\(3\\leq b\\leq 5\\) gives \\((a,b) = (3,4)\\) only. \n\nOnce \\(p\\geq 5\\) , if \\(b! = p^{p} - p = p(p^{p - 1} - 1)\\) then applying Zsigmondy gets a prime factor \\(q\\equiv 1\\) (mod \\(p - 1\\) ) which divides \\(p^{p - 1} - 1\\) . Yet \\(q\\leq b\\leq 2p - 2\\) and \\(q\\neq p\\) , contradiction.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2022-notes.jsonl", "problem_match": "5. ", "solution_match": "## \\(\\S 2.2\\) IMO 2022/5, proposed by Tijs Buggenhout (BEL) \n"}}
{"year": "2022", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "Let \\(n\\) be a positive integer. A Nordic square is an \\(n\\times n\\) board containing all the integers from 1 to \\(n^{2}\\) so that each cell contains exactly one number. An uphill path is a sequence of one or more cells such that: \n\na) the first cell in the sequence is a valley, meaning the number written is less than all its orthogonal neighbors, \nb) each subsequent cell in the sequence is orthogonally adjacent to the previous cell, and \nc) the numbers written in the cells in the sequence are in increasing order. \n\nFind, as a function of \\(n\\) , the smallest possible total number of uphill paths in a Nordic square.", "solution": "Lre. \n\nAnswer: \\(2n^{2} - 2n + 1\\) \n\nBound. The lower bound is the \"obvious\" one: \n\n- For any pair of adjacent cells, say \\(a > b\\) , one can extend it to a downhill path (the reverse of an uphill path) by walking downwards until one reaches a valley. This gives \\(2n(n - 1) = 2n^{2} - 2n\\) uphill paths of length \\(\\geq 2\\) . \n\n- There is always at least one uphill path of length 1, namely the single cell \\(\\{1\\}\\) (or indeed any valley). \n\nConstruction. For the construction, the ideas it build a tree \\(T\\) on the grid such that no two cells not in \\(T\\) are adjacent. \n\nAn example of such a grid is shown below for \\(n = 15\\) with \\(T\\) in yellow and cells not in \\(T\\) in black; it generalizes to any \\(3\\mid n\\) , and then to any \\(n\\) by deleting the last \\(n\\) mod 3 rows and either/both of the leftmost/rightmost column.\n\n\n\n\n| 1 | 2 | 3 | 32 | | 36 | 62 | 63 | 64 | 93 | | 97 | 123 | 124 | 125 |
| 4 | | 33 | 34 | 35 | | 65 | | 94 | 95 | 96 | | 126 | |
| 6 | 5 | 7 | | 37 | | 67 | 66 | 68 | | 98 | | 128 | 127 | 129 |
| 8 | | 39 | 38 | 40 | | 69 | | 100 | 99 | 101 | | 130 | |
| 10 | 9 | 11 | | 41 | | 71 | 70 | 72 | | 102 | | 132 | 131 | 133 |
| 12 | | 43 | 42 | 44 | | 73 | | 104 | 103 | 105 | | 134 | |
| 14 | 13 | 15 | | 45 | | 75 | 74 | 76 | | 106 | | 136 | 135 | 137 |
| 16 | | 47 | 46 | 48 | | 77 | | 108 | 107 | 109 | | 138 | |
| 18 | 17 | 19 | | 49 | | 79 | 78 | 80 | | 110 | | 140 | 139 | 141 |
| 20 | | 51 | 50 | 52 | | 81 | | 112 | 111 | 113 | | 142 | |
| 22 | 21 | 23 | | 53 | | 83 | 82 | 84 | | 114 | | 144 | 143 | 145 |
| 24 | | 55 | 54 | 56 | | 85 | | 116 | 115 | 117 | | 146 | |
| 26 | 25 | 27 | | 57 | | 87 | 86 | 88 | | 118 | | 148 | 147 | 149 |
| 28 | | 59 | 58 | 60 | | 89 | | 120 | 119 | 121 | | 150 | |
| 30 | 29 | 31 | | 61 | | 91 | 90 | 92 | | 122 | | 152 | 151 | 153 |
\n\nPlace 1 anywhere in \\(T\\) and then place all the small numbers at most \\(|T|\\) adjacent to previously placed numbers (example above). Then place the remaining numbers outside \\(T\\) arbitrarily. \n\nBy construction, as 1 is the only valley, any uphill path must start from 1. And by construction, it may only reach a given pair of terminal cells in one way, i.e. the downhill paths we mentioned are the only one. End proof.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2022-notes.jsonl", "problem_match": "6. ", "solution_match": "## \\(\\S 2.3\\) IMO 2022/6, proposed by Nikola Petrovic (SRB) \n"}}
diff --git a/IMO/segmented/en-IMO-2023-notes.jsonl b/IMO/segmented/en-IMO-2023-notes.jsonl
index ca34055b08d7d2ebe78146d14330817addd85290..c7e9352f8d2a8ab97909b1a9cf31bcf673673d1f 100644
--- a/IMO/segmented/en-IMO-2023-notes.jsonl
+++ b/IMO/segmented/en-IMO-2023-notes.jsonl
@@ -1,6 +1,6 @@
{"year": "2023", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "Determine all composite integers \\(n > 1\\) that satisfy the following property: if \\(d_{1}< d_{2}< \\dots < d_{k}\\) are all the positive divisors of \\(n\\) with then \\(d_{i}\\) divides \\(d_{i + 1} + d_{i + 2}\\) for every \\(1\\leq i\\leq k - 2\\)", "solution": "Problem statement . \n\nThe answer is prime powers. \n\nVerification that these work. When \\(n = p^{e}\\) , we get \\(d_{i} = p^{i - 1}\\) . The \\(i^{\\mathrm{th}}\\) relationship reads \n\n\\[p^{i - 1}\\mid p^{i} + p^{i + 1}\\] \n\nwhich is obviously true. \n\nProof that these are the only answers. Conversely, suppose \\(n\\) has at least two distinct prime divisors. Let \\(p< q\\) denote the two smallest ones, and let \\(p^{e}\\) be the largest power of \\(p\\) which both divides \\(n\\) and is less than \\(q\\) , hence \\(e\\geq 1\\) . Then the smallest factors of \\(n\\) are 1, \\(p\\) , ..., \\(p^{e}\\) , \\(q\\) . So we are supposed to have \n\n\\[\\frac{n}{q}\\mid \\frac{n}{p^{e}} +\\frac{n}{p^{e - 1}} = \\frac{(p + 1)n}{p^{e}}\\] \n\nwhich means that the ratio \n\n\\[\\frac{q(p + 1)}{p^{e}}\\] \n\nneeds to be an integer, which is obviously not possible.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2023-notes.jsonl", "problem_match": "1. ", "solution_match": "## \\(\\S 1.1\\) IMO 2023/1, proposed by Santiago Rodriguez (COL) \n"}}
-{"year": "2023", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "Let \\(A B C\\) be an acute-angled triangle with \\(A B< A C\\) . Let \\(\\Omega\\) be the circumcircle of \\(A B C\\) . Let \\(S\\) be the midpoint of the arc \\(C B\\) of \\(\\Omega\\) containing \\(A\\) . The perpendicular from \\(A\\) to \\(B C\\) meets \\(B S\\) at \\(D\\) and meets \\(\\Omega\\) again at \\(E\\neq A\\) . The line through \\(D\\) parallel to \\(B C\\) meets line \\(B E\\) at \\(L\\) . Denote the circumcircle of triangle \\(B D L\\) by \\(\\omega\\) . Let \\(\\omega\\) meet \\(\\Omega\\) again at \\(P\\neq B\\) . Prove that the line tangent to \\(\\omega\\) at \\(P\\) meets line \\(B S\\) on the internal angle bisector of \\(\\angle B A C\\) .", "solution": "## Problem statemlaim — We have \\(LPS\\) collinear. \n\nProof. Because \\(\\angle LPB = \\angle LDB = \\angle CBD = \\angle CBS = \\angle SCB = \\angle SPB\\) . \\(\\square\\) \n\nLet \\(F\\) be the antipode of \\(A\\) , so \\(AMFS\\) is a rectangle. \n\nClaim — We have \\(PDF\\) collinear. (This lets us erase \\(L\\) .) \n\nProof. Because \\(\\angle SPD = \\angle LPD = \\angle LBD = \\angle SBE = \\angle FCS = \\angle FPS\\) . \\(\\square\\) \n\nLet us define \\(X = \\overline{AM} \\cap \\overline{BS}\\) and complete chord \\(\\overline{PXQ}\\) . We aim to show that \\(\\overline{PXQ}\\) is tangent to \\((PDLB)\\) . \n\n\n \n\nClaim (Main projective claim) — We have \\(XP = XA\\) . \n\nProof. Introduce \\(Y = \\overline{PDF} \\cap \\overline{AM}\\) . Note that \n\n\\[-1 = (SM;EF) \\stackrel{A}{=} (S,X;D,\\overline{AF} \\cap \\overline{ES}) \\stackrel{E}{=} (\\infty X;YA)\\] \n\nwhere \\(\\infty = \\overline{AM} \\cap \\overline{SF}\\) is at infinity (because \\(AMSF\\) is a rectangle). Thus, \\(XY = XA\\) .\n\n\n\n \n\nSince \\(\\triangle APY\\) is also right, we get \\(XP = XA\\) . \n\nAlternative proof of claim without harmonic bundles, from Solution 9 of the marking scheme. With \\(Y = \\overline{PDF} \\cap \\overline{AM}\\) defined as before, note that \\(\\overline{AE} \\parallel \\overline{SM}\\) and \\(\\overline{AM} \\parallel \\overline{SF}\\) (as AMFS is a rectangle) gives respectively the similar triangles \n\n\\[\\triangle AXD\\sim \\triangle M X S,\\qquad \\triangle XDY\\sim \\triangle SDF.\\] \n\nFrom this we conclude \n\n\\[\\frac{AX}{XD} = \\frac{AX + XM}{XD + SX} = \\frac{AM}{SD} = \\frac{SF}{SD} = \\frac{XY}{XD}.\\] \n\nSo \\(AX = XY\\) and as before we conclude \\(XP = XA\\) . \n\nFrom \\(XP = XA\\) , we conclude that \\(\\widehat{PM}\\) and \\(\\widehat{AQ}\\) have the same measure. Since \\(\\widehat{AS}\\) and \\(\\widehat{EM}\\) have the same measure, it follows \\(\\widehat{PE}\\) and \\(\\widehat{SQ}\\) have the same measure. The desired tangency then follows from \n\n\\[\\angle QPL = \\angle QPS = \\angle PQE = \\angle PFE = \\angle PDL.\\] \n\nRemark (Logical ordering). This solution is split into two phases: the \"synthetic phase\" where we do a bunch of angle chasing, and the \"projective phase\" where we use cross- ratios because I like projective. For logical readability (so we write in only one logical direction), the projective phase is squeezed in two halves of the synthetic phase, but during an actual solve it's expected to complete the whole synthetic phase first (i.e. to reduce the problem to show \\(XP = XA\\) ). \n\nRemark. There are quite a multitude of approaches for this problem; the marking scheme for this problem at the actual IMO had 13 different solutions.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2023-notes.jsonl", "problem_match": "2. ", "solution_match": "## \\(\\S 1.2\\) IMO 2023/2, proposed by Tiago Mourão and Nuno Arala (POR) \n"}}
+{"year": "2023", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "Let \\(A B C\\) be an acute-angled triangle with \\(A B< A C\\) . Let \\(\\Omega\\) be the circumcircle of \\(A B C\\) . Let \\(S\\) be the midpoint of the arc \\(C B\\) of \\(\\Omega\\) containing \\(A\\) . The perpendicular from \\(A\\) to \\(B C\\) meets \\(B S\\) at \\(D\\) and meets \\(\\Omega\\) again at \\(E\\neq A\\) . The line through \\(D\\) parallel to \\(B C\\) meets line \\(B E\\) at \\(L\\) . Denote the circumcircle of triangle \\(B D L\\) by \\(\\omega\\) . Let \\(\\omega\\) meet \\(\\Omega\\) again at \\(P\\neq B\\) . Prove that the line tangent to \\(\\omega\\) at \\(P\\) meets line \\(B S\\) on the internal angle bisector of \\(\\angle B A C\\) .", "solution": "## Problem statemlaim — We have \\(LPS\\) collinear. \n\nProof. Because \\(\\angle LPB = \\angle LDB = \\angle CBD = \\angle CBS = \\angle SCB = \\angle SPB\\) . \\(\\square\\) \n\nLet \\(F\\) be the antipode of \\(A\\) , so \\(AMFS\\) is a rectangle. \n\nClaim — We have \\(PDF\\) collinear. (This lets us erase \\(L\\) .) \n\nProof. Because \\(\\angle SPD = \\angle LPD = \\angle LBD = \\angle SBE = \\angle FCS = \\angle FPS\\) . \\(\\square\\) \n\nLet us define \\(X = \\overline{AM} \\cap \\overline{BS}\\) and complete chord \\(\\overline{PXQ}\\) . We aim to show that \\(\\overline{PXQ}\\) is tangent to \\((PDLB)\\) . \n\n\n \n\nClaim (Main projective claim) — We have \\(XP = XA\\) . \n\nProof. Introduce \\(Y = \\overline{PDF} \\cap \\overline{AM}\\) . Note that \n\n\\[-1 = (SM;EF) \\stackrel{A}{=} (S,X;D,\\overline{AF} \\cap \\overline{ES}) \\stackrel{E}{=} (\\infty X;YA)\\] \n\nwhere \\(\\infty = \\overline{AM} \\cap \\overline{SF}\\) is at infinity (because \\(AMSF\\) is a rectangle). Thus, \\(XY = XA\\) .\n\n\n\n \n\nSince \\(\\triangle APY\\) is also right, we get \\(XP = XA\\) . \n\nAlternative proof of claim without harmonic bundles, from Solution 9 of the marking scheme. With \\(Y = \\overline{PDF} \\cap \\overline{AM}\\) defined as before, note that \\(\\overline{AE} \\parallel \\overline{SM}\\) and \\(\\overline{AM} \\parallel \\overline{SF}\\) (as AMFS is a rectangle) gives respectively the similar triangles \n\n\\[\\triangle AXD\\sim \\triangle M X S,\\qquad \\triangle XDY\\sim \\triangle SDF.\\] \n\nFrom this we conclude \n\n\\[\\frac{AX}{XD} = \\frac{AX + XM}{XD + SX} = \\frac{AM}{SD} = \\frac{SF}{SD} = \\frac{XY}{XD}.\\] \n\nSo \\(AX = XY\\) and as before we conclude \\(XP = XA\\) . \n\nFrom \\(XP = XA\\) , we conclude that \\(\\widehat{PM}\\) and \\(\\widehat{AQ}\\) have the same measure. Since \\(\\widehat{AS}\\) and \\(\\widehat{EM}\\) have the same measure, it follows \\(\\widehat{PE}\\) and \\(\\widehat{SQ}\\) have the same measure. The desired tangency then follows from \n\n\\[\\angle QPL = \\angle QPS = \\angle PQE = \\angle PFE = \\angle PDL.\\] \n\nRemark (Logical ordering). This solution is split into two phases: the \"synthetic phase\" where we do a bunch of angle chasing, and the \"projective phase\" where we use cross- ratios because I like projective. For logical readability (so we write in only one logical direction), the projective phase is squeezed in two halves of the synthetic phase, but during an actual solve it's expected to complete the whole synthetic phase first (i.e. to reduce the problem to show \\(XP = XA\\) ). \n\nRemark. There are quite a multitude of approaches for this problem; the marking scheme for this problem at the actual IMO had 13 different solutions.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2023-notes.jsonl", "problem_match": "2. ", "solution_match": "## \\(\\S 1.2\\) IMO 2023/2, proposed by Tiago Mourão and Nuno Arala (POR) \n"}}
{"year": "2023", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "For each integer \\(k\\geq 2\\) , determine all infinite sequences of positive integers \\(a_{1}\\) , \\(a_{2}\\) , ... for which there exists a polynomial \\(P\\) of the form \n\n\\[P(x) = x^{k} + c_{k - 1}x^{k - 1} + \\dots +c_{1}x + c_{0},\\] \n\nwhere \\(c_{0}\\) , \\(c_{1}\\) , ..., \\(c_{k - 1}\\) are non-negative integers, such that \n\n\\[P(a_{n}) = a_{n + 1}a_{n + 2}\\cdot \\cdot \\cdot a_{n + k}\\] \n\nfor every integer \\(n\\geq 1\\)", "solution": "## Problem stateme \n\nThe answer is \\(a_{n}\\) being an arithmetic progression. Indeed, if \\(a_{n} = d(n - 1) + a_{1}\\) for \\(d \\geq 0\\) and \\(n \\geq 1\\) , then \n\n\\[a_{n + 1}a_{n + 2}\\dots a_{n + k} = (a_{n} + d)(a_{n} + 2d)\\dots (a_{n} + kd)\\] \n\nso we can just take \\(P(x) = (x + d)(x + 2d)\\dots (x + kd)\\) . \n\nThe converse direction takes a few parts. \n\nClaim — Either \\(a_{1} < a_{2} < \\dots\\) or the sequence is constant. \n\nProof. Note that \n\n\\[P(a_{n - 1}) = a_{n}a_{n + 1}\\dots a_{n + k - 1\\] \\[P(a_{n}) = a_{n + 1}a_{n + 2}\\dots a_{n + k\\] \\[\\implies a_{n + k} = \\frac{P(a_{n})}{P(a_{n - 1})}\\cdot a_{n}.\\] \n\nNow the polynomial \\(P\\) is strictly increasing over \\(\\mathbb{N}\\) . \n\nSo assume for contradiction there's an index \\(n\\) such that \\(a_{n} < a_{n - 1}\\) . Then in fact the above equation shows \\(a_{n + k} < a_{n} < a_{n - 1}\\) . Then there's an index \\(\\ell \\in [n + 1, n + k]\\) such that \\(a_{\\ell} < a_{\\ell - 1}\\) , and also \\(a_{\\ell} < a_{n}\\) . Continuing in this way, we can an infinite descending subsequence of \\((a_{n})\\) , but that's impossible because we assumed integers. \n\nHence we have \\(a_{1} \\leq a_{2} \\leq \\dots\\) . Now similarly, if \\(a_{n} = a_{n - 1}\\) for any index \\(n\\) , then \\(a_{n + k} = a_{n}\\) , ergo \\(a_{n - 1} = a_{n} = a_{n + 1} = \\dots = a_{n + k}\\) . So the sequence is eventually constant, and then by downwards induction, it is fully constant. \\(\\square\\) \n\nClaim — There exists a constant \\(C\\) (depending only \\(P\\) , \\(k\\) ) such that we have \\(a_{n + 1} \\leq a_{n} + C\\) . \n\nProof. Let \\(C\\) be a constant such that \\(P(x) < x^{k} + Cx^{k - 1}\\) for all \\(x \\in \\mathbb{N}\\) (for example \\(C = c_{0} + c_{1} + \\dots + c_{k - 1} + 1\\) works). We have \n\n\\[a_{n + k} = \\frac{P(a_{n})}{a_{n + 1}a_{n + 2}\\dots a_{n + k - 1}}\\]\n\n\n\n\\[< \\frac{P(a_{n})}{(a_{n} + 1)(a_{n} + 2)\\ldots(a_{n} + k - 1)}\\] \\[< \\frac{a_{n}^{k} + C\\cdot a_{n}^{k - 1}}{(a_{n} + 1)(a_{n} + 2)\\ldots(a_{n} + k - 1)}\\] \\[< a_{n} + C + 1.\\] \n\nAssume henceforth \\(a_{n}\\) is nonconstant, and hence unbounded. For each index \\(n\\) and term \\(a_{n}\\) in the sequence, consider the associated differences \\(d_{1} = a_{n + 1} - a_{n}\\) , \\(d_{2} = a_{n + 2} - a_{n + 1}\\) , ..., \\(d_{k} = a_{n + k} - a_{n + k - 1}\\) , which we denote by \n\n\\[\\Delta (n):= (d_{1},\\ldots ,d_{k}).\\] \n\nThis \\(\\Delta\\) can only take up to \\(C^{k}\\) different values. So in particular, some tuple \\((d_{1},\\ldots ,d_{n})\\) must appear infinitely often as \\(\\Delta (n)\\) ; for that tuple, we obtain \n\n\\[P(a_{N}) = (a_{N} + d_{1})(a_{N} + d_{1} + d_{2})\\ldots (a_{N} + d_{1} + \\dots +d_{k})\\] \n\nfor infinitely many \\(N\\) . But because of that, we actually must have \n\n\\[P(X) = (X + d_{1})(X + d_{1} + d_{2})\\ldots (X + d_{1} + \\dots +d_{k}).\\] \n\nHowever, this also means that exactly one output to \\(\\Delta\\) occurs infinitely often (because that output is determined by \\(P\\) ). Consequently, it follows that \\(\\Delta\\) is eventually constant. For this to happen, \\(a_{n}\\) must eventually coincide with an arithmetic progression of some common difference \\(d\\) , and \\(P(X) = (X + d)(X + 2d)\\ldots (X + kd)\\) . Finally, this implies by downwards induction that \\(a_{n}\\) is an arithmetic progression on all inputs.\n\n\n\n## \\(\\S 2\\) Solutions to Day 2", "metadata": {"resource_path": "IMO/segmented/en-IMO-2023-notes.jsonl", "problem_match": "3. ", "solution_match": "## \\(\\S 1.3\\) IMO 2023/3, proposed by Ivan Chan (MAS) \n"}}
{"year": "2023", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "Let \\(x_{1}\\) , \\(x_{2}\\) , ..., \\(x_{2023}\\) be pairwise different positive real numbers such that \n\n\\[a_{n} = \\sqrt{(x_{1} + x_{2} + \\cdot \\cdot \\cdot + x_{n})\\left(\\frac{1}{x_{1}} +\\frac{1}{x_{2}} +\\cdot \\cdot \\cdot +\\frac{1}{x_{n}}\\right)}\\] \n\nis an integer for every \\(n = 1,2,\\ldots ,2023\\) . Prove that \\(a_{2023}\\geq 3034\\)", "solution": "Note that \\(a_{n + 1} > \\sqrt{\\sum_{1}^{n}x_{i}\\sum_{1}^{n}\\frac{1}{x_{i}}} = a_{n}\\) for all \\(n\\) , so that \\(a_{n + 1}\\geq a_{n} + 1\\) . Observe \\(a_{1} = 1\\) . We are going to prove that \n\n\\[a_{2m + 1}\\geq 3m + 1\\qquad \\mathrm{for~all~}m\\geq 0\\] \n\nby induction on \\(m\\) , with the base case being clear. \n\nWe now present two variations of the induction. The first shorter solution compares \\(a_{n + 2}\\) directly to \\(a_{n}\\) , showing it increases by at least 3. Then we give a longer approach that compares \\(a_{n + 1}\\) to \\(a_{n}\\) , and shows it cannot increase by 1 twice in a row. \n\n\\(\\P\\) Induct- by- two solution. Let \\(u = \\sqrt{\\frac{x_{n + 1}}{x_{n + 2}}}\\neq 1\\) . Note that by using Cauchy- Schwarz with three terms: \n\n\\[a_{n + 2}^{2} = \\left[(x_{1} + \\dots +x_{n}) + x_{n + 1} + x_{n + 2}\\right]\\left[\\left(\\frac{1}{x_{1}} +\\dots +\\frac{1}{x_{n}}\\right) + \\frac{1}{x_{n + 2}} +\\frac{1}{x_{n + 1}}\\right]\\] \\[\\qquad \\geq \\left(\\sqrt{(x_{1} + \\dots +x_{n})\\left(\\frac{1}{x_{1}} +\\dots +\\frac{1}{x_{n}}\\right)} +\\sqrt{\\frac{x_{n + 1}}{x_{n + 2}}} +\\sqrt{\\frac{x_{n + 2}}{x_{n + 1}}}\\right)^{2}\\] \\[\\qquad = \\left(a_{n} + u + \\frac{1}{u}\\right)^{2}.\\] \\[\\Rightarrow a_{n + 2}\\geq a_{n} + u + \\frac{1}{u} >a_{n} + 2\\] \n\nwhere the last equality \\(u + \\frac{1}{u} >2\\) is by AM- GM, strict as \\(u\\neq 1\\) . It follows that \\(a_{n + 2}\\geq a_{n} + 3\\) , completing the proof. \n\n\\(\\P\\) Induct- by- one solution. The main claim is: \n\nClaim — It's impossible to have \\(a_{n} = c\\) , \\(a_{n + 1} = c + 1\\) , \\(a_{n + 2} = c + 2\\) for any \\(c\\) and \\(n\\) . \n\nProof. Let \\(p = x_{n + 1}\\) and \\(q = x_{n + 2}\\) for brevity. Let \\(s = \\sum_{1}^{n}x_{i}\\) and \\(t = \\sum_{1}^{n}\\frac{1}{x_{n}}\\) , so \\(c^{2} = a_{n}^{2} = st\\) . \n\nFrom \\(a_{n} = c\\) and \\(a_{n + 1} = c + 1\\) we have \n\n\\[(c + 1)^{2} = a_{n + 1}^{2} = (p + s)\\left(\\frac{1}{p} +t\\right)\\]\n\n\n\n\\[= s t + p t + \\frac{1}{p} s + 1 = c^{2} + p t + \\frac{1}{p} s + 1\\] \\[\\overset {\\mathrm{AM - GM}}{\\geq} c^{2} + 2\\sqrt{s t} + 1 = c^{2} + 2\\sqrt{c^{2}} + 1 = (c + 1)^{2}.\\] \n\nHence, equality must hold in the AM- GM we must have exactly \n\n\\[p t = \\frac{1}{p} s = c.\\] \n\nIf we repeat the argument again on \\(a_{n + 1} = c + 1\\) and \\(a_{n + 2} = c + 2\\) , then \n\n\\[q\\left(\\frac{1}{p} +t\\right) = \\frac{1}{q} (p + s) = c + 1.\\] \n\nHowever this forces \\(\\frac{p}{q} = \\frac{q}{p} = 1\\) which is impossible.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2023-notes.jsonl", "problem_match": "4. ", "solution_match": "## \\(\\S 2.1\\) IMO 2023/4, proposed by Merlijn Staps (NLD) \n"}}
-{"year": "2023", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "Let \\(n\\) be a positive integer. A Japanese triangle consists of \\(1 + 2 + \\dots +n\\) circles arranged in an equilateral triangular shape such that for each \\(1\\leq i\\leq n\\) , the \\(i^{\\mathrm{th}}\\) row contains exactly \\(i\\) circles, exactly one of which is colored red. A ninja path in a Japanese triangle is a sequence of \\(n\\) circles obtained by starting in the top row, then repeatedly going from a circle to one of the two circles immediately below it and finishing in the bottom row. Here is an example of a Japanese triangle with \\(n = 6\\) , along with a ninja path in that triangle containing two red circles. \n\n\n \n\nIn terms of \\(n\\) , find the greatest \\(k\\) such that in each Japanese triangle there is a ninja path containing at least \\(k\\) red circles.", "solution": "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)\n \n\nIn terms of \\(n\\) , find the greatest \\(k\\) such that in each Japanese triangle there is a ninja path containing at least \\(k\\) red circles. \n\nThe answer is \n\n\\[k = \\lfloor \\log_2(n)\\rfloor +1.\\] \n\nConstruction. It suffices to find a Japanese triangle for \\(n = 2^{e} - 1\\) with the property that at most \\(e\\) red circles in any ninja path. The construction shown below for \\(e = 4\\) obviously generalizes, and works because in each of the sets \\(\\{1\\}\\) , \\(\\{2,3\\}\\) , \\(\\{4,5,6,7\\}\\) , ..., \\(\\{2^{e - 1}, \\ldots , 2^{e} - 1\\}\\) , at most one red circle can be taken. (These sets are colored in different shades of red for visual clarity). \n\n\n\n\n\n\nBound. Conversely, we show that in any Japanese triangle, one can find a ninja path containing at least \n\n\\[k = \\lfloor \\log_2(n)\\rfloor +1.\\] \n\nThe following short solution was posted at https://aops.com/community/p28134004, apparently first found by the team leader for Iran. \n\nWe construct a rooted binary tree \\(T_{1}\\) on the set of all circles as follows. For each row, other than the bottom row: \n\nConnect the red circle to both circles under it; White circles to the left of the red circle in its row are connected to the left; White circles to the right of the red circle in its row are connected to the right. \n\nThe circles in the bottom row are all leaves of this tree. For example, the \\(n = 6\\) construction in the beginning gives the tree shown on the left half of the figure below: \n\n\n \n\nNow focus on only the red circles, as shown in the right half of the figure. We build a new rooted tree \\(T_{2}\\) where each red circle is joined to the red circle below it if there was a path of (zero or more) white circles in \\(T_{1}\\) between them. Then each red circle has at most 2 direct descendants in \\(T_{2}\\) . Hence the depth of the new tree \\(T_{2}\\) exceeds \\(\\log_2(n)\\) , which produces the desired path. \n\nAnother recursive proof of bound, communicated by Helio Ng. We give another proof that \\(\\lfloor \\log_2n\\rfloor +1\\) is always achievable. Define \\(f(i,j)\\) to be the maximum number of red circles contained in the portion of a ninja path from \\((1,1)\\) to \\((i,j)\\) , including the endpoints \\((1,1)\\) and \\((i,j)\\) . (If \\((i,j)\\) is not a valid circle in the triangle, define \\(f(i,j) = 0\\) for convenience.) An example is shown below with the values of \\(f(i,j)\\) drawn in the circles. \n\n\n\n\n\n\nWe have that \n\n\\[f(i,j) = \\max \\left\\{f(i - 1,j - 1),f(i,j - 1)\\right\\} +\\left\\{ \\begin{array}{l l}{1} & {\\mathrm{if~}(i,j)\\mathrm{~is~red~}}\\\\ {0} & {\\mathrm{otherwise}} \\end{array} \\right.\\] \n\nsince every ninja path passing through \\((i,j)\\) also passes through either \\((i - 1,j - 1)\\) or \\((i,j - 1)\\) . Now consider the quantity \\(S_{j} = f(0,j) + \\dots +f(j,j)\\) . We obtain the following recurrence: \n\n\\[\\mathrm{Claim} - S_{j + 1}\\geq S_{j} + \\left\\lceil \\frac{S_{j}}{j}\\right\\rceil +1.\\] \n\nProof. Consider a maximal element \\(f(m,j)\\) of \\(\\{f(0,j),\\ldots ,f(j,j)\\}\\) . We perform the following manipulations: \n\n\\[S_{j + 1} = \\sum_{i = 0}^{j + 1}\\max \\left\\{f(i - 1,j),f(i,j)\\right\\} +\\sum_{i = 0}^{j + 1}\\left\\{ \\begin{array}{l l}{1} & {\\mathrm{if~}(i,j + 1)\\mathrm{~is~red~}}\\\\ {0} & {\\mathrm{otherwise}} \\end{array} \\right.\\] \\[\\quad = \\sum_{i = 0}^{m}\\max \\left\\{f(i - 1,j),f(i,j)\\right\\} +\\sum_{i = m + 1}^{j}\\max \\left\\{f(i - 1,j),f(i,j)\\right\\} +1\\] \\[\\quad \\geq \\sum_{i = 0}^{m}f(i,j) + \\sum_{i = m + 1}^{j}f(i - 1,j) + 1\\] \\[\\quad = S_{j} + f(m,j) + 1\\] \\[\\quad \\geq S_{j} + \\left\\lceil \\frac{S_{j}}{j}\\right\\rceil +1\\] \n\nwhere the last inequality is due to Pigeonhole. \n\nThis is actually enough to solve the problem. Write \\(n = 2^{c} + r\\) , where \\(0 \\leq r \\leq 2^{c} - 1\\) . \n\nClaim — \\(S_{n} \\geq cn + 2r + 1\\) . In particular, \\(\\left\\lceil \\frac{S_{n}}{n} \\right\\rceil \\geq c + 1\\) . \n\nProof. First note that \\(S_{n} \\geq cn + 2r + 1\\) implies \\(\\left\\lceil \\frac{S_{n}}{n} \\right\\rceil \\geq c + 1\\) because \n\n\\[\\left\\lceil \\frac{S_{n}}{n}\\right\\rceil \\geq \\left\\lceil \\frac{cn + 2r + 1}{n}\\right\\rceil = c + \\left\\lceil \\frac{2r + 1}{n}\\right\\rceil = c + 1.\\] \n\nWe proceed by induction on \\(n\\) . The base case \\(n = 1\\) is clearly true as \\(S_{1} = 1\\) . Assuming that the claim holds for some \\(n = j\\) , we have \n\n\\[S_{j + 1}\\geq S_{j} + \\left\\lceil \\frac{S_{j}}{j}\\right\\rceil +1\\] \\[\\qquad \\geq cj + 2r + 1 + c + 1 + 1\\] \\[\\qquad = c(j + 1) + 2(r + 1) + 1\\] \n\nso the claim is proved for \\(n = j + 1\\) if \\(j + 1\\) is not a power of 2. If \\(j + 1 = 2^{c + 1}\\) , then by writing \\(c(j + 1) + 2(r + 1) + 1 = c(j + 1) + (j + 1) + 1 = (c + 2)(j + 1) + 1\\) , the claim is also proved. \\(\\square\\) \n\nNow \\(\\left\\lceil \\frac{S_{n}}{n}\\right\\rceil \\geq c + 1\\) implies the existence of some ninja path containing at least \\(c + 1\\) red circles, and we are done.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2023-notes.jsonl", "problem_match": "5. ", "solution_match": "## \\(\\S 2.2\\) IMO 2023/5, proposed by Merlijn Staps and Daniël Kroes (NLD) \n"}}
-{"year": "2023", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "Let \\(A B C\\) be an equilateral triangle. Let \\(A_{1}\\) , \\(B_{1}\\) , \\(C_{1}\\) be interior points of \\(A B C\\) such that \\(B A_{1} = A_{1}C\\) , \\(C B_{1} = B_{1}A\\) , \\(A C_{1} = C_{1}B\\) , and \n\n\\[\\angle B A_{1}C + \\angle C B_{1}A + \\angle A C_{1}B = 480^{\\circ}.\\]\n\n\n\nLet \\(A_{2} = \\overline{BC_{1}} \\cap \\overline{CB_{1}}\\) , \\(B_{2} = \\overline{CA_{1}} \\cap \\overline{AC_{1}}\\) , \\(C_{2} = \\overline{AB_{1}} \\cap \\overline{BA_{1}}\\) . Prove that if triangle \\(A_{1}B_{1}C_{1}\\) is scalene, then the circumcircles of triangles \\(AA_{1}A_{2}\\) , \\(BB_{1}B_{2}\\) , and \\(CC_{1}C_{2}\\) all pass through two common points.", "solution": "Lts. \n\nThis is the second official solution from the marking scheme, also communicated to me by Michael Ren. Define \\(O\\) as the center of \\(A B C\\) and set the angles \n\n\\[\\alpha := \\angle A_{1}C B = \\angle C B A_{1}\\] \\[\\beta := \\angle A C B_{1} = \\angle B_{1}A C\\] \\[\\gamma := \\angle C_{1}A B = \\angle C_{1}B A\\] \n\nso that \n\n\\[\\alpha +\\beta +\\gamma = 30^{\\circ}.\\] \n\nIn particular, \\(\\max (\\alpha ,\\beta ,\\gamma)< 30^{\\circ}\\) , so it follows that \\(A_{1}\\) lies inside \\(\\triangle O B C\\) , and similarly for the others. This means for example that \\(C_{1}\\) lies between \\(B\\) and \\(A_{2}\\) , and so on. Therefore the polygon \\(A_{2}C_{1}B_{2}A_{1}C_{2}B_{1}\\) is convex.\n\n\n\n \n\nWe start by providing the \"interpretation\" for the \\(480^{\\circ}\\) angle in the statement: \n\nClaim — Point \\(A_{1}\\) is the circumcenter of \\(\\triangle A_{2}BC\\) , and similarly for the others. \n\nProof. We have \\(\\angle B A_{1}C = 180^{\\circ} - 2\\alpha\\) , and \n\n\\[\\angle B A_{2}C = 180^{\\circ} - \\angle C B C_{1} - \\angle B_{1}C B\\] \\[\\qquad = 180^{\\circ} - (60^{\\circ} - \\gamma) - (60^{\\circ} - \\beta)\\] \\[\\qquad = 60^{\\circ} + \\beta +\\gamma = 90^{\\circ} - \\alpha = \\frac{1}{2}\\angle B A_{1}C.\\] \n\nSince \\(A_{1}\\) lies inside \\(\\triangle B A_{2}C\\) , it follows \\(A_{1}\\) is exactly the circumcenter. \n\nClaim — Quadrilateral \\(B_{2}C_{1}B_{1}C\\) can be inscribed in a circle, say \\(\\gamma_{a}\\) . Circles \\(\\gamma_{b}\\) and \\(\\gamma_{c}\\) can be defined similarly. Finally, these three circles are pairwise distinct. \n\nProof. Using directed angles now, we have \n\n\\[\\angle B_{2}B_{1}C_{2} = 180^{\\circ} - \\angle A B_{1}B_{2} = 180^{\\circ} - 2\\angle A C B = 180^{\\circ} - 2(60^{\\circ} - \\alpha) = 60^{\\circ} + 2\\alpha .\\] \n\nBy the same token, \\(\\angle B_{2}C_{1}C_{2} = 60^{\\circ} + 2\\alpha\\) . This establishes the existence of \\(\\gamma_{a}\\) . \n\nThe proof for \\(\\gamma_{b}\\) and \\(\\gamma_{c}\\) is the same. Finally, to show the three circles are distinct, it would be enough to verify that the convex hexagon \\(A_{2}C_{1}B_{2}A_{1}C_{2}B_{1}\\) is not cyclic.\n\n\n\nAssume for contradiction it was cyclic. Then \n\n\\[360^{\\circ} = \\angle C_{2}A_{1}B_{2} + \\angle B_{2}C_{1}A_{2} + \\angle A_{2}B_{1}C_{2} = \\angle B A_{1}C + \\angle C B_{1}A + \\angle A C_{1}B = 480^{\\circ}\\] \n\nwhich is absurd. This contradiction eliminates the degenerate case, so the three circles are distinct. \\(\\square\\) \n\nFor the remainder of the solution, let \\(\\mathrm{Pow}(P,\\omega)\\) denote the power of a point \\(P\\) with respect to a circle \\(\\omega\\) . \n\nLet line \\(A A_{1}\\) meet \\(\\gamma_{b}\\) and \\(\\gamma_{c}\\) again at \\(X\\) and \\(Y\\) , and set \\(k_{a}:= \\frac{A X}{A Y}\\) . Consider the locus of all points \\(P\\) such that \n\n\\[\\mathcal{C}_{a}:= \\Big\\{\\mathrm{points} P\\mathrm{in}\\mathrm{the}\\mathrm{plane}\\mathrm{satisfying}\\mathrm{Pow}(P,\\gamma_{b}) = k_{a}\\mathrm{Pow}(P,\\gamma_{c})\\Big\\} .\\] \n\nWe recall the coaxiality lemma \\(^{1}\\) , which states that (given \\(\\gamma_{b}\\) and \\(\\gamma_{c}\\) are not concentric) the locus \\(\\mathcal{C}_{a}\\) must be either a circle (if \\(k_{a}\\neq 1\\) ) or a line (if \\(k_{a} = 1\\) ). \n\nOn the other hand, \\(A_{1}\\) , \\(A_{2}\\) , and \\(A\\) all obviously lie on \\(\\mathcal{C}_{a}\\) . (For \\(A_{1}\\) and \\(A_{2}\\) , the powers are both zero, and for the point \\(A\\) , we have \\(\\mathrm{Pow}(P,\\gamma_{b}) = AX\\cdot AA_{1}\\) and \\(\\mathrm{Pow}(P,\\gamma_{c}) = AY\\cdot AA_{1}\\) .) So \\(\\mathcal{C}_{a}\\) must be exactly the circumcircle of \\(\\triangle AA_{1}A_{2}\\) from the problem statement. \n\nWe turn to evaluating \\(k_{a}\\) more carefully. First, note that \n\n\\[\\angle A_{1}X B_{1} = \\angle A_{1}B_{2}B_{1} = \\angle C B_{2}B_{1} = 90^{\\circ} - \\angle B_{2}A C = 90^{\\circ} - (60^{\\circ} - \\gamma) = 30^{\\circ} + \\gamma .\\] \n\nNow using the law of sines, we derive \n\n\\[\\frac{A X}{A B_{1}} = \\frac{\\sin\\angle A B_{1}X}{\\sin\\angle A X B_{1}} = \\frac{\\sin(\\angle A_{1}X B_{1} - \\angle X A B_{1})}{\\sin\\angle A_{1}X B_{1}}\\] \\[\\qquad = \\frac{\\sin((30^{\\circ} + \\gamma) - (30^{\\circ} - \\beta))}{\\sin(30^{\\circ} + \\gamma)} = \\frac{\\sin(\\beta + \\gamma)}{\\sin(30^{\\circ} + \\gamma)}.\\] \n\nSimilarly, \\(A Y = A C_{1}\\cdot \\frac{\\sin(\\beta + \\gamma)}{\\sin(30^{\\circ} + \\beta)}\\) , so \n\n\\[k_{a} = \\frac{A X}{A Y} = \\frac{A B_{1}}{A C_{1}}\\cdot \\frac{\\sin(30^{\\circ} + \\beta)}{\\sin(30^{\\circ} + \\gamma)}.\\] \n\nNow define analogous constants \\(k_{b}\\) and \\(k_{c}\\) and circles \\(\\mathcal{C}_{b}\\) and \\(\\mathcal{C}_{c}\\) . Owing to the symmetry of the expressions, we have the key relation \n\n\\[k_{a}k_{b}k_{c} = 1.\\] \n\nIn summary, the three circles in the problem statement may be described as \n\n\\[\\mathcal{C}_{a} = (A A_{1}A_{2}) = \\{\\mathrm{points} P\\mathrm{such}\\mathrm{that}\\mathrm{Pow}(P,\\gamma_{b}) = k_{a}\\mathrm{Pow}(P,\\gamma_{c})\\}\\] \\[\\mathcal{C}_{b} = (B B_{1}B_{2}) = \\{\\mathrm{points} P\\mathrm{such}\\mathrm{that}\\mathrm{Pow}(P,\\gamma_{c}) = k_{b}\\mathrm{Pow}(P,\\gamma_{a})\\}\\] \\[\\mathcal{C}_{c} = (C C_{1}C_{2}) = \\{\\mathrm{points} P\\mathrm{such}\\mathrm{that}\\mathrm{Pow}(P,\\gamma_{a}) = k_{c}\\mathrm{Pow}(P,\\gamma_{b})\\} .\\] \n\nSince \\(k_{a}\\) , \\(k_{b}\\) , \\(k_{c}\\) have product 1, it follows that any point on at least two of the circles must lie on the third circle as well. The convexity of hexagon \\(A_{2}C_{1}B_{2}A_{1}C_{2}B_{1}\\) mentioned earlier ensures these any two of these circles do intersect at two different points, completing the solution.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2023-notes.jsonl", "problem_match": "6. ", "solution_match": "## \\(\\S 2.3\\) IMO 2023/6, proposed by Ankan Bhattacharya, Luke Robitaille (USA) \n"}}
+{"year": "2023", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "Let \\(n\\) be a positive integer. A Japanese triangle consists of \\(1 + 2 + \\dots +n\\) circles arranged in an equilateral triangular shape such that for each \\(1\\leq i\\leq n\\) , the \\(i^{\\mathrm{th}}\\) row contains exactly \\(i\\) circles, exactly one of which is colored red. A ninja path in a Japanese triangle is a sequence of \\(n\\) circles obtained by starting in the top row, then repeatedly going from a circle to one of the two circles immediately below it and finishing in the bottom row. Here is an example of a Japanese triangle with \\(n = 6\\) , along with a ninja path in that triangle containing two red circles. \n\n\n \n\nIn terms of \\(n\\) , find the greatest \\(k\\) such that in each Japanese triangle there is a ninja path containing at least \\(k\\) red circles.", "solution": "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)\n \n\nIn terms of \\(n\\) , find the greatest \\(k\\) such that in each Japanese triangle there is a ninja path containing at least \\(k\\) red circles. \n\nThe answer is \n\n\\[k = \\lfloor \\log_2(n)\\rfloor +1.\\] \n\nConstruction. It suffices to find a Japanese triangle for \\(n = 2^{e} - 1\\) with the property that at most \\(e\\) red circles in any ninja path. The construction shown below for \\(e = 4\\) obviously generalizes, and works because in each of the sets \\(\\{1\\}\\) , \\(\\{2,3\\}\\) , \\(\\{4,5,6,7\\}\\) , ..., \\(\\{2^{e - 1}, \\ldots , 2^{e} - 1\\}\\) , at most one red circle can be taken. (These sets are colored in different shades of red for visual clarity). \n\n\n\n\n\n\nBound. Conversely, we show that in any Japanese triangle, one can find a ninja path containing at least \n\n\\[k = \\lfloor \\log_2(n)\\rfloor +1.\\] \n\nThe following short solution was posted at https://aops.com/community/p28134004, apparently first found by the team leader for Iran. \n\nWe construct a rooted binary tree \\(T_{1}\\) on the set of all circles as follows. For each row, other than the bottom row: \n\nConnect the red circle to both circles under it; White circles to the left of the red circle in its row are connected to the left; White circles to the right of the red circle in its row are connected to the right. \n\nThe circles in the bottom row are all leaves of this tree. For example, the \\(n = 6\\) construction in the beginning gives the tree shown on the left half of the figure below: \n\n\n \n\nNow focus on only the red circles, as shown in the right half of the figure. We build a new rooted tree \\(T_{2}\\) where each red circle is joined to the red circle below it if there was a path of (zero or more) white circles in \\(T_{1}\\) between them. Then each red circle has at most 2 direct descendants in \\(T_{2}\\) . Hence the depth of the new tree \\(T_{2}\\) exceeds \\(\\log_2(n)\\) , which produces the desired path. \n\nAnother recursive proof of bound, communicated by Helio Ng. We give another proof that \\(\\lfloor \\log_2n\\rfloor +1\\) is always achievable. Define \\(f(i,j)\\) to be the maximum number of red circles contained in the portion of a ninja path from \\((1,1)\\) to \\((i,j)\\) , including the endpoints \\((1,1)\\) and \\((i,j)\\) . (If \\((i,j)\\) is not a valid circle in the triangle, define \\(f(i,j) = 0\\) for convenience.) An example is shown below with the values of \\(f(i,j)\\) drawn in the circles. \n\n\n\n\n\n\nWe have that \n\n\\[f(i,j) = \\max \\left\\{f(i - 1,j - 1),f(i,j - 1)\\right\\} +\\left\\{ \\begin{array}{l l}{1} & {\\mathrm{if~}(i,j)\\mathrm{~is~red~}}\\\\ {0} & {\\mathrm{otherwise}} \\end{array} \\right.\\] \n\nsince every ninja path passing through \\((i,j)\\) also passes through either \\((i - 1,j - 1)\\) or \\((i,j - 1)\\) . Now consider the quantity \\(S_{j} = f(0,j) + \\dots +f(j,j)\\) . We obtain the following recurrence: \n\n\\[\\mathrm{Claim} - S_{j + 1}\\geq S_{j} + \\left\\lceil \\frac{S_{j}}{j}\\right\\rceil +1.\\] \n\nProof. Consider a maximal element \\(f(m,j)\\) of \\(\\{f(0,j),\\ldots ,f(j,j)\\}\\) . We perform the following manipulations: \n\n\\[S_{j + 1} = \\sum_{i = 0}^{j + 1}\\max \\left\\{f(i - 1,j),f(i,j)\\right\\} +\\sum_{i = 0}^{j + 1}\\left\\{ \\begin{array}{l l}{1} & {\\mathrm{if~}(i,j + 1)\\mathrm{~is~red~}}\\\\ {0} & {\\mathrm{otherwise}} \\end{array} \\right.\\] \\[\\quad = \\sum_{i = 0}^{m}\\max \\left\\{f(i - 1,j),f(i,j)\\right\\} +\\sum_{i = m + 1}^{j}\\max \\left\\{f(i - 1,j),f(i,j)\\right\\} +1\\] \\[\\quad \\geq \\sum_{i = 0}^{m}f(i,j) + \\sum_{i = m + 1}^{j}f(i - 1,j) + 1\\] \\[\\quad = S_{j} + f(m,j) + 1\\] \\[\\quad \\geq S_{j} + \\left\\lceil \\frac{S_{j}}{j}\\right\\rceil +1\\] \n\nwhere the last inequality is due to Pigeonhole. \n\nThis is actually enough to solve the problem. Write \\(n = 2^{c} + r\\) , where \\(0 \\leq r \\leq 2^{c} - 1\\) . \n\nClaim — \\(S_{n} \\geq cn + 2r + 1\\) . In particular, \\(\\left\\lceil \\frac{S_{n}}{n} \\right\\rceil \\geq c + 1\\) . \n\nProof. First note that \\(S_{n} \\geq cn + 2r + 1\\) implies \\(\\left\\lceil \\frac{S_{n}}{n} \\right\\rceil \\geq c + 1\\) because \n\n\\[\\left\\lceil \\frac{S_{n}}{n}\\right\\rceil \\geq \\left\\lceil \\frac{cn + 2r + 1}{n}\\right\\rceil = c + \\left\\lceil \\frac{2r + 1}{n}\\right\\rceil = c + 1.\\] \n\nWe proceed by induction on \\(n\\) . The base case \\(n = 1\\) is clearly true as \\(S_{1} = 1\\) . Assuming that the claim holds for some \\(n = j\\) , we have \n\n\\[S_{j + 1}\\geq S_{j} + \\left\\lceil \\frac{S_{j}}{j}\\right\\rceil +1\\] \\[\\qquad \\geq cj + 2r + 1 + c + 1 + 1\\] \\[\\qquad = c(j + 1) + 2(r + 1) + 1\\] \n\nso the claim is proved for \\(n = j + 1\\) if \\(j + 1\\) is not a power of 2. If \\(j + 1 = 2^{c + 1}\\) , then by writing \\(c(j + 1) + 2(r + 1) + 1 = c(j + 1) + (j + 1) + 1 = (c + 2)(j + 1) + 1\\) , the claim is also proved. \\(\\square\\) \n\nNow \\(\\left\\lceil \\frac{S_{n}}{n}\\right\\rceil \\geq c + 1\\) implies the existence of some ninja path containing at least \\(c + 1\\) red circles, and we are done.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2023-notes.jsonl", "problem_match": "5. ", "solution_match": "## \\(\\S 2.2\\) IMO 2023/5, proposed by Merlijn Staps and Daniël Kroes (NLD) \n"}}
+{"year": "2023", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "Let \\(A B C\\) be an equilateral triangle. Let \\(A_{1}\\) , \\(B_{1}\\) , \\(C_{1}\\) be interior points of \\(A B C\\) such that \\(B A_{1} = A_{1}C\\) , \\(C B_{1} = B_{1}A\\) , \\(A C_{1} = C_{1}B\\) , and \n\n\\[\\angle B A_{1}C + \\angle C B_{1}A + \\angle A C_{1}B = 480^{\\circ}.\\]\n\n\n\nLet \\(A_{2} = \\overline{BC_{1}} \\cap \\overline{CB_{1}}\\) , \\(B_{2} = \\overline{CA_{1}} \\cap \\overline{AC_{1}}\\) , \\(C_{2} = \\overline{AB_{1}} \\cap \\overline{BA_{1}}\\) . Prove that if triangle \\(A_{1}B_{1}C_{1}\\) is scalene, then the circumcircles of triangles \\(AA_{1}A_{2}\\) , \\(BB_{1}B_{2}\\) , and \\(CC_{1}C_{2}\\) all pass through two common points.", "solution": "Lts. \n\nThis is the second official solution from the marking scheme, also communicated to me by Michael Ren. Define \\(O\\) as the center of \\(A B C\\) and set the angles \n\n\\[\\alpha := \\angle A_{1}C B = \\angle C B A_{1}\\] \\[\\beta := \\angle A C B_{1} = \\angle B_{1}A C\\] \\[\\gamma := \\angle C_{1}A B = \\angle C_{1}B A\\] \n\nso that \n\n\\[\\alpha +\\beta +\\gamma = 30^{\\circ}.\\] \n\nIn particular, \\(\\max (\\alpha ,\\beta ,\\gamma)< 30^{\\circ}\\) , so it follows that \\(A_{1}\\) lies inside \\(\\triangle O B C\\) , and similarly for the others. This means for example that \\(C_{1}\\) lies between \\(B\\) and \\(A_{2}\\) , and so on. Therefore the polygon \\(A_{2}C_{1}B_{2}A_{1}C_{2}B_{1}\\) is convex.\n\n\n\n \n\nWe start by providing the \"interpretation\" for the \\(480^{\\circ}\\) angle in the statement: \n\nClaim — Point \\(A_{1}\\) is the circumcenter of \\(\\triangle A_{2}BC\\) , and similarly for the others. \n\nProof. We have \\(\\angle B A_{1}C = 180^{\\circ} - 2\\alpha\\) , and \n\n\\[\\angle B A_{2}C = 180^{\\circ} - \\angle C B C_{1} - \\angle B_{1}C B\\] \\[\\qquad = 180^{\\circ} - (60^{\\circ} - \\gamma) - (60^{\\circ} - \\beta)\\] \\[\\qquad = 60^{\\circ} + \\beta +\\gamma = 90^{\\circ} - \\alpha = \\frac{1}{2}\\angle B A_{1}C.\\] \n\nSince \\(A_{1}\\) lies inside \\(\\triangle B A_{2}C\\) , it follows \\(A_{1}\\) is exactly the circumcenter. \n\nClaim — Quadrilateral \\(B_{2}C_{1}B_{1}C\\) can be inscribed in a circle, say \\(\\gamma_{a}\\) . Circles \\(\\gamma_{b}\\) and \\(\\gamma_{c}\\) can be defined similarly. Finally, these three circles are pairwise distinct. \n\nProof. Using directed angles now, we have \n\n\\[\\angle B_{2}B_{1}C_{2} = 180^{\\circ} - \\angle A B_{1}B_{2} = 180^{\\circ} - 2\\angle A C B = 180^{\\circ} - 2(60^{\\circ} - \\alpha) = 60^{\\circ} + 2\\alpha .\\] \n\nBy the same token, \\(\\angle B_{2}C_{1}C_{2} = 60^{\\circ} + 2\\alpha\\) . This establishes the existence of \\(\\gamma_{a}\\) . \n\nThe proof for \\(\\gamma_{b}\\) and \\(\\gamma_{c}\\) is the same. Finally, to show the three circles are distinct, it would be enough to verify that the convex hexagon \\(A_{2}C_{1}B_{2}A_{1}C_{2}B_{1}\\) is not cyclic.\n\n\n\nAssume for contradiction it was cyclic. Then \n\n\\[360^{\\circ} = \\angle C_{2}A_{1}B_{2} + \\angle B_{2}C_{1}A_{2} + \\angle A_{2}B_{1}C_{2} = \\angle B A_{1}C + \\angle C B_{1}A + \\angle A C_{1}B = 480^{\\circ}\\] \n\nwhich is absurd. This contradiction eliminates the degenerate case, so the three circles are distinct. \\(\\square\\) \n\nFor the remainder of the solution, let \\(\\mathrm{Pow}(P,\\omega)\\) denote the power of a point \\(P\\) with respect to a circle \\(\\omega\\) . \n\nLet line \\(A A_{1}\\) meet \\(\\gamma_{b}\\) and \\(\\gamma_{c}\\) again at \\(X\\) and \\(Y\\) , and set \\(k_{a}:= \\frac{A X}{A Y}\\) . Consider the locus of all points \\(P\\) such that \n\n\\[\\mathcal{C}_{a}:= \\Big\\{\\mathrm{points} P\\mathrm{in}\\mathrm{the}\\mathrm{plane}\\mathrm{satisfying}\\mathrm{Pow}(P,\\gamma_{b}) = k_{a}\\mathrm{Pow}(P,\\gamma_{c})\\Big\\} .\\] \n\nWe recall the coaxiality lemma \\(^{1}\\) , which states that (given \\(\\gamma_{b}\\) and \\(\\gamma_{c}\\) are not concentric) the locus \\(\\mathcal{C}_{a}\\) must be either a circle (if \\(k_{a}\\neq 1\\) ) or a line (if \\(k_{a} = 1\\) ). \n\nOn the other hand, \\(A_{1}\\) , \\(A_{2}\\) , and \\(A\\) all obviously lie on \\(\\mathcal{C}_{a}\\) . (For \\(A_{1}\\) and \\(A_{2}\\) , the powers are both zero, and for the point \\(A\\) , we have \\(\\mathrm{Pow}(P,\\gamma_{b}) = AX\\cdot AA_{1}\\) and \\(\\mathrm{Pow}(P,\\gamma_{c}) = AY\\cdot AA_{1}\\) .) So \\(\\mathcal{C}_{a}\\) must be exactly the circumcircle of \\(\\triangle AA_{1}A_{2}\\) from the problem statement. \n\nWe turn to evaluating \\(k_{a}\\) more carefully. First, note that \n\n\\[\\angle A_{1}X B_{1} = \\angle A_{1}B_{2}B_{1} = \\angle C B_{2}B_{1} = 90^{\\circ} - \\angle B_{2}A C = 90^{\\circ} - (60^{\\circ} - \\gamma) = 30^{\\circ} + \\gamma .\\] \n\nNow using the law of sines, we derive \n\n\\[\\frac{A X}{A B_{1}} = \\frac{\\sin\\angle A B_{1}X}{\\sin\\angle A X B_{1}} = \\frac{\\sin(\\angle A_{1}X B_{1} - \\angle X A B_{1})}{\\sin\\angle A_{1}X B_{1}}\\] \\[\\qquad = \\frac{\\sin((30^{\\circ} + \\gamma) - (30^{\\circ} - \\beta))}{\\sin(30^{\\circ} + \\gamma)} = \\frac{\\sin(\\beta + \\gamma)}{\\sin(30^{\\circ} + \\gamma)}.\\] \n\nSimilarly, \\(A Y = A C_{1}\\cdot \\frac{\\sin(\\beta + \\gamma)}{\\sin(30^{\\circ} + \\beta)}\\) , so \n\n\\[k_{a} = \\frac{A X}{A Y} = \\frac{A B_{1}}{A C_{1}}\\cdot \\frac{\\sin(30^{\\circ} + \\beta)}{\\sin(30^{\\circ} + \\gamma)}.\\] \n\nNow define analogous constants \\(k_{b}\\) and \\(k_{c}\\) and circles \\(\\mathcal{C}_{b}\\) and \\(\\mathcal{C}_{c}\\) . Owing to the symmetry of the expressions, we have the key relation \n\n\\[k_{a}k_{b}k_{c} = 1.\\] \n\nIn summary, the three circles in the problem statement may be described as \n\n\\[\\mathcal{C}_{a} = (A A_{1}A_{2}) = \\{\\mathrm{points} P\\mathrm{such}\\mathrm{that}\\mathrm{Pow}(P,\\gamma_{b}) = k_{a}\\mathrm{Pow}(P,\\gamma_{c})\\}\\] \\[\\mathcal{C}_{b} = (B B_{1}B_{2}) = \\{\\mathrm{points} P\\mathrm{such}\\mathrm{that}\\mathrm{Pow}(P,\\gamma_{c}) = k_{b}\\mathrm{Pow}(P,\\gamma_{a})\\}\\] \\[\\mathcal{C}_{c} = (C C_{1}C_{2}) = \\{\\mathrm{points} P\\mathrm{such}\\mathrm{that}\\mathrm{Pow}(P,\\gamma_{a}) = k_{c}\\mathrm{Pow}(P,\\gamma_{b})\\} .\\] \n\nSince \\(k_{a}\\) , \\(k_{b}\\) , \\(k_{c}\\) have product 1, it follows that any point on at least two of the circles must lie on the third circle as well. The convexity of hexagon \\(A_{2}C_{1}B_{2}A_{1}C_{2}B_{1}\\) mentioned earlier ensures these any two of these circles do intersect at two different points, completing the solution.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2023-notes.jsonl", "problem_match": "6. ", "solution_match": "## \\(\\S 2.3\\) IMO 2023/6, proposed by Ankan Bhattacharya, Luke Robitaille (USA) \n"}}
diff --git a/IMO/segmented/en-IMO-2024-notes.jsonl b/IMO/segmented/en-IMO-2024-notes.jsonl
index 5b5f7d045007b646a8866ffc1b3b636e346e2a0b..9b40d1e57b3c5411696e569c6bf82085b383c972 100644
--- a/IMO/segmented/en-IMO-2024-notes.jsonl
+++ b/IMO/segmented/en-IMO-2024-notes.jsonl
@@ -1,6 +1,6 @@
{"year": "2024", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "Find all real numbers \\(\\alpha\\) so that, for every positive integer \\(n\\) , the integer \n\n\\[|\\alpha | + |2\\alpha | + |3\\alpha | + \\dots +|n\\alpha |\\] \n\nis divisible by \\(n\\)", "solution": ". \n\nThe answer is that \\(\\alpha\\) must be an even integer. Let \\(S(n,\\alpha)\\) denote the sum in question. \n\n\\(\\P\\) Analysis for \\(\\alpha\\) an integer. If \\(\\alpha\\) is an integer, then the sum equals \n\n\\[S(n,\\alpha) = (1 + 2 + \\cdot \\cdot \\cdot +n)\\alpha = \\frac{n(n + 1)}{2}\\cdot \\alpha\\] \n\nwhich is obviously a multiple of \\(n\\) if \\(2\\mid \\alpha\\) ; meanwhile, if \\(\\alpha\\) is an odd integer then \\(n = 2\\) gives a counterexample. \n\n\\(\\P\\) Main case. Suppose \\(\\alpha\\) is not an integer; we show the desired condition can never be true. Note that replacing \\(\\alpha\\) with \\(\\alpha \\pm 2\\) changes by \n\n\\[S(n\\pm 2,\\alpha) - S(n,\\alpha) = 2(1 + 2 + \\cdot \\cdot \\cdot +n) = n(n + 1)\\equiv 0\\pmod {n}\\] \n\nfor every \\(n\\) . Thus, by shifting appropriately we may assume \\(- 1< \\alpha < 1\\) and \\(\\alpha \\notin \\mathbb{Z}\\) \n\n. If \\(0< \\alpha < 1\\) , then let \\(m\\geq 2\\) be the smallest integer such that \\(m\\alpha \\geq 1\\) . Then \n\n\\[S(m,\\alpha) = \\underbrace{0 + \\cdot \\cdot \\cdot +0 + 1}_{m - 1\\mathrm{~terms}} = 1\\] \n\nis not a multiple of \\(m\\) \n\n. If \\(- 1< \\alpha < 0\\) , then let \\(m\\geq 2\\) be the smallest integer such that \\(m\\alpha \\leq - 1\\) . Then \n\n\\[S(m,\\alpha) = \\underbrace{(-1) + \\cdot \\cdot \\cdot +(-1)}_{m - 1\\mathrm{~terms}} + 0 = -(m - 1)\\] \n\nis not a multiple of \\(m\\)", "metadata": {"resource_path": "IMO/segmented/en-IMO-2024-notes.jsonl", "problem_match": "1. ", "solution_match": "## \\(\\S 1.1\\) IMO 2024/1, proposed by Santiago Rodriguez (COL) \n"}}
{"year": "2024", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "For which pairs of positive integers \\((a,b)\\) is the sequence \n\n\\[\\gcd (a^{n} + b,b^{n} + a)\\qquad n = 1,2,\\ldots\\] \n\neventually constant?", "solution": "The answer is \\((a,b) = (1,1)\\) only, which obviously works since the sequence is always 2. Conversely, assume the sequence \n\n\\[x_{n}:= \\gcd (a^{n} + b,b^{n} + a)\\] \n\nis eventually constant. The main crux of the other direction is to consider \n\n\\[M:= ab + 1.\\] \n\nRemark (Motivation). The reason to consider the number is the same technique used in IMO 2005/4, namely the idea to consider \" \\(n = - 1\\) \". The point is that the two rational numbers \n\n\\[\\frac{1}{a} +b = \\frac{ab + 1}{b},\\qquad \\frac{1}{b} +a = \\frac{ab + 1}{a}\\] \n\nhave a large common factor: we could write \" \\(x_{- 1} = ab + 1\\) \", loosely speaking. \n\nNow, the sequence is really only defined for \\(n\\geq 1\\) , so one should instead take \\(n\\equiv - 1\\) (mod \\(\\phi (M)\\) ) — and this is exactly what we do. \n\nObviously \\(\\gcd (a,M) = \\gcd (b,M) = 1\\) . Let \\(n\\) be a sufficiently large multiple of \\(\\phi (M)\\) so that \n\n\\[x_{n - 1} = x_{n} = x_{n + 1} = \\dots .\\] \n\nWe consider the first three terms: the first one is the \"key\" one that gets the bulk of the work, and the rest is bookkeeping and extraction. \n\nConsider \\(x_{n - 1}\\) . Note that \n\n\\[a(a^{n - 1} + b) = a^{n} + ab\\equiv 1 + (-1)\\equiv 0\\pmod {M}\\] \n\nand similarly \\(b(b^{n} + a)\\equiv 0\\) (mod \\(M\\) ). Hence \\(M\\mid x_{n - 1}\\) \n\nConsider \\(x_{n}\\) , which is now known to be divisible by \\(M\\) . Note that \n\n\\[0\\equiv a^{n} + b\\equiv 1 + b\\pmod {M\\] \\[0\\equiv b^{n} + a\\equiv 1 + a\\pmod {M\\] \n\nSo \\(a\\equiv b\\equiv - 1\\) (mod \\(M\\) ). \n\nConsider \\(x_{n + 1}\\) , which is now known to be divisible by \\(M\\) . Note that \n\n\\[0\\equiv a^{n + 1} + b\\equiv b^{n + 1} + a\\equiv a + b\\pmod {M}.\\] \n\nWe knew \\(a\\equiv b\\equiv - 1\\) (mod \\(M\\) ), hence this means \\(0\\equiv 2\\) (mod \\(M\\) ), so \\(M = 2\\) . From \\(M = 2\\) we then conclude \\(a = b = 1\\) , as desired.\n\n\n\nRemark (No alternate solutions known). At the time nobody seems to know any solution not depending critically on \\(M = ab + 1\\) (or prime numbers dividing \\(M\\) , etc.). They vary in execution once some term of the form \\(x_{k \\varphi (n) - 1}\\) is taken, but avoiding the key idea altogether does not currently seem possible. \n\nA good example to consider for ruling out candidate ideas is \\((a, b) = (18, 9)\\) .", "metadata": {"resource_path": "IMO/segmented/en-IMO-2024-notes.jsonl", "problem_match": "2. ", "solution_match": "## \\(\\S 1.2\\) IMO 2024/2, proposed by Valentino Iverson (IDN) \n"}}
-{"year": "2024", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "Let \\(a_{1}\\) , \\(a_{2}\\) , \\(a_{3}\\) , ... be an infinite sequence of positive integers, and let \\(N\\) be a positive integer. Suppose that, for each \\(n > N\\) , the number \\(a_{n}\\) is equal to the number of times \\(a_{n - 1}\\) appears in the list \\((a_{1},a_{2},\\ldots ,a_{n - 1})\\) . Prove that at least one of the sequences \\(a_{1}\\) , \\(a_{3}\\) , \\(a_{5}\\) , ... and \\(a_{2}\\) , \\(a_{4}\\) , \\(a_{6}\\) , ... is eventually periodic.", "solution": "We present the solution from \"gigamilkmen'tgeg\" in https://aops.com/community/p31224483, with some adaptation from the first shortlist official solution as well. Set \\(M:= \\max (a_{1},\\ldots ,a_{N})\\) . \n\n\\(\\P\\) Setup. We will visualize the entire process as follows. We draw a stack of towers labeled 1, 2, ..., each initially empty. For \\(i = 1,2,\\ldots\\) , we imagine the term \\(a_{i}\\) as adding a block \\(B_{i}\\) to tower \\(a_{i}\\) . \n\nThen there are \\(N\\) initial blocks placed, colored red. The rest of the blocks are colored yellow: if the last block \\(B_{i}\\) was added to a tower that then reaches height \\(a_{i + 1}\\) , the next block \\(B_{i + 1}\\) is added to tower \\(a_{i + 1}\\) . We'll say \\(B_{i}\\) contributes to the tower containing \\(B_{i + 1}\\) . \n\nIn other words, the yellow blocks \\(B_{i}\\) for \\(i > N\\) are given coordinates \\(B_{i} = (a_{i},a_{i + 1})\\) for \\(i > N\\) . Note in particular that in towers \\(M + 1\\) , \\(M + 2\\) , ..., the blocks are all yellow. \n\n\n \n\nWe let \\(h_{\\ell}\\) denote the height of the \\(\\ell^{\\mathrm{th}}\\) tower at a given time \\(n\\) . (This is an abuse of notation and we should write \\(h_{\\ell}(n)\\) at time \\(n\\) , but \\(n\\) will always be clear from context.) \n\n\\(\\P\\) Up to alternating up and down. We start with two independent easy observations: the set of numbers that occur infinitely often is downwards closed, and consecutive terms cannot both be huge.\n\n\n\n \n\nClaim — If the \\((k + 1)^{\\mathrm{st}}\\) tower grows arbitrarily high, so does tower \\(k\\) . In fact, there exists a constant \\(C\\) such that \\(h_{k} \\geq h_{k + 1} - C\\) at all times. \n\nProof. Suppose \\(B_{n}\\) is a yellow block in tower \\(k + 1\\) . Then with at most finitely many exceptions, \\(B_{n - 1}\\) is a yellow block at height \\(k + 1\\) , and the block \\(B_{r}\\) right below \\(B_{n - 1}\\) is also yellow; then \\(B_{r + 1}\\) is in tower \\(k\\) . Hence, with at most finitely many exceptions, the map \n\n\\[B_{n}\\mapsto B_{n - 1}\\mapsto B_{r}\\mapsto B_{r + 1}\\] \n\nprovides an injective map taking each yellow block in tower \\(k + 1\\) to a yellow block in tower \\(k\\) . (The figure above shows \\(B_{32} \\to B_{31} \\to B_{19} \\to B_{20}\\) as an example.) \\(\\square\\) \n\nClaim — If \\(a_{n} > M\\) then \\(a_{n + 1} \\leq M\\) . \n\nProof. Assume for contradiction there's a first moment where \\(a_{n} > M\\) and \\(a_{n + 1} > M\\) , meaning the block \\(B_{n}\\) was added to an all- yellow tower past \\(M\\) that has height exceeding \\(M\\) . (This is the X'ed out region in the figure above.) In \\(B_{n}\\) 's tower, every (yellow) block (including \\(B_{n}\\) ) was contributed by a block placed in different towers at height \\(a_{n} > M\\) . So before \\(B_{n}\\) , there were already \\(a_{n + 1} > M\\) towers of height more than \\(M\\) . This contradicts minimality of \\(n\\) . \\(\\square\\) \n\nIt follows that the set of indices with \\(a_{n} \\leq M\\) has arithmetic density at least half, so certainly at least some of the numbers must occur infinitely often. Of the numbers in \\(\\{1, 2, \\ldots , M\\}\\) , define \\(L\\) such that towers 1 through \\(L\\) grow unbounded but towers \\(L + 1\\) through \\(M\\) do not. Then we can pick a larger threshold \\(N' > N\\) such that\n\n\n\n- Towers 1 through \\(L\\) have height greater than \\((M, N)\\) ; \n\n- Towers \\(L + 1\\) through \\(M\\) will receive no further blocks; \n\n- \\(a_{N^{\\prime}} \\leq L\\) . \n\nAfter this threshold, the following statement is true: \n\nClaim (Alternating small and big) — The terms \\(a_{N^{\\prime}}\\) , \\(a_{N^{\\prime} + 2}\\) , \\(a_{N^{\\prime} + 4}\\) , ... are all at most \\(L\\) while the terms \\(a_{N^{\\prime} + 1}\\) , \\(a_{N^{\\prime} + 3}\\) , \\(a_{N^{\\prime} + 5}\\) , ... are all greater than \\(M\\) . \n\n\\(\\P\\) Automaton for \\(n \\equiv N^{\\prime} \\pmod {2}\\) . From now on we always assume \\(n > N^{\\prime}\\) . When \\(n \\equiv N^{\\prime} \\pmod {2}\\) , i.e., when \\(a_{n}\\) is small, we define the state \n\n\\[S(n) = (h_{1}, h_{2}, \\ldots , h_{L}; a_{n}).\\] \n\nFor example, in the figure below, we illustrate how \n\n\\[S(34) = (9, 11; a_{34} = 1) \\longrightarrow S(36) = (9, 12; a_{36} = 2)\\] \n\n\n \n\nThe final element \\(a_{n}\\) simply reminds us which tower was most recently incremented. At this point we can give a complete description of how to move from \\(S(n)\\) to \\(S(n + 2)\\) : \n\n- The intermediate block \\(B_{n + 1}\\) is placed in the tower corresponding to the height \\(a_{n + 1}\\) of \\(B_{n}\\) ; \n\n- That tower will have height \\(a_{n + 2}\\) equal to the number of towers with height at least \\(a_{n + 1}\\) ; that is, it equals the cardinality of the set \n\n\\[\\{i: h_{i} \\geq h_{a_{n}}\\}\\]\n\n\n\n- We increment \\(h_{a_{n + 2}}\\) by 1 and update \\(a_{n}\\) . \n\nFor example, the illustrated \\(S(34) \\to S(36)\\) corresponds to the block \\(B_{34}\\) at height \\(h_{1}\\) in tower 1 giving the block \\(B_{35}\\) at height 2 in tower \\(h_{1}\\) , then block \\(B_{36}\\) at height \\(h_{2} + 1\\) being placed in tower 2. \n\nPigeonhole periodicity argument. Because only the relative heights matter in the automata above, if we instead define \n\n\\[T(n) = (h_{1} - h_{2},h_{2} - h_{3},\\ldots ,h_{L - 1} - h_{L};a_{n}).\\] \n\nthen \\(T(n + 2)\\) can be determined from just \\(T(n)\\) . \n\nSo it would be sufficient to show \\(T(n)\\) only takes on finitely many values to show that \\(T(n)\\) (and hence \\(a_{n}\\) ) is eventually periodic. \n\nSince we have the bound \\(h_{k + 1} \\leq h_{k} + C\\) , we are done upon proving the following lower bound: \n\nClaim — For every \\(1 \\leq \\ell < L\\) and \\(n > N'\\) , we have \\(h_{\\ell} \\leq h_{\\ell + 1} + C \\cdot (L - 1)\\) . \n\nProof. Assume for contradiction that there is some moment \\(n > N'\\) such that \n\n\\[h_{\\ell} > h_{\\ell +1} + C\\cdot (L - 1)\\] \n\nand WLOG assume that \\(h_{\\ell}\\) was just updated at the moment \\(n\\) . Together with \\(h_{k + 1} \\leq h_{k} + C\\) for all \\(k\\) and triangle inequality, we conclude \n\n\\[\\min (h_{1},\\ldots ,h_{\\ell}) > q:= \\max (h_{\\ell +1},\\ldots ,h_{L}).\\] \n\nWe find that the blocks now in fact alternate between being placed among the first \\(\\ell\\) towers and in towers with indices greater than \\(q\\) thereafter. Hence the heights \\(h_{\\ell +1}\\) , ..., \\(h_{L}\\) never grow after this moment. This contradicts the definition of \\(L\\) . \\(\\square\\) \n\nRemark. In fact, it can be shown that the period is actually exactly \\(L\\) , meaning the periodic part will be exactly a permutation of \\((1,2,\\ldots ,L)\\) . For any \\(L\\) , it turns out there is indeed a permutation achieving that periodic part.\n\n\n\n## \\(\\S 2\\) Solutions to Day 2", "metadata": {"resource_path": "IMO/segmented/en-IMO-2024-notes.jsonl", "problem_match": "3. ", "solution_match": "## \\(\\S 1.3\\) IMO 2024/3, proposed by William Steinberg (AUS) \n"}}
-{"year": "2024", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "Let triangle \\(A B C\\) with incenter \\(I\\) satisfying \\(A B< A C< B C\\) . Let \\(X\\) be a point on line \\(B C\\) , different from \\(C\\) , such that the line through \\(X\\) and parallel to \\(A C\\) is tangent to the incircle. Similarly, let \\(Y\\) be a point on line \\(B C\\) , different from \\(B\\) , such that the line through \\(Y\\) and parallel to \\(A B\\) is tangent to the incircle. Line \\(A I\\) intersects the circumcircle of triangle \\(A B C\\) again at \\(P\\) . Let \\(K\\) and \\(L\\) be the midpoints of \\(A C\\) and \\(A B\\) , respectively. Prove that \\(\\angle K I L + \\angle Y P X = 180^{\\circ}\\) .", "solution": "be the reflection of \\(A\\) over \\(I\\) , the most important point to add since it gets rid of \\(K\\) and \\(L\\) as follows. \n\nClaim — We have \\(\\angle KIL = \\angle BTC\\) , and lines \\(TX\\) and \\(TY\\) are tangent to the incircle. \n\nProof. The first part is true since \\(\\triangle BTC\\) is the image of \\(\\triangle KIL\\) under a homothety of ratio 2. The second part is true because lines \\(AB\\) , \\(AC\\) , \\(TX\\) , \\(TY\\) determine a rhombus with center \\(I\\) . \\(\\square\\) \n\nWe thus delete \\(K\\) and \\(L\\) from the picture altogether; they aren't needed anymore. \n\n\n\n\n\n\nClaim — We have \\(B X P T\\) and \\(C Y P T\\) are cyclic. \n\nProof. \\(\\angle T Y C = \\angle T Y B = \\angle A B C = \\angle A P C = \\angle T P C\\) and similarly. (Some people call this Reim's theorem.) \\(\\square\\) \n\nTo finish, observe that \n\n\\[\\angle C T B = \\angle C T P + \\angle P T B = \\angle C Y P + \\angle P X B = \\angle X Y P + \\angle X Y P = \\angle X P Y\\] \n\nas desired. (The length conditions \\(A C > A B > B C\\) ensure that \\(B\\) , \\(X\\) , \\(Y\\) , \\(C\\) are collinear in that order, and that \\(T\\) lies on the opposite side of \\(\\overline{B C}\\) as \\(A\\) . Hence the directed equality \\(\\angle C T B = \\angle X P Y\\) translates to the undirected \\(\\angle B T C + \\angle X P Y = 180^{\\circ}\\) .)", "metadata": {"resource_path": "IMO/segmented/en-IMO-2024-notes.jsonl", "problem_match": "4. ", "solution_match": "## \\(\\S 2.1\\) IMO 2024/4, proposed by Dominik Burek (POL) \n"}}
-{"year": "2024", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "Turbo the snail is in the top row of a grid with 2024 rows and 2023 columns and wants to get to the bottom row. However, there are 2022 hidden monsters, one in every row except the first and last, with no two monsters in the same column. \n\nTurbo makes a series of attempts to go from the first row to the last row. On each attempt, he chooses to start on any cell in the first row, then repeatedly moves to an orthogonal neighbor. (He is allowed to return to a previously visited cell.) If Turbo reaches a cell with a monster, his attempt ends and he is transported back to the first row to start a new attempt. The monsters do not move between attempts, and Turbo remembers whether or not each cell he has visited contains a monster. If he reaches any cell in the last row, his attempt ends and Turbo wins. \n\nFind the smallest integer \\(n\\) such that Turbo has a strategy which guarantees being able to reach the bottom row in at most \\(n\\) attempts, regardless of how the monsters are placed.", "solution": "Surprisingly the answer is \\(n = 3\\) for any grid size \\(s\\times (s - 1)\\) when \\(s\\geq 4\\) . We prove this in that generality. \n\n\\(\\P\\) Proof that at least three attempts are needed. When Turbo first moves into the second row, Turbo could encounter a monster \\(M_{1}\\) right away. Then on the next attempt, Turbo must enter the third row in different column as \\(M_{1}\\) , and again could encounter a monster \\(M_{2}\\) right after doing so. This means no strategy can guarantee fewer than three attempts. \n\n\\(\\P\\) Strategy with three attempts. On the first attempt, we have Turbo walk through the entire second row until he finds the monster \\(M_{1}\\) in it. Then we get two possible cases. \n\nCase where \\(M_{1}\\) is not on the edge. In the first case, if that monster \\(M_{1}\\) is not on the edge of the row, then Turbo can trace two paths below it as shown below. At least one of these paths works, hence three attempts is sufficient. \n\n\n \n\nCase where \\(M_{1}\\) is on the edge. WLOG, \\(M_{1}\\) is in the leftmost cell. Then Turbo follows the green staircase pattern shown in the left figure below. If the staircase is free\n\n\n\nof monsters, then Turbo wins on the second attempt. Otherwise, if a monster \\(M_{2}\\) is encountered on the staircase, Turbo has found a safe path to the left of \\(M_{2}\\) ; then Turbo can use this to reach the column \\(M_{1}\\) is in, and escape from there. This is shown in purple in the center and right figure (there are two slightly different cases depending on whether \\(M_{2}\\) was encountered going east or south). \n\n\n \n\nThus the problem is solved in three attempts, as promised. \n\n\\(\\P\\) Extended remark: all working strategies look similar to this. As far as we know, all working strategies are variations of the above. In fact, we will try to give a description of the space of possible strategies, although this needs a bit of notation. \n\nDefinition. For simplicity, we only use \\(s\\) even in the figures below. We define the happy triangle as the following cells: \n\nAll \\(s - 1\\) cells in the first row (which has no monsters). The center \\(s - 3\\) cells in the second row. The center \\(s - 5\\) cells in the third row. The center cell in the \\(\\frac{s}{2}\\) th row. \n\nFor \\(s = 12\\) , the happy triangle is the region shaded in the thick border below. \n\n\n\n\n\n\nDefinition. Given a cell, define a shoulder to be the cell directly northwest or northeast of it. Hence there are two shoulders of cells outside the first and last column, and one shoulder otherwise. \n\nThen solutions roughly must distinguish between these two cases: \n\n- Inside happy triangle: If the first monster \\(M_{1}\\) is found in the happy triangle, and there is a safe path found by Turbo to the two shoulders (marked \\(\\bigstar\\) in the figure), then one can finish in two more moves by considering the two paths from \\(\\bigstar\\) that cut under the monster \\(M_{1}\\) ; one of them must work. This slightly generalizes the easier case in the solution above (which focuses only on the case where \\(M_{1}\\) is in the first row). \n\n\n \n\n- Outside happy triangle: Now suppose the first monster \\(M_{1}\\) is outside the happy triangle. Of the two shoulders, take the one closer to the center (if in the center column, either one works; if only one shoulder, use it). If there is a safe path to that shoulder, then one can take a staircase pattern towards the center, as shown in the figure. In that case, the choice of shoulder and position guarantees the staircase reaches the bottom row, so that if no monster is along this path, the algorithm ends. Otherwise, if one encounters a second monster along the staircase, then one can use the third trial to cut under the monster \\(M_{1}\\) . \n\n\n \n\nWe now prove the following proposition: in any valid strategy for Turbo, in the case where Turbo first encounters a monster upon leaving the happy triangle, the second path must outline the same staircase shape.\n\n\n\nThe monsters pre- commit to choosing their pattern to be either a NW- SE diagonal or NE- SW diagonal, with a single one- column gap; see figure below for an example. Note that this forces any valid path for Turbo to pass through the particular gap. \n\n\n \n\nWe may assume without loss of generality that Turbo first encounters a monster \\(M_{1}\\) when Turbo first leaves the happy triangle, and that this forces an NW- SE configuration. \n\n\n \n\nThen the following is true:\n\n\n\n## Proposition \n\nPropositionThe strategy of Turbo on the second path must visit every cell in \"slightly raised diagonal\" marked with \\(\\clubsuit\\) in the figure above in order from top to bottom, until it encounters a second Monster \\(M_{2}\\) (or reaches the bottom row and wins anyway). It's both okay and irrelevant if Turbo visits other cells above this diagonal, but the marked cells must be visited from top to bottom in that order. \n\nProof. If Turbo tries to sidestep by visiting the cell southeast of \\(M_{1}\\) (marked \\(X\\) in the Figure), then Turbo clearly cannot finish after this (for \\(s\\) large enough). Meanwhile, suppose Turbo tries to \"skip\" one of the \\(\\clubsuit\\) , say in column \\(C\\) , then the gap could equally well be in the column to the left of \\(C\\) . This proves the proposition. \\(\\square\\) \n\nRemark (Memories of safe cells are important, not just monster cells). Here is one additional observation that one can deduce from this. We say a set \\(S\\) of revealed monsters is called obviously winnable if, based on only the positions of the monsters (and not the moves or algorithm that were used to obtain them), one can identify a guaranteed winning path for Turbo using only \\(S\\) . For example, two monsters in adjacent columns which are not diagonally adjacent is obviously winnable. \n\nThen no strategy can guarantee obtaining an obviously winnable set in 2 moves (or even \\(k\\) moves for any constant \\(k\\) , if \\(s\\) is large enough in terms of \\(k\\) ). So any valid strategy must also use the memory of identified safe cells that do not follow just from the revealed monster positions.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2024-notes.jsonl", "problem_match": "5. ", "solution_match": "## \\(\\S 2.2\\) IMO 2024/5, proposed by Chu Cheuk Hei (HKG) \n"}}
+{"year": "2024", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "Let \\(a_{1}\\) , \\(a_{2}\\) , \\(a_{3}\\) , ... be an infinite sequence of positive integers, and let \\(N\\) be a positive integer. Suppose that, for each \\(n > N\\) , the number \\(a_{n}\\) is equal to the number of times \\(a_{n - 1}\\) appears in the list \\((a_{1},a_{2},\\ldots ,a_{n - 1})\\) . Prove that at least one of the sequences \\(a_{1}\\) , \\(a_{3}\\) , \\(a_{5}\\) , ... and \\(a_{2}\\) , \\(a_{4}\\) , \\(a_{6}\\) , ... is eventually periodic.", "solution": "We present the solution from \"gigamilkmen'tgeg\" in https://aops.com/community/p31224483, with some adaptation from the first shortlist official solution as well. Set \\(M:= \\max (a_{1},\\ldots ,a_{N})\\) . \n\n\\(\\P\\) Setup. We will visualize the entire process as follows. We draw a stack of towers labeled 1, 2, ..., each initially empty. For \\(i = 1,2,\\ldots\\) , we imagine the term \\(a_{i}\\) as adding a block \\(B_{i}\\) to tower \\(a_{i}\\) . \n\nThen there are \\(N\\) initial blocks placed, colored red. The rest of the blocks are colored yellow: if the last block \\(B_{i}\\) was added to a tower that then reaches height \\(a_{i + 1}\\) , the next block \\(B_{i + 1}\\) is added to tower \\(a_{i + 1}\\) . We'll say \\(B_{i}\\) contributes to the tower containing \\(B_{i + 1}\\) . \n\nIn other words, the yellow blocks \\(B_{i}\\) for \\(i > N\\) are given coordinates \\(B_{i} = (a_{i},a_{i + 1})\\) for \\(i > N\\) . Note in particular that in towers \\(M + 1\\) , \\(M + 2\\) , ..., the blocks are all yellow. \n\n\n \n\nWe let \\(h_{\\ell}\\) denote the height of the \\(\\ell^{\\mathrm{th}}\\) tower at a given time \\(n\\) . (This is an abuse of notation and we should write \\(h_{\\ell}(n)\\) at time \\(n\\) , but \\(n\\) will always be clear from context.) \n\n\\(\\P\\) Up to alternating up and down. We start with two independent easy observations: the set of numbers that occur infinitely often is downwards closed, and consecutive terms cannot both be huge.\n\n\n\n \n\nClaim — If the \\((k + 1)^{\\mathrm{st}}\\) tower grows arbitrarily high, so does tower \\(k\\) . In fact, there exists a constant \\(C\\) such that \\(h_{k} \\geq h_{k + 1} - C\\) at all times. \n\nProof. Suppose \\(B_{n}\\) is a yellow block in tower \\(k + 1\\) . Then with at most finitely many exceptions, \\(B_{n - 1}\\) is a yellow block at height \\(k + 1\\) , and the block \\(B_{r}\\) right below \\(B_{n - 1}\\) is also yellow; then \\(B_{r + 1}\\) is in tower \\(k\\) . Hence, with at most finitely many exceptions, the map \n\n\\[B_{n}\\mapsto B_{n - 1}\\mapsto B_{r}\\mapsto B_{r + 1}\\] \n\nprovides an injective map taking each yellow block in tower \\(k + 1\\) to a yellow block in tower \\(k\\) . (The figure above shows \\(B_{32} \\to B_{31} \\to B_{19} \\to B_{20}\\) as an example.) \\(\\square\\) \n\nClaim — If \\(a_{n} > M\\) then \\(a_{n + 1} \\leq M\\) . \n\nProof. Assume for contradiction there's a first moment where \\(a_{n} > M\\) and \\(a_{n + 1} > M\\) , meaning the block \\(B_{n}\\) was added to an all- yellow tower past \\(M\\) that has height exceeding \\(M\\) . (This is the X'ed out region in the figure above.) In \\(B_{n}\\) 's tower, every (yellow) block (including \\(B_{n}\\) ) was contributed by a block placed in different towers at height \\(a_{n} > M\\) . So before \\(B_{n}\\) , there were already \\(a_{n + 1} > M\\) towers of height more than \\(M\\) . This contradicts minimality of \\(n\\) . \\(\\square\\) \n\nIt follows that the set of indices with \\(a_{n} \\leq M\\) has arithmetic density at least half, so certainly at least some of the numbers must occur infinitely often. Of the numbers in \\(\\{1, 2, \\ldots , M\\}\\) , define \\(L\\) such that towers 1 through \\(L\\) grow unbounded but towers \\(L + 1\\) through \\(M\\) do not. Then we can pick a larger threshold \\(N' > N\\) such that\n\n\n\n- Towers 1 through \\(L\\) have height greater than \\((M, N)\\) ; \n\n- Towers \\(L + 1\\) through \\(M\\) will receive no further blocks; \n\n- \\(a_{N^{\\prime}} \\leq L\\) . \n\nAfter this threshold, the following statement is true: \n\nClaim (Alternating small and big) — The terms \\(a_{N^{\\prime}}\\) , \\(a_{N^{\\prime} + 2}\\) , \\(a_{N^{\\prime} + 4}\\) , ... are all at most \\(L\\) while the terms \\(a_{N^{\\prime} + 1}\\) , \\(a_{N^{\\prime} + 3}\\) , \\(a_{N^{\\prime} + 5}\\) , ... are all greater than \\(M\\) . \n\n\\(\\P\\) Automaton for \\(n \\equiv N^{\\prime} \\pmod {2}\\) . From now on we always assume \\(n > N^{\\prime}\\) . When \\(n \\equiv N^{\\prime} \\pmod {2}\\) , i.e., when \\(a_{n}\\) is small, we define the state \n\n\\[S(n) = (h_{1}, h_{2}, \\ldots , h_{L}; a_{n}).\\] \n\nFor example, in the figure below, we illustrate how \n\n\\[S(34) = (9, 11; a_{34} = 1) \\longrightarrow S(36) = (9, 12; a_{36} = 2)\\] \n\n\n \n\nThe final element \\(a_{n}\\) simply reminds us which tower was most recently incremented. At this point we can give a complete description of how to move from \\(S(n)\\) to \\(S(n + 2)\\) : \n\n- The intermediate block \\(B_{n + 1}\\) is placed in the tower corresponding to the height \\(a_{n + 1}\\) of \\(B_{n}\\) ; \n\n- That tower will have height \\(a_{n + 2}\\) equal to the number of towers with height at least \\(a_{n + 1}\\) ; that is, it equals the cardinality of the set \n\n\\[\\{i: h_{i} \\geq h_{a_{n}}\\}\\]\n\n\n\n- We increment \\(h_{a_{n + 2}}\\) by 1 and update \\(a_{n}\\) . \n\nFor example, the illustrated \\(S(34) \\to S(36)\\) corresponds to the block \\(B_{34}\\) at height \\(h_{1}\\) in tower 1 giving the block \\(B_{35}\\) at height 2 in tower \\(h_{1}\\) , then block \\(B_{36}\\) at height \\(h_{2} + 1\\) being placed in tower 2. \n\nPigeonhole periodicity argument. Because only the relative heights matter in the automata above, if we instead define \n\n\\[T(n) = (h_{1} - h_{2},h_{2} - h_{3},\\ldots ,h_{L - 1} - h_{L};a_{n}).\\] \n\nthen \\(T(n + 2)\\) can be determined from just \\(T(n)\\) . \n\nSo it would be sufficient to show \\(T(n)\\) only takes on finitely many values to show that \\(T(n)\\) (and hence \\(a_{n}\\) ) is eventually periodic. \n\nSince we have the bound \\(h_{k + 1} \\leq h_{k} + C\\) , we are done upon proving the following lower bound: \n\nClaim — For every \\(1 \\leq \\ell < L\\) and \\(n > N'\\) , we have \\(h_{\\ell} \\leq h_{\\ell + 1} + C \\cdot (L - 1)\\) . \n\nProof. Assume for contradiction that there is some moment \\(n > N'\\) such that \n\n\\[h_{\\ell} > h_{\\ell +1} + C\\cdot (L - 1)\\] \n\nand WLOG assume that \\(h_{\\ell}\\) was just updated at the moment \\(n\\) . Together with \\(h_{k + 1} \\leq h_{k} + C\\) for all \\(k\\) and triangle inequality, we conclude \n\n\\[\\min (h_{1},\\ldots ,h_{\\ell}) > q:= \\max (h_{\\ell +1},\\ldots ,h_{L}).\\] \n\nWe find that the blocks now in fact alternate between being placed among the first \\(\\ell\\) towers and in towers with indices greater than \\(q\\) thereafter. Hence the heights \\(h_{\\ell +1}\\) , ..., \\(h_{L}\\) never grow after this moment. This contradicts the definition of \\(L\\) . \\(\\square\\) \n\nRemark. In fact, it can be shown that the period is actually exactly \\(L\\) , meaning the periodic part will be exactly a permutation of \\((1,2,\\ldots ,L)\\) . For any \\(L\\) , it turns out there is indeed a permutation achieving that periodic part.\n\n\n\n## \\(\\S 2\\) Solutions to Day 2", "metadata": {"resource_path": "IMO/segmented/en-IMO-2024-notes.jsonl", "problem_match": "3. ", "solution_match": "## \\(\\S 1.3\\) IMO 2024/3, proposed by William Steinberg (AUS) \n"}}
+{"year": "2024", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "Let triangle \\(A B C\\) with incenter \\(I\\) satisfying \\(A B< A C< B C\\) . Let \\(X\\) be a point on line \\(B C\\) , different from \\(C\\) , such that the line through \\(X\\) and parallel to \\(A C\\) is tangent to the incircle. Similarly, let \\(Y\\) be a point on line \\(B C\\) , different from \\(B\\) , such that the line through \\(Y\\) and parallel to \\(A B\\) is tangent to the incircle. Line \\(A I\\) intersects the circumcircle of triangle \\(A B C\\) again at \\(P\\) . Let \\(K\\) and \\(L\\) be the midpoints of \\(A C\\) and \\(A B\\) , respectively. Prove that \\(\\angle K I L + \\angle Y P X = 180^{\\circ}\\) .", "solution": "be the reflection of \\(A\\) over \\(I\\) , the most important point to add since it gets rid of \\(K\\) and \\(L\\) as follows. \n\nClaim — We have \\(\\angle KIL = \\angle BTC\\) , and lines \\(TX\\) and \\(TY\\) are tangent to the incircle. \n\nProof. The first part is true since \\(\\triangle BTC\\) is the image of \\(\\triangle KIL\\) under a homothety of ratio 2. The second part is true because lines \\(AB\\) , \\(AC\\) , \\(TX\\) , \\(TY\\) determine a rhombus with center \\(I\\) . \\(\\square\\) \n\nWe thus delete \\(K\\) and \\(L\\) from the picture altogether; they aren't needed anymore. \n\n\n\n\n\n\nClaim — We have \\(B X P T\\) and \\(C Y P T\\) are cyclic. \n\nProof. \\(\\angle T Y C = \\angle T Y B = \\angle A B C = \\angle A P C = \\angle T P C\\) and similarly. (Some people call this Reim's theorem.) \\(\\square\\) \n\nTo finish, observe that \n\n\\[\\angle C T B = \\angle C T P + \\angle P T B = \\angle C Y P + \\angle P X B = \\angle X Y P + \\angle X Y P = \\angle X P Y\\] \n\nas desired. (The length conditions \\(A C > A B > B C\\) ensure that \\(B\\) , \\(X\\) , \\(Y\\) , \\(C\\) are collinear in that order, and that \\(T\\) lies on the opposite side of \\(\\overline{B C}\\) as \\(A\\) . Hence the directed equality \\(\\angle C T B = \\angle X P Y\\) translates to the undirected \\(\\angle B T C + \\angle X P Y = 180^{\\circ}\\) .)", "metadata": {"resource_path": "IMO/segmented/en-IMO-2024-notes.jsonl", "problem_match": "4. ", "solution_match": "## \\(\\S 2.1\\) IMO 2024/4, proposed by Dominik Burek (POL) \n"}}
+{"year": "2024", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "Turbo the snail is in the top row of a grid with 2024 rows and 2023 columns and wants to get to the bottom row. However, there are 2022 hidden monsters, one in every row except the first and last, with no two monsters in the same column. \n\nTurbo makes a series of attempts to go from the first row to the last row. On each attempt, he chooses to start on any cell in the first row, then repeatedly moves to an orthogonal neighbor. (He is allowed to return to a previously visited cell.) If Turbo reaches a cell with a monster, his attempt ends and he is transported back to the first row to start a new attempt. The monsters do not move between attempts, and Turbo remembers whether or not each cell he has visited contains a monster. If he reaches any cell in the last row, his attempt ends and Turbo wins. \n\nFind the smallest integer \\(n\\) such that Turbo has a strategy which guarantees being able to reach the bottom row in at most \\(n\\) attempts, regardless of how the monsters are placed.", "solution": "Surprisingly the answer is \\(n = 3\\) for any grid size \\(s\\times (s - 1)\\) when \\(s\\geq 4\\) . We prove this in that generality. \n\n\\(\\P\\) Proof that at least three attempts are needed. When Turbo first moves into the second row, Turbo could encounter a monster \\(M_{1}\\) right away. Then on the next attempt, Turbo must enter the third row in different column as \\(M_{1}\\) , and again could encounter a monster \\(M_{2}\\) right after doing so. This means no strategy can guarantee fewer than three attempts. \n\n\\(\\P\\) Strategy with three attempts. On the first attempt, we have Turbo walk through the entire second row until he finds the monster \\(M_{1}\\) in it. Then we get two possible cases. \n\nCase where \\(M_{1}\\) is not on the edge. In the first case, if that monster \\(M_{1}\\) is not on the edge of the row, then Turbo can trace two paths below it as shown below. At least one of these paths works, hence three attempts is sufficient. \n\n\n \n\nCase where \\(M_{1}\\) is on the edge. WLOG, \\(M_{1}\\) is in the leftmost cell. Then Turbo follows the green staircase pattern shown in the left figure below. If the staircase is free\n\n\n\nof monsters, then Turbo wins on the second attempt. Otherwise, if a monster \\(M_{2}\\) is encountered on the staircase, Turbo has found a safe path to the left of \\(M_{2}\\) ; then Turbo can use this to reach the column \\(M_{1}\\) is in, and escape from there. This is shown in purple in the center and right figure (there are two slightly different cases depending on whether \\(M_{2}\\) was encountered going east or south). \n\n\n \n\nThus the problem is solved in three attempts, as promised. \n\n\\(\\P\\) Extended remark: all working strategies look similar to this. As far as we know, all working strategies are variations of the above. In fact, we will try to give a description of the space of possible strategies, although this needs a bit of notation. \n\nDefinition. For simplicity, we only use \\(s\\) even in the figures below. We define the happy triangle as the following cells: \n\nAll \\(s - 1\\) cells in the first row (which has no monsters). The center \\(s - 3\\) cells in the second row. The center \\(s - 5\\) cells in the third row. The center cell in the \\(\\frac{s}{2}\\) th row. \n\nFor \\(s = 12\\) , the happy triangle is the region shaded in the thick border below. \n\n\n\n\n\n\nDefinition. Given a cell, define a shoulder to be the cell directly northwest or northeast of it. Hence there are two shoulders of cells outside the first and last column, and one shoulder otherwise. \n\nThen solutions roughly must distinguish between these two cases: \n\n- Inside happy triangle: If the first monster \\(M_{1}\\) is found in the happy triangle, and there is a safe path found by Turbo to the two shoulders (marked \\(\\bigstar\\) in the figure), then one can finish in two more moves by considering the two paths from \\(\\bigstar\\) that cut under the monster \\(M_{1}\\) ; one of them must work. This slightly generalizes the easier case in the solution above (which focuses only on the case where \\(M_{1}\\) is in the first row). \n\n\n \n\n- Outside happy triangle: Now suppose the first monster \\(M_{1}\\) is outside the happy triangle. Of the two shoulders, take the one closer to the center (if in the center column, either one works; if only one shoulder, use it). If there is a safe path to that shoulder, then one can take a staircase pattern towards the center, as shown in the figure. In that case, the choice of shoulder and position guarantees the staircase reaches the bottom row, so that if no monster is along this path, the algorithm ends. Otherwise, if one encounters a second monster along the staircase, then one can use the third trial to cut under the monster \\(M_{1}\\) . \n\n\n \n\nWe now prove the following proposition: in any valid strategy for Turbo, in the case where Turbo first encounters a monster upon leaving the happy triangle, the second path must outline the same staircase shape.\n\n\n\nThe monsters pre- commit to choosing their pattern to be either a NW- SE diagonal or NE- SW diagonal, with a single one- column gap; see figure below for an example. Note that this forces any valid path for Turbo to pass through the particular gap. \n\n\n \n\nWe may assume without loss of generality that Turbo first encounters a monster \\(M_{1}\\) when Turbo first leaves the happy triangle, and that this forces an NW- SE configuration. \n\n\n \n\nThen the following is true:\n\n\n\n## Proposition \n\nPropositionThe strategy of Turbo on the second path must visit every cell in \"slightly raised diagonal\" marked with \\(\\clubsuit\\) in the figure above in order from top to bottom, until it encounters a second Monster \\(M_{2}\\) (or reaches the bottom row and wins anyway). It's both okay and irrelevant if Turbo visits other cells above this diagonal, but the marked cells must be visited from top to bottom in that order. \n\nProof. If Turbo tries to sidestep by visiting the cell southeast of \\(M_{1}\\) (marked \\(X\\) in the Figure), then Turbo clearly cannot finish after this (for \\(s\\) large enough). Meanwhile, suppose Turbo tries to \"skip\" one of the \\(\\clubsuit\\) , say in column \\(C\\) , then the gap could equally well be in the column to the left of \\(C\\) . This proves the proposition. \\(\\square\\) \n\nRemark (Memories of safe cells are important, not just monster cells). Here is one additional observation that one can deduce from this. We say a set \\(S\\) of revealed monsters is called obviously winnable if, based on only the positions of the monsters (and not the moves or algorithm that were used to obtain them), one can identify a guaranteed winning path for Turbo using only \\(S\\) . For example, two monsters in adjacent columns which are not diagonally adjacent is obviously winnable. \n\nThen no strategy can guarantee obtaining an obviously winnable set in 2 moves (or even \\(k\\) moves for any constant \\(k\\) , if \\(s\\) is large enough in terms of \\(k\\) ). So any valid strategy must also use the memory of identified safe cells that do not follow just from the revealed monster positions.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2024-notes.jsonl", "problem_match": "5. ", "solution_match": "## \\(\\S 2.2\\) IMO 2024/5, proposed by Chu Cheuk Hei (HKG) \n"}}
{"year": "2024", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "A function \\(f\\colon \\mathbb{Q}\\to \\mathbb{Q}\\) is called aquaesulian if the following property holds: for every \\(x,y\\in \\mathbb{Q}\\) \n\n\\[f(x + f(y)) = f(x) + y\\quad \\mathrm{or}\\quad f(f(x) + y) = x + f(y).\\] \n\nShow that there exists an integer \\(c\\) such that for any aquaesulian function \\(f\\) there are at most \\(c\\) different rational numbers of the form \\(f(r) + f(- r)\\) for some rational number \\(r\\) , and find the smallest possible value of \\(c\\) .", "solution": "We will prove that \n\n\\[\\{f(x) + f(-x)\\mid x\\in \\mathbb{Q}\\}\\] \n\ncontains at most 2 elements and give an example where there are indeed 2 elements. \n\nWe fix the notation \\(x\\to y\\) to mean that \\(f(x + f(y)) = f(x) + y\\) . So the problem statement means that either \\(x\\to y\\) or \\(y\\to x\\) for all \\(x,y\\) . In particular, we always have \\(x\\to x\\) , and hence \n\n\\[f(x + f(x)) = x + f(x)\\] \n\nfor every \\(x\\) \n\nConstruction. The function \n\n\\[f(x) = \\lfloor 2x\\rfloor -x\\] \n\ncan be seen to satisfy the problem conditions. Moreover, \\(f(0) + f(0) = 0\\) but \\(f(1 / 3) + f(- 1 / 3) = - 1\\) . \n\nRemark. Here is how I (Evan) found the construction. Let \\(h(x):= x + f(x)\\) , and let \\(S:= h(\\mathbb{Q}) = \\{h(x)\\mid x\\in \\mathbb{Q}\\}\\) . Hence \\(f\\) is the identity on all of \\(S\\) . If we rewrite the problem condition in terms of \\(h\\) instead of \\(f\\) , it asserts that at least one of the equations \n\n\\[h(x + h(y) - y) = h(x) + h(y)\\] \\[h(y + h(x) - x) = h(x) + h(y)\\] \n\nis true. In particular, \\(S\\) is closed under addition. \n\nNow, the two trivial solutions for \\(h\\) are \\(h(x) = 2x\\) and \\(h(x) = 0\\) . To get a nontrivial construction, we must also have \\(S\\neq \\{0\\}\\) and \\(S\\neq \\mathbb{Q}\\) . So a natural guess is to take \\(S = \\mathbb{Z}\\) . And indeed \\(h(x) = \\lfloor 2x\\rfloor\\) works fine. \n\nRemark. This construction is far from unique. For example, \\(f(x) = 2\\lfloor x\\rfloor - x = \\lfloor x\\rfloor - \\{x\\}\\) seems to have been more popular to find. \n\nProof (communicated by Abel George Mathew). We start by proving: \n\nClaim — \\(f\\) is injective.\n\n\n\nProof. Suppose \\(f(a) = f(b)\\) . WLOG \\(a \\to b\\) . Then \n\n\\[f(a) + a = f(a + f(a)) = f(a + f(b)) = f(a) + b \\Rightarrow a = b. \\quad (□)\\] \n\nClaim — Suppose \\(s \\to r\\) . Then either \\(f(r) + f(- r) = 0\\) or \\(f(f(s)) = s + f(r) + f(- r)\\) . \n\nProof. Take the given statement with \\(x = s + f(r)\\) and \\(y = - r\\) ; then \n\n\\[x + f(y) = s + f(r) + f(-r)\\] \\[y + f(x) = f(s + f(r)) - r = f(s).\\] \n\nBecause \\(f\\) is injective, if \\(x \\to y\\) then \\(f(r) + f(- r) = 0\\) . Meanwhile, if \\(y \\to x\\) then indeed \\(f(f(s)) = s + f(r) + f(- r)\\) . \\(\\square\\) \n\nFinally, suppose \\(a\\) and \\(b\\) are different numbers for which \\(f(a) + f(- a)\\) and \\(f(b) + f(- b)\\) are both nonzero. Again, WLOG \\(a \\to b\\) . Then \n\n\\[f(a) + f(-a) \\stackrel{a \\Rightarrow a}{=} f(f(a)) - a \\stackrel{a \\Rightarrow b}{=} f(b) + f(-b).\\] \n\nThis shows at most two values can occur. \n\nRemark. The above solution works equally well for \\(f: \\mathbb{R} \\to \\mathbb{R}\\) . But the choice of \\(\\mathbb{Q}\\) permits some additional alternate solutions. \n\nRemark. After showing \\(f\\) injective, a common lemma proved is that \\(- f(- f(x)) = x\\) , i.e. \\(f\\) is an involution. This provides some alternative paths for solutions.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2024-notes.jsonl", "problem_match": "6. ", "solution_match": "## \\(\\S 2.3\\) IMO 2024/6, proposed by Japan \n"}}
diff --git a/IMO/segmented/en-IMO-2025-notes.jsonl b/IMO/segmented/en-IMO-2025-notes.jsonl
index dfad1bef8d38b24af59216fafdd17d0f0271a65c..f3c9647e5a0a32dc583dbd6ae5fb52a8b1bf4c9e 100644
--- a/IMO/segmented/en-IMO-2025-notes.jsonl
+++ b/IMO/segmented/en-IMO-2025-notes.jsonl
@@ -1,6 +1,6 @@
-{"year": "2025", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "A line in the plane is called sunny if it is not parallel to any of the \\(x\\) -axis, the \\(y\\) -axis, or the line \\(x + y = 0\\) . \n\nLet \\(n \\geq 3\\) be a given integer. Determine all nonnegative integers \\(k\\) such that there exist \\(n\\) distinct lines in the plane satisfying both of the following: \n\n- for all positive integers \\(a\\) and \\(b\\) with \\(a + b \\leq n + 1\\) , the point \\((a, b)\\) lies on at least one of the lines; and \n\n- exactly \\(k\\) of the \\(n\\) lines are sunny.", "solution": "y. \n\nThe answer is 0, 1, or 3 sunny lines. \n\nIn what follows, we draw the grid as equilateral instead of a right triangle; this has no effect on the problem statement but is more symmetric. \n\nWe say a long line is one of the three lines at the edge of the grid, i.e. one of the (non- sunny) lines passing through \\(n\\) points. The main claim is the following. \n\nClaim — If \\(n \\geq 4\\) , any set of \\(n\\) lines must have at least one long line. \n\nProof. Consider the \\(3(n - 1)\\) points on the outer edge of the grid. If there was no long line, each of the \\(n\\) lines passes through at most two such points. So we obtain \\(2n \\geq 3(n - 1)\\) , which forces \\(n \\leq 3\\) . \\(\\square\\) \n\nHence, by induction we may repeatedly delete a long line without changing the number of sunny lines until \\(n = 3\\) (and vice- versa: given a construction for smaller \\(n\\) we can increase \\(n\\) by one and add a long line). \n\nWe now classify all the ways to cover the \\(1 + 2 + 3 = 6\\) points in an \\(n = 3\\) grid with 3 lines. \n\n\n\nLong line present \n\n\n\nNo long line \n\n- If there is a long line (say, the red one in the figure), the remaining \\(1 + 2 = 3\\) points (circled in blue) are covered with two lines. One of the lines passes through 2 points\n\n\n\nand must not be sunny; the other line may or may not be sunny. Hence in this case the possible counts of sunny lines are 0 or 1. \n\n- If there is no long line, each of the three lines passes through at most 2 points. But there are 6 total lines, so in fact each line must pass through exactly two points. The only way to do this is depicted in the figure in the right. In this case there are 3 sunny lines. \n\nThis proves that 0, 1, 3 are the only possible answers. \n\nRemark. The concept of a sunny line is not that important to the problem. The proof above essentially classifies all the ways to cover the \\(1 + 2 + \\dots + n\\) points with exactly \\(n\\) lines. Namely, one should repeatedly take a long line and decrease \\(n\\) until \\(n = 3\\) , and then pick one of the finitely many cases for \\(n = 3\\) . The count of sunny lines just happens to be whatever is possible for \\(n = 3\\) , since long lines are not sunny.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2025-notes.jsonl", "problem_match": "1. ", "solution_match": "## \\(\\S 1.1\\) IMO 2025/1, proposed by Linus Tang (USA) \n"}}
-{"year": "2025", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "Let \\(\\Omega\\) and \\(\\Gamma\\) be circles with centres \\(M\\) and \\(N\\) , respectively, such that the radius of \\(\\Omega\\) is less than the radius of \\(\\Gamma\\) . Suppose \\(\\Omega\\) and \\(\\Gamma\\) intersect at two distinct points \\(A\\) and \\(B\\) . Line \\(MN\\) intersects \\(\\Omega\\) at \\(C\\) and \\(\\Gamma\\) at \\(D\\) , so that \\(C\\) , \\(M\\) , \\(N\\) , \\(D\\) lie on \\(MN\\) in that order. Let \\(P\\) be the circumcenter of triangle \\(ACD\\) . Line \\(AP\\) meets \\(\\Omega\\) again at \\(E \\neq A\\) and meets \\(\\Gamma\\) again at \\(F \\neq A\\) . Let \\(H\\) be the orthocenter of triangle \\(PMN\\) . Prove that the line through \\(H\\) parallel to \\(AP\\) is tangent to the circumcircle of triangle \\(BEF\\) .", "solution": "L . \n\nThroughout the solution, we define \n\n\\[\\alpha := \\angle DCA = \\angle BCD \\Rightarrow \\angle PAD = \\angle CAB = 90^{\\circ} - \\alpha\\] \\[\\beta := \\angle ADC = \\angle CDB \\Rightarrow \\angle CAP = \\angle BAD = 90^{\\circ} - \\beta .\\] \n\nIgnore the points \\(H\\) , \\(M\\) , \\(N\\) for now and focus on the remaining ones. \n\nClaim — We have \\(\\overline{CE} \\parallel \\overline{AD}\\) and \\(\\overline{DF} \\parallel \\overline{AC}\\) . \n\nProof. \\(\\angle AEC = \\angle ABC = \\angle CAB = 90^{\\circ} - \\alpha\\) . \n\nHence, if we let \\(A' := \\overline{CE} \\cap \\overline{DF}\\) , we have a parallelogram \\(ACA'D\\) . Note in particular that \\(\\overline{BA'} \\parallel \\overline{CD}\\) .\n\n\n\n \n\nNext, let \\(T\\) denote the circumcenter of \\(\\triangle A^{\\prime}EF\\) . (This will be the tangency point later in the problem.) \n\nClaim — Point \\(T\\) also lies on \\(\\overline{BA^{\\prime}}\\) and is also the arc midpoint of \\(\\overline{EF}\\) on \\((BEF)\\) . \n\nProof. We compute the angles of \\(\\triangle A^{\\prime}EF\\) : \n\n\\[\\angle FEA^{\\prime} = \\angle AEC = \\angle ABC = \\angle CAB = 90^{\\circ} - \\alpha\\] \\[\\angle A^{\\prime}FE = \\angle DFA = \\angle DBA = \\angle BAD = 90^{\\circ} - \\beta\\] \\[\\angle EA^{\\prime}F = \\alpha +\\beta .\\] \n\nThen, since \\(T\\) is the circumcenter, it follows that: \n\n\\[\\angle EA^{\\prime}T = \\angle 90^{\\circ} - \\angle A^{\\prime}FE = \\beta = \\angle A^{\\prime}CD = \\angle CA^{\\prime}B.\\] \n\nThis shows that \\(T\\) lies on \\(\\overline{BA^{\\prime}}\\) . \n\nAlso, we have \\(\\angle ETF = 2\\angle EA^{\\prime}F = 2(\\alpha +\\beta)\\) and \n\n\\[\\angle EBF = \\angle EBA + \\angle ABF = \\angle ECA + \\angle ADF\\]\n\n\n\n\\[= \\angle A^{\\prime}C A + \\angle A D A^{\\prime} = (\\alpha +\\beta) + (\\alpha +\\beta) = 2(\\alpha +\\beta)\\] \n\nwhich proves that \\(T\\) also lies on \\(\\overline{AB^{\\prime}}\\) . \n\nWe then bring \\(M\\) and \\(N\\) into the picture as follows: \n\nClaim — Point \\(T\\) lies on both lines \\(ME\\) and \\(NF\\) . \n\nProof. To show that \\(F\\) , \\(T\\) , \\(N\\) are collinear, note that \\(\\triangle F E A^{\\prime} \\sim \\triangle F A D\\) via a homothety at \\(F\\) . This homothety maps \\(T\\) to \\(N\\) . \\(\\square\\) \n\nWe now deal with point \\(H\\) using two claims. \n\nClaim — We have \\(\\overline{M H} \\parallel \\overline{A D}\\) and \\(\\overline{N H} \\parallel \\overline{A C}\\) . \n\nProof. Note that \\(\\overline{M H} \\perp \\overline{P N}\\) , but \\(\\overline{P N}\\) is the perpendicular bisector of \\(\\overline{A D}\\) , so in fact \\(\\overline{M H} \\parallel \\overline{A D}\\) . Similarly, \\(\\overline{N H} \\parallel \\overline{A C}\\) . \\(\\square\\) \n\nClaim — Lines \\(\\overline{M H}\\) and \\(\\overline{N H}\\) bisect \\(\\angle N M T\\) and \\(\\angle M N T\\) . In fact, point \\(H\\) is the incenter of \\(\\triangle T M N\\) , and \\(\\angle N T H = \\angle H T M = 90^{\\circ} - (\\alpha +\\beta)\\) . \n\nProof. Hence, \\(\\angle H M N = \\angle A^{\\prime}C D = \\angle A D C = \\beta\\) . But \\(\\angle T M N = \\angle C M E = 2\\angle C A E =\\) \\(- 2(90^{\\circ} - 2\\beta) = 2\\beta\\) . That proves \\(\\overline{M H}\\) bisects \\(\\angle N M T\\) ; the other one is similar. \n\nTo show that \\(H\\) is an incenter (rather than an excenter) and get the last angle equality, we need to temporarily undirect our angles. Assume WLOG that \\(\\triangle A C D\\) is directed counterclockwise. The problem condition that \\(C\\) and \\(D\\) are the farther intersections of line \\(M N\\) mean that \\(\\angle N H M = \\angle C A D > 90^{\\circ}\\) . We are also promised \\(C\\) , \\(M\\) , \\(N\\) , \\(D\\) are collinear in that order. Hence the reflections of line \\(M N\\) over lines \\(M H\\) and \\(N H\\) , which meet at \\(T\\) , should meet at a point for which \\(T\\) lies on the same side as \\(H\\) . In other words, \\(\\triangle M T N\\) is oriented counterclockwise and contains \\(H\\) . \n\nWorking with undirected \\(\\alpha = \\angle D C A\\) and \\(\\beta = \\angle A D C\\) with \\(\\alpha + \\beta < 90^{\\circ}\\) , \n\n\\[\\angle N T H = \\angle H T M = \\frac{1}{2}\\angle N T M = \\frac{1}{2} (180^{\\circ} - 2(\\alpha +\\beta)) = 90^{\\circ} - (\\alpha +\\beta).\\] \n\nThis matches the claim and finishes the result. \\(\\square\\) \n\nNow \n\n\\[\\angle N F A = 90^{\\circ} - \\angle A D F = 90^{\\circ} - (\\alpha +\\beta) = \\angle N T H\\] \n\nso \\(\\overline{H T} \\parallel \\overline{A P}\\) . And since \\(T E = T F\\) , we have the tangency requested too now, as desired. \n\nRemark. There are many other ways to describe the point \\(T\\) . For example, \\(A M T N\\) is a parallelogram and \\(M B T N\\) is an isosceles trapezoid. In coordination, we joked that it was impossible to write a false conjecture.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2025-notes.jsonl", "problem_match": "2. ", "solution_match": "## \\(\\S 1.2\\) IMO 2025/2, proposed by Tran Quang Hung (VNM) \n"}}
+{"year": "2025", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "A line in the plane is called sunny if it is not parallel to any of the \\(x\\) -axis, the \\(y\\) -axis, or the line \\(x + y = 0\\) . \n\nLet \\(n \\geq 3\\) be a given integer. Determine all nonnegative integers \\(k\\) such that there exist \\(n\\) distinct lines in the plane satisfying both of the following: \n\n- for all positive integers \\(a\\) and \\(b\\) with \\(a + b \\leq n + 1\\) , the point \\((a, b)\\) lies on at least one of the lines; and \n\n- exactly \\(k\\) of the \\(n\\) lines are sunny.", "solution": "y. \n\nThe answer is 0, 1, or 3 sunny lines. \n\nIn what follows, we draw the grid as equilateral instead of a right triangle; this has no effect on the problem statement but is more symmetric. \n\nWe say a long line is one of the three lines at the edge of the grid, i.e. one of the (non- sunny) lines passing through \\(n\\) points. The main claim is the following. \n\nClaim — If \\(n \\geq 4\\) , any set of \\(n\\) lines must have at least one long line. \n\nProof. Consider the \\(3(n - 1)\\) points on the outer edge of the grid. If there was no long line, each of the \\(n\\) lines passes through at most two such points. So we obtain \\(2n \\geq 3(n - 1)\\) , which forces \\(n \\leq 3\\) . \\(\\square\\) \n\nHence, by induction we may repeatedly delete a long line without changing the number of sunny lines until \\(n = 3\\) (and vice- versa: given a construction for smaller \\(n\\) we can increase \\(n\\) by one and add a long line). \n\nWe now classify all the ways to cover the \\(1 + 2 + 3 = 6\\) points in an \\(n = 3\\) grid with 3 lines. \n\n\n\nLong line present \n\n\n\nNo long line \n\n- If there is a long line (say, the red one in the figure), the remaining \\(1 + 2 = 3\\) points (circled in blue) are covered with two lines. One of the lines passes through 2 points\n\n\n\nand must not be sunny; the other line may or may not be sunny. Hence in this case the possible counts of sunny lines are 0 or 1. \n\n- If there is no long line, each of the three lines passes through at most 2 points. But there are 6 total lines, so in fact each line must pass through exactly two points. The only way to do this is depicted in the figure in the right. In this case there are 3 sunny lines. \n\nThis proves that 0, 1, 3 are the only possible answers. \n\nRemark. The concept of a sunny line is not that important to the problem. The proof above essentially classifies all the ways to cover the \\(1 + 2 + \\dots + n\\) points with exactly \\(n\\) lines. Namely, one should repeatedly take a long line and decrease \\(n\\) until \\(n = 3\\) , and then pick one of the finitely many cases for \\(n = 3\\) . The count of sunny lines just happens to be whatever is possible for \\(n = 3\\) , since long lines are not sunny.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2025-notes.jsonl", "problem_match": "1. ", "solution_match": "## \\(\\S 1.1\\) IMO 2025/1, proposed by Linus Tang (USA) \n"}}
+{"year": "2025", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "Let \\(\\Omega\\) and \\(\\Gamma\\) be circles with centres \\(M\\) and \\(N\\) , respectively, such that the radius of \\(\\Omega\\) is less than the radius of \\(\\Gamma\\) . Suppose \\(\\Omega\\) and \\(\\Gamma\\) intersect at two distinct points \\(A\\) and \\(B\\) . Line \\(MN\\) intersects \\(\\Omega\\) at \\(C\\) and \\(\\Gamma\\) at \\(D\\) , so that \\(C\\) , \\(M\\) , \\(N\\) , \\(D\\) lie on \\(MN\\) in that order. Let \\(P\\) be the circumcenter of triangle \\(ACD\\) . Line \\(AP\\) meets \\(\\Omega\\) again at \\(E \\neq A\\) and meets \\(\\Gamma\\) again at \\(F \\neq A\\) . Let \\(H\\) be the orthocenter of triangle \\(PMN\\) . Prove that the line through \\(H\\) parallel to \\(AP\\) is tangent to the circumcircle of triangle \\(BEF\\) .", "solution": "L . \n\nThroughout the solution, we define \n\n\\[\\alpha := \\angle DCA = \\angle BCD \\Rightarrow \\angle PAD = \\angle CAB = 90^{\\circ} - \\alpha\\] \\[\\beta := \\angle ADC = \\angle CDB \\Rightarrow \\angle CAP = \\angle BAD = 90^{\\circ} - \\beta .\\] \n\nIgnore the points \\(H\\) , \\(M\\) , \\(N\\) for now and focus on the remaining ones. \n\nClaim — We have \\(\\overline{CE} \\parallel \\overline{AD}\\) and \\(\\overline{DF} \\parallel \\overline{AC}\\) . \n\nProof. \\(\\angle AEC = \\angle ABC = \\angle CAB = 90^{\\circ} - \\alpha\\) . \n\nHence, if we let \\(A' := \\overline{CE} \\cap \\overline{DF}\\) , we have a parallelogram \\(ACA'D\\) . Note in particular that \\(\\overline{BA'} \\parallel \\overline{CD}\\) .\n\n\n\n \n\nNext, let \\(T\\) denote the circumcenter of \\(\\triangle A^{\\prime}EF\\) . (This will be the tangency point later in the problem.) \n\nClaim — Point \\(T\\) also lies on \\(\\overline{BA^{\\prime}}\\) and is also the arc midpoint of \\(\\overline{EF}\\) on \\((BEF)\\) . \n\nProof. We compute the angles of \\(\\triangle A^{\\prime}EF\\) : \n\n\\[\\angle FEA^{\\prime} = \\angle AEC = \\angle ABC = \\angle CAB = 90^{\\circ} - \\alpha\\] \\[\\angle A^{\\prime}FE = \\angle DFA = \\angle DBA = \\angle BAD = 90^{\\circ} - \\beta\\] \\[\\angle EA^{\\prime}F = \\alpha +\\beta .\\] \n\nThen, since \\(T\\) is the circumcenter, it follows that: \n\n\\[\\angle EA^{\\prime}T = \\angle 90^{\\circ} - \\angle A^{\\prime}FE = \\beta = \\angle A^{\\prime}CD = \\angle CA^{\\prime}B.\\] \n\nThis shows that \\(T\\) lies on \\(\\overline{BA^{\\prime}}\\) . \n\nAlso, we have \\(\\angle ETF = 2\\angle EA^{\\prime}F = 2(\\alpha +\\beta)\\) and \n\n\\[\\angle EBF = \\angle EBA + \\angle ABF = \\angle ECA + \\angle ADF\\]\n\n\n\n\\[= \\angle A^{\\prime}C A + \\angle A D A^{\\prime} = (\\alpha +\\beta) + (\\alpha +\\beta) = 2(\\alpha +\\beta)\\] \n\nwhich proves that \\(T\\) also lies on \\(\\overline{AB^{\\prime}}\\) . \n\nWe then bring \\(M\\) and \\(N\\) into the picture as follows: \n\nClaim — Point \\(T\\) lies on both lines \\(ME\\) and \\(NF\\) . \n\nProof. To show that \\(F\\) , \\(T\\) , \\(N\\) are collinear, note that \\(\\triangle F E A^{\\prime} \\sim \\triangle F A D\\) via a homothety at \\(F\\) . This homothety maps \\(T\\) to \\(N\\) . \\(\\square\\) \n\nWe now deal with point \\(H\\) using two claims. \n\nClaim — We have \\(\\overline{M H} \\parallel \\overline{A D}\\) and \\(\\overline{N H} \\parallel \\overline{A C}\\) . \n\nProof. Note that \\(\\overline{M H} \\perp \\overline{P N}\\) , but \\(\\overline{P N}\\) is the perpendicular bisector of \\(\\overline{A D}\\) , so in fact \\(\\overline{M H} \\parallel \\overline{A D}\\) . Similarly, \\(\\overline{N H} \\parallel \\overline{A C}\\) . \\(\\square\\) \n\nClaim — Lines \\(\\overline{M H}\\) and \\(\\overline{N H}\\) bisect \\(\\angle N M T\\) and \\(\\angle M N T\\) . In fact, point \\(H\\) is the incenter of \\(\\triangle T M N\\) , and \\(\\angle N T H = \\angle H T M = 90^{\\circ} - (\\alpha +\\beta)\\) . \n\nProof. Hence, \\(\\angle H M N = \\angle A^{\\prime}C D = \\angle A D C = \\beta\\) . But \\(\\angle T M N = \\angle C M E = 2\\angle C A E =\\) \\(- 2(90^{\\circ} - 2\\beta) = 2\\beta\\) . That proves \\(\\overline{M H}\\) bisects \\(\\angle N M T\\) ; the other one is similar. \n\nTo show that \\(H\\) is an incenter (rather than an excenter) and get the last angle equality, we need to temporarily undirect our angles. Assume WLOG that \\(\\triangle A C D\\) is directed counterclockwise. The problem condition that \\(C\\) and \\(D\\) are the farther intersections of line \\(M N\\) mean that \\(\\angle N H M = \\angle C A D > 90^{\\circ}\\) . We are also promised \\(C\\) , \\(M\\) , \\(N\\) , \\(D\\) are collinear in that order. Hence the reflections of line \\(M N\\) over lines \\(M H\\) and \\(N H\\) , which meet at \\(T\\) , should meet at a point for which \\(T\\) lies on the same side as \\(H\\) . In other words, \\(\\triangle M T N\\) is oriented counterclockwise and contains \\(H\\) . \n\nWorking with undirected \\(\\alpha = \\angle D C A\\) and \\(\\beta = \\angle A D C\\) with \\(\\alpha + \\beta < 90^{\\circ}\\) , \n\n\\[\\angle N T H = \\angle H T M = \\frac{1}{2}\\angle N T M = \\frac{1}{2} (180^{\\circ} - 2(\\alpha +\\beta)) = 90^{\\circ} - (\\alpha +\\beta).\\] \n\nThis matches the claim and finishes the result. \\(\\square\\) \n\nNow \n\n\\[\\angle N F A = 90^{\\circ} - \\angle A D F = 90^{\\circ} - (\\alpha +\\beta) = \\angle N T H\\] \n\nso \\(\\overline{H T} \\parallel \\overline{A P}\\) . And since \\(T E = T F\\) , we have the tangency requested too now, as desired. \n\nRemark. There are many other ways to describe the point \\(T\\) . For example, \\(A M T N\\) is a parallelogram and \\(M B T N\\) is an isosceles trapezoid. In coordination, we joked that it was impossible to write a false conjecture.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2025-notes.jsonl", "problem_match": "2. ", "solution_match": "## \\(\\S 1.2\\) IMO 2025/2, proposed by Tran Quang Hung (VNM) \n"}}
{"year": "2025", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "A function \\(f: \\mathbb{N} \\to \\mathbb{N}\\) is said to be bonza if \n\n\\[f(a) \\quad \\text{divides} \\quad b^{a} - f(b)^{f(a)}\\] \n\nfor all positive integers \\(a\\) and \\(b\\) . \n\nDetermine the smallest real constant \\(c\\) such that \\(f(n) \\leq cn\\) for all bonza functions \\(f\\) and all positive integers \\(n\\) .", "solution": "The answer is \\(c = 4\\) \n\nLet \\(P(a,b)\\) denote the given statement \\(f(a)\\mid b^{a} - f(b)^{f(a)}\\) \n\nClaim — We have \\(f(n)\\mid n^{n}\\) for all \\(n\\) \n\nProof. Take \\(P(n,n)\\) \n\nClaim — Unless \\(f = \\mathrm{id}\\) , we have \\(f(p) = 1\\) for all odd primes \\(p\\) \n\nProof. Consider any prime \\(q\\) with \\(f(q) > 1\\) . Then \\(f(q)\\) is a power of \\(q\\) , and for each \\(n\\) we get \n\n\\[P(q,n)\\Rightarrow q\\mid f(q)\\mid n^{q} - f(n)^{f(q)}.\\] \n\nFermat's little theorem now gives \\(n^{q}\\equiv n\\) (mod \\(q\\) ) and \\(f(n)^{f(q)}\\equiv f(n)\\) (mod \\(q\\) ) (since \\(f(q)\\) is a power of \\(q\\) ), and therefore \\(q\\mid n - f(n)\\) . Hence, unless \\(f\\) is the identity function, only finitely many \\(q\\) could have \\(f(q) > 1\\) . \n\nNow let \\(p\\) be any odd prime, and let \\(q\\) be a large prime such that \\(q\\not\\equiv 1\\) (mod \\(p\\) ) (possible for all \\(p > 2\\) , say by Dirichlet). Then \n\n\\[P(p,q)\\Rightarrow f(p)\\mid q^{p} - 1^{p}.\\] \n\nThe RHS is \\(q^{p} - 1\\equiv q - 1\\not\\equiv 0\\) (mod \\(p\\) ), so \\(f(p) = 1\\) \n\nClaim — We have \\(f(n)\\mid 2^{\\infty}\\) for all \\(n\\) \n\nProof. If \\(p\\mid f(n)\\) is odd then \\(P(n,p)\\) gives \\(p\\mid f(n)\\mid p^{n} - 1^{n}\\) , contradiction. \n\n(In particular, we now know \\(f(n) = 1\\) for all odd \\(n\\) , though we don't use this.) \n\nClaim — We have \\(f(n)\\leq 2^{\\nu_{2}(n) + 2}\\) for all \\(n\\) \n\nProof. Consider \\(P(n,5)\\Rightarrow f(n)\\mid 5^{n} - 1^{n}\\) . It's well- known that \\(\\nu_{2}(5^{n} - 1) = \\nu_{2}(n) + 2\\) for all \\(n\\) .\n\n\n\nThis immediately shows \\(f(n) \\leq 4n\\) for all \\(n\\) , hence \\(c = 4\\) in the problem statement works. For the construction, the simplest one seems to be \n\n\\[f(n) = \\left\\{ \\begin{array}{ll}1 & n \\mathrm{odd} \\\\ 16 & n = 4 \\\\ 2 & n \\mathrm{even}, n \\neq 4 \\end{array} \\right.\\] \n\nwhich is easily checked to work and has \\(f(4) = 16\\) . \n\nRemark. With a little more case analysis we can classify all functions \\(f\\) . The two trivial solutions are \\(f(n) = n\\) and \\(f(n) = 1\\) ; the others are described by writing \\(f(n) = 2^{e(n)}\\) for any function \\(e\\) satisfying \n\n\\(\\cdot e(n) = 0\\) for odd \\(n\\) \n\n\\(\\cdot 1 \\leq e(2) \\leq 2\\) \n\n\\(\\cdot 1 \\leq e(n) \\leq \\nu_{2}(n) + 2\\) for even \\(n > 2\\) \n\nThis basically means that there are almost no additional constraints beyond what is suggested by the latter two claims.\n\n\n\n## \\(\\S 2\\) Solutions to Day 2", "metadata": {"resource_path": "IMO/segmented/en-IMO-2025-notes.jsonl", "problem_match": "3. ", "solution_match": "## \\(\\S 1.3\\) IMO 2025/3, proposed by Lorenzo Sarria (COL) \n"}}
{"year": "2025", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "An infinite sequence \\(a_{1}, a_{2}, \\ldots\\) consists of positive integers has each of which has at least three proper divisors. Suppose that for each \\(n \\geq 1\\) , \\(a_{n+1}\\) is the sum of the three largest proper divisors of \\(a_{n}\\) . Determine all possible values of \\(a_{1}\\) .", "solution": "An) . \n\nThe answer is \\(a_{1} = 12^{e} \\cdot 6 \\cdot \\ell\\) for any \\(e, \\ell \\geq 0\\) with \\(\\gcd (\\ell , 10) = 1\\) . \n\nLet \\(\\mathbf{S}\\) denote the set of positive integers with at least three divisors. For \\(x \\in \\mathbf{S}\\) , let \\(\\psi (x)\\) denote the sum of the three largest ones, so that \\(\\psi (a_{n}) = a_{n + 1}\\) . \n\nProof that all such \\(a_{1}\\) work. Let \\(x = 12^{e} \\cdot 6 \\cdot \\ell \\in \\mathbf{S}\\) with \\(\\gcd (\\ell , 10) = 1\\) . As \\(\\frac{1}{2} + \\frac{1}{3} + \\frac{1}{4} = \\frac{13}{12}\\) , we get \n\n\\[\\psi (x) = \\left\\{ \\begin{array}{ll}x & e = 0\\] \\[\\frac{13}{12} x & e > 0 \\end{array} \\right.\\] \n\nso by induction on the value of \\(e\\) we see that \\(\\psi (x) \\in \\mathbf{S}\\) (the base \\(e = 0\\) coming from \\(\\psi\\) fixing \\(x\\) ). \n\nProof that all \\(a_{1}\\) are of this form. In what follows \\(x\\) is always an element of \\(\\mathbf{S}\\) , not necessarily an element of the sequence. \n\nClaim — Let \\(x \\in \\mathbf{S}\\) . If \\(2 \\mid \\psi (x)\\) then \\(2 \\mid x\\) . \n\nProof. If \\(x\\) is odd then every divisor of \\(x\\) is odd, so \\(\\psi (x)\\) is the sum of three odd numbers. \n\nClaim — Let \\(x \\in \\mathbf{S}\\) . If \\(6 \\mid \\psi (x)\\) then \\(6 \\mid x\\) . \n\nProof. We consider only \\(x\\) even because of the previous claim. We prove the contrapositive that \\(3 \\nmid x \\implies 6 \\nmid \\psi (x)\\) (for even \\(x\\) ). \n\n- If \\(4 \\mid x\\) , then letting \\(d\\) be the third largest proper divisor of \\(x\\) , \n\n\\[\\psi (x) = \\frac{x}{2} + \\frac{x}{4} +d = \\frac{3}{4} x + d \\equiv d \\not\\equiv 0 \\pmod {3}.\\] \n\n- Otherwise, let \\(p \\mid x\\) be the smallest prime dividing \\(x\\) , with \\(p > 3\\) . If the third-smallest nontrivial divisor of \\(x\\) is \\(2p\\) , then \n\n\\[\\psi (x) = \\frac{x}{2} + \\frac{x}{p} + \\frac{x}{2p} = \\frac{3}{2p} x + \\frac{x}{2} \\equiv \\frac{x}{2} \\not\\equiv 0 \\pmod {3}.\\] \n\nIf the third- smallest nontrivial divisor of \\(x\\) is instead an odd prime \\(q\\) , then \n\n\\[\\psi (x) = \\frac{x}{2} + \\frac{x}{p} + \\frac{x}{q} \\equiv 1 + 0 + 0 \\equiv 1 \\pmod {2}.\\] \n\nTo tie these two claims into the problem, we assert:\n\n\n\nClaim — Every \\(a_{i}\\) must be divisible by 6. \n\nProof. The idea is to combine the previous two claims (which have no dependence on the sequence) with a size argument. \n\n- For odd \\(x \\in \\mathbf{S}\\) note that \\(\\psi (x) < \\left(\\frac{1}{3} + \\frac{1}{5} + \\frac{1}{7}\\right)x < x\\) and \\(\\psi (x)\\) is still odd. So if any \\(a_{i}\\) is odd the sequence is strictly decreasing and that's impossible. Hence, we may assume \\(a_{1}, a_{2}, \\ldots\\) are all even. \n\n- If \\(x \\in \\mathbf{S}\\) is even but \\(3 \\nmid x\\) then \\(\\psi (x) < \\left(\\frac{1}{2} + \\frac{1}{4} + \\frac{1}{5}\\right)x < x\\) and \\(\\psi (x)\\) is still not a multiple of 3. So if any \\(a_{i}\\) is not divisible by 3 the sequence is again strictly decreasing. \\(\\square\\) \n\nOn the other hand, if \\(x\\) is a multiple of 6, we have the following formula for \\(\\psi (x)\\) : \n\n\\[\\psi (x) = \\left\\{ \\begin{array}{ll}\\frac{13}{12} x & 4\\mid x\\] \\[\\frac{31}{30} x & 4\\mid x\\mathrm{~but~}5\\mid x\\] \\[x & 4\\mid x\\mathrm{~and~}5\\mid x. \\end{array} \\right.\\] \n\nLooking back on our sequence of \\(a_{i}\\) (which are all multiples of 6), the center case cannot happen with our \\(a_{i}\\) , because \\(\\frac{31}{30} x\\) is odd when \\(x \\equiv 2\\) (mod 4). Hence in actuality \n\n\\[a_{n + 1} = \\frac{13}{12} a_{n}\\quad \\mathrm{or}\\quad a_{n + 1} = a_{n}\\] \n\nfor every \\(n\\) . \n\nLet \\(T\\) be the smallest index such that \\(a_{T} = a_{T + 1} = a_{T + 2} = \\dots\\) (it must exist because we cannot multiply by \\(\\frac{13}{12}\\) forever). Then we can exactly describe the sequence by \n\n\\[a_{n} = a_{1}\\cdot \\left(\\frac{13}{12}\\right)^{\\min (n,T) - 1}.\\] \n\nHence \\(a_{1} = \\left(\\frac{12}{13}\\right)^{T - 1}a_{T}\\) , and since \\(a_{T}\\) is a multiple of 6 not divisible by 4 or 5, it follows \\(a_{1}\\) has the required form.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2025-notes.jsonl", "problem_match": "4. ", "solution_match": "## \\(\\S 2.1\\) IMO 2025/4, proposed by Paulius Aleknavičius (LIT) \n"}}
{"year": "2025", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "Alice and Bazza are playing the inekolaty game, a two-player game whose rules depend on a positive real number \\(\\lambda\\) which is known to both players. On the \\(n\\) th turn of the game (starting with \\(n = 1\\) ) the following happens: \n\n- If \\(n\\) is odd, Alice chooses a nonnegative real number \\(x_{n}\\) such that \n\n\\[x_{1} + x_{2} + \\dots +x_{n}\\leq \\lambda n.\\] \n\n- If \\(n\\) is even, Bazza chooses a nonnegative real number \\(x_{n}\\) such that \n\n\\[x_{1}^{2} + x_{2}^{2} + \\dots +x_{n}^{2}\\leq n.\\] \n\nIf a player cannot choose a suitable \\(x_{n}\\) , the game ends and the other player wins. If the game goes on forever, neither player wins. All chosen numbers are known to both players. \n\nDetermine all values of \\(\\lambda\\) for which Alice has a winning strategy and all those for which Bazza has a winning strategy.", "solution": "he answer is that Alice has a winning strategy for \\(\\lambda >1 / \\sqrt{2}\\) , and Bazz a has a winning strategy for \\(\\lambda < 1 / \\sqrt{2}\\) . (Neither player can guarantee winning for \\(\\lambda = 1 / \\sqrt{2}\\) .) \n\nWe divide the proof into two parts. \n\n\\(\\P\\) Alice's strategy when \\(\\lambda \\geq 1 / \\sqrt{2}\\) . Consider the strategy where Alice always plays \\(x_{2i + 1} = 0\\) for \\(i = 0,\\ldots ,k - 1\\) \n\nIn this situation, when \\(n = 2k + 1\\) we have \n\n\\[\\sum_{1}^{2k}x_{i} = 0 + x_{2} + 0 + x_{4} + \\dots +0 + x_{2k}\\] \\[\\qquad \\leq k\\cdot \\sqrt{\\frac{x_{2}^{2} + \\dots + x_{2k}^{2}}{k}} = \\sqrt{2}\\cdot k< \\lambda \\cdot (2k + 1)\\] \n\nand so the choices for \\(x_{2k + 1}\\) are \n\n\\[x_{2k + 1}\\in [0,\\lambda \\cdot (2k + 1) - \\sqrt{2} k]\\] \n\nwhich is nonempty. Hence Alice can't ever lose with this strategy. \n\nBut suppose further \\(\\lambda >\\frac{1}{\\sqrt{2}}\\) ; we show Alice can win. Choose \\(k\\) large enough that \n\n\\[\\sqrt{2}\\cdot k< \\lambda \\cdot (2k + 1) - \\sqrt{2k + 2}.\\] \n\nThen on the \\((2k + 1)\\) st turn, Alice can (after playing 0 on all earlier turns) play a number greater than \\(\\sqrt{2k + 2}\\) and cause Bazz a to lose.\n\n\n\n\\(\\parallel\\) Baza strategy when \\(\\lambda \\leq 1 / \\sqrt{2}\\) . Consider the strategy where Baza always plays \\(x_{2i + 2} = \\sqrt{2 - x_{2i + 1}^{2}}\\) for all \\(i = 0,\\ldots ,k - 1\\) (i.e. that is, the largest possible value Baza can play). \n\nTo analyze Baza's choices on each of his turns, we first need to estimate \\(x_{2k + 1}\\) . We do this by writing \n\n\\[\\lambda \\cdot (2k + 1)\\geq x_{1} + x_{2} + \\cdot \\cdot \\cdot +x_{2k + 1}\\] \\[\\qquad = \\left(x_{1} + \\sqrt{2 - x_{1}^{2}}\\right) + \\left(x_{3} + \\sqrt{2 - x_{3}^{2}}\\right)\\] \\[\\qquad \\qquad +\\cdot \\cdot \\cdot +\\left(x_{2k - 1} + \\sqrt{2 - x_{2k}^{2}}\\right) + x_{2k + 1}\\] \\[\\qquad \\geq \\underbrace{\\sqrt{2} + \\cdot \\cdot \\cdot + \\sqrt{2}}_{k\\mathrm{~times}} + x_{2k + 1} = \\sqrt{2}\\cdot k + x_{2k + 1}\\] \n\nwhere we have used the fact that \\(t + \\sqrt{2 - t^{2}}\\geq 2\\) for all \\(t\\geq 0\\) . This means that \n\n\\[x_{2k + 1}\\leq \\lambda \\cdot (2k + 1) - \\sqrt{2} k< \\sqrt{2}.\\] \n\nAnd \\(x_{1}^{2} + x_{2}^{2} + \\cdot \\cdot \\cdot +x_{2k}^{2} + x_{2k + 1}^{2} = (2 + \\cdot \\cdot \\cdot +2) + x_{2k + 1}^{2}\\) , Baza can indeed choose \\(x_{2k + 2} = \\sqrt{2 - x_{2k + 1}^{2}}\\) and always has a move. \n\nBut suppose further \\(\\lambda < 1 / \\sqrt{2}\\) . Then the above calculation also shows that Alice couldn't have made a valid choice for large enough \\(k\\) , since \\(\\lambda \\cdot (2k + 1) - \\sqrt{2} k< 0\\) for large \\(k\\) . \n\nRemark. In the strategies above, we saw that Alice prefers to always play 0 and Baza prefers to always play as large as possible. One could consider what happens in the opposite case: \n\n- If Alice tries to always play the largest number possible, her strategy still wins for \\(\\lambda >1\\) . \n\n- If Baza tries to always play 0, Alice can win no matter the value for \\(\\lambda >0\\) .", "metadata": {"resource_path": "IMO/segmented/en-IMO-2025-notes.jsonl", "problem_match": "5. ", "solution_match": "## \\(\\S 2.2\\) IMO 2025/5, proposed by Massimiliano Foschi, Leonardo Franchi (ITA) \n"}}
-{"year": "2025", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "Consider a \\(2025 \\times 2025\\) grid of unit squares. Matilda wishes to place on the grid some rectangular tiles, possibly of different sizes, such that each side of every tile lies on a grid line and every unit square is covered by at most one tile. \n\nDetermine the minimum number of tiles Matilda needs to place so that each row and each column of the grid has exactly one unit square that is not covered by any tile.", "solution": "The answer is \\(2112 = 2025 + 2 \\cdot 45 - 3\\) . In general, the answer turns out to be \\([n + 2 \\sqrt{n} - 3]\\) , but when \\(n\\) is not a perfect square the solution is more complicated. \n\nRemark. The 2017 Romanian Masters in Math asked the same problem where the tiles are replaced by sticks, i.e. \\(1 \\times k\\) tiles. The answer to that problem is completely different and as far as I know there is no connection at all to the present IMO problem. \n\nConstruction. We show a general construction when \\(n = k^{2}\\) illustrated below for \\(k = 5\\) ; it generalizes readily. There are a total of \\((k - 1)^{2}\\) tiles which are \\(k \\times k\\) squares and another \\(4(k - 1)\\) tiles on the boundary, giving a total of \n\n\\[(k - 1)^{2} + 4(k - 1) = k^{2} + 2k - 3\\] \n\ntiles as promised. \n\n\n\n\n\n\nRemark. Ironically, the construction obtaining the answer in the floor pattern at Sunshine Coast airport, the closest airport to the site of the exam. See this image or this image. \n\nBound. There are several approaches (all hard), but the shortest proof seems to be the following one that exploits the Erdős- Szekeres theorem; it is Solution 6 in the shortlist. The theorem being quoted is: \n\n## Theorem (Erdős- Szekeres) \n\nLet \\(n \\geq 1\\) be an integer. Given a permutation of \\((1, \\ldots , n)\\) , if a longest increasing subsequence (LIS) has length \\(a\\) and a longest decreasing subsequence (LDS) has length \\(b\\) , then \\(ab \\geq n\\) . \n\nThis is a stronger version of the theorem compared to another version which instead just asserts that \\(\\max (a, b) \\geq \\sqrt{n}\\) . \n\nTo apply this to the present problem, take the \\(n\\) uncovered squares which we henceforth call \"black\" as a permutation. Then consider both an LIS of length \\(a\\) and an LDS of length \\(b\\) . We do the following artistic illustration: \n\n- Draw the LIS as a broken line, then connect it to the southwest and northwest corner of the board.- Similarly, draw the LDS as a broken line, then connect it to the northwest and southeast corner of the board.- These two steps partition the board into four quadrants, which we call north, east, south, west.- For each black cell in the north quadrant, write an \\(\\mathbf{N}\\) in the cell above it (for the cell in the first row, this will be off the board). Do the same for \\(\\mathbf{E}\\) (east), \\(\\mathbf{S}\\) (south), \\(\\mathbf{W}\\) (west).- Some black cells are in multiple quadrants (i.e. part of the LIS/LDS). Write all letters in that case. \n\nThe figure below shows two examples of the process, each for a board with \\(n = 9\\) , for two choices of LIS and LDS. The cells in the LIS and LDS have been marked with green circles, and the boundaries of the quadrants are drawn in green lines. In the left example, the LIS and LDS have a black square in common (that cell has all four directions labeled). In the right example, the LIS and do not have common squares.\n\n\n\n \n\nWe observe that: \n\nClaim — In this algorithm, the total number of letters written is exactly \n\n\\[C:= \\left\\{ \\begin{array}{ll}n + a + b + 1 & \\mathrm{if~the~LIS~and~LDS~intersect}\\\\ n + a + b & \\mathrm{otherwise.} \\end{array} \\right.\\] \n\nProof. This is obvious. Each black square contributes at least one letter. Each black square on exactly one of the LIS and LDS contributes one extra letter. And a black square on both contributes 4 letters instead of \\(1 + 1 + 1\\) . \\(\\square\\) \n\nNote by AM- GM we have \\(a + b \\geq 2\\sqrt{ab} \\geq 2\\sqrt{n}\\) , so we have a bound of \n\n\\[C\\geq n + 2\\sqrt{n} +\\epsilon \\quad \\mathrm{where}\\quad \\epsilon := \\left\\{ \\begin{array}{ll}1 & \\mathrm{if~the~LIS~and~LDS~intersect}\\\\ 0 & \\mathrm{otherwise.} \\end{array} \\right.\\] \n\nTo relate \\(C\\) to the number of tiles, the critical claim is the following, which is by construction: \n\nClaim — None of Matilda's tiles can have more than one letter written in any cell. \n\nProof. This follows from the construction. \n\nWe split into two cases based on \\(\\epsilon\\) . \n\n- When \\(\\epsilon = 1\\) , at most four letters go off the grid (one for each direction), the number of tiles is at least \\(C - 4 \\geq n + 2\\sqrt{n} - 3\\) . \n\n- Suppose \\(\\epsilon = 0\\) . Then \\(C - 4 \\geq n + 2\\sqrt{n} - 4\\) . However, we make the additional observation here that the tile where the LIS and LDS meet has no letters on it either; hence there are at least \\(1 + (C - 4) \\geq n + 2\\sqrt{n} - 3\\) tiles. \n\nRemark. USJL mentions that when \\(n = ab\\) , it is in fact always possible to guarantee \\(\\epsilon = 1\\) . Moreover, when \\(a \\neq b\\) , the AM- GM inequality is strict. This gives a way to avoid the additional observation needed for \\(\\epsilon = 0\\) above.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2025-notes.jsonl", "problem_match": "6. ", "solution_match": "## \\(\\S 2.3\\) IMO 2025/6, proposed by Zhao Yu Ma and David Lin Kewei (SGP) \n"}}
+{"year": "2025", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "Consider a \\(2025 \\times 2025\\) grid of unit squares. Matilda wishes to place on the grid some rectangular tiles, possibly of different sizes, such that each side of every tile lies on a grid line and every unit square is covered by at most one tile. \n\nDetermine the minimum number of tiles Matilda needs to place so that each row and each column of the grid has exactly one unit square that is not covered by any tile.", "solution": "The answer is \\(2112 = 2025 + 2 \\cdot 45 - 3\\) . In general, the answer turns out to be \\([n + 2 \\sqrt{n} - 3]\\) , but when \\(n\\) is not a perfect square the solution is more complicated. \n\nRemark. The 2017 Romanian Masters in Math asked the same problem where the tiles are replaced by sticks, i.e. \\(1 \\times k\\) tiles. The answer to that problem is completely different and as far as I know there is no connection at all to the present IMO problem. \n\nConstruction. We show a general construction when \\(n = k^{2}\\) illustrated below for \\(k = 5\\) ; it generalizes readily. There are a total of \\((k - 1)^{2}\\) tiles which are \\(k \\times k\\) squares and another \\(4(k - 1)\\) tiles on the boundary, giving a total of \n\n\\[(k - 1)^{2} + 4(k - 1) = k^{2} + 2k - 3\\] \n\ntiles as promised. \n\n\n\n\n\n\nRemark. Ironically, the construction obtaining the answer in the floor pattern at Sunshine Coast airport, the closest airport to the site of the exam. See this image or this image. \n\nBound. There are several approaches (all hard), but the shortest proof seems to be the following one that exploits the Erdős- Szekeres theorem; it is Solution 6 in the shortlist. The theorem being quoted is: \n\n## Theorem (Erdős- Szekeres) \n\nLet \\(n \\geq 1\\) be an integer. Given a permutation of \\((1, \\ldots , n)\\) , if a longest increasing subsequence (LIS) has length \\(a\\) and a longest decreasing subsequence (LDS) has length \\(b\\) , then \\(ab \\geq n\\) . \n\nThis is a stronger version of the theorem compared to another version which instead just asserts that \\(\\max (a, b) \\geq \\sqrt{n}\\) . \n\nTo apply this to the present problem, take the \\(n\\) uncovered squares which we henceforth call \"black\" as a permutation. Then consider both an LIS of length \\(a\\) and an LDS of length \\(b\\) . We do the following artistic illustration: \n\n- Draw the LIS as a broken line, then connect it to the southwest and northwest corner of the board.- Similarly, draw the LDS as a broken line, then connect it to the northwest and southeast corner of the board.- These two steps partition the board into four quadrants, which we call north, east, south, west.- For each black cell in the north quadrant, write an \\(\\mathbf{N}\\) in the cell above it (for the cell in the first row, this will be off the board). Do the same for \\(\\mathbf{E}\\) (east), \\(\\mathbf{S}\\) (south), \\(\\mathbf{W}\\) (west).- Some black cells are in multiple quadrants (i.e. part of the LIS/LDS). Write all letters in that case. \n\nThe figure below shows two examples of the process, each for a board with \\(n = 9\\) , for two choices of LIS and LDS. The cells in the LIS and LDS have been marked with green circles, and the boundaries of the quadrants are drawn in green lines. In the left example, the LIS and LDS have a black square in common (that cell has all four directions labeled). In the right example, the LIS and do not have common squares.\n\n\n\n \n\nWe observe that: \n\nClaim — In this algorithm, the total number of letters written is exactly \n\n\\[C:= \\left\\{ \\begin{array}{ll}n + a + b + 1 & \\mathrm{if~the~LIS~and~LDS~intersect}\\\\ n + a + b & \\mathrm{otherwise.} \\end{array} \\right.\\] \n\nProof. This is obvious. Each black square contributes at least one letter. Each black square on exactly one of the LIS and LDS contributes one extra letter. And a black square on both contributes 4 letters instead of \\(1 + 1 + 1\\) . \\(\\square\\) \n\nNote by AM- GM we have \\(a + b \\geq 2\\sqrt{ab} \\geq 2\\sqrt{n}\\) , so we have a bound of \n\n\\[C\\geq n + 2\\sqrt{n} +\\epsilon \\quad \\mathrm{where}\\quad \\epsilon := \\left\\{ \\begin{array}{ll}1 & \\mathrm{if~the~LIS~and~LDS~intersect}\\\\ 0 & \\mathrm{otherwise.} \\end{array} \\right.\\] \n\nTo relate \\(C\\) to the number of tiles, the critical claim is the following, which is by construction: \n\nClaim — None of Matilda's tiles can have more than one letter written in any cell. \n\nProof. This follows from the construction. \n\nWe split into two cases based on \\(\\epsilon\\) . \n\n- When \\(\\epsilon = 1\\) , at most four letters go off the grid (one for each direction), the number of tiles is at least \\(C - 4 \\geq n + 2\\sqrt{n} - 3\\) . \n\n- Suppose \\(\\epsilon = 0\\) . Then \\(C - 4 \\geq n + 2\\sqrt{n} - 4\\) . However, we make the additional observation here that the tile where the LIS and LDS meet has no letters on it either; hence there are at least \\(1 + (C - 4) \\geq n + 2\\sqrt{n} - 3\\) tiles. \n\nRemark. USJL mentions that when \\(n = ab\\) , it is in fact always possible to guarantee \\(\\epsilon = 1\\) . Moreover, when \\(a \\neq b\\) , the AM- GM inequality is strict. This gives a way to avoid the additional observation needed for \\(\\epsilon = 0\\) above.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2025-notes.jsonl", "problem_match": "6. ", "solution_match": "## \\(\\S 2.3\\) IMO 2025/6, proposed by Zhao Yu Ma and David Lin Kewei (SGP) \n"}}
diff --git a/NewZealand_MO/md/en-nzmo1_2019_solutions.md b/NewZealand_MO/md/en-nzmo1_2019_solutions.md
index 30401bd59d045d03ce22ad90c211524aa68c94af..28dcd616dc7d5cf9886a053d9c1223ba762dd99b 100644
--- a/NewZealand_MO/md/en-nzmo1_2019_solutions.md
+++ b/NewZealand_MO/md/en-nzmo1_2019_solutions.md
@@ -34,7 +34,7 @@ Remark: In order to receive full marks, a student would have to demonstrate that
Solution: First notice that the sidelengths of \(\triangle ABC\) are 3, 4 and 5. By Pythagoras this implies that triangle \(ABC\) is right- angled at \(B\) . Now we can put the diagram on coordinate axes such that \(B = (0,0)\) and \(A = (0,3)\) and \(C = (4,0)\) . Furthermore we get \(D = (0,2)\) and since \(E\) divides \(CA\) into the ratio \(2:3\) we get \(E = (2.4,1.2)\) , as shown in the diagram.
-
+
Now we can calculate the slope of the line \(D E\) to be \(\frac{- 0.8}{2.4} = - \frac{1}{3}\) . This means that the equation of line \(D E\) is given by \(y = - \frac{x}{3} +2\) . Therefore the \(x\) - intercept of this line is the solution to \(0 = - \frac{x}{3} +2\) . The solution is when \(x = 6\) , and thus \(F = (6,0)\) . Hence \(C F = 2\) . \(\square\)
@@ -75,7 +75,7 @@ which is a perfect square. Therefore the only solution is \(n = 2\) . \(\square\
Solution: To simplify notation, we will let \(m\) be \(m(U)\) and let \(n\) be \(|U|\) . First note that there are 10 different diagonal-lengths in a regular 21- gon. Now consider the following set of 5 vertices.
-
+
Note that each of the 10 different diagonal- lengths appear (exactly once each). So for this set of 5 vertices we have \(\frac{m}{n} = \frac{10}{5} = 2\) . We will now show that this is the maximum possible value for \(\frac{m}{n}\) . If \(U\) is an arbitrary non- empty set of vertices, then there are two cases:
@@ -99,7 +99,7 @@ Remark: The construction given is unique up to rotations and reflections. I.e. a
Solution: Let \(M\) be the midpoint of \(AF\) and let \(O\) be the circumcentre of triangle \(CPD\) . Now construct \(Q\) to be the point such that \(CPDQ\) is a parallelogram, and let \(R\) be the centre of this parallelogram (i.e. \(R\) is the intersection of \(PQ\) with \(CD\) , and also \(R\) is the midpoint of \(PQ\) ).
-
+
Note that \(Q D = C P = F P\) and \(D P = P A\) and \(\angle Q D P = 180^{\circ} - \angle D P C = \angle F P A\) Therefore (by SAS) we have a pair of congruent triangles:
@@ -112,7 +112,7 @@ Therefore \(\angle M A P = \angle R P D\) and \(A F = P Q\) . Thus \(A M = \frac
Therefore \(\angle A P M = \angle R D P\) . Let \(x = \angle A P M\) so that \(\angle C D P = \angle R D P = x\) also.
-
+
Since the angle subtended at the circumcentre is double the angle subtended at the circumference, we get \(\angle C O P = 2x\) (recall that \(O\) is the circumcentre of \(\triangle P C D\) ). Finally we get \(\angle O P C = 90^{\circ} - x\) because \(\triangle C O P\) is isosceles. Putting this all together, we get
diff --git a/NewZealand_MO/md/en-nzmo1_2020_solutions.md b/NewZealand_MO/md/en-nzmo1_2020_solutions.md
index 69156acf0f70e2de493a62dc1efc62e55166c2fd..77f30335805493511ab94d5fda4e36bc11acfc53 100644
--- a/NewZealand_MO/md/en-nzmo1_2020_solutions.md
+++ b/NewZealand_MO/md/en-nzmo1_2020_solutions.md
@@ -17,7 +17,7 @@ numbers between 1 and 50 which are divisible by 2 exactly \(k\) times. Hence the
Solution:
-
+
Construct point \(E\) on diagonal \(BD\) so that \(EX\) is parallel to \(CD\) . So \(\angle BXE = \angle BCD = 90^\circ\) . Also \(\angle XBE = 45^\circ\) because \(E\) is on diagonal \(BD\) . Therefore triangle \(\triangle BEX\) is an isosceles right- angled triangle. Hence
@@ -48,7 +48,7 @@ Solution: All the angles in squares and equilateral triangles are multiples of \
Hence \(n \leq 12\) . Also all polygons have at least 3 sides so \(3 \leq n \leq 12\) . Finally we demonstrate that it is possible for any \(3 \leq n \leq 12\) using the following illustrations.
-
+
@@ -137,7 +137,7 @@ Hence \(f(z) = 2z + 4\) is the unique function \(f\) satisfying the given equati
Solution: Let \(Z\) be the point where \(PY\) intersects \(AX\) . The problem asks us to prove that \(AZ = ZX\) .
-
+
@@ -222,7 +222,7 @@ To simplify notation, we define a \(k\) - strip to be a \(1 \times 8\) rectangle
Start by finding 44 disjoint \(1 \times 9\) rectangles on the board.
-
+
- Josie places a white stone in one end of each of these 44 rectangles on her first 22 turns.
diff --git a/NewZealand_MO/md/en-nzmo1_2021_solutions.md b/NewZealand_MO/md/en-nzmo1_2021_solutions.md
index 0221e21500190db3efd5386ea10a47905dec41fa..480f10fdc33094e1963673d6964f67b1395a8bce 100644
--- a/NewZealand_MO/md/en-nzmo1_2021_solutions.md
+++ b/NewZealand_MO/md/en-nzmo1_2021_solutions.md
@@ -37,7 +37,7 @@ First note that if \(x\geq 0.8n\) then \(\frac{x}{x + c}\geq \frac{x}{n}\geq 0.8
Solution: First note that triangle \(ABE\) is isosceles because \(AB = BE\).
-
+
Let \(x = \angle DEC\) . Angle chasing gives:
@@ -106,7 +106,7 @@ Therefore the only solutions for \((x,p,n)\) are \((2,5,2)\) and \((0,3,1)\) .
Solution: Let \(E^{\prime}\) be the point on line \(BC\) such that \(DE^{\prime}\) is parallel to \(AB\) . We know that \(E^{\prime}\) lies between \(B\) and \(C\) because \(D\) lies between \(A\) and \(C\) . So it suffices for us to prove that \(BE^{\prime} = CD\) .
-
+
\[\angle A C B = \angle C B A\] \[= \angle C E^{\prime}D.\]
diff --git a/NewZealand_MO/md/en-nzmo1_2022_solutions.md b/NewZealand_MO/md/en-nzmo1_2022_solutions.md
index f2ea1d6b6c4803b319ae012ac55da379c9d02920..7b3b5b7ebf7b0248c952ea162143680bae93b5fa 100644
--- a/NewZealand_MO/md/en-nzmo1_2022_solutions.md
+++ b/NewZealand_MO/md/en-nzmo1_2022_solutions.md
@@ -9,7 +9,7 @@ Solution: (Kevin Shen)
\(M\) is the midpoint of \(AD\) , by symmetry \(MB = MC\) . The side lengths of an equilateral triangle are all equal, so \(MB = MP\) .
-
+
As \(MB = MC = MP\) , \(M\) is the circumcenter of triangle \(BCP\) . For any chord of any circle, the angle subtended at the center is always double the angle subtended at the circumference. Since \(\angle PCB\) and \(\angle PMB\) are both subtended by arc \(BP\) , we get
@@ -268,7 +268,7 @@ Solution A: (Ross Atkins)
Let \(F\) be the point on the circle such that \(AF\) is a diameter. So \(M\) is the midpoint of \(AF\) . Now let \(E \neq F\) be the point on the circle such that \(D, E, F\) are colinear. This means \(\angle AEF = 90^\circ\) by Thales' theorem.
-
+
By the Power of a Point Theorem we get \(DP^2 = DE \times DF = DQ^2\) . Therefore we have two pairs of similar triangles: \(EPD \sim PFD\) and \(EQD \sim QFD\) . Hence
diff --git a/NewZealand_MO/md/en-nzmo1_2023_solutions.md b/NewZealand_MO/md/en-nzmo1_2023_solutions.md
index 3aea6734d1b4ccbfaf33fbab0a7f024de6b62c68..2206ae07eae812ed94f76b853167563222c0bace 100644
--- a/NewZealand_MO/md/en-nzmo1_2023_solutions.md
+++ b/NewZealand_MO/md/en-nzmo1_2023_solutions.md
@@ -37,7 +37,7 @@ where \(|X Y Z|\) denotes the area of triangle \(X Y Z\) .
Solution: (Kevin Shen)
-
+
Since \(PQ\) is parallel to \(BC\) , there is a dilation (centred at \(A\) ) of factor \(AB / AP\) that sends triangle \(APQ\) to triangle \(ABC\) . Thus
@@ -156,7 +156,7 @@ Solution: (Kevin Shen)
Wlog let \(AB = 1\) and \(BC = a\) . Also let \(BD = x\) . We will try to find all the lengths in the diagram in terms of \(a\) .
-
+
By Pythagoras in \(\triangle ABC\) we get \(AC = \sqrt{a^2 - 1}\) . By the angle- bisector theorem we get \(\frac{AD}{DC} = \frac{AB}{BC} = \frac{1}{a}\) , and so \(DC = a \times AD\) . This can be substituted into \(AD + DC = AC = \sqrt{a^2 - 1}\) to get \(AD(1 + a) = \sqrt{a^2 - 1}\) . Therefore
@@ -213,7 +213,7 @@ Let \(E\) be the foot of the perpendicular from \(D\) to \(BC\) . Note that \(E\
Therefore \(AB = EB = EN\) and hence \(2AB = BE + EN = BN\) . We also get \(BD = DN\) . Let \(F\) be the point on line \(BC\) such \(BD = BF\) .
-
+
Assume \(BC - BD = 2AB\) . Since \(BD = DN\) and \(2AB = BN\) we can get \(BC - DN = BN\) . Therefore
diff --git a/NewZealand_MO/md/en-nzmo1_2024_solutions.md b/NewZealand_MO/md/en-nzmo1_2024_solutions.md
index c500a0b5064ef0dd4a1ce7e9f68b9f5372b00bae..7c6b5c9e119cc27def40135f9d50071c0fc17334 100644
--- a/NewZealand_MO/md/en-nzmo1_2024_solutions.md
+++ b/NewZealand_MO/md/en-nzmo1_2024_solutions.md
@@ -47,7 +47,7 @@ Solution: (Ross Atkins)
Let the original rectangle be \(ABCD\) . Let the fold line be \(XY\) with point \(X\) lying on side \(AB\) and point \(Y\) lying on side \(CD\) . Consider the fold such that point \(D\) lands on point \(B\) . Let \(P\) be the location of corner \(A\) after the fold.
-
+
The two-sheets-thick area is \(\triangle BXY\) which should be \(80\%\) . By symmetry, triangles \(\triangle PXB\) and \(\triangle CYB\) are congruent with equal hypotenuses
@@ -72,7 +72,7 @@ Combining this with \(CD = CY + YD = CY + YB = 9CY\) we get
4. Problem: A dot-trapezium consists of several rows of dots such that each row contains one more dot than the row immediately above (apart from the top row). For example here is a dot-trapezium consisting of 15 dots, having 3 rows and 4 dot in the top row.
-
+
A positive integer \(n\) is called a trapezium- number if there exists a dot- trapezium consisting of exactly \(n\) dots, with at least two rows and at least two dots in the top row. How many trapezium- numbers are there less than 100?
@@ -109,7 +109,7 @@ However this doesn't work because we require \(x > x + 1\) . (when \(y = 1\) the
The only remaining possibility is when \(m = 2^{k + 1} \pm 1\) and \(m\) is composite. If \(k > 2\) then \(m = 2^{k + 1} \pm 1 \geq 15\) , so \(n = 2^{k} m \geq 8 \times 15 > 100\) and we don’t need to consider it. For \(k \leq 2\) we get \(m = 1, 3, 5, 7, 9\) . We can’t have \(m = 1\) because then \(n\) would be a power of 2. We also can’t have \(m = 3, 5, 7\) because they are prime. Finally we consider \(m = 9\) and so \(k = 2\) . In this case we get \(n = 2^{k} m = 36\) , which is a trapezium number as seen here:
-
+
Therefore all non- trapezium- numbers less than 100, are the powers of two and numbers of the form \(n = 2^{k}(2^{k + 1} \pm 1)\) where \((2^{k + 1} \pm 1)\) is prime (and \(k \leq 2\) ). The powers of two are: \(\{1, 2, 4, 8, 16, 32, 64\}\) . The numbers of the form \(2^{k}(2^{k + 1} \pm 1)\) with \(k \leq 2\) and \((2^{k + 1} \pm 1)\) being prime are
@@ -176,7 +176,7 @@ Solution: (Ross Atkins)
Let \(a, b, c\) be the sidelengths \(BC, AC, AB\) respectively, and let \(s\) be the semiperimeter of triangle \(ABC\) (i.e. let \(s = \frac{a + b + c}{2}\) ). Since \(X\) and \(Y\) are the points of contact of the incircle we get \(AX = AY\) . Similarly \(BX = BZ\) and \(CY = CZ\) where \(Z\) is the point of tangency between \(\omega\) and side \(BC\) . Let \(p = AX = AY\) and \(q = BX = BZ\) and \(r = CY + CZ\) as in the diagram.
-
+
We have the system of equations:
@@ -201,7 +201,7 @@ Adding these together gives us \(C D + C E = a + b + (A F + B F) = a + b + c\) .
Therefore \(C D = C E = s\) . This means \(B F = C D - a = (s - a)\) and thus \(X^{\prime}\) and \(F\) are the same point.
-
+
Now consider the homothety (centred at \(C\) ) that carries \(\omega\) to \(\omega_{C}\) . This homothety sends point \(P\) to point \(X^{\prime}\) (since \(C, P, X^{\prime}\) are colinear). It also carries the tangency point \(Y\) to \(E\) . It follows that
diff --git a/NewZealand_MO/md/en-nzmo1_2025_solutions.md b/NewZealand_MO/md/en-nzmo1_2025_solutions.md
index 4a662b696f92aafb572b309df186e4a697f5521a..289b9ba5eff4cce1a890d7016ebb9c14a4ff0252 100644
--- a/NewZealand_MO/md/en-nzmo1_2025_solutions.md
+++ b/NewZealand_MO/md/en-nzmo1_2025_solutions.md
@@ -31,7 +31,7 @@ Solution: (Nico McKinlay)
Construct point \(E\) so that \(ABEC\) is a rectangle. The diagonals of any rectangle bisect each other, that is, they meet at each other's midpoints. Hence \(AE\) and \(BC\) meet at \(M\) , i.e. \(E\) lies on line \(AM\) .
-
+
By symmetry in rectangle \(ABEC\) , we have
@@ -178,7 +178,7 @@ and we simply choose \(n > \frac{2}{\epsilon}\) to complete the proof.
Solution 1: (Nico McKinlay)
-
+
Assume \(T\) lies on \(AD\) . Then by the angle bisector theorem in triangles \(AED\) and \(AFD\) , we have
diff --git a/NewZealand_MO/md/en-nzmo2_2019_solutions.md b/NewZealand_MO/md/en-nzmo2_2019_solutions.md
index 4f7655731b20146015e69f7329233fc8769ca5cb..dba4a56ceb68334109256f83ded3d5cf69cf1f65 100644
--- a/NewZealand_MO/md/en-nzmo2_2019_solutions.md
+++ b/NewZealand_MO/md/en-nzmo2_2019_solutions.md
@@ -65,13 +65,13 @@ Solution: For the purpose of this proof, we will consider a vertex to be any poi
There can be no other types of vertex, because all the angles must be \(60^{\circ}\) or \(180^{\circ}\) (or \(300^{\circ}\) in the corners of the original large triangle). We will now colour each edge- end either green or blue, as shown in the diagram (the green edge- ends are also a bit thicker).
-
+
-
+
-
+
@@ -82,7 +82,7 @@ An edge- end is coloured green if it touches a Type \(B\) vertex along the \(180
Note also that the 6 edges with an endpoint at a type \(A\) vertex must all have their other end being green. There are more blue edge- ends than green edge- ends, so it can't be the case that every blue edge- end is connected to a green edge- end. There must therefore be an edge such that both of its ends are blue. Since neither endpoint of such an edge is a type \(A\) vertex, we can conclude that all four of the angles at the endpoints of this double blue- ended edge must be \(60^{\circ}\) .
-
+
Therefore this edge is a shared side of two triangular pieces. These two triangular pieces must therefore be the same size. \(\square\)
diff --git a/NewZealand_MO/md/en-nzmo2_2020_solutions.md b/NewZealand_MO/md/en-nzmo2_2020_solutions.md
index 4d98789bde6b4e9cf63e9c806997a0e00b0d6f91..76a8095067c2cfdad5bf820c79651c6b54ee23b6 100644
--- a/NewZealand_MO/md/en-nzmo2_2020_solutions.md
+++ b/NewZealand_MO/md/en-nzmo2_2020_solutions.md
@@ -49,7 +49,7 @@ Therefore the smallest positive possibility for \(N + 2\) is 504. Thus \(N = 502
Solution: Divide the circle into 6 equal ( \(60^{\circ}\) ) arcs. By the pigeon-hole principle (since \(13 > 2 \times 6\) ) there exists at least one arc which contains at least three of the marked points. Let this \(60^{\circ}\) arc be \(AB\) , and let the three marked points be \(X\) , \(Y\) and \(Z\) in that order.
-
+
Let \(M\) be the midpoint of arc \(AB\) . Points \(X\) and \(Z\) both lie within the minor arc \(AB\) so the base of triangle \(\triangle XYZ\) is less than or equal to the base of triangle \(\triangle AMB\) . Also the distance from \(Y\) to \(XZ\) is less than or equal to the distance from \(Y\) to \(AB\) , which is at most the distance from \(M\) to \(AB\) . Hence: the base and height of triangle \(\triangle XYZ\) are less than or equal to the base and height of triangle \(\triangle AMB\) respectively. Therefore
@@ -78,7 +78,7 @@ as required.
Solution: Let \(\lambda\) be the common tangent of \(\Gamma_{1}\) and \(\Gamma_{2}\) at point \(A\) . Let \(P\) be a point on \(\lambda\) such that \(P\) and \(C\) are on opposite sides of line \(AB\) . Let \(Q\) and \(R\) be the points of intersection of \(\Gamma_{1}\) with \(AB\) and \(AC\) respectively.
-
+
\[\angle BCA = \angle BAP\] \[\qquad = \angle QAP\] \[\qquad = \angle QRA\]
diff --git a/NewZealand_MO/md/en-nzmo2_2021_solutions.md b/NewZealand_MO/md/en-nzmo2_2021_solutions.md
index 0beae3be7db3e0ef338dd901da3b522ce46a20d3..2d448921ac94fa659ce91661b698d12fa647cc66 100644
--- a/NewZealand_MO/md/en-nzmo2_2021_solutions.md
+++ b/NewZealand_MO/md/en-nzmo2_2021_solutions.md
@@ -11,7 +11,7 @@ Determine the length of the diagonal \(BD\) .
Solution: Since \(AD = DC\) and \(\angle ADC = 90^{\circ}\) , we can fit four copies of quadrilateral \(ABCD\) around vertex \(D\) as shown in the diagram.
-
+
The outer shape is a quadrilateral because \(\angle DAB + \angle BCD = 180^{\circ}\) . Moreover it is a rectangle because \(\angle ABC = 90^{\circ}\) . In fact it is a square with side- length 2021 because of rotational symmetry and \(AB + BC = 2021\) . Also \(D\) is the centre of the square because it is the centre of the rotational symmetry. So \(BD\) is the distance from a vertex to the centre of the square, which is half the length of the diagonal of the square. Thus
@@ -102,7 +102,7 @@ Comment: The example \(\{1,3,7,9\}\) is not the only example that satisfies the
Solution: Construct the tangent line to \(\Gamma\) at \(P\) . Note that this line is also tangent to the circle through points \(C\) and \(P\) with centre \(O\) . Also construct point \(E\) on this tangent line to the right of \(P\) . Note that \(\angle EPO = 90^{\circ}\) and \(\angle OCA = 90^{\circ}\) because the radii and tangents are perpendicular.
-
+
Let \(x = \angle PBQ\)
diff --git a/NewZealand_MO/md/en-nzmo2_2022_solutions.md b/NewZealand_MO/md/en-nzmo2_2022_solutions.md
index d8d653f6a57731dc1ea416ee32422a0b0611b5e0..d4d56313b35340e411e61cbf8d168384ee192c64 100644
--- a/NewZealand_MO/md/en-nzmo2_2022_solutions.md
+++ b/NewZealand_MO/md/en-nzmo2_2022_solutions.md
@@ -95,7 +95,7 @@ Solution: (Kevin Shen)
First note that \(ID = IE = IF\) because they are all radii of the incircle, and \(\angle BFI = \angle BDI = 90^{\circ}\) because tangents are perpendicular to radii. Since \(\angle ABC = 90^{\circ}\) we have \(BFID\) a square and so \(BD = BF = ID\) too. Thus \(\triangle EIF\) is isosceles and so \(\angle IFE = \angle FEI\) .
-
+
Since \(CI\) is the bisector of \(\angle DCE\) , we see that \(D\) and \(E\) are reflections of each other over line \(CIP\) . Therefore \(\angle IDP = \angle PEI\) . Hence
diff --git a/NewZealand_MO/md/en-nzmo2_2023_solutions.md b/NewZealand_MO/md/en-nzmo2_2023_solutions.md
index 8feb724fdf52a67d5fb611af1fcb43cdd1d7e206..569074f25d19dcc639371304adf792711694981f 100644
--- a/NewZealand_MO/md/en-nzmo2_2023_solutions.md
+++ b/NewZealand_MO/md/en-nzmo2_2023_solutions.md
@@ -63,7 +63,7 @@ Solution: (Kevin Shen)
Since \(Z\) lies on diagonal \(AC\) , we have \(\angle DAZ = 45^{\circ}\) and \(\angle ZAB = 45^{\circ}\) . Therefore \(B\) lies on diagonal \(AY\) of square \(AXYZ\) and \(\angle BAX = 45^{\circ}\) .
-
+
Since \(AB = AD\) and \(AX = AZ\) and \(\angle BAX = 45^{\circ} = \angle DAZ\) , we have congruent triangles
diff --git a/NewZealand_MO/md/en-nzmo2_2024_solutions.md b/NewZealand_MO/md/en-nzmo2_2024_solutions.md
index 8770b59c8a4c67c0bd6925b3c60bbea5307a1f0c..cbca4e84bcf12d7e988d67f3a509a211c1836e51 100644
--- a/NewZealand_MO/md/en-nzmo2_2024_solutions.md
+++ b/NewZealand_MO/md/en-nzmo2_2024_solutions.md
@@ -49,7 +49,7 @@ Let \(x = \angle ABF\) and let \(y = \angle BCA\) . Since \(\triangle ABF\) is i
\[\Rightarrow \angle BCA = \angle BDA = \angle ACE = \angle ADE = y.\]
-
+
Since opposite angles in a cyclic quadrilateral (ABDE) are supplementary, we get \(\angle ABD + \angle DEA = 180^{\circ}\) . Therefore \(\angle DEA = 180^{\circ} - x\) . Now consider triangles \(AED\) and \(AFD\) . We have
diff --git a/NewZealand_MO/md/en-nzmo2_2025_solutions.md b/NewZealand_MO/md/en-nzmo2_2025_solutions.md
index f1e287679cb807ed71dd3b79e368151e9a280ca9..dd834f3c65a82dda46c51976ccbbffa1e0832557 100644
--- a/NewZealand_MO/md/en-nzmo2_2025_solutions.md
+++ b/NewZealand_MO/md/en-nzmo2_2025_solutions.md
@@ -47,7 +47,7 @@ Solution: (Nico McKinlay & George Zhu)
Let \(\alpha = \angle B A C\) . Let \(B B^{\prime}\) and \(C C^{\prime}\) be altitudes in triangle \(A B C\) , as shown.
-
+
Claim. Triangle \(C D E\) is isosceles with \(C E = D E\) .
diff --git a/NewZealand_MO/segmented/en-nzmo1_2019_solutions.jsonl b/NewZealand_MO/segmented/en-nzmo1_2019_solutions.jsonl
index 1ff1a347a374707924fe66323c64cf93f77c0ef8..5847ce4e99ec548dd7b718b2bc27e56f08df2a17 100644
--- a/NewZealand_MO/segmented/en-nzmo1_2019_solutions.jsonl
+++ b/NewZealand_MO/segmented/en-nzmo1_2019_solutions.jsonl
@@ -1,8 +1,8 @@
{"year": "2019", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "NewZealand_MO", "problem": "How many positive integers less than 2019 are divisible by either 18 or 21, but not both?", "solution": ": For any positive integer \\(n\\) , the number of multiples of \\(n\\) less than or equal to 2019 is given by \n\n\\[\\left\\lfloor \\frac{2019}{n}\\right\\rfloor .\\] \n\nSo there are \\(\\left\\lfloor \\frac{2019}{18}\\right\\rfloor = 112\\) multiples of 18, and \\(\\left\\lfloor \\frac{2019}{21}\\right\\rfloor = 96\\) multiples of 21. Moreover, since \\(lcm(18,21) = 126\\) there are \\(\\left\\lfloor \\frac{2019}{126}\\right\\rfloor = 16\\) positive integers less than 2019 which are a multiple of both 18 and 21. Therefore the final answer is \n\n\\[\\left\\lfloor \\frac{2019}{18}\\right\\rfloor +\\left\\lfloor \\frac{2019}{21}\\right\\rfloor -2\\left\\lfloor \\frac{2019}{126}\\right\\rfloor = 112 + 96 - 2\\times 16 = 176.\\]", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2019_solutions.jsonl", "problem_match": "\n1.", "solution_match": "\nSolution"}}
{"year": "2019", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "NewZealand_MO", "problem": "Find all real solutions to the equation \n\n\\[(x^{2} + 3x + 1)^{x^{2} - x - 6} = 1.\\]", "solution": ": Let \\(a = x^{2} + 3x + 1\\) and let \\(b = x^{2} - x - 6\\) . The only way to have \\(a^{b} = 1\\) , is if \\(a = \\pm 1\\) or \\(b = 0\\) . \n\n- If \\(b = 0\\) , then we solve the quadratic \\(x^{2} - x - 6 = 0\\) which has solutions \\(x = -2,3\\) (we would also have to check that \\(a \\neq 0\\) in this case) \n\n- If \\(a = 1\\) , then we solve the quadratic \\(x^{2} + 3x + 1 = 1\\) which has solutions \\(x = 0, -3\\) . \n\n- If \\(a = -1\\) , then we solve the quadratic \\(x^{2} + 3x + 1 = -1\\) which has solutions \\(x = -1, -2\\) . (we also have to check that \\(b\\) is an even integer in this case) \n\nTherefore there are a total of 5 candidate solutions: \\(x = - 3, - 2, - 1,0,3\\) . \n\nRemark: In order to receive full marks, a student would have to demonstrate that \\(x = - 3, - 2, - 1,0,3\\) are actually all solutions, by substituting each of these values into the expression, and verify that the result is indeed 1.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2019_solutions.jsonl", "problem_match": "\n2.", "solution_match": "\nSolution"}}
-{"year": "2019", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "NewZealand_MO", "problem": "In triangle \\(ABC\\) , points \\(D\\) and \\(E\\) lie on the interior of segments \\(AB\\) and \\(AC\\) , respectively, such that \\(AD = 1\\) , \\(DB = 2\\) , \\(BC = 4\\) , \\(CE = 2\\) and \\(EA = 3\\) . Let \\(DE\\) intersect \\(BC\\) at \\(F\\) . Determine the length of \\(CF\\) .", "solution": ": First notice that the sidelengths of \\(\\triangle ABC\\) are 3, 4 and 5. By Pythagoras this implies that triangle \\(ABC\\) is right- angled at \\(B\\) . Now we can put the diagram on coordinate axes such that \\(B = (0,0)\\) and \\(A = (0,3)\\) and \\(C = (4,0)\\) . Furthermore we get \\(D = (0,2)\\) and since \\(E\\) divides \\(CA\\) into the ratio \\(2:3\\) we get \\(E = (2.4,1.2)\\) , as shown in the diagram.\n\n\n\n \n\nNow we can calculate the slope of the line \\(D E\\) to be \\(\\frac{- 0.8}{2.4} = - \\frac{1}{3}\\) . This means that the equation of line \\(D E\\) is given by \\(y = - \\frac{x}{3} +2\\) . Therefore the \\(x\\) - intercept of this line is the solution to \\(0 = - \\frac{x}{3} +2\\) . The solution is when \\(x = 6\\) , and thus \\(F = (6,0)\\) . Hence \\(C F = 2\\) . \\(\\square\\)", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2019_solutions.jsonl", "problem_match": "\n3.", "solution_match": "\nSolution"}}
+{"year": "2019", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "NewZealand_MO", "problem": "In triangle \\(ABC\\) , points \\(D\\) and \\(E\\) lie on the interior of segments \\(AB\\) and \\(AC\\) , respectively, such that \\(AD = 1\\) , \\(DB = 2\\) , \\(BC = 4\\) , \\(CE = 2\\) and \\(EA = 3\\) . Let \\(DE\\) intersect \\(BC\\) at \\(F\\) . Determine the length of \\(CF\\) .", "solution": ": First notice that the sidelengths of \\(\\triangle ABC\\) are 3, 4 and 5. By Pythagoras this implies that triangle \\(ABC\\) is right- angled at \\(B\\) . Now we can put the diagram on coordinate axes such that \\(B = (0,0)\\) and \\(A = (0,3)\\) and \\(C = (4,0)\\) . Furthermore we get \\(D = (0,2)\\) and since \\(E\\) divides \\(CA\\) into the ratio \\(2:3\\) we get \\(E = (2.4,1.2)\\) , as shown in the diagram.\n\n\n\n \n\nNow we can calculate the slope of the line \\(D E\\) to be \\(\\frac{- 0.8}{2.4} = - \\frac{1}{3}\\) . This means that the equation of line \\(D E\\) is given by \\(y = - \\frac{x}{3} +2\\) . Therefore the \\(x\\) - intercept of this line is the solution to \\(0 = - \\frac{x}{3} +2\\) . The solution is when \\(x = 6\\) , and thus \\(F = (6,0)\\) . Hence \\(C F = 2\\) . \\(\\square\\)", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2019_solutions.jsonl", "problem_match": "\n3.", "solution_match": "\nSolution"}}
{"year": "2019", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "NewZealand_MO", "problem": "Show that the number \\(122^{n} - 102^{n} - 21^{n}\\) is always one less than a multiple of 2020, for any positive integer \\(n\\) .", "solution": ": Let \\(f(n) = 122^{n} - 102^{n} - 21^{n}\\) . We consider \\(f(n)\\) in mod 101 and in mod 20 separately. \n\nConsider \\(f(n)\\) mod 101. \n\n\\[f(n) = 122^{n} - 102^{n} - 21^{n\\] \\[\\equiv 21^{n} - 1^{n} - 21^{n} \\pmod {101\\] \\[= -1\\] \n\nConsider \\(f(n)\\) mod 20. \n\n\\[f(n) = 122^{n} - 102^{n} - 21^{n\\] \\[\\equiv 2^{n} - 2^{n} - 1^{n} \\pmod {20\\] \\[= -1\\] \n\nTherefore \\(f(n) \\equiv - 1\\) both in mod 20 and in mod 101. Since 20 and 101 are relatively prime, this means \\(f(n) \\equiv - 1\\) (mod 2020). As required. \\(\\square\\)", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2019_solutions.jsonl", "problem_match": "\n4.", "solution_match": "\nSolution"}}
{"year": "2019", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "NewZealand_MO", "problem": "Find all positive integers \\(n\\) such that \\(n^{4} - n^{3} + 3n^{2} + 5\\) is a perfect square.", "solution": ": Let \\(f(n) = 4n^{4} - 4n^{3} + 12n^{2} + 20 = 4(n^{4} - n^{3} + 3n^{2} + 5)\\) and note that \\((n^{4} - n^{3} + 3n^{2} + 5)\\) is a perfect square if and only if \\(f(n)\\) is. First note that: \n\n\\[(2n^{2} - n + 5)^{2} - f(n) = 9n^{2} - 10n + 5 = 4n^{2} + 5(n - 1)^{2} > 0.\\] \n\nAlso note that \n\n\\[f(n) - (2n^{2} - n + 2)^{2} = 3n^{2} + 4n + 16 = 2n^{2} + (n + 2)^{2} + 12 > 0.\\] \n\nTherefore \\((2n^{2} - n + 2)^{2}< f(n)< (2n^{2} - n + 5)^{2}\\) , so the only way \\(f(n)\\) could be a perfect square is if it is \\((2n^{2} - n + 3)^{2}\\) or \\((2n^{2} - n + 4)^{2}\\) . Solving \\(f(n) = (2n^{2} - n + 3)^{2}\\) gives us the quadratic \\(n^{2} - 6n - 11 = 0\\) which has no integer solutions. Solving \\(f(n) = (2n^{2} - n + 4)^{2}\\) gives us \\(5n^{2} - 8n - 4 = (5n + 2)(n - 2) = 0\\) . which has only one integer solution \\(n = 2\\) . Checking \n\n\\[(2)^{4} - (2)^{3} + 3(2)^{2} + 5 = 25\\] \n\nwhich is a perfect square. Therefore the only solution is \\(n = 2\\) . \\(\\square\\)", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2019_solutions.jsonl", "problem_match": "\n5.", "solution_match": "\nSolution"}}
-{"year": "2019", "tier": "T1", "problem_label": "6", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(V\\) be the set of vertices of a regular 21-gon. Given a non-empty subset \\(U\\) of \\(V\\) , let \\(m(U)\\) be the number of distinct lengths that occur between two distinct vertices in \\(U\\) . What is the maximum value of \\(\\frac{m(U)}{|U|}\\) as \\(U\\) varies over all non-empty subsets of \\(V\\) ?", "solution": ": To simplify notation, we will let \\(m\\) be \\(m(U)\\) and let \\(n\\) be \\(|U|\\) . First note that there are 10 different diagonal-lengths in a regular 21- gon. Now consider the following set of 5 vertices. \n\n\n \n\nNote that each of the 10 different diagonal- lengths appear (exactly once each). So for this set of 5 vertices we have \\(\\frac{m}{n} = \\frac{10}{5} = 2\\) . We will now show that this is the maximum possible value for \\(\\frac{m}{n}\\) . If \\(U\\) is an arbitrary non- empty set of vertices, then there are two cases: \n\n- Case 1: \\(n < 5\\) . The total number of pairs of vertices in \\(U\\) is given by \\(\\frac{1}{2} n(n - 1)\\) . Since \\(n - 1 < 4\\) this gives us the bound: \n\n\\[m \\leq \\frac{n(n - 1)}{2} < \\frac{n \\times 4}{2} = 2n.\\] \n\nThus \\(\\frac{m}{n} < 2\\) in this case. \n\n- Case 1: \\(n \\geq 5\\) . The total number of distances in \\(U\\) is at most 10 because there are only 10 different diagonal lengths in the 21-gon. Therefore \n\n\\[\\frac{m}{n} \\leq \\frac{10}{n} \\leq \\frac{10}{5} = 2\\] \n\nas required. \n\nRemark: The construction given is unique up to rotations and reflections. I.e. all sets that achieve the value \\(\\frac{m}{n} = 2\\) are congruent to the example given here. \\(\\square\\)", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2019_solutions.jsonl", "problem_match": "\n6.", "solution_match": "\nSolution"}}
-{"year": "2019", "tier": "T1", "problem_label": "7", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(ABCDEF\\) be a convex hexagon containing a point \\(P\\) in its interior such that \\(PABC\\) and \\(PDEF\\) are congruent rectangles with \\(PA = BC = PD = EF\\) (and \\(AB = PC = DE = PF\\) ). Let \\(\\ell\\) be the line through the midpoint of \\(AF\\) and the circumcentre of \\(PCD\\) . Prove that \\(\\ell\\) passes through \\(P\\) .", "solution": ": Let \\(M\\) be the midpoint of \\(AF\\) and let \\(O\\) be the circumcentre of triangle \\(CPD\\) . Now construct \\(Q\\) to be the point such that \\(CPDQ\\) is a parallelogram, and let \\(R\\) be the centre of this parallelogram (i.e. \\(R\\) is the intersection of \\(PQ\\) with \\(CD\\) , and also \\(R\\) is the midpoint of \\(PQ\\) ).\n\n\n\n \n\nNote that \\(Q D = C P = F P\\) and \\(D P = P A\\) and \\(\\angle Q D P = 180^{\\circ} - \\angle D P C = \\angle F P A\\) Therefore (by SAS) we have a pair of congruent triangles: \n\n\\[\\triangle Q D P\\cong \\triangle F P A.\\] \n\nTherefore \\(\\angle M A P = \\angle R P D\\) and \\(A F = P Q\\) . Thus \\(A M = \\frac{1}{2} A F = \\frac{1}{2} P Q = P R\\) . Therefore (by SAS) we have another pair of congruent triangles: \n\n\\[\\triangle M A P\\cong \\triangle R P D.\\] \n\nTherefore \\(\\angle A P M = \\angle R D P\\) . Let \\(x = \\angle A P M\\) so that \\(\\angle C D P = \\angle R D P = x\\) also. \n\n\n \n\nSince the angle subtended at the circumcentre is double the angle subtended at the circumference, we get \\(\\angle C O P = 2x\\) (recall that \\(O\\) is the circumcentre of \\(\\triangle P C D\\) ). Finally we get \\(\\angle O P C = 90^{\\circ} - x\\) because \\(\\triangle C O P\\) is isosceles. Putting this all together, we get \n\n\\[\\angle O P M = \\angle O P C + \\angle C P A + \\angle A P M = (90^{\\circ} - x) + 90^{\\circ} + x = 180^{\\circ}.\\] \n\nTherefore \\(\\angle O P M\\) is a straight line.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2019_solutions.jsonl", "problem_match": "\n7.", "solution_match": "\nSolution"}}
+{"year": "2019", "tier": "T1", "problem_label": "6", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(V\\) be the set of vertices of a regular 21-gon. Given a non-empty subset \\(U\\) of \\(V\\) , let \\(m(U)\\) be the number of distinct lengths that occur between two distinct vertices in \\(U\\) . What is the maximum value of \\(\\frac{m(U)}{|U|}\\) as \\(U\\) varies over all non-empty subsets of \\(V\\) ?", "solution": ": To simplify notation, we will let \\(m\\) be \\(m(U)\\) and let \\(n\\) be \\(|U|\\) . First note that there are 10 different diagonal-lengths in a regular 21- gon. Now consider the following set of 5 vertices. \n\n\n \n\nNote that each of the 10 different diagonal- lengths appear (exactly once each). So for this set of 5 vertices we have \\(\\frac{m}{n} = \\frac{10}{5} = 2\\) . We will now show that this is the maximum possible value for \\(\\frac{m}{n}\\) . If \\(U\\) is an arbitrary non- empty set of vertices, then there are two cases: \n\n- Case 1: \\(n < 5\\) . The total number of pairs of vertices in \\(U\\) is given by \\(\\frac{1}{2} n(n - 1)\\) . Since \\(n - 1 < 4\\) this gives us the bound: \n\n\\[m \\leq \\frac{n(n - 1)}{2} < \\frac{n \\times 4}{2} = 2n.\\] \n\nThus \\(\\frac{m}{n} < 2\\) in this case. \n\n- Case 1: \\(n \\geq 5\\) . The total number of distances in \\(U\\) is at most 10 because there are only 10 different diagonal lengths in the 21-gon. Therefore \n\n\\[\\frac{m}{n} \\leq \\frac{10}{n} \\leq \\frac{10}{5} = 2\\] \n\nas required. \n\nRemark: The construction given is unique up to rotations and reflections. I.e. all sets that achieve the value \\(\\frac{m}{n} = 2\\) are congruent to the example given here. \\(\\square\\)", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2019_solutions.jsonl", "problem_match": "\n6.", "solution_match": "\nSolution"}}
+{"year": "2019", "tier": "T1", "problem_label": "7", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(ABCDEF\\) be a convex hexagon containing a point \\(P\\) in its interior such that \\(PABC\\) and \\(PDEF\\) are congruent rectangles with \\(PA = BC = PD = EF\\) (and \\(AB = PC = DE = PF\\) ). Let \\(\\ell\\) be the line through the midpoint of \\(AF\\) and the circumcentre of \\(PCD\\) . Prove that \\(\\ell\\) passes through \\(P\\) .", "solution": ": Let \\(M\\) be the midpoint of \\(AF\\) and let \\(O\\) be the circumcentre of triangle \\(CPD\\) . Now construct \\(Q\\) to be the point such that \\(CPDQ\\) is a parallelogram, and let \\(R\\) be the centre of this parallelogram (i.e. \\(R\\) is the intersection of \\(PQ\\) with \\(CD\\) , and also \\(R\\) is the midpoint of \\(PQ\\) ).\n\n\n\n \n\nNote that \\(Q D = C P = F P\\) and \\(D P = P A\\) and \\(\\angle Q D P = 180^{\\circ} - \\angle D P C = \\angle F P A\\) Therefore (by SAS) we have a pair of congruent triangles: \n\n\\[\\triangle Q D P\\cong \\triangle F P A.\\] \n\nTherefore \\(\\angle M A P = \\angle R P D\\) and \\(A F = P Q\\) . Thus \\(A M = \\frac{1}{2} A F = \\frac{1}{2} P Q = P R\\) . Therefore (by SAS) we have another pair of congruent triangles: \n\n\\[\\triangle M A P\\cong \\triangle R P D.\\] \n\nTherefore \\(\\angle A P M = \\angle R D P\\) . Let \\(x = \\angle A P M\\) so that \\(\\angle C D P = \\angle R D P = x\\) also. \n\n\n \n\nSince the angle subtended at the circumcentre is double the angle subtended at the circumference, we get \\(\\angle C O P = 2x\\) (recall that \\(O\\) is the circumcentre of \\(\\triangle P C D\\) ). Finally we get \\(\\angle O P C = 90^{\\circ} - x\\) because \\(\\triangle C O P\\) is isosceles. Putting this all together, we get \n\n\\[\\angle O P M = \\angle O P C + \\angle C P A + \\angle A P M = (90^{\\circ} - x) + 90^{\\circ} + x = 180^{\\circ}.\\] \n\nTherefore \\(\\angle O P M\\) is a straight line.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2019_solutions.jsonl", "problem_match": "\n7.", "solution_match": "\nSolution"}}
{"year": "2019", "tier": "T1", "problem_label": "8", "problem_type": null, "exam": "NewZealand_MO", "problem": "Suppose that \\(x_{1},x_{2},x_{3},\\ldots x_{n}\\) are real numbers between 0 and 1 with sum \\(s\\) . Prove that \n\n\\[\\sum_{i = 1}^{n}{\\frac{x_{i}}{s + 1 - x_{i}}} + \\prod_{i = 1}^{n}(1 - x_{i})\\leq 1.\\]", "solution": ": Let \\(i\\) be arbitrary and consider the set \\(A = \\{a_{1},a_{2},\\ldots ,a_{n}\\}\\) defined by \\(a_{i} = s + 1 - x_{i}\\) and let \\(a_{j} = 1 - x_{j}\\) for all \\(j\\neq i\\) . For example, if \\(i = 2\\) then \\(A\\) would be \\(\\{1 - x_{1},s + 1 - x_{2},1 - x_{3},\\ldots ,1 - x_{n}\\}\\) . The AM- GM inequality on \\(A\\) tells us \n\n\\[1 = \\frac{(s + 1 - x_{i}) + \\sum_{j\\neq i}(1 - x_{j})}{n}\\geq \\left((1 + s - x_{i})\\prod_{j\\neq i}(1 - x_{j})\\right)^{\\frac{1}{n}}\\] \n\nWhich rearranges to give us \n\n\\[1 - (s + 1 - x_{i})\\prod_{j\\neq i}(1 - x_{j})\\geq 0.\\] \n\nFrom here we can multiply both sides by \\((1 - x_{i})\\) , then add \\(s\\) to both sides and factorise the LHS to get: \n\n\\[(s + 1 - x_{i})\\left(1 - \\prod_{j = 1}^{n}(1 - x_{j})\\right)\\geq s.\\] \n\nNow multipy both sides by \\(\\frac{x_{i}}{s(s + 1 - x_{i})}\\) to get the following equation. \n\n\\[\\left(1 - \\prod_{j = 1}^{n}(1 - x_{j})\\right)\\frac{x_{i}}{s}\\geq \\frac{x_{i}}{s + 1 - x_{i}} \\quad (1)\\] \n\nNote that this equation holds for all \\(i\\) . Now consider the sum of Equation 1 over all \\(1\\leq i\\leq n\\) . Since \\((1 - \\prod (1 - x_{j}))\\) is constant and \\(\\sum \\frac{x_{i}}{s} = 1\\) , the sum of all the LHS equals \\(\\left(1 - \\prod (1 - x_{j})\\right)\\) . So we get \n\n\\[1 - \\prod_{j = 1}^{n}(1 - x_{j})\\geq \\sum_{i = 1}^{n}\\frac{x_{i}}{s + 1 - x_{i}}\\] \n\nas required.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2019_solutions.jsonl", "problem_match": "\n8.", "solution_match": "\nSolution"}}
diff --git a/NewZealand_MO/segmented/en-nzmo1_2020_solutions.jsonl b/NewZealand_MO/segmented/en-nzmo1_2020_solutions.jsonl
index 1fb7ad3210190a93ef011b2ca41ed2dfd98da398..747b349f692a4e853d8818cba440943b7dfbe2f3 100644
--- a/NewZealand_MO/segmented/en-nzmo1_2020_solutions.jsonl
+++ b/NewZealand_MO/segmented/en-nzmo1_2020_solutions.jsonl
@@ -1,14 +1,14 @@
{"year": "2020", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "NewZealand_MO", "problem": "What is the maximum integer \\(n\\) such that \\(\\frac{50!}{2^n}\\) is an integer?", "solution": ": \\(50! = 1 \\times 2 \\times 3 \\times \\dots \\times 50\\) . Of the numbers up to 50, we need to find how many of them are divisible by \\(2^k\\) for each \\(k = 1, 2, 3, 4, 5, 6\\) . There are \\(\\left\\lfloor \\frac{50}{2^k} \\right\\rfloor\\) numbers which are divisible by \\(2^k\\) and there are \\(\\left\\lfloor \\frac{50}{2^{k+1}} \\right\\rfloor\\) which are divisible by \\(2^{k+1}\\) . Therefore there are \n\n\\[\\left\\lfloor \\frac{50}{2^k}\\right\\rfloor -\\left\\lfloor \\frac{50}{2^{k + 1}}\\right\\rfloor\\] \n\nnumbers between 1 and 50 which are divisible by 2 exactly \\(k\\) times. Hence the answer we are looking for is: \n\n\\[\\begin{aligned} & \\left(\\left\\lfloor \\frac{50}{2}\\right\\rfloor -\\left\\lfloor \\frac{50}{4}\\right\\rfloor\\right) + 2\\left(\\left\\lfloor \\frac{50}{4}\\right\\rfloor -\\left\\lfloor \\frac{50}{8}\\right\\rfloor\\right) + 3\\left(\\left\\lfloor \\frac{50}{8}\\right\\rfloor -\\left\\lfloor \\frac{50}{16}\\right\\rfloor\\right) + 4\\left(\\left\\lfloor \\frac{50}{16}\\right\\rfloor -\\left\\lfloor \\frac{50}{32}\\right\\rfloor\\right) + 5\\left(\\left\\lfloor \\frac{50}{32}\\right\\rfloor -\\left\\lfloor \\frac{50}{64}\\right\\rfloor\\right) \\\\ & = 13 + 2 \\times 6 + 3 \\times 3 + 4 \\times 2 + 5 \\times 1 \\\\ & = 47. \\end{aligned}\\]", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2020_solutions.jsonl", "problem_match": "\n1. Problem:", "solution_match": "\nSolution"}}
-{"year": "2020", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(ABCD\\) be a square and let \\(X\\) be any point on side \\(BC\\) between \\(B\\) and \\(C\\) . Let \\(Y\\) be the point on line \\(CD\\) such that \\(BX = YD\\) and \\(D\\) is between \\(C\\) and \\(Y\\) . Prove that the midpoint of \\(XY\\) lies on diagonal \\(BD\\) .", "solution": ": \n\n\n \n\nConstruct point \\(E\\) on diagonal \\(BD\\) so that \\(EX\\) is parallel to \\(CD\\) . So \\(\\angle BXE = \\angle BCD = 90^\\circ\\) . Also \\(\\angle XBE = 45^\\circ\\) because \\(E\\) is on diagonal \\(BD\\) . Therefore triangle \\(\\triangle BEX\\) is an isosceles right- angled triangle. Hence \n\n\\[EX = BX = YD.\\] \n\nSince segments \\(EX\\) and \\(YD\\) are parallel and equal in length, this implies that \\(EXDY\\) is a parallelogram. Since the diagonals of a parallelogram bisect each other, we deduce that the intersection of \\(XY\\) and \\(DE\\) is the midpoint of \\(XY\\) . Therefore the midpoint of \\(XY\\) lies on line \\(DE\\) (which is a segment of diagonal \\(BD\\) ).", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2020_solutions.jsonl", "problem_match": "\n2. Problem:", "solution_match": "\nSolution"}}
+{"year": "2020", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(ABCD\\) be a square and let \\(X\\) be any point on side \\(BC\\) between \\(B\\) and \\(C\\) . Let \\(Y\\) be the point on line \\(CD\\) such that \\(BX = YD\\) and \\(D\\) is between \\(C\\) and \\(Y\\) . Prove that the midpoint of \\(XY\\) lies on diagonal \\(BD\\) .", "solution": ": \n\n\n \n\nConstruct point \\(E\\) on diagonal \\(BD\\) so that \\(EX\\) is parallel to \\(CD\\) . So \\(\\angle BXE = \\angle BCD = 90^\\circ\\) . Also \\(\\angle XBE = 45^\\circ\\) because \\(E\\) is on diagonal \\(BD\\) . Therefore triangle \\(\\triangle BEX\\) is an isosceles right- angled triangle. Hence \n\n\\[EX = BX = YD.\\] \n\nSince segments \\(EX\\) and \\(YD\\) are parallel and equal in length, this implies that \\(EXDY\\) is a parallelogram. Since the diagonals of a parallelogram bisect each other, we deduce that the intersection of \\(XY\\) and \\(DE\\) is the midpoint of \\(XY\\) . Therefore the midpoint of \\(XY\\) lies on line \\(DE\\) (which is a segment of diagonal \\(BD\\) ).", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2020_solutions.jsonl", "problem_match": "\n2. Problem:", "solution_match": "\nSolution"}}
{"year": "2020", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(ABCD\\) be a square and let \\(X\\) be any point on side \\(BC\\) between \\(B\\) and \\(C\\) . Let \\(Y\\) be the point on line \\(CD\\) such that \\(BX = YD\\) and \\(D\\) is between \\(C\\) and \\(Y\\) . Prove that the midpoint of \\(XY\\) lies on diagonal \\(BD\\) .", "solution": "Let \\(A B C D\\) be the unit square, with \\(A = (0,1)\\) , \\(B = (1,1)\\) , \\(C = (1,0)\\) and \\(D = (0,0)\\) . Now let \\(a = B X = D Y\\) . This means that \\(X = (1,1 - a)\\) and \\(Y = (- a,0)\\) . Now let \\(Z\\) be the midpoint of \\(X Y\\) . We compute the co- ordinates of \\(Z\\) to be \n\n\\[Z = \\left(\\frac{1 + (-a)}{2}, \\frac{(1 - a) + 0}{2}\\right).\\] \n\nThe \\(x\\) and \\(y\\) coordinates of \\(Z\\) are equal, therefore \\(Z\\) lies on diagonal \\(B D\\) .", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2020_solutions.jsonl", "problem_match": "\n2. Problem:", "solution_match": "\nAlternative Solution A:"}}
{"year": "2020", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(ABCD\\) be a square and let \\(X\\) be any point on side \\(BC\\) between \\(B\\) and \\(C\\) . Let \\(Y\\) be the point on line \\(CD\\) such that \\(BX = YD\\) and \\(D\\) is between \\(C\\) and \\(Y\\) . Prove that the midpoint of \\(XY\\) lies on diagonal \\(BD\\) .", "solution": "Let \\(s\\) be the side- length of the square and let \\(a = B X = D Y\\) . Let \\(Z\\) be the midpoint of \\(X Y\\) . Now we apply the converse of Menelaus' Theorem to traversal \\(B Z D\\) of \\(\\triangle X Y C\\) . \n\n\\[\\frac{X Z}{Z Y} \\times \\frac{Y D}{D C} \\times \\frac{C B}{B X} = 1 \\times \\frac{a}{s} \\times \\frac{s}{a} = 1.\\] \n\nTherefore \\(Z\\) , \\(D\\) and \\(B\\) are colinear.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2020_solutions.jsonl", "problem_match": "\n2. Problem:", "solution_match": "\nAlternative Solution B:"}}
-{"year": "2020", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "NewZealand_MO", "problem": "You have an unlimited supply of square tiles with side length 1 and equilateral triangle tiles with side length 1. For which \\(n\\) can you use these tiles to create a convex \\(n\\) -sided polygon? The tiles must fit together without gaps and may not overlap.", "solution": ": All the angles in squares and equilateral triangles are multiples of \\(30^{\\circ}\\) . So all the external angles of the \\(n\\) - sided polygon are multiples of \\(30^{\\circ}\\) . Since the polygon is convex, this implies that all external angles are greater than or equal to \\(30^{\\circ}\\) . However, the sum of the external angles is \\(360^{\\circ}\\) , therefore \n\n\\[n \\times 30^{\\circ} \\leq 360^{\\circ}.\\] \n\nHence \\(n \\leq 12\\) . Also all polygons have at least 3 sides so \\(3 \\leq n \\leq 12\\) . Finally we demonstrate that it is possible for any \\(3 \\leq n \\leq 12\\) using the following illustrations. \n\n", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2020_solutions.jsonl", "problem_match": "\n3. Problem:", "solution_match": "\nSolution"}}
+{"year": "2020", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "NewZealand_MO", "problem": "You have an unlimited supply of square tiles with side length 1 and equilateral triangle tiles with side length 1. For which \\(n\\) can you use these tiles to create a convex \\(n\\) -sided polygon? The tiles must fit together without gaps and may not overlap.", "solution": ": All the angles in squares and equilateral triangles are multiples of \\(30^{\\circ}\\) . So all the external angles of the \\(n\\) - sided polygon are multiples of \\(30^{\\circ}\\) . Since the polygon is convex, this implies that all external angles are greater than or equal to \\(30^{\\circ}\\) . However, the sum of the external angles is \\(360^{\\circ}\\) , therefore \n\n\\[n \\times 30^{\\circ} \\leq 360^{\\circ}.\\] \n\nHence \\(n \\leq 12\\) . Also all polygons have at least 3 sides so \\(3 \\leq n \\leq 12\\) . Finally we demonstrate that it is possible for any \\(3 \\leq n \\leq 12\\) using the following illustrations. \n\n", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2020_solutions.jsonl", "problem_match": "\n3. Problem:", "solution_match": "\nSolution"}}
{"year": "2020", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "NewZealand_MO", "problem": "Determine all prime numbers \\(p\\) such that \\(p^{2} - 6\\) and \\(p^{2} + 6\\) are both prime numbers.", "solution": ": If \\(p > 5\\) then the units digit of \\(p\\) must be 1, 3, 7 or 9. \n\n- If the units digit of \\(p\\) is 1 or 9 then the units digit of \\(p^{2}\\) is 1. Therefore the units digit of \\(p^{2} - 6\\) is 5. Since \\(p^{2} - 6 > 5\\) this means that \\(p^{2} - 6\\) is not prime. \n\n- If the units digit of \\(p\\) is 3 or 7 then the units digit of \\(p^{2}\\) is 9. Therefore the units digit of \\(p^{2} + 6\\) is 5. Since \\(p^{2} + 6 > 5\\) this means that \\(p^{2} + 6\\) is not prime. \n\nTherefore we must have \\(p \\leq 5\\) . Hence \\(p\\) must be 2, 3 or 5. \n\n- If \\(p = 2\\) then \\(p^{2} + 6 = 10\\) is not prime. \n\n- If \\(p = 3\\) then \\(p^{2} + 6 = 15\\) is not prime. \n\n- If \\(p = 5\\) then \\(p^{2} \\pm 6\\) are 19 and 31 which are both prime. \n\nTherefore the only answer is \\(p = 5\\) .", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2020_solutions.jsonl", "problem_match": "\n4. Problem:", "solution_match": "\nSolution"}}
{"year": "2020", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "NewZealand_MO", "problem": "Determine all prime numbers \\(p\\) such that \\(p^{2} - 6\\) and \\(p^{2} + 6\\) are both prime numbers.", "solution": "Consider the following product modulo 5. \n\n\\[p(p^{2} - 6)(p^{2} + 6) = p^{5} - 36p \\equiv p^{5} - p \\pmod {5}\\] \n\nBy Fermat's Little Theorem, this product is 0 (mod 5). So if \\(p\\) , \\(p^{2} - 6\\) and \\(p^{2} + 6\\) are all prime numbers then at least one of them must be equal to 5. \n\n- If \\(p = 5\\) then \\(6p^{2} - 1 = 29\\) and \\(6p^{2} + 1 = 31\\) . This is one solution. \n\n- If \\(6p^{2} - 1 = 5\\) then \\(p = \\pm 1\\) . Neither 1 nor \\(-1\\) is prime, so this case leads to no solutions. \n\n- If \\(6p^{2} + 1 = 5\\) then \\(p\\) is not an integer. No solutions in this case. \n\nTherefore the only solution is \\(p = 5\\) .", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2020_solutions.jsonl", "problem_match": "\n4. Problem:", "solution_match": "\nAlternative Solution:"}}
{"year": "2020", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "NewZealand_MO", "problem": "Find all functions \\(f: \\mathbb{R} \\to \\mathbb{R}\\) that satisfy \n\n\\[f(x + f(y)) = 2x + 2f(y + 1)\\] \n\nfor all real numbers \\(x\\) and \\(y\\) .", "solution": ": Substitute \\(x = - f(y)\\) into the given equation to get \n\n\\[f(0) = -2f(y) + 2f(y + 1) \\quad (1)\\] \n\nWe can then substitute \\(y = - 1\\) into (1) to get \\(f(0) = - 2f(- 1) + 2f(0)\\) . Hence \n\n\\[f(-1) = \\frac{f(0)}{2}. \\quad (2)\\] \n\nAlso we can substitute \\(y = - 2\\) into (1) to get \\(f(0) = - 2f(- 2) + 2f(- 1)\\) . Hence \n\n\\[f(-1) = f(-2) + \\frac{f(0)}{2}. \\quad (3)\\] \n\nEquations (2) and (3) tell us that \\(f(- 2) = 0\\) . Now substitute \\(y = - 2\\) into the original equation to get \n\n\\[f(x + f(-2)) = 2x + 2f(-1)\\quad \\Longrightarrow \\quad f(x) = 2x + 2f(-1).\\]\n\n\n\nTherefore our function can be written in the form \\(f(x) = 2x + c\\) , where \\(c = 2f(- 1)\\) is a constant. We can substitute this into the original equation to get \n\n\\[f(x + f(y)) = 2x + 2f(y + 1)\\] \\[f(x + (2y + c)) = 2x + 2(2(y + 1) + c)\\] \\[2(x + (2y + c)) + c = 2x + 4y + 4 + 2c\\] \\[2x + 4y + 3c = 2x + 4y + 2c + 4\\] \\[c = 4.\\] \n\nTherefore the function \\(f\\) works if and only if \\(c = 4\\) . Hence \\(f(x) = 2x + 4\\) (for all \\(x\\) ) is the only solution.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2020_solutions.jsonl", "problem_match": "\n5. Problem:", "solution_match": "\nSolution"}}
{"year": "2020", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "NewZealand_MO", "problem": "Find all functions \\(f: \\mathbb{R} \\to \\mathbb{R}\\) that satisfy \n\n\\[f(x + f(y)) = 2x + 2f(y + 1)\\] \n\nfor all real numbers \\(x\\) and \\(y\\) .", "solution": "Substitute \\(x = z - f(0)\\) and \\(y = 0\\) into the given equation to get \n\n\\[f((z - f(0)) + f(0)) = 2(z - f(0)) + 2f(1)\\] \\[f(z) = 2z + (2f(1) - 2f(0)).\\] \n\nHence \\(f(z) = 2z + c\\) for any real number \\(z\\) , where \\(c = 2f(1) - 2f(0)\\) is constant. Now \n\n\\[f(1) - f(0) = (2\\cdot 1 + c) - (2\\cdot 0 + c) = 2.\\] \n\nHence \\(c = 2f(1) - 2f(0) = 4\\) . Therefore \\(f(z) = 2z + 4\\) is the only possible function satisfying the given equation. \n\nNow need to check that \\(f(z) = 2z + 4\\) does indeed satisfy the given equation \n\n\\[f(x + f(y)) = f(x + 2y + 4) = 2(x + 2y + 4) + 4 = 2x + 4y + 12\\] \\[\\qquad = 2x + 2(2(y + 1) + 4) = 2x + 2f(y + 1).\\] \n\nHence \\(f(z) = 2z + 4\\) is the unique function \\(f\\) satisfying the given equation.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2020_solutions.jsonl", "problem_match": "\n5. Problem:", "solution_match": "# Alternative Solution:"}}
-{"year": "2020", "tier": "T1", "problem_label": "6", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(\\triangle ABC\\) be an acute triangle with \\(AB > AC\\) . Let \\(P\\) be the foot of the altitude from \\(C\\) to \\(AB\\) and let \\(Q\\) be the foot of the altitude from \\(B\\) to \\(AC\\) . Let \\(X\\) be the intersection of \\(PQ\\) and \\(BC\\) . Let the intersection of the circumcircles of triangle \\(\\triangle AXC\\) and triangle \\(\\triangle PQC\\) be distinct points: \\(C\\) and \\(Y\\) . Prove that \\(PY\\) bisects \\(AX\\) .", "solution": ": Let \\(Z\\) be the point where \\(PY\\) intersects \\(AX\\) . The problem asks us to prove that \\(AZ = ZX\\) . \n\n\n\n\n\n\nSince \\(\\angle BPC = \\angle BQC = 90^{\\circ}\\) we conclude that \\(BPQC\\) is a cyclic quadrilateral. Hence \\(BPYQC\\) is a cyclic pentagon. A quick angle chase gives us: \n\n\\[\\angle ZPA = 180^{\\circ} - \\angle BPY\\] \\[\\qquad = \\angle BCY\\] \\[\\qquad = 180^{\\circ} - \\angle XCY\\] \\[\\qquad = \\angle XAY\\] \\[\\qquad = \\angle ZAY.\\] \n\nHence triangles \\(\\triangle ZPA\\) and \\(\\triangle ZAY\\) are similar ( \\(\\angle Z\\) is shared). Therefore \\(\\frac{ZP}{ZA} = \\frac{ZA}{ZY}\\) , and hence \\(ZA^{2} = ZP \\times ZY\\) . Another angle chase gives us: \n\n\\[\\angle XPZ = \\angle QPY\\] \\[\\qquad = \\angle QCY\\] \\[\\qquad = \\angle ACY\\] \\[\\qquad = \\angle AXY\\] \\[\\qquad = \\angle ZXY.\\] \n\nHence triangles \\(\\triangle ZPX\\) and \\(\\triangle ZXY\\) are similar ( \\(\\angle Z\\) is shared). Therefore \\(\\frac{ZP}{ZX} = \\frac{ZX}{ZY}\\) and hence \\(ZX^{2} = ZP \\times ZY\\) . Putting this together gives us: \n\n\\[ZA^{2} = ZP \\times ZY = ZX^{2}.\\] \n\nHence \\(ZA = ZX\\) , and therefore \\(Z\\) is the midpoint of \\(AX\\) .", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2020_solutions.jsonl", "problem_match": "\n6. Problem:", "solution_match": "\nSolution"}}
+{"year": "2020", "tier": "T1", "problem_label": "6", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(\\triangle ABC\\) be an acute triangle with \\(AB > AC\\) . Let \\(P\\) be the foot of the altitude from \\(C\\) to \\(AB\\) and let \\(Q\\) be the foot of the altitude from \\(B\\) to \\(AC\\) . Let \\(X\\) be the intersection of \\(PQ\\) and \\(BC\\) . Let the intersection of the circumcircles of triangle \\(\\triangle AXC\\) and triangle \\(\\triangle PQC\\) be distinct points: \\(C\\) and \\(Y\\) . Prove that \\(PY\\) bisects \\(AX\\) .", "solution": ": Let \\(Z\\) be the point where \\(PY\\) intersects \\(AX\\) . The problem asks us to prove that \\(AZ = ZX\\) . \n\n\n\n\n\n\nSince \\(\\angle BPC = \\angle BQC = 90^{\\circ}\\) we conclude that \\(BPQC\\) is a cyclic quadrilateral. Hence \\(BPYQC\\) is a cyclic pentagon. A quick angle chase gives us: \n\n\\[\\angle ZPA = 180^{\\circ} - \\angle BPY\\] \\[\\qquad = \\angle BCY\\] \\[\\qquad = 180^{\\circ} - \\angle XCY\\] \\[\\qquad = \\angle XAY\\] \\[\\qquad = \\angle ZAY.\\] \n\nHence triangles \\(\\triangle ZPA\\) and \\(\\triangle ZAY\\) are similar ( \\(\\angle Z\\) is shared). Therefore \\(\\frac{ZP}{ZA} = \\frac{ZA}{ZY}\\) , and hence \\(ZA^{2} = ZP \\times ZY\\) . Another angle chase gives us: \n\n\\[\\angle XPZ = \\angle QPY\\] \\[\\qquad = \\angle QCY\\] \\[\\qquad = \\angle ACY\\] \\[\\qquad = \\angle AXY\\] \\[\\qquad = \\angle ZXY.\\] \n\nHence triangles \\(\\triangle ZPX\\) and \\(\\triangle ZXY\\) are similar ( \\(\\angle Z\\) is shared). Therefore \\(\\frac{ZP}{ZX} = \\frac{ZX}{ZY}\\) and hence \\(ZX^{2} = ZP \\times ZY\\) . Putting this together gives us: \n\n\\[ZA^{2} = ZP \\times ZY = ZX^{2}.\\] \n\nHence \\(ZA = ZX\\) , and therefore \\(Z\\) is the midpoint of \\(AX\\) .", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2020_solutions.jsonl", "problem_match": "\n6. Problem:", "solution_match": "\nSolution"}}
{"year": "2020", "tier": "T1", "problem_label": "6", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(\\triangle ABC\\) be an acute triangle with \\(AB > AC\\) . Let \\(P\\) be the foot of the altitude from \\(C\\) to \\(AB\\) and let \\(Q\\) be the foot of the altitude from \\(B\\) to \\(AC\\) . Let \\(X\\) be the intersection of \\(PQ\\) and \\(BC\\) . Let the intersection of the circumcircles of triangle \\(\\triangle AXC\\) and triangle \\(\\triangle PQC\\) be distinct points: \\(C\\) and \\(Y\\) . Prove that \\(PY\\) bisects \\(AX\\) .", "solution": "First we will embed the diagram in the Argand plane, such that point \\(B\\) is represented by the complex number \\(b = - 1\\) and point \\(C\\) is represented by the complex number \\(c = 1\\) . Lower case letters will always denote the complex number representing the corresponding upper case letter (so \\(a\\) is the complex number representing point \\(A\\) and \\(x\\) is the complex number representing point \\(X\\) , etc). We will endeavour to find expressions for all the points in the diagram in terms of \\(p\\) and \\(q\\) . \n\nSince \\(\\angle BPC = \\angle BQC = 90^{\\circ}\\) , we know that points \\(P\\) and \\(Q\\) both lie on the unit circle. So \\(p\\bar{p} = q\\bar{q} = 1\\) . Therefore the circumcircle of triangle \\(PQC\\) is the unit circle and thus \\(y\\bar{y} = 1\\) too. Since point \\(A\\) is the intersection of chords \\(BP\\) and \\(CQ\\) , we can compute \\(a\\) using the formula for the intersection of two chords. \n\n\\[a = \\frac{cq(b + p) - bp(c + q)}{cq - bp} = \\frac{q(-1 + p) - (-1)p(1 + q)}{q - (-1)p} = \\frac{2pq + p - q}{p + q}\\] \\[\\qquad \\Rightarrow \\qquad \\bar{a} = \\frac{2\\bar{p}\\bar{q} + \\bar{p} - \\bar{q}}{\\bar{p} + \\bar{q}} = \\frac{(2\\bar{p}\\bar{q} + \\bar{p} - \\bar{q})pq}{(\\bar{p} + \\bar{q})pq} = \\frac{2 + q - p}{q + p}\\] \\[\\qquad \\mathrm{note:}\\qquad \\frac{1 - a}{1 - \\bar{a}} = \\frac{1 - \\frac{2pq + p - q}{p + q}}{1 - \\frac{2 + q - p}{q + p}} = \\frac{(p + q) - (2pq + p - q)}{(p + q) - (2 + q - p)} = \\frac{2q - 2pq}{2p - 2} = -q.\\] \n\nSimilarly \\(X\\) is the intersection of chords \\(PQ\\) and \\(BC\\) , so \n\n\\[x = \\frac{pq(b + c) - bc(p + q)}{pq - bc} = \\frac{pq(0) - (-1)(p + q)}{pq - (-1)} = \\frac{p + q}{pq + 1}\\]\n\n\n\nNow we have formulas for \\(a\\) and \\(x\\) in terms of \\(p\\) and \\(q\\) . Next we will use the fact that \\(AYCX\\) is cyclic to find a formula for \\(y\\) in terms of \\(p\\) and \\(q\\) . \\(AYCX\\) being cyclic is equivalent to \\(\\angle CAY = \\angle CXY\\) . This is equivalent to \n\n\\[\\left(\\frac{c - a}{\\bar{c} - \\bar{a}}\\right) / \\left(\\frac{y - a}{\\bar{y} - \\bar{a}}\\right) = \\left(\\frac{c - x}{\\bar{c} - \\bar{x}}\\right) / \\left(\\frac{y - x}{\\bar{y} - \\bar{x}}\\right)\\] \n\nTo simplify this, first recall that \\(\\frac{c - a}{\\bar{c} - \\bar{a}} = \\frac{1 - a}{1 - \\bar{a}} = - q\\) . Also, since \\(c = \\bar{c}\\) and \\(x = \\bar{x}\\) ( \\(c\\) and \\(x\\) are real numbers) the \\(\\frac{c - x}{\\bar{c} - \\bar{x}}\\) factor is 1. Furthermore, since \\(y\\bar{y} = 1\\) we can replace \\(\\bar{y}\\) with \\(y^{- 1}\\) . Thus the equation for \\(AYCX\\) being cyclic becomes: \n\n\\[(-q) / \\left(\\frac{y - a}{y^{-1} - \\bar{a}}\\right) = 1 / \\left(\\frac{y - x}{y^{-1} - x}\\right).\\] \n\nFrom here, We can multiply out the denominators, expand the brackets and collect like terms to get a quadratic in \\(y\\) . \n\n\\[(q\\bar{a} +x)y^{2} - (q\\bar{a}x + q + xa + 1)y + (qx + a) = 0.\\] \n\nNow (using \\(\\frac{1 - a}{1 - \\bar{a}} = - q\\) ) we can get \\((q\\bar{a} +x + qx + a) = (q\\bar{a}x + q + xa + 1)\\) . Thus the quadratic factorises as: \n\n\\[(y - 1)\\Big((q\\bar{a} +x)y - (qx + a)\\Big) = 0.\\] \n\nSince \\(Y\\) and \\(C\\) are distinct points, we know \\(y\\neq 1\\) and so we finally get a formula for \\(y\\) \n\n\\[(q\\bar{a} +x)y - (qx + a) = 0\\qquad \\Longrightarrow \\qquad y = \\frac{qx + a}{q\\bar{a} + x}\\] \n\nWe can now subsitute our formulas for \\(a\\) and \\(x\\) ( \\(a = \\frac{2pq + p - q}{p + q}\\) and \\(x = \\frac{p + q}{pq + 1}\\) ) into this expression to find \\(y\\) in terms of \\(p\\) and \\(q\\) . After some algebraic simplification this yields: \n\n\\[y = \\frac{2p^{2}q + (q + 1)p + q^{2} - q}{(1 - q)p^{2} + (q + q^{2})p + 2q}.\\] \n\nLet \\(M\\) be the midpoint of \\(AX\\) . So \n\n\\[m = \\frac{a + x}{2} = \\frac{\\frac{2pq + p - q}{p + q} + \\frac{p + q}{pq + 1}}{2} = \\frac{(pq + 1)(2pq + p - q) + (p + q)^{2}}{2(p + q)(pq + 1)}.\\] \n\n\\[\\overline{m} = \\frac{(pq + 1)(2 + q - p) + (p + q)^{2}}{2(p + q)(pq + 1)}.\\] \n\n\\[p\\overline{m} -1 = \\frac{(p - 1)((1 - q)p^{2} + (q + q^{2})p + 2q)}{2(p + q)(pq + 1)}\\] \n\n\\[y(p\\overline{m} -1) = \\frac{(p - 1)(2p^{2}q + (q + 1)p + q^{2} - q)}{2(p + q)(pq + 1)}\\] \\[y(p\\overline{m} -1) + m = \\frac{(p - 1)(2p^{2}q + (q + 1)p + q^{2} - q)}{2(p + q)(pq + 1)} +\\frac{(pq + 1)(2pq + p - q) + (p + q)^{2}}{2(p + q)(pq + 1)}\\] \\[= \\frac{(p - 1)(2p^{2}q + (q + 1)p + q^{2} - q) + (pq + 1)(2pq + p - q) + (p + q)^{2}}{2(p + q)(pq + 1)}\\] \\[= \\frac{2p^{3}q + 2p^{2}q^{2} + 2p^{2} + 2pq}{2(p + q)(pq + 1)}\\] \\[= p.\\] \n\nWe have shown that \\(y(p\\overline{m} -1) + m = p\\) . Hence \n\n\\[m = p + y - py\\overline{m}.\\] \n\nWhich interpeted geometrically means that point \\(M\\) lies on chord \\(PY\\) of the unit circle.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2020_solutions.jsonl", "problem_match": "\n6. Problem:", "solution_match": "# Alternative Solution (outline):"}}
-{"year": "2020", "tier": "T1", "problem_label": "7", "problem_type": null, "exam": "NewZealand_MO", "problem": "Josie and Ross are playing a game on a \\(20 \\times 20\\) chessboard. Initially the chessboard is empty. The two players alternately take turns, with Josie going first. On Josie's turn, she selects any two different empty cells, and places one white stone in each of them. On Ross' turn, he chooses any one white stone currently on the board, and replaces it with a black stone. If at any time there are 8 consecutive cells in a line (horizontally or vertically) all of which contain a white stone, Josie wins. Is it possible that Ross can stop Josie winning — regardless of how Josie plays?", "solution": ": Ross can't stop Josie winning — Josie has a strategy in which she can ensure that there will be 8 white stones in a row. We will give an explicit example of such a strategy. \n\nTo simplify notation, we define a \\(k\\) - strip to be a \\(1 \\times 8\\) rectangle, in which the first \\(k\\) cells are filled with white stones and the other \\(8 - k\\) cells are empty. \n\n- Step One. Josie creates 32 disjoint 1-strips using the following technique. \n\nStart by finding 44 disjoint \\(1 \\times 9\\) rectangles on the board. \n\n\n \n\n- Josie places a white stone in one end of each of these 44 rectangles on her first 22 turns. \n\n- Ross \"ruins\" 22 of them. Each of the other 22 rectangles are 1-strips (by ignoring the empty end cell). \n\n- On Josie's next 10 turns she then chooses 20 of the ruined \\(1 \\times 9\\) rectangles, and \"unruins\" them by placing a white stone in the other end. These are now 1-strips (by ignoring the cell containing the black stone). \n\n- However Ross \"ruins\" a further 10 of them. So in total we have \n\n\\[22 + 20 - 10 = 32 \\text{ disjoint 1-strips.}\\] \n\n- Step Two. Repeat the following operation for \\(k = 1, 2, 3, 4, 5\\) . \n\nStarting with \\(2^{6 - k}\\) disjoint \\(k\\) - strips. Josie can use her next \\(2^{5 - k}\\) turns to add one white stone to each of these strips. On Ross' next \\(2^{5 - k}\\) turns he can spoil at most \\(2^{5 - k}\\) of them. So we are left with at least \\(2^{5 - k}\\) disjoint \\((k + 1)\\) - strips. \n\n- Step Three. Now we have at least one 6-strip. Josie wins immediately by placing both her white stones into the empty cells in the 6-strip.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2020_solutions.jsonl", "problem_match": "\n7. Problem:", "solution_match": "\nSolution"}}
+{"year": "2020", "tier": "T1", "problem_label": "7", "problem_type": null, "exam": "NewZealand_MO", "problem": "Josie and Ross are playing a game on a \\(20 \\times 20\\) chessboard. Initially the chessboard is empty. The two players alternately take turns, with Josie going first. On Josie's turn, she selects any two different empty cells, and places one white stone in each of them. On Ross' turn, he chooses any one white stone currently on the board, and replaces it with a black stone. If at any time there are 8 consecutive cells in a line (horizontally or vertically) all of which contain a white stone, Josie wins. Is it possible that Ross can stop Josie winning — regardless of how Josie plays?", "solution": ": Ross can't stop Josie winning — Josie has a strategy in which she can ensure that there will be 8 white stones in a row. We will give an explicit example of such a strategy. \n\nTo simplify notation, we define a \\(k\\) - strip to be a \\(1 \\times 8\\) rectangle, in which the first \\(k\\) cells are filled with white stones and the other \\(8 - k\\) cells are empty. \n\n- Step One. Josie creates 32 disjoint 1-strips using the following technique. \n\nStart by finding 44 disjoint \\(1 \\times 9\\) rectangles on the board. \n\n\n \n\n- Josie places a white stone in one end of each of these 44 rectangles on her first 22 turns. \n\n- Ross \"ruins\" 22 of them. Each of the other 22 rectangles are 1-strips (by ignoring the empty end cell). \n\n- On Josie's next 10 turns she then chooses 20 of the ruined \\(1 \\times 9\\) rectangles, and \"unruins\" them by placing a white stone in the other end. These are now 1-strips (by ignoring the cell containing the black stone). \n\n- However Ross \"ruins\" a further 10 of them. So in total we have \n\n\\[22 + 20 - 10 = 32 \\text{ disjoint 1-strips.}\\] \n\n- Step Two. Repeat the following operation for \\(k = 1, 2, 3, 4, 5\\) . \n\nStarting with \\(2^{6 - k}\\) disjoint \\(k\\) - strips. Josie can use her next \\(2^{5 - k}\\) turns to add one white stone to each of these strips. On Ross' next \\(2^{5 - k}\\) turns he can spoil at most \\(2^{5 - k}\\) of them. So we are left with at least \\(2^{5 - k}\\) disjoint \\((k + 1)\\) - strips. \n\n- Step Three. Now we have at least one 6-strip. Josie wins immediately by placing both her white stones into the empty cells in the 6-strip.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2020_solutions.jsonl", "problem_match": "\n7. Problem:", "solution_match": "\nSolution"}}
{"year": "2020", "tier": "T1", "problem_label": "8", "problem_type": null, "exam": "NewZealand_MO", "problem": "For a positive integer \\(x\\) , define a sequence \\(a_{0}, a_{1}, a_{2}, \\ldots\\) according to the following rules: \\(a_{0} = 1\\) , \\(a_{1} = x + 1\\) and \n\n\\[a_{n + 2} = x a_{n + 1} - a_{n}\\qquad \\mathrm{for~all~}n\\geq 0.\\] \n\nProve that there exist infinitely many positive integers \\(x\\) such that this sequence does not contain a prime number.", "solution": ": For each integer \\(n \\geq 0\\) and \\(x > 2\\) , we recursively define integers \\(a_{n}(x)\\) and \\(b_{n}(x)\\) . Let \\(a_{0}(x) = b_{0}(x) = 1\\) and \\(a_{1}(x) = x + 1\\) and \\(b_{1}(x) = x - 1\\) . For all \\(n \\geq 0\\) let \n\n\\[a_{n + 2}(x) = x a_{n + 1}(x) - a_{n}(x)\\quad \\mathrm{and}\\quad b_{n + 2}(x) = x b_{n + 1}(x) - b_{n}(x).\\] \n\nSo for constant \\(x\\) , the sequences \\((a_{n}\\) and \\(b_{n}\\) ), both satisfy the same recurrence but have different initial conditions. The chracteristic polynomial for this recurrence is \\(\\lambda^{2} - x\\lambda +1\\) Now let \n\n\\[\\beta = \\beta (x) = \\frac{x + \\sqrt{x^{2} - 4}}{2}\\] \n\nand note that the roots of the characteristic polynomial are \\(\\beta\\) and \\(\\beta^{- 1}\\) (the roots are reciprocals because the constant term is 1). \n\nLemma: \n\n\\[a_{n}(x) = \\frac{\\beta^{n + 1} - \\beta^{-n}}{\\beta - 1}\\qquad \\mathrm{and}\\qquad b_{n}(x) = \\frac{\\beta^{n + 1} + \\beta^{-n}}{\\beta + 1}.\\] \n\nProof of Lemma: \n\nSince these formalas for \\(a_{n}(x)\\) and \\(b_{n}(x)\\) are both linear combinations of \\(\\beta^{n}\\) and \\(\\beta^{- n}\\) , we simply need to verify them for \\(n = 0\\) and \\(n = 1\\) . \n\n\\[\\frac{\\beta^{0 + 1} - \\beta^{-0}}{\\beta - 1} = 1 = a_{0}(x)\\qquad \\mathrm{and}\\qquad \\frac{\\beta^{0 + 1} + \\beta^{-0}}{\\beta + 1} = 1 = b_{0}(x)\\] \n\nSince \\(\\beta\\) is a root of \\(\\lambda^{2} - x\\lambda + 1\\) , this means that \\(\\beta^{2} + 1 = x\\beta\\) and thus \\(\\beta + \\beta^{- 1} = x\\) . So \n\n\\[\\frac{\\beta^{2} - \\beta^{-1}}{\\beta - 1} = \\beta +1 + \\beta^{-1}\\] \\[\\qquad = x + 1\\] \\[\\qquad = a_{1}(x).\\] \n\nThus the Lemma is now proven. Also note that \\((\\beta (x))^{2} = \\beta (y)\\) for \\(y = x^{2} - 2\\) , because \n\n\\[(\\beta (x))^{2} = \\left(\\frac{x + \\sqrt{x^{2} - 4}}{2}\\right)^{2} = \\frac{(x^{2} - 2) + \\sqrt{(x^{2} - 2)^{2} - 4}}{2} = \\frac{y + \\sqrt{y^{2} - 4}}{2} = \\beta (y).\\] \n\nNow we use \\(\\beta^{2} = (\\beta (x))^{2} = \\beta (y)\\) to show that \\(a_{n}(x)b_{n}(x) = a_{n}(y)\\) . \n\n\\[a_{n}(x)b_{n}(x) = \\left(\\frac{\\beta^{n + 1} - \\beta^{-n}}{\\beta - 1}\\right)\\left(\\frac{\\beta^{n + 1} + \\beta^{-n}}{\\beta + 1}\\right)\\] \\[\\qquad = \\frac{(\\beta^{2})^{n + 1} - (\\beta^{2})^{-n}}{(\\beta^{2} - 1)}\\] \\[\\qquad = \\frac{(\\beta(y))^{n + 1} - (\\beta(y))^{-n}}{(\\beta(y) - 1)}\\] \\[\\qquad = a_{n}(y).\\]\n\n\n\nThis shows that the sequence \\(a_{1}(y), a_{2}(y), a_{3}(y), \\ldots\\) cannot contain any prime numbers. Since there are infinitely many integers of the form \\(y = x^{2} - 2\\) , we are done.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2020_solutions.jsonl", "problem_match": "\n8. Problem:", "solution_match": "\nSolution"}}
{"year": "2020", "tier": "T1", "problem_label": "8", "problem_type": null, "exam": "NewZealand_MO", "problem": "For a positive integer \\(x\\) , define a sequence \\(a_{0}, a_{1}, a_{2}, \\ldots\\) according to the following rules: \\(a_{0} = 1\\) , \\(a_{1} = x + 1\\) and \n\n\\[a_{n + 2} = x a_{n + 1} - a_{n}\\qquad \\mathrm{for~all~}n\\geq 0.\\] \n\nProve that there exist infinitely many positive integers \\(x\\) such that this sequence does not contain a prime number.", "solution": "First note that (for \\(x \\geq 3\\) ) each term in the sequence is more than double the previous term, because \n\n\\[a_{n + 1} = x a_{n} - a_{n - 1} = (x - 1)a_{n} + (a_{n} - a_{n - 1})\\geq (x - 1)a_{n}\\geq 2a_{n}.\\] \n\nMoreover we can easily verify that \\(a_{n} > x\\) for all \\(n \\geq 1\\) because \\(a_{n} \\geq a_{1} = x + 1\\) . Now we prove the following Lemma. \n\nLemma: \n\n\\[a_{n}^{2} - a_{n + 1}a_{n - 1} = x + 2\\] \n\nProof of Lemma: (by Induction) \n\nFirst note that \\(a_{2} = x a_{1} - a_{0} = x(x + 1) - 1\\) . Now for the base case, we check this expression for \\(n = 1\\) . \n\n\\[a_{1}^{2} - a_{2}a_{0} = (x + 1)^{2} - (x(x + 1) - 1)(1) = x + 2.\\] \n\nThis proves the Lemma when \\(n = 0\\) . Now for the inductive step, we assume the Lemma is true when \\(n = k\\) . Assume: \\(a_{k}^{2} - a_{k + 1}a_{k - 1} = x + 2\\) . Now we will use \\(x a_{k} = a_{k + 1} + a_{k - 1}\\) to simplify \\(a_{k + 1}^{2} - a_{k + 2}a_{k}\\) . \n\n\\[a_{k + 1}^{2} - a_{k + 2}a_{k} = a_{k + 1}^{2} - (x a_{k + 1} - a_{k})a_{k}\\] \\[\\qquad = a_{k + 1}^{2} - x a_{k}a_{k + 1} + a_{k}^{2}\\] \\[\\qquad = a_{k + 1}^{2} - (a_{k + 1} + a_{k - 1})a_{k + 1} + a_{k}^{2}\\] \\[\\qquad = a_{k + 1}^{2} - a_{k + 1}^{2} - a_{k - 1}a_{k + 1} + a_{k}^{2}\\] \\[\\qquad = a_{k}^{2} - a_{k - 1}a_{k + 1}\\] \\[\\qquad = x + 2.\\] \n\nThus that the Lemma is proven. Now lets assume that \\((x + 2)\\) is a perfect square; let \\(x + 2 = c^{2}\\) for some integer \\(c \\geq 3\\) . This implies that \\(a_{n}^{2} - a_{n + 1}a_{n - 1} = c^{2}\\) . Therefore \n\n\\[a_{n + 1}a_{n - 1} = a_{n}^{2} - c^{2} = (a_{n} + c)(a_{n} - c)\\] \n\nand thus \\((a_{n} + c)(a_{n} - c)\\) is a multiple of \\(a_{n + 1}\\) . We know that both factors \\((a_{n} + c)\\) and \\((a_{n} - c)\\) are positive because \\(a_{n} > x = c^{2} - 2 > c\\) for all \\(c \\geq 3\\) . \n\nNow suppose for the sake of contradiction that \\(a_{n + 1}\\) was prime. This would mean that either \\((a_{n} + c)\\) or \\((a_{n} - c)\\) is a multiple of \\(a_{n + 1}\\) . Therefore either \n\n\\[a_{n + 1}\\leq a_{n} + c\\qquad \\mathrm{or}\\qquad a_{n + 1}\\leq a_{n} - c.\\] \n\nEither way we get \\(a_{n + 1} \\leq a_{n} + c\\) . Now using \\(a_{n} > c\\) we obtain \n\n\\[a_{n + 1}\\leq a_{n} + c< a_{n} + a_{n} = 2a_{n}.\\] \n\nHowever this contradicts the fact that each term in the sequence is double the previous term. Therefore \\(a_{n + 1}\\) cannot be prime (when \\(x = c^{2} - 2\\) ) for any \\(n \\geq 1\\) . Finally we notice that \\(a_{0} = 1\\) is not prime, and \\(a_{1} = x + 1 = (c^{2} - 2) + 1 = (c + 1)(c - 1)\\) is clearly composite for all \\(c \\geq 3\\) . Thus we have shown that the sequence \\(a_{0}, a_{1}, a_{2}, \\ldots\\) contains no prime numbers whenever \\(x\\) is in the form \\(x = c^{2} - 2\\) for any integer \\(c \\geq 3\\) .", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2020_solutions.jsonl", "problem_match": "\n8. Problem:", "solution_match": "\nAlternative Solution:"}}
diff --git a/NewZealand_MO/segmented/en-nzmo1_2021_solutions.jsonl b/NewZealand_MO/segmented/en-nzmo1_2021_solutions.jsonl
index 5045257a962831a8e6b894638afb7e85d5aaeab8..146ceae1599ad63125a01cd582c8a83a0db0c374 100644
--- a/NewZealand_MO/segmented/en-nzmo1_2021_solutions.jsonl
+++ b/NewZealand_MO/segmented/en-nzmo1_2021_solutions.jsonl
@@ -1,8 +1,8 @@
{"year": "2021", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "NewZealand_MO", "problem": "A school offers three subjects: Mathematics, Art and Science. At least \\(80\\%\\) of students study both Mathematics and Art. At least \\(80\\%\\) of students study both Mathematics and Science. Prove that at least \\(80\\%\\) of students who study both Art and Science, also study Mathematics.", "solution": ": Let \\(n\\) be the total number of students. Let \\(x\\) be the number of students that study all three subjects. Let \\(a\\) be the number of students that study Maths and Art but not Science. Let \\(b\\) be the number of students that study Maths and Science but not Art. Let \\(c\\) be the number of students that study Art and Science but not Maths. \n\nSo \\(a,b,c,x,n\\) are non- negative real numbers with \n\n\\(\\cdot n\\geq a + b + c + x\\) (total number of students). \n\n\\(\\cdot x + a\\geq 0.8\\times n\\) (studying both Mathematics and Art), \n\n\\(\\cdot x + b\\geq 0.8\\times n\\) (studying both Mathematics and Science). \n\nAdding all these together gives us \\(2x + a + b\\geq 1.6\\times n\\) and since \\(n\\geq a + b + c + x\\) we get: \n\n\\[2x + a + b\\geq 1.6\\times (a + b + c + x).\\] \n\nFinally multiply both sides by 0.5 and rearrange to get: \n\n\\[x\\geq 0.8\\times (x + c) + 0.3\\times (a + b)\\geq 0.8\\times (x + c).\\] \n\nTherefore \\(\\frac{x}{x + c}\\geq 0.8\\) as required.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2021_solutions.jsonl", "problem_match": "\n1. Problem:", "solution_match": "\nSolution"}}
{"year": "2021", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "NewZealand_MO", "problem": "A school offers three subjects: Mathematics, Art and Science. At least \\(80\\%\\) of students study both Mathematics and Art. At least \\(80\\%\\) of students study both Mathematics and Science. Prove that at least \\(80\\%\\) of students who study both Art and Science, also study Mathematics.", "solution": "First note that if \\(x\\geq 0.8n\\) then \\(\\frac{x}{x + c}\\geq \\frac{x}{n}\\geq 0.8\\) . Otherwise we have \\(x< 0.8n\\) and so \\(n > \\frac{5x}{4}\\) . Hence \n\n\\[\\frac{x}{x + c} = \\frac{x}{(a + b + c + x) - (a + x) - (b + x) + 2x\\] \\[\\geq \\frac{x}{n - 0.8n - 0.8n + 2x\\] \\[= \\frac{x}{2x - 0.6n\\] \\[>\\frac{x}{2x - 0.6\\times\\frac{5x}{4\\] \\[= 0.8.\\]", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2021_solutions.jsonl", "problem_match": "\n1. Problem:", "solution_match": "# Alternative Solution:"}}
-{"year": "2021", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(ABCD\\) be a trapezium such that \\(AB \\parallel CD\\). Let \\(E\\) be the intersection of diagonals \\(AC\\) and \\(BD\\). Suppose that \\(AB = BE\\) and \\(AC = DE\\). Prove that the internal angle bisector of \\(\\angle BAC\\) is perpendicular to \\(AD\\).", "solution": ": First note that triangle \\(ABE\\) is isosceles because \\(AB = BE\\). \n\n\n \n\nLet \\(x = \\angle DEC\\) . Angle chasing gives: \n\n\\[x = \\angle DEC\\] \\[\\quad = \\angle BEA\\] \\[\\quad = \\angle EAB\\] \\[\\quad = \\angle ACD.\\] \n\nTherefore triangle \\(CDE\\) is isosceles. Hence \\(DE = DC\\) . Since we are also given \\(AC = DE\\) this implies \\(AC = DC\\) . Therefore \\(\\triangle ACD\\) is isosceles. Since \\(x = \\angle ACD\\) this gives us \n\n\\[\\angle CDA = \\angle DAC = 90^{\\circ} - \\frac{x}{2}\\] \n\nNow let \\(\\lambda\\) be the angle bisector of \\(\\angle BAC\\) . Since \\(x = \\angle BAC\\) we know that \\(\\lambda\\) makes an angle of \\(\\frac{1}{2}\\angle BAC = \\frac{x}{2}\\) with line \\(AC\\) . Therefore the angle between \\(\\lambda\\) and \\(AD\\) is \n\n\\[\\frac{x}{2} +\\angle DAC = \\frac{x}{2} +\\left(90^{\\circ} - \\frac{x}{2}\\right) = 90^{\\circ}\\] \n\nas required.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2021_solutions.jsonl", "problem_match": "\n2. Problem:", "solution_match": "\nSolution"}}
+{"year": "2021", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(ABCD\\) be a trapezium such that \\(AB \\parallel CD\\). Let \\(E\\) be the intersection of diagonals \\(AC\\) and \\(BD\\). Suppose that \\(AB = BE\\) and \\(AC = DE\\). Prove that the internal angle bisector of \\(\\angle BAC\\) is perpendicular to \\(AD\\).", "solution": ": First note that triangle \\(ABE\\) is isosceles because \\(AB = BE\\). \n\n\n \n\nLet \\(x = \\angle DEC\\) . Angle chasing gives: \n\n\\[x = \\angle DEC\\] \\[\\quad = \\angle BEA\\] \\[\\quad = \\angle EAB\\] \\[\\quad = \\angle ACD.\\] \n\nTherefore triangle \\(CDE\\) is isosceles. Hence \\(DE = DC\\) . Since we are also given \\(AC = DE\\) this implies \\(AC = DC\\) . Therefore \\(\\triangle ACD\\) is isosceles. Since \\(x = \\angle ACD\\) this gives us \n\n\\[\\angle CDA = \\angle DAC = 90^{\\circ} - \\frac{x}{2}\\] \n\nNow let \\(\\lambda\\) be the angle bisector of \\(\\angle BAC\\) . Since \\(x = \\angle BAC\\) we know that \\(\\lambda\\) makes an angle of \\(\\frac{1}{2}\\angle BAC = \\frac{x}{2}\\) with line \\(AC\\) . Therefore the angle between \\(\\lambda\\) and \\(AD\\) is \n\n\\[\\frac{x}{2} +\\angle DAC = \\frac{x}{2} +\\left(90^{\\circ} - \\frac{x}{2}\\right) = 90^{\\circ}\\] \n\nas required.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2021_solutions.jsonl", "problem_match": "\n2. Problem:", "solution_match": "\nSolution"}}
{"year": "2021", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "NewZealand_MO", "problem": "In a sequence of numbers, a term is called golden if it is divisible by the term immediately before it. What is the maximum possible number of golden terms in a permutation of \\(1,2,3,\\ldots ,2021?\\)", "solution": ": Let \\(k\\) be the number of golden terms. We claim that \\(k\\leq 1010\\) \n\nProof: Define the term immediately before a golden term to be a silver term. The number of silver terms is also \\(k\\) . If \\(a\\) is any silver term and \\(b\\) is the corresponding golden term then we must have \n\n\\[a\\leq \\frac{b}{2}\\leq \\frac{2021}{2} < 1011.\\] \n\nTherefore \\(a\\leq 1010\\) . Hence \\(a\\in \\{1,2,3,\\ldots ,1010\\}\\) and therefore there can be at most 1010 different silver terms. Therefore \\(k\\leq 1010\\) \n\nNow we need to show that \\(k = 1010\\) is indeed possible. To do this we first partition the set \\(\\{1,2,3,\\ldots ,2021\\}\\) into the following distinct parts: \n\n\\[\\{1,2,4,8,16,32,64,128,256,512,1024\\},\\] \\[\\{3,6,12,24,48,96,192,384,768,1536\\},\\] \\[\\{5,10,20,40,80,160,320,640,1280\\},\\] \\[\\{7,14,28,56,112,224,448,896,1792\\},\\] \\[\\ldots\\] \\[\\{2019\\},\\] \\[\\{2021\\} .\\] \n\nEach part starts with an odd number, which is then doubled until the result is larger than 2021. \n\nPutting each part in the above sequence in order is our example. In this example every even term, \\(2n\\) , is golden because it occurs immediately after the term \\(n\\) . There are 1010 even numbers, so \\(k = 1010\\) is possible. \n\n## Comment: \n\nA full solution needs to show that any permutation cannot have 1011 (or more) golden terms. Simply claiming that a particular permutation is \"optimal\" or \"best\" would not receive full marks for this problem. There are many permutations with 1010 golden terms such that not every golden term is double it's predecessor. For example: \n\n\\[4,8,16,32,64,128,256,512,1024,\\] \\[1,2,6,12,24,48,96,192,384,768,1536,\\] \\[5,10,20,40,80,160,320,640,1280,\\] \\[7,14,28,56,112,224,448,896,1792,\\] \\[3,9,18,36,72,144,288,576,1152,\\] \\[11,\\ldots\\] \\[2021.\\]", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2021_solutions.jsonl", "problem_match": "\n3. Problem:", "solution_match": "\nSolution"}}
{"year": "2021", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "NewZealand_MO", "problem": "Find all triples \\((x,p,n)\\) of non-negative integers such that \\(p\\) is prime and \n\n\\[2x(x + 5) = p^{n} + 3(x - 1).\\]", "solution": ": The equation rearranges to be \n\n\\[p^{n} = 2x(x + 5) - 3(x - 1) = 2x^{2} + 7x + 3 = (2x + 1)(x + 3).\\] \n\nSince \\(x\\) is a non- negative integer, both factors \\((2x + 1)\\) and \\((x + 3)\\) must be positive integers. Therefore both \\((2x + 1)\\) and \\((x + 3)\\) are both powers of \\(p\\) . Let \n\n\\[2x + 1 = p^{a}\\] \\[x + 3 = p^{b}.\\] \n\nNow note that \\(\\gcd (2x + 1,x + 3) = \\gcd (x - 2,x + 3) = \\gcd (5,x + 3)\\) which must equal either 1 or 5 (because 5 is prime). We consider each case individually: \n\nIf \\(\\gcd (2x + 1,x + 3) = 5\\) then \\(p = 5\\) and \\(\\min (a,b) = 1\\) . Thus either \\(2x + 1 = 5\\) or \\(x + 3 = 5\\) . Either way we get \\(x = 2\\) . This leads to \\(p^{n} = 25\\) and so \\((x,p,n) = (2,5,2)\\) . If \\(\\gcd (2x + 1,x + 3) = 1\\) then \\(\\min (a,b) = 0\\) . Thus \\(2x + 1 = 1\\) or \\(x + 3 = 1\\) . So either \\(x = 0\\) or \\(x = - 2\\) . We can't have \\(x< 0\\) so we must have \\(x = 0\\) . This leads to \\(p^{n} = 3\\) and so \\((x,p,n) = (0,3,1)\\) . \n\nTherefore the only solutions for \\((x,p,n)\\) are \\((2,5,2)\\) and \\((0,3,1)\\) .", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2021_solutions.jsonl", "problem_match": "\n4. Problem:", "solution_match": "\nSolution"}}
-{"year": "2021", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(ABC\\) be an isosceles triangle with \\(AB = AC\\) . Point \\(D\\) lies on side \\(AC\\) such that \\(BD\\) is the angle bisector of \\(\\angle ABC\\) . Point \\(E\\) lies on side \\(BC\\) between \\(B\\) and \\(C\\) such that \\(BE = CD\\) . Prove that \\(DE\\) is parallel to \\(AB\\) .", "solution": ": Let \\(E^{\\prime}\\) be the point on line \\(BC\\) such that \\(DE^{\\prime}\\) is parallel to \\(AB\\) . We know that \\(E^{\\prime}\\) lies between \\(B\\) and \\(C\\) because \\(D\\) lies between \\(A\\) and \\(C\\) . So it suffices for us to prove that \\(BE^{\\prime} = CD\\) . \n\n\n \n\n\\[\\angle A C B = \\angle C B A\\] \\[= \\angle C E^{\\prime}D.\\] \n\nTherefore triangle \\(CDE^{\\prime}\\) is isosceles with \\(CD = DE^{\\prime}\\) . \n\n\\[\\angle E^{\\prime}BD = \\angle DBA\\] \\[= \\angle ADE^{\\prime}.\\] \n\nTherefore triangle \\(BE^{\\prime}D\\) is isosceles with \\(BE^{\\prime} = DE^{\\prime}\\) . Hence \\(CD = DE^{\\prime} = BE^{\\prime}\\) as required.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2021_solutions.jsonl", "problem_match": "\n5. Problem:", "solution_match": "\nSolution"}}
+{"year": "2021", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(ABC\\) be an isosceles triangle with \\(AB = AC\\) . Point \\(D\\) lies on side \\(AC\\) such that \\(BD\\) is the angle bisector of \\(\\angle ABC\\) . Point \\(E\\) lies on side \\(BC\\) between \\(B\\) and \\(C\\) such that \\(BE = CD\\) . Prove that \\(DE\\) is parallel to \\(AB\\) .", "solution": ": Let \\(E^{\\prime}\\) be the point on line \\(BC\\) such that \\(DE^{\\prime}\\) is parallel to \\(AB\\) . We know that \\(E^{\\prime}\\) lies between \\(B\\) and \\(C\\) because \\(D\\) lies between \\(A\\) and \\(C\\) . So it suffices for us to prove that \\(BE^{\\prime} = CD\\) . \n\n\n \n\n\\[\\angle A C B = \\angle C B A\\] \\[= \\angle C E^{\\prime}D.\\] \n\nTherefore triangle \\(CDE^{\\prime}\\) is isosceles with \\(CD = DE^{\\prime}\\) . \n\n\\[\\angle E^{\\prime}BD = \\angle DBA\\] \\[= \\angle ADE^{\\prime}.\\] \n\nTherefore triangle \\(BE^{\\prime}D\\) is isosceles with \\(BE^{\\prime} = DE^{\\prime}\\) . Hence \\(CD = DE^{\\prime} = BE^{\\prime}\\) as required.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2021_solutions.jsonl", "problem_match": "\n5. Problem:", "solution_match": "\nSolution"}}
{"year": "2021", "tier": "T1", "problem_label": "6", "problem_type": null, "exam": "NewZealand_MO", "problem": "Is it possible to place a positive integer in every cell of a \\(10 \\times 10\\) array in such a way that both the following conditions are satisfied? \n\n- Each number (not in the top row) is a proper divisor of the number immediately below. \n\n- Each row consists of 10 consecutive positive integers (but not necessarily in order).", "solution": ": Answer: Yes. In fact it is even possible to achieve such an array where each row consists of ten consecutive positive integers in increasing order. We shall construct an example explicitly. \n\nInitially let the top row be \\((1,2,3,\\ldots ,10)\\) in this order. Then iteratively if the contents of a particular row are \n\n\\[(n,n + 1,n + 2,\\ldots ,n + 9)\\] \n\nthen construct the next row to be \n\n\\[\\left((n + 9)! + n,(n + 9)! + n + 1,(n + 9)! + n + 2,\\ldots ,(n + 9)! + n + 9\\right).\\] \n\nSo the completed array will look like this: \n\n| 1 | 2 | 3 | ... | 10 |
| 10! + 1 | 10! + 2 | 10! + 3 | ... | 10! + 10 |
| : | : | : | : | : |
| n | n + 1 | n + 2 | ... | n + 9 |
| (n + 9)! + n | (n + 9)! + n + 1 | (n + 9)! + n + 2 | ... | (n + 9)! + n + 9 |
| : | : | : | : | : |
\n\nTo prove that this construction works, we simply need to verify that \\(\\left((n + 9)! + n + i\\right)\\) is always a multiple of \\((n + i)\\) for each \\(i = 0,1,2,3,\\ldots ,9\\) . We prove this by combining these two trivial divisibilities: \n\n\\[(n + i)\\mid (n + 9)! \\qquad \\mathrm{and} \\qquad (n + i)\\mid (n + i).\\] \n\nThus this construction satisfies the required conditions. \n\n## Comment: \n\nThe construction given here can be generalised in the following way. Let the first row be any arrangement of 10 consecutive positive integers. Then iteratively if the contents of a particular row is \\((a_{1},a_{2},\\ldots ,a_{10})\\) , then choose some positive integer \\(x\\) which is a common multiple of each of \\(a_{1},a_{2},\\ldots ,a_{10}\\) and construct the next row to be \\((b_{1},b_{2},\\ldots ,b_{10})\\) , where \\(b_{i} = x + a_{i}\\) for all \\(i\\) . Note that we cannot simply let \\(x = 0\\) because a number is not a proper divisor of itself.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2021_solutions.jsonl", "problem_match": "\n6. Problem:", "solution_match": "\nSolution"}}
{"year": "2021", "tier": "T1", "problem_label": "7", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(a, b, c, d\\) be integers such that \\(a > b > c > d \\geq -2021\\) and \n\n\\[\\frac{a + b}{b + c} = \\frac{c + d}{d + a}\\] \n\n(and \\(b + c \\neq 0 \\neq d + a\\) ). What is the maximum possible value of \\(ac\\) ?", "solution": ": We claim that the maximum value of \\(ac\\) is \\(2 \\times 505^{2} = 510050\\) , and this is uniquely achieved by \\((a, b, c, d) = (1010, 506, 505, - 2021)\\) . \n\nTo prove this we start by rearranging the expression \\(\\frac{a + b}{b + c} = \\frac{c + d}{d + a}\\) to get \\((c + b)(c + d) = (a + b)(a + d)\\) . Now expand the brackets and move all the terms to one side to get \\((c - a)(a + b + c + d) = 0\\) . Since \\(c > a\\) we can divide by \\(c - a\\) to get \n\n\\[a + b + c + d = 0.\\] \n\nNote that this is equivalent to the \\(\\frac{a + b}{b + c} = \\frac{c + d}{d + a}\\) condition. Now assume that \\((a, b, c, d)\\) is the tuple such that \\(ac\\) is maximised. Since \\(4a > a + b + c + d = 0\\) we trivially have \\(a > 0\\) . Furthermore if \\(c \\leq 0\\) then we would have \\(ac \\leq 0\\) which would contradict the definition that \\(ac\\) is maximal. Therefore \n\n\\[a > b > c > 0 \\geq d \\geq -2021.\\] \n\nNow we perform two little proofs by contradiction: \n\n- If \\(d > -2021\\) then we could consider \\((a', b', c', d') = (a + 1, b, c, d - 1)\\) so that \n\n\\[a^{\\prime}c^{\\prime} = (a + 1)c > ac.\\] \n\nHowever this would contradict the definition that \\(ac\\) was maximal. \n\nTherefore \\(d = - 2021\\) \n\n- If \\(b > c + 1\\) then we could consider \\((a', b', c', d') = (a + 1, b - 1, c, d)\\) so that \n\n\\[a^{\\prime}c^{\\prime} = (a + 1)c > ac.\\] \n\nHowever this would contradict the definition that \\(ac\\) was maximal. \n\nTherefore \\(b = c + 1\\) \n\nTherefore \\((a, b, c, d) = (a, c + 1, c, - 2021)\\) , and since \\(a + b + c + d = 0\\) , this means \\(a = 2020 - 2c\\) . So the final expression we are trying to maximise is \n\n\\[a c = (2020 - 2c)c\\] \\[\\qquad = 2\\times 505^{2} - 2(c - 505)^{2}\\] \\[\\qquad \\geq 2\\times 505^{2}.\\] \n\nwith equality iff \\(c = 505\\) . Therefore the maximum value of \\(ac\\) is \\(2 \\times 505^{2} = 510050\\) which is achieved by \n\n\\[(a, b, c, d) = (1010, 506, 505, -2021).\\]", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2021_solutions.jsonl", "problem_match": "\n7. Problem:", "solution_match": "\nSolution"}}
diff --git a/NewZealand_MO/segmented/en-nzmo1_2022_solutions.jsonl b/NewZealand_MO/segmented/en-nzmo1_2022_solutions.jsonl
index 3595985d408322828da171d5ad1debd32e7eea5a..db5232e5cdb12e241d30e4f368eb844077345e0d 100644
--- a/NewZealand_MO/segmented/en-nzmo1_2022_solutions.jsonl
+++ b/NewZealand_MO/segmented/en-nzmo1_2022_solutions.jsonl
@@ -1,10 +1,10 @@
-{"year": "2022", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "NewZealand_MO", "problem": "\\(ABCD\\) is a rectangle with side lengths \\(AB = CD = 1\\) and \\(BC = DA = 2\\) . Let \\(M\\) be the midpoint of \\(AD\\) . Point \\(P\\) lies on the opposite side of line \\(MB\\) to \\(A\\) , such that triangle \\(MBP\\) is equilateral. Find the value of \\(\\angle PCB\\) .", "solution": ": (Kevin Shen) \n\n\\(M\\) is the midpoint of \\(AD\\) , by symmetry \\(MB = MC\\) . The side lengths of an equilateral triangle are all equal, so \\(MB = MP\\) . \n\n\n \n\nAs \\(MB = MC = MP\\) , \\(M\\) is the circumcenter of triangle \\(BCP\\) . For any chord of any circle, the angle subtended at the center is always double the angle subtended at the circumference. Since \\(\\angle PCB\\) and \\(\\angle PMB\\) are both subtended by arc \\(BP\\) , we get \n\n\\[\\angle PCB = \\frac{1}{2}\\angle PMB = \\frac{1}{2} 60^{\\circ} = 30^{\\circ}\\] \n\nas required.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2022_solutions.jsonl", "problem_match": "\n1. Problem:", "solution_match": "\nSolution"}}
+{"year": "2022", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "NewZealand_MO", "problem": "\\(ABCD\\) is a rectangle with side lengths \\(AB = CD = 1\\) and \\(BC = DA = 2\\) . Let \\(M\\) be the midpoint of \\(AD\\) . Point \\(P\\) lies on the opposite side of line \\(MB\\) to \\(A\\) , such that triangle \\(MBP\\) is equilateral. Find the value of \\(\\angle PCB\\) .", "solution": ": (Kevin Shen) \n\n\\(M\\) is the midpoint of \\(AD\\) , by symmetry \\(MB = MC\\) . The side lengths of an equilateral triangle are all equal, so \\(MB = MP\\) . \n\n\n \n\nAs \\(MB = MC = MP\\) , \\(M\\) is the circumcenter of triangle \\(BCP\\) . For any chord of any circle, the angle subtended at the center is always double the angle subtended at the circumference. Since \\(\\angle PCB\\) and \\(\\angle PMB\\) are both subtended by arc \\(BP\\) , we get \n\n\\[\\angle PCB = \\frac{1}{2}\\angle PMB = \\frac{1}{2} 60^{\\circ} = 30^{\\circ}\\] \n\nas required.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2022_solutions.jsonl", "problem_match": "\n1. Problem:", "solution_match": "\nSolution"}}
{"year": "2022", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "NewZealand_MO", "problem": "Is it possible to pair up the numbers \\(0, 1, 2, 3, \\ldots , 61\\) in such a way that when we sum each pair, the product of the 31 numbers we get is a perfect fifth power?", "solution": ": (Jamie Craik) \n\nClaim: It can be achieved. One way to achieve it is: \n\nto pair 0 with 1, and \n\nto pair \\(k\\) with \\((63 - k)\\) for \\(k = 2,3,4,\\ldots ,31\\) \n\nThis would result in a product equal to: \n\n\\[(0 + 1)\\prod_{k = 2}^{31}\\left(k + (63 - k)\\right) = 1\\times 63^{30} = \\left(63^{6}\\right)^{5}\\] \n\nwhich is a perfect \\(5^{\\mathrm{th}}\\) power of \\(63^{6}\\) \n\nNote that there are many ways to solve this problem, for example: \n\n\\[(0 + 1)\\times (2 + 11)(3 + 10)(4 + 9)(5 + 8)(6 + 7)\\] \\[\\qquad \\times (12 + 21)(13 + 20)(14 + 19)(15 + 18)(16 + 17)\\] \\[\\qquad \\times (22 + 31)(23 + 30)(24 + 29)(25 + 28)(26 + 27)\\] \\[\\qquad \\times (32 + 41)(33 + 40)(34 + 39)(35 + 38)(36 + 37)\\] \\[\\qquad \\times (42 + 51)(43 + 50)(44 + 49)(45 + 48)(46 + 47)\\] \\[\\qquad \\times (52 + 61)(53 + 60)(54 + 59)(55 + 58)(56 + 57)\\] \\[\\qquad = 1\\times 13^{5}\\times 33^{5}\\times 53^{5}\\times 73^{5}\\times 93^{5}\\times 113^{5}.\\]", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2022_solutions.jsonl", "problem_match": "\n2. Problem:", "solution_match": "\nSolution"}}
{"year": "2022", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "NewZealand_MO", "problem": "Find all real numbers \\(x\\) and \\(y\\) such that \n\n\\[\\frac{x^{2} + y^{2} = 2,}{2 - y} +\\frac{y^{2}}{2 - x} = 2.\\]", "solution": ": (Viet Hoang) \n\nFrom the second equation, after a few steps of algebraic manipulation, one has \n\n\\[x^{2}(2 - x) + y^{2}(2 - y) = 2(2 - x)(2 - y)\\] \\[2(x^{2} + y^{2}) - (x + y)(x^{2} + y^{2} - x y) = 8 - 4(x + y) + 2x y\\] \\[4 - (x + y)(2 - x y) = 8 - 4(x + y) + 2x y.\\] \\[-2(x + y) + x y(x + y) = 4 - 4(x + y) + 2x y\\] \\[2(x + y) + x y(x + y) = 4 + 2x y\\] \\[(x + y - 2)(2 + x y) = 0\\] \n\nIn the case \\(x y = - 2\\) , substituting this into the first equation, we have \\(x^{2} + x y + y^{2} = 0\\) However, this means \n\n\\[\\left(x + \\frac{y}{2}\\right)^{2} + \\frac{3y^{2}}{4} = 0\\Longrightarrow x = y = 0\\] \n\nSubstituting these values into the second equation, it yields a contradiction. For the second case, using the substitution \\(y = 2 - x\\) , we get \n\n\\[x^{2} + y^{2} = 2,\\] \\[x + y = 2.\\] \n\nFrom here we get: \\((x - 1)^{2} + (y - 1)^{2} = (x^{2} + y^{2}) - 2(x + y) + 2 = 2 - 4 + 2 = 0\\) Therefore \\(x = y = 1\\) .", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2022_solutions.jsonl", "problem_match": "\n3. Problem:", "solution_match": "\nSolution"}}
{"year": "2022", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "NewZealand_MO", "problem": "On a table, there is an empty bag and a chessboard containing exactly one token on each square. Next to the table is a large pile that contains an unlimited supply of tokens. Using only the following types of moves what is the maximum possible number of tokens that can be in the bag? \n\n- Type 1: Choose a non-empty square on the chessboard that is not in the rightmost column. Take a token from this square and place it, along with one token from the pile, on the square immediately to its right. \n\n- Type 2: Choose a non-empty square on the chessboard that is not in the bottommost row. Take a token from this square and place it, along with one token from the pile, on the square immediately below it. \n\n- Type 3: Choose two adjacent non-empty squares. Remove a token from each and put them both into the bag.", "solution": "A: (Ishan Nath) \n\nLet \\(a_{i,j}\\) be the number of tokens in the square in the \\(i^{\\mathrm{th}}\\) row and \\(j^{\\mathrm{th}}\\) column, where the first row is the topmost row and the first column is the leftmost column. Furthermore let \\(b\\) denote the number of tokens in the bag. We define a monovariant as follows. \n\n\\[m = \\frac{3b}{2} +\\sum_{i = 1}^{8}\\sum_{j = 1}^{8}2^{16 - i - j}a_{i,j}\\] \n\nThis either stays the same, or decreases after each move, because: \n\n- If Ross plays a type 1 move, then \\(a_{i,j}\\) decreases by 1 and \\(a_{i,j + 1}\\) increases by 2 for some \\(i,j\\) . Thus \\(m\\) changes by \n\n\\[m^{\\prime} - m = 2^{16 - i - j}(-1) + 2^{16 - i - (j + 1)}(+2)\\] \\[\\qquad = 0.\\] \n\nHence \\(m\\) stays the same after type 1 moves. \n\n- If Ross plays a type 2 move, then \\(a_{i,j}\\) decreases by 1 and \\(a_{i + 1,j}\\) increases by 2 for some \\(i,j\\) . Thus \\(m\\) changes by \n\n\\[m^{\\prime} - m = 2^{16 - i - j}(-1) + 2^{16 - (i + 1) - j}(+2)\\] \\[\\qquad = 0.\\] \n\nHence \\(m\\) stays the same after type 2 moves. \n\n- If Ross plays a type 3 move, then \\(b\\) increases by 2 while \\(a_{i,j}\\) and \\(a_{i^{\\prime},j^{\\prime}}\\) both decrease by 1 for some \\(i,j,i^{\\prime},j^{\\prime}\\) . \n\n\\[m^{\\prime} - m = \\frac{3}{2} (+2) + 2^{16 - i - j}(-1) + 2^{16 - i^{\\prime} - j^{\\prime}}(-1)\\] \\[\\qquad \\leq \\frac{3}{2} (+2) + 2^{0}(-1) + 2^{1}(-1)\\] \\[\\qquad = 0.\\] \n\nHence \\(m\\) stays the same or decreases after these moves.\n\n\n\nAt the beginning, \n\n\\[m = \\frac{3b}{2} +\\sum_{i = 1}^{8}\\sum_{j = 1}^{8}2^{16 - i - j}a_{i,j}\\] \\[\\qquad = 0 + \\sum_{i = 1}^{8}\\sum_{j = 1}^{8}2^{16 - i - j}\\] \\[\\qquad = 0 + 2^{16}\\sum_{i = 1}^{8}2^{-i}\\sum_{j = 1}^{8}2^{-j}\\] \\[\\qquad = 0 + 2^{16}\\sum_{i = 1}^{8}2^{-i}(1 - 2^{-8})\\] \\[\\qquad = 0 + 2^{16}(1 - 2^{-8})(1 - 2^{-8})\\] \\[\\qquad = 65025.\\] \n\nTherefore at the end we must have \\(m \\leq 65025\\) . Thus we obtain \n\n\\[\\frac{3}{2} b \\leq m \\leq 65025 \\iff b \\leq 43350.\\] \n\nTo obtain equality we simply need to perform any strategy in which: \n\n- Moves of type 3 are only ever performed on either \\(a_{7,8}\\) and \\(a_{8,8}\\) , or \\(a_{8,7}\\) and \\(a_{8,8}\\) . \n\n- The chessboard is empty at the end of the game. \n\nNow consider a strategy in which we start by applying all moves of type 1 and type 2, until \\(a_{i,j} = 0\\) for all squares except for \\(a_{7,8}\\) , \\(a_{8,7}\\) and \\(a_{8,8}\\) . At this point we will have \\(a_{8,8} = 1\\) and since \\(m\\) is invariant, \n\n\\[65025 = 0 + a_{8,8} + 2a_{7,8} + 2a_{8,7} = 1 + 2(a_{7,8} + a_{8,7}).\\] \n\nThus \\(a_{7,8} + a_{8,7} = 32512\\) . Now perform 10837 further moves (either type 1 on \\(a_{7,8}\\) or type 2 on \\(a_{8,7}\\) ) so that we end up with a state in which: \n\n\\[\\cdot a_{7,8} + a_{8,7} = 32512 - 10837 = 21675,\\mathrm{and}\\] \n\n\\[\\cdot a_{8,8} = 1 + 2\\times 10837 = 21675.\\] \n\nFrom here we can finish by performing 21675 type 3 moves and obtain \\(b = 2 \\times 21675 = 43350\\) tokens in the bag.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2022_solutions.jsonl", "problem_match": "\n4. Problem:", "solution_match": "\nSolution"}}
{"year": "2022", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "NewZealand_MO", "problem": "On a table, there is an empty bag and a chessboard containing exactly one token on each square. Next to the table is a large pile that contains an unlimited supply of tokens. Using only the following types of moves what is the maximum possible number of tokens that can be in the bag? \n\n- Type 1: Choose a non-empty square on the chessboard that is not in the rightmost column. Take a token from this square and place it, along with one token from the pile, on the square immediately to its right. \n\n- Type 2: Choose a non-empty square on the chessboard that is not in the bottommost row. Take a token from this square and place it, along with one token from the pile, on the square immediately below it. \n\n- Type 3: Choose two adjacent non-empty squares. Remove a token from each and put them both into the bag.", "solution": "B: (Michael Albert) \n\nFirst note that we may assume that all type 3 moves are performed after all the type 1 and type 2 moves have been made. WLOG We may also assume that all moves (type 1 and 2) into square \\((8,8)\\) occur all other type 1 and type 2 moves. \n\nClaim: In an optimal solution no type 3 move is made except from a pair including the lower right corner. \n\nProof: If we make one elsewhere, then we could first double each of those coins and move them to another adjacent pair of squares. Then we could perform two type 3 moves (in this new adjacent pair of squares) and get result in \\(+2\\) more coins in the bag. \\(\\square\\) \n\nFrom here it is clear that all the coins (except for the one that starts in \\((8,8)\\) ) must eventually move to either \\((7,8)\\) or \\((8,7)\\) . So there will be a point in which there is: \n\none coin on \\((8,8)\\) , \\(A\\) coins on \\((7,8)\\) , \\(B\\) coins on \\((8,7)\\) , and all other squares are empty. \n\nA coin starting in square \\((x,y)\\) will need to double \\((8 + 7 - x - y)\\) times on it's way to either \\((7,8)\\) or \\((8,7)\\) . Therefore \n\n\\[A + B = \\sum_{(x,y)\\neq (8,8)}2^{8 + 7 - x - y} = 2^{15}\\sum_{x = 1}^{8}2^{-x}\\sum_{y = 1}^{8}2^{-y} - 2^{-1} = 2^{-15}\\left(\\frac{2^{-9} - 2^{1}}{2^{-1} - 1}\\right)^{2} - \\frac{1}{2} = 32512.\\] \n\nNow let \\(m\\) be the number of type 1 and type 2 moves we make before we finish the game by making as many type 3 moves as possible. \n\nAfter these \\(m\\) moves there will be \\(2m + 1\\) coins on square \\((8,8)\\) . \n\nAfter these \\(m\\) moves there will be a total of \\(A + B - m = 32512 - m\\) coins on squares \\((7,8)\\) and \\((8,7)\\) . \n\nSo the maximum number of type 3 moves we can make is \n\n\\[M = \\min \\left(2m + 1,32512 - m\\right).\\] \n\nIf \\(m \\leq 10837\\) then \\(M = 2m + 1 \\leq 21675\\) . If \\(m \\geq 10837\\) then \\(M = 32512 - m \\leq 21675\\) . Hence the maximum occurs when \\(m = 10837\\) and so \\(M = 21675\\) type 3 moves are made.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2022_solutions.jsonl", "problem_match": "\n4. Problem:", "solution_match": "\nSolution"}}
{"year": "2022", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "NewZealand_MO", "problem": "A round-robin tournament is one where each team plays every other team exactly once. Five teams take part in such a tournament getting: 3 points for a win, 1 point for a draw and 0 points for a loss. At the end of the tournament the teams are ranked from first to last according to the number of points. \n\n(a) Is it possible that at the end of the tournament, each team has a different number of points, and each team except for the team ranked last has exactly two more points than the next-ranked team? \n\n(b) Is this possible if there are six teams in the tournament instead?", "solution": ": (Ishan Nath) \n\nWe show the answer is no for five teams, and yes for six. \n\nFirst, we show five teams cannot have this property. Suppose the teams had points \\(x, x + 2, x + 4, x + 6, x + 8\\) in the final ranking. If there were \\(d\\) draws and \\(10 - d\\) decisive games, then the total number of points is \n\n\\[5x + 20 = x + (x + 2) + \\dots +(x + 8) = 2d + 3(10 - d) = 30 - d.\\] \n\nSince \\(0 \\leq d \\leq 10\\) , we have \n\n\\[20 \\leq 5x + 20 \\leq 30.\\] \n\nHence \\(x = 0,1\\) or 2. \n\n- If \\(x = 0\\) , then there are 20 points in total. Hence \\(d = 10\\) , so every match was a draw. However then every team would have 4 points, contradiction. \n\n- If \\(x = 2\\) , then there are 30 points in total. Hence \\(d = 0\\) , so every match was decisive. But this means every team should have a point total which is a multiple of 3, since they can earn either 0 or 3 points per game. However the lowest team has 2 points, contradiction. \n\n- If \\(x = 1\\) , then there are 25 points in total. Hence \\(d = 5\\) , meaning there are 5 draws and 5 decisive games. In this case, the teams had points \\(1,3,5,7,9\\) . \n\nConsider the total number of draws. For each team, the number of points satisfies \\(p = 3w + d\\) , where \\(p\\) is the number of points, \\(w\\) is the number of wins, and \\(d\\) is the number of draws. Hence the number of draws is equal to the number of points, modulo 3. \n\n- The team with 1 points has drawn at most once \n\n- The team with 3 points has drawn at most 3 times. \n\n- The team with 5 points has drawn at most 2 times, since they play 4 times. \n\n- The team with 7 points has drawn at most once. They cannot draw 4 times, otherwise they would have drawn every game, ending up with 4 points. \n\n- The team with 9 points cannot have drawn. Otherwise, they would have drawn at least 3 times, requiring 6 points from the remaining game, which is impossible. \n\nHence there are at most 7 draws. However, if 5 matches ended in draws, then there should be at least 10 draws, contradiction. \n\nAll of these options lead to contradictions, so the rankings cannot have each team having exactly two more points than the least.\n\n\n\nWith six teams it is possible: Take 6 teams, labeled 1 to 6. We assign the following matches: \n\nTeams 1 and 2 draw, teams 1 and 4 draw, and teams 4 and 5 draw. \n\nTeam 2 beats team 6. \n\nOtherwise, if \\(1 \\leq x < y \\leq 6\\) , then team \\(y\\) beats team \\(x\\) . \n\nThen team \\(i\\) has \\(2i\\) points, satisfying the condition.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2022_solutions.jsonl", "problem_match": "\n5. Problem:", "solution_match": "\nSolution"}}
{"year": "2022", "tier": "T1", "problem_label": "6", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let a positive integer \\(n\\) be given. Determine, in terms of \\(n\\) , the least positive integer \\(k\\) such that among any \\(k\\) positive integers, it is always possible to select a positive even number of them having sum divisible by \\(n\\) .", "solution": ": (Ethan Ng) \n\nFirst we consider the \\(n\\) odd case. If \\(n\\) is odd we claim the answer is \\(k = 2n\\) . \n\nLet the numbers be \\(x_{1}, x_{2}, x_{3}, \\ldots , x_{2n}\\) . Consider the following partial sums: \n\n\\[s_{1} = x_{1} + x_{2}\\] \\[s_{2} = x_{1} + x_{2} + x_{3} + x_{4}\\] \\[\\vdots\\] \\[s_{i} = x_{1} + x_{2} + x_{3} + \\dots +x_{2i}\\] \\[\\vdots\\] \\[s_{n} = x_{1} + x_{2} + x_{3} + x_{4} + \\dots +x_{2n}.\\] \n\nIf we have \\(s_{i} \\equiv 0\\) (mod \\(n\\) ) for any \\(i\\) , then we are done. Otherwise, by the Pigeonhole Principle we must have a pair of distinct indices \\(1 \\leq i < j \\leq n\\) such that \\(s_{i} \\equiv s_{j}\\) (mod \\(n\\) ). Thus we can construct: \n\n\\[x_{2i + 1} + x_{2i + 2} + \\dots +x_{2j} = s_{j} - s_{i} \\equiv 0 \\pmod {n}\\] \n\nas required. \n\nHowever if \\(k < 2n\\) then we could choose: \\(x_{1} = x_{2} = \\dots = x_{k} = 1\\) and so the sum of any non- empty subset would be a positive integer less than \\(2n\\) . So if this sum is a multiple of \\(n\\) then it must be equal to \\(n\\) , and so we must have selected \\(n\\) of them. But \\(n\\) is odd, so it is impossible to select an even number of them having sum divisible by \\(n\\) . \n\nNow we consider the \\(n\\) even case. If \\(n\\) is even we claim the answer is \\(k = n + 1\\) . \n\nLet \\(n = 2m\\) . By the Pigeonhole Principle, amongst any three positive integers we can choose two of them with the same parity. Therefore we can order the \\(k = 2m + 1\\) integers: \\(x_{1}, x_{2}, x_{3}, \\ldots , x_{2m + 1}\\) such that \n\n\\[a_{1} = \\frac{x_{1} + x_{2}}{2}, a_{2} = \\frac{x_{3} + x_{4}}{2}, \\ldots a_{m} = \\frac{x_{2m - 1} + x_{2m}}{2}\\] \n\nare all integers. This can be done by iteratively choosing \\(x_{2i - 1}\\) and \\(x_{2i}\\) so that they have the same parity. Now consider the partial sums: \n\n\\[t_{1} = a_{1\\] \\[t_{2} = a_{1} + a_{2\\] \\[t_{3} = a_{1} + a_{2} + a_{3\\] \\[\\qquad \\vdots\\] \\[t_{m} = a_{1} + a_{2} + a_{3} + \\dots +a_{m}.\\]\n\n\n\nIf any \\(t_{i}\\) were a multiple of \\(m\\) then we would have \\(2t_{i} = x_{1} + x_{2} + x_{3} + \\dots + x_{2i}\\) being a multiple of \\(2m\\) . Otherwise consider the \\(m\\) pigeons \\((t_{1},\\ldots t_{m})\\) and the \\(m - 1\\) pigeonholes (the non- zero residues mod \\(m\\) ). By the Pigeonhole Principle we can find indices \\(1\\leq i<\\) \\(j\\leq m\\) such that \\(t_{i}\\equiv t_{j}\\) (mod \\(m\\) ). Therefore \n\n\\[a_{i + 1} + a_{i + 2} + \\dots +a_{j} = t_{j} - t_{i}\\equiv 0(\\mathrm{mod} m).\\] \n\nThus \\((x_{2i + 1} + \\dots +x_{2j}) / 2\\) is a multiple of \\(m\\) and hence \\(x_{2i + 1} + \\dots +x_{2j}\\) is a multiple of \\(2m = n\\) . \n\nHowever if \\(k\\leq n\\) then we could choose: \\(x_{1} = n\\) and \\(x_{2} = \\dots = x_{k} = 1\\) and so the sum of any non- empty subset would be a positive integer at most \\(2n - 1\\) . So if this sum is a multiple of \\(n\\) then it must be equal to \\(n\\) . However the only way to achieve a sum of \\(n\\) is to only select \\(x_{1}\\) which is not selecting an even number of numbers. \n\nComment It is possible to come up with optimal constructions in which the \\((x_{i})\\) are all different. For example when \\(n\\) is odd choose: \n\n\\[(x_{1},x_{2},x_{3},\\ldots ,x_{k}) = (n + 1,2n + 1,3n + 1,\\ldots ,kn + 1).\\] \n\nAnd when \\(n\\) is even we can choose: \n\n\\[(x_{1},x_{2},x_{3},\\ldots ,x_{k}) = (n,2n + 1,3n + 1,\\ldots ,kn + 1).\\]", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2022_solutions.jsonl", "problem_match": "\n6. Problem:", "solution_match": "\nSolution"}}
-{"year": "2022", "tier": "T1", "problem_label": "7", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(M\\) be the midpoint of side \\(BC\\) of acute triangle \\(ABC\\) . The circle centered at \\(M\\) passing through \\(A\\) intersects the lines \\(AB\\) and \\(AC\\) again at \\(P\\) and \\(Q\\) , respectively. The tangents to this circle at \\(P\\) and \\(Q\\) meet at \\(D\\) . Prove that the perpendicular bisector of \\(BC\\) bisects segment \\(AD\\) .", "solution": "A: (Ross Atkins) \n\nLet \\(F\\) be the point on the circle such that \\(AF\\) is a diameter. So \\(M\\) is the midpoint of \\(AF\\) . Now let \\(E \\neq F\\) be the point on the circle such that \\(D, E, F\\) are colinear. This means \\(\\angle AEF = 90^\\circ\\) by Thales' theorem. \n\n\n \n\nBy the Power of a Point Theorem we get \\(DP^2 = DE \\times DF = DQ^2\\) . Therefore we have two pairs of similar triangles: \\(EPD \\sim PFD\\) and \\(EQD \\sim QFD\\) . Hence \n\n\\[\\frac{EP}{PF} = \\frac{PD}{FD} = \\frac{QD}{FD} = \\frac{EQ}{QF}.\\] \n\ni.e. \\(EPFQ\\) is a harmonic quadrilateral. Now apply the extended sine rule to get: \n\n\\[\\frac{\\sin\\angle BAM}{\\sin\\angle MAC} = \\frac{\\sin\\angle PAF}{\\sin\\angle FAQ} = \\frac{PF}{FQ} = \\frac{PE}{EQ} = \\frac{\\sin\\angle PAE}{\\sin\\angle QAE} = \\frac{\\sin\\angle BAE}{\\sin\\angle CAE}.\\] \n\nSince \\(\\frac{\\sin\\angle BAM}{\\sin\\angle MAC} = \\frac{\\sin\\angle BAE}{\\sin\\angle CAE}\\) , and \\(M\\) is the midpoint of \\(BC\\) , this means \\(AE\\) is parallel to \\(BC\\) . Hence the perpendicular bisector of \\(BC\\) is parallel to line \\(DEF\\) . \n\nNow let \\(N\\) be the midpoint of \\(AD\\) . By midpoint theorem in triangle \\(AFD\\) we get \\(NM \\parallel DF\\) and therefore \\(N\\) lies on the perpendicular bisector of \\(BC\\) .", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2022_solutions.jsonl", "problem_match": "\n7. Problem:", "solution_match": "\nSolution"}}
+{"year": "2022", "tier": "T1", "problem_label": "7", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(M\\) be the midpoint of side \\(BC\\) of acute triangle \\(ABC\\) . The circle centered at \\(M\\) passing through \\(A\\) intersects the lines \\(AB\\) and \\(AC\\) again at \\(P\\) and \\(Q\\) , respectively. The tangents to this circle at \\(P\\) and \\(Q\\) meet at \\(D\\) . Prove that the perpendicular bisector of \\(BC\\) bisects segment \\(AD\\) .", "solution": "A: (Ross Atkins) \n\nLet \\(F\\) be the point on the circle such that \\(AF\\) is a diameter. So \\(M\\) is the midpoint of \\(AF\\) . Now let \\(E \\neq F\\) be the point on the circle such that \\(D, E, F\\) are colinear. This means \\(\\angle AEF = 90^\\circ\\) by Thales' theorem. \n\n\n \n\nBy the Power of a Point Theorem we get \\(DP^2 = DE \\times DF = DQ^2\\) . Therefore we have two pairs of similar triangles: \\(EPD \\sim PFD\\) and \\(EQD \\sim QFD\\) . Hence \n\n\\[\\frac{EP}{PF} = \\frac{PD}{FD} = \\frac{QD}{FD} = \\frac{EQ}{QF}.\\] \n\ni.e. \\(EPFQ\\) is a harmonic quadrilateral. Now apply the extended sine rule to get: \n\n\\[\\frac{\\sin\\angle BAM}{\\sin\\angle MAC} = \\frac{\\sin\\angle PAF}{\\sin\\angle FAQ} = \\frac{PF}{FQ} = \\frac{PE}{EQ} = \\frac{\\sin\\angle PAE}{\\sin\\angle QAE} = \\frac{\\sin\\angle BAE}{\\sin\\angle CAE}.\\] \n\nSince \\(\\frac{\\sin\\angle BAM}{\\sin\\angle MAC} = \\frac{\\sin\\angle BAE}{\\sin\\angle CAE}\\) , and \\(M\\) is the midpoint of \\(BC\\) , this means \\(AE\\) is parallel to \\(BC\\) . Hence the perpendicular bisector of \\(BC\\) is parallel to line \\(DEF\\) . \n\nNow let \\(N\\) be the midpoint of \\(AD\\) . By midpoint theorem in triangle \\(AFD\\) we get \\(NM \\parallel DF\\) and therefore \\(N\\) lies on the perpendicular bisector of \\(BC\\) .", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2022_solutions.jsonl", "problem_match": "\n7. Problem:", "solution_match": "\nSolution"}}
{"year": "2022", "tier": "T1", "problem_label": "7", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(M\\) be the midpoint of side \\(BC\\) of acute triangle \\(ABC\\) . The circle centered at \\(M\\) passing through \\(A\\) intersects the lines \\(AB\\) and \\(AC\\) again at \\(P\\) and \\(Q\\) , respectively. The tangents to this circle at \\(P\\) and \\(Q\\) meet at \\(D\\) . Prove that the perpendicular bisector of \\(BC\\) bisects segment \\(AD\\) .", "solution": "B: (Ishan Nath) \n\nAs in the previous solution, let \\(F\\) be the point on the circle such that \\(AF\\) is a diameter. Notice that \n\n\\[\\angle B P F = 90^{\\circ} = \\angle M P D,\\] \n\nwhere the first equality follows from Thales' theorem, and the second follows as \\(PD\\) is tangent to a circle with center \\(M\\) . We also have \n\n\\[\\angle P B F = \\angle B A C = \\angle P A Q = \\frac{\\angle P M Q}{2} = \\angle P M D,\\] \n\nsince \\(C A B F\\) is a parallelogram (because the diagonals bisect each other at \\(M\\) ) and \\(\\angle P M Q = 2\\angle P M D\\) (because \\(P M Q D\\) is a kite). This implies that the triangles \\(P B F\\) and \\(P M D\\) are similar. Since they are also similarly oriented, we can use similar switch to deduce that triangles \\(P B M\\) and \\(P F D\\) are similar. \n\nConsider rotation at \\(P\\) , followed by scaling, that takes triangle \\(P B M\\) to \\(P F D\\) . Since \\(\\angle B P F = 90^{\\circ}\\) , the rotation must be \\(90^{\\circ}\\) . However this means the transformation taking \\(B M\\) to \\(F D\\) must rotate \\(B M\\) by \\(90^{\\circ}\\) , implying \\(B M\\) and \\(F D\\) are perpendicular. \n\nFinally, letting \\(N\\) be the midpoint of \\(A D\\) , we get \\(N M \\parallel D F \\perp B C\\) by the midpoint theorem. So \\(N\\) lies on the perpendicular bisector of \\(B C\\) .", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2022_solutions.jsonl", "problem_match": "\n7. Problem:", "solution_match": "\nSolution"}}
{"year": "2022", "tier": "T1", "problem_label": "8", "problem_type": null, "exam": "NewZealand_MO", "problem": "Find all real numbers \\(x\\) such that \\(-1 < x \\leq 2\\) and \n\n\\[\\sqrt{2 - x} + \\sqrt{2 + 2x} = \\sqrt{\\frac{x^{4} + 1}{x^{2} + 1}} +\\frac{x + 3}{x + 1}.\\]", "solution": ": (Viet Hoang) \n\nNotice that we are solving for \\(x\\) in the domain \\(- 1 < x \\leq 2\\) . Using Cauchy- Schwarz Inequality on the left hand side, one has \n\n\\[\\sqrt{2 - x} + \\sqrt{2 + 2x} = \\sqrt{\\frac{1}{2}\\cdot(4 - 2x) + \\sqrt{2 + 2x}}\\leq \\sqrt{\\left(\\frac{1}{2} + 1\\right)(2 + 2x + 4 - 2x)} = 3.\\] \n\nUsing the fact that \\(x^{4} + 1 \\geq 2x^{2}\\) (because \\((x^{2} - 1)^{2} \\geq 0\\) ) and \\(x^{2} + 1 \\geq 2x\\) (because \\((x - 1)^{2} \\geq 0\\) we can get: \n\n\\[2(x^{4} + 1) \\geq (x^{2} + 1)^{2} \\geq (x^{2} + 1) \\frac{(x + 1)^{2}}{2}.\\] \n\nIf we substitute this into the right- hand side we get \n\n\\[\\sqrt{\\frac{x^{4} + 1}{x^{2} + 1}} +\\frac{x + 3}{x + 1} \\geq \\frac{x + 1}{2} +\\frac{x + 3}{x + 1} = \\frac{x + 1}{2} +\\frac{2}{x + 1} +1.\\] \n\nFinally applying the AM- GM again for 2 positive numbers \\(\\frac{x + 1}{2}\\) and \\(\\frac{2}{x + 1}\\) we obtain \\(\\frac{x + 1}{2} + \\frac{2}{x + 1} \\geq 2\\) . Thus the right- hand side is \\(\\geq 3\\) . \n\nTherefore, we can see that equality must occur in our AM- GM inequality and hence, \\(\\frac{x + 1}{2} = \\frac{2}{x + 1} \\Rightarrow x = 1\\) .", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2022_solutions.jsonl", "problem_match": "\n8. Problem:", "solution_match": "\nSolution"}}
diff --git a/NewZealand_MO/segmented/en-nzmo1_2023_solutions.jsonl b/NewZealand_MO/segmented/en-nzmo1_2023_solutions.jsonl
index ddf1da7f8fb04a943f5cd9da3212fc05b5ae88f4..6c94b1a9dd3fe654a89521c94c1711fec4c770a1 100644
--- a/NewZealand_MO/segmented/en-nzmo1_2023_solutions.jsonl
+++ b/NewZealand_MO/segmented/en-nzmo1_2023_solutions.jsonl
@@ -1,9 +1,9 @@
{"year": "2023", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "NewZealand_MO", "problem": "There are 2023 employees in the office, each of them knowing exactly 1686 of the others. For any pair of employees they either both know each other or both don't know each other. Prove that we can find 7 employees each of them knowing all 6 others.", "solution": "A: (Ross Atkins) \n\nIf every person knows 1686 others then for each person, there are \\(2023 - 1686 - 1 = 336\\) people that they don't know. Now consider any group of \\(p\\) people from the office. There will be at most \\(336p\\) people who don't know someone in the group (336 for each person in the group). Therefore there are at least \\((2023 - p) - 336p\\) people not in the group who know everyone in the group. If \\(p \\leq 6\\) then \n\n\\[(2023 - p) - 336p \\geq (2023 - 6) - 336 \\times 6 = 1.\\] \n\nSo for any group of at most 6 people, there exists at least one person not in the group that knows everyone in the group. Hence we perform the following process. \n\n- Start with a random group of \\(p = 2\\) people who know each other, and then- while \\(p \\leq 6\\) choose a person who knows all the current members of the group (at random) and add them to the group. \n\nThis process ends with a group of 7 people each knowing everyone in the group.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2023_solutions.jsonl", "problem_match": "\n1. Problem:", "solution_match": "\nSolution"}}
{"year": "2023", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "NewZealand_MO", "problem": "There are 2023 employees in the office, each of them knowing exactly 1686 of the others. For any pair of employees they either both know each other or both don't know each other. Prove that we can find 7 employees each of them knowing all 6 others.", "solution": "B: (Ishan Nath) \n\nWe will show by induction that, for non- negative integers \\(k\\) and \\(n > 0\\) , if there are \\(nk + 1\\) people such that any of them know at least \\(n(k - 1) + 1\\) of the others, then there are \\(k + 1\\) people who all know each other. For \\(k = 0\\) this is true, as we have one person. \n\nNow assume this is true for some integer \\(k\\) . Among any \\(n(k + 1) + 1\\) people who all know at least \\(nk + 1\\) of the others, pick an arbitrary person \\(P\\) , and a set \\(\\mathcal{S}\\) of \\(nk + 1\\) people that \\(P\\) knows. \n\nFor any person in \\(\\mathcal{S}\\) , they must know at least \\(n(k - 1) + 1\\) others in \\(\\mathcal{S}\\) , as there are exactly \\(n\\) people outside of \\(\\mathcal{S}\\) , and they know at least \\(nk + 1\\) people in total. Hence by our induction hypothesis, \\(\\mathcal{S}\\) contains \\(k + 1\\) people who all know each other. \n\nAs \\(P\\) knows everyone in \\(\\mathcal{S}\\) , including \\(P\\) gives a group of \\(k + 2\\) people who all know each other, proving our inductive result. \n\nNow letting \\(n = 337\\) and \\(k = 6\\) , we get the desired result.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2023_solutions.jsonl", "problem_match": "\n1. Problem:", "solution_match": "\nSolution"}}
-{"year": "2023", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(ABCD\\) be a parallelogram, and let \\(P\\) be a point on the side \\(AB\\) . Let the line through \\(P\\) parallel to \\(BC\\) intersect the diagonal \\(AC\\) at point \\(Q\\) . Prove that \n\n\\[|D A Q|^{2} = |P A Q|\\times |B C D|,\\] \n\nwhere \\(|X Y Z|\\) denotes the area of triangle \\(X Y Z\\) .", "solution": ": (Kevin Shen) \n\n\n \n\nSince \\(PQ\\) is parallel to \\(BC\\) , there is a dilation (centred at \\(A\\) ) of factor \\(AB / AP\\) that sends triangle \\(APQ\\) to triangle \\(ABC\\) . Thus \n\n\\[|P A Q| = |A B C|\\times \\left(\\frac{A P}{A B}\\right)^{2}.\\] \n\nSince \\(AD\\) is parallel to \\(BC\\) , triangle \\(ABC\\) has the same height as triangle \\(BCD\\) . Triangles \\(ABC\\) and \\(BCD\\) also have a common base, so \\(|ABC| = |BCD|\\) . Similarly \\(|DAQ| = |DAP|\\) (because \\(PQ||AD\\) and \\(AD\\) is a common base). \n\n\\[\\therefore |D A P| = |D A B|\\times \\frac{A P}{A B}\\] \\[\\qquad = |B C D|\\times \\frac{A P}{A B}\\] \\[\\qquad |D A P|^{2} = |B C D|\\times |B C D|\\times \\left(\\frac{A P}{A B}\\right)^{2}\\] \\[\\qquad = |B C D|\\times |P A Q|\\qquad \\mathrm{(diagonal~splits~parallelograms~in~half)}\\] \\[\\qquad = |B C D|\\times |P A Q|\\qquad \\mathrm{(because~}|P A Q| = |A B C|(A P / A B)^{2})\\] \n\nas required.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2023_solutions.jsonl", "problem_match": "\n2. Problem:", "solution_match": "\nSolution"}}
+{"year": "2023", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(ABCD\\) be a parallelogram, and let \\(P\\) be a point on the side \\(AB\\) . Let the line through \\(P\\) parallel to \\(BC\\) intersect the diagonal \\(AC\\) at point \\(Q\\) . Prove that \n\n\\[|D A Q|^{2} = |P A Q|\\times |B C D|,\\] \n\nwhere \\(|X Y Z|\\) denotes the area of triangle \\(X Y Z\\) .", "solution": ": (Kevin Shen) \n\n\n \n\nSince \\(PQ\\) is parallel to \\(BC\\) , there is a dilation (centred at \\(A\\) ) of factor \\(AB / AP\\) that sends triangle \\(APQ\\) to triangle \\(ABC\\) . Thus \n\n\\[|P A Q| = |A B C|\\times \\left(\\frac{A P}{A B}\\right)^{2}.\\] \n\nSince \\(AD\\) is parallel to \\(BC\\) , triangle \\(ABC\\) has the same height as triangle \\(BCD\\) . Triangles \\(ABC\\) and \\(BCD\\) also have a common base, so \\(|ABC| = |BCD|\\) . Similarly \\(|DAQ| = |DAP|\\) (because \\(PQ||AD\\) and \\(AD\\) is a common base). \n\n\\[\\therefore |D A P| = |D A B|\\times \\frac{A P}{A B}\\] \\[\\qquad = |B C D|\\times \\frac{A P}{A B}\\] \\[\\qquad |D A P|^{2} = |B C D|\\times |B C D|\\times \\left(\\frac{A P}{A B}\\right)^{2}\\] \\[\\qquad = |B C D|\\times |P A Q|\\qquad \\mathrm{(diagonal~splits~parallelograms~in~half)}\\] \\[\\qquad = |B C D|\\times |P A Q|\\qquad \\mathrm{(because~}|P A Q| = |A B C|(A P / A B)^{2})\\] \n\nas required.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2023_solutions.jsonl", "problem_match": "\n2. Problem:", "solution_match": "\nSolution"}}
{"year": "2023", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "NewZealand_MO", "problem": "Find the sum of the smallest and largest possible values for \\(x\\) which satisfy the following equation. \n\n\\[9^{x + 1} + 2187 = 3^{6x - x^{2}}\\]", "solution": ": (Viet Hoang) \n\nFirst we prove that the given equation has at least one root. To do this we consider the following function. \n\n\\[f(x) = 9^{x + 1} + 2187 - 3^{6x - x^{2}}\\] \n\nNote that this function is continuous. Note also that \\(f(0) = 9 + 2197 - 1 > 0\\) and \\(f(3) = 9^{4} + 2187 - 3^{9}< 0\\) . Therefore by the intermediate value theorem there exists a real number \\(0< z< 3\\) such that \\(f(z) = 0\\) . Therefore there is at least one value \\(x\\) which satisfies the equation. \n\nNext, we can rewrite the equation in the following form \n\n\\[3^{x} + 3^{5 - x} = 3^{x(5 - x) - 2} \\quad (1)\\] \n\nLet \\(r\\) and \\(s\\) be any two numbers such that \\(r + s = 5\\) . Notice that if \\(r\\) is a root of 1 then \\(s\\) must also be a root (and vice versa). This is because \n\n\\[3^{r} + 3^{5 - r} = 3^{5 - s} + 3^{s}\\qquad \\mathrm{and}\\qquad 3^{r(5 - r) - 2} = 3^{(5 - s)s - 2}.\\] \n\nNow we claim that if \\(x_{0}\\) is the smallest root of 1, then \\((5 - x_{0})\\) must be the largest root. Indeed if \\(x_{1}\\) were a root of 1 greater than \\((5 - x_{0})\\) , then \\((5 - x_{1})\\) would also have to be a root, but \\((5 - x_{1})< (5 - (5 - x_{0})) = x_{0}\\) and this would contradict the fact that \\(x_{0}\\) is the smallest root. \n\nTherefore, the sum of the largest and the smallest roots of 1 is \\(x_{0} + (5 - x_{0}) = 5\\)", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2023_solutions.jsonl", "problem_match": "\n3. Problem:", "solution_match": "\nSolution"}}
{"year": "2023", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(p\\) be a prime and let \\(f(x) = ax^{2} + bx + c\\) be a quadratic polynomial with integer coefficients such that \\(0 < a, b, c \\leq p\\) . Suppose \\(f(x)\\) is divisible by \\(p\\) whenever \\(x\\) is a positive integer. Find all possible values of \\(a + b + c\\) .", "solution": ": (James Xu) \n\nFirst substitute \\(x = p\\) , to get \\(p \\mid ap^{2} + bp + c\\) so \\(p \\mid c\\) . Therefore \\(c = p\\) . Next substitute \\(x = 1\\) , to get \\(p \\mid a + b + c\\) . Since \\(c = p\\) , this gives us \\(p \\mid a + b\\) . Finally substitute \\(x = p - 1\\) , to get \\((p - 1)^{2}a - (p - 1)b + c \\equiv 0\\) (mod \\(p\\) ). Thus \\(a - b + c \\equiv 0\\) (mod \\(p\\) ), and since \\(c = p\\) , we get \\(p \\mid a - b\\) . Adding and subtracting \\(p \\mid a + b\\) and \\(p \\mid a - b\\) we get \n\n\\[p \\mid (a + b) + (a - b) = 2a \\qquad \\text{and} \\qquad p \\mid (a + b) - (a - b) = 2b.\\] \n\nNow there are two cases: either \\(p = 2\\) or \\(p\\) is odd. \n\nCase \\(1 p = 2\\) \n\n2 \\(|a - b\\) means that either \\(a = b = 1\\) or \\(a = b = 2\\) , i.e. the only two possibilities for \\((a,b,c)\\) are \\((1,1,2)\\) and \\((2,2,2)\\) . So the only candidate polynomials are \n\n\\[x^{2} + x + 2\\qquad \\mathrm{and}\\qquad 2x^{2} + 2x + 2.\\] \n\nNow we check that these work. If \\(f(x) = x^{2} + x + 2 = 2 + x(x + 1)\\) which is always even because both 2 and \\(x(x + 1)\\) are even for all integers \\(x\\) . If \\(f(x) = 2x^{2} + 2x + 2 = 2(x^{2} + x + 1)\\) which is always even because it has a factor of 2. \n\nCase \\(2 p\\) is an odd prime: \n\nSince \\((2,p) = 1\\) , \\(p \\mid a\\) so \\(a = p\\) . Also note that \\(p \\mid 2b\\) by subtracting \\(a - b\\) from \\(a + b\\) , so by a similar argument, \\(b = p\\) . Thus, \\(a + b + c = p + p + p = 3p\\) \n\nCombining the two cases, \\(a + b + c = 3p\\) for all \\(p\\) and \\(a + b + c = 4\\) when \\(p = 2\\) are the only possible values of \\(a + b + c\\) .", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2023_solutions.jsonl", "problem_match": "\n4. Problem:", "solution_match": "\nSolution"}}
{"year": "2023", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "NewZealand_MO", "problem": "Find all triples \\((a,b,n)\\) of positive integers such that \\(a\\) and \\(b\\) are both divisors of \\(n\\) , and \\(a + b = \\frac{n}{2}\\) .", "solution": ": (James Xu) \n\nSince \\(a\\) and \\(b\\) are both factors of \\(n\\) , we can find positive integers \\(x\\) and \\(y\\) such that \\(a = \\frac{n}{x}\\) and \\(b = \\frac{n}{y}\\) . Then \\(\\frac{n}{x} +\\frac{n}{y} = \\frac{n}{2}\\) so \n\n\\[\\frac{1}{x} +\\frac{1}{y} = \\frac{1}{2}.\\] \n\nWithout loss of generality assume \\(a\\leqslant b\\) . So \\(x\\geqslant y\\) and \\(\\frac{1}{x}\\leqslant \\frac{1}{y}\\) and thus \n\n\\[\\frac{2}{y}\\geqslant \\frac{1}{x} +\\frac{1}{y} = \\frac{1}{2}.\\] \n\nTherefore \\(y\\leqslant 4\\) . So we try \\(y = 1,2,3,4\\) one- by- one. \n\nCase 1 \\(y = 1\\) : \n\nThis implies \\(\\frac{1}{x} = \\frac{1}{2} - \\frac{1}{y} = 0\\) thus \\(x = - 2\\) . This doesn't work because \\(a\\) (and therefore \\(x\\) ) must be positive. \n\nCase 2 \\(y = 2\\) : \n\nThis implies \\(\\frac{1}{x} = \\frac{1}{2} - \\frac{1}{y} = 0\\) which is not possible. \n\nCase 3 \\(y = 3\\) : \n\nThis implies \\(\\frac{1}{x} = \\frac{1}{2} - \\frac{1}{y} = \\frac{1}{6}\\) and thus \\(x = 6\\) . This means that \\(n\\) must be a multiple of 6. Now let \\(n = 6k\\) , and we get \n\n\\[a = \\frac{n}{x} = \\frac{6k}{6} = k\\qquad \\mathrm{and}\\qquad b = \\frac{n}{y} = \\frac{6k}{3} = 2k.\\] \n\nThis yields the family of solutions \\((a,b,n) = (k,2k,6k)\\) where \\(k\\) is any positive integer. \n\nCase 4 \\(y = 4\\) : \n\nThis implies \\(\\frac{1}{x} = \\frac{1}{2} - \\frac{1}{y} = \\frac{1}{4}\\) and thus \\(x = 4\\) . This means that \\(n\\) must be a multiple of 4. Now let \\(n = 4k\\) , and we get \n\n\\[a = \\frac{n}{x} = \\frac{4k}{4} = k\\qquad \\mathrm{and}\\qquad b = \\frac{n}{y} = \\frac{4k}{4} = k.\\] \n\nThis yields the family of solutions \\((a,b,n) = (k,k,4k)\\) where \\(k\\) is any positive integer. \n\nRemember that we assumed \\(a\\leqslant b\\) so we still need to swap the roles of \\(a\\) and \\(b\\) for our final answer. So in summary, the triples \\((a,b,n)\\) which satisfy the given equation are \n\n\\[(k,2k,6k),(2k,k,6k)\\mathrm{~and~}(k,k,4k)\\] \n\nwhere \\(k\\) is any positive integer.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2023_solutions.jsonl", "problem_match": "\n5. Problem:", "solution_match": "\nSolution"}}
-{"year": "2023", "tier": "T1", "problem_label": "6", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let triangle \\(ABC\\) be right-angled at \\(A\\) . Let \\(D\\) be the point on \\(AC\\) such that \\(BD\\) bisects angle \\(\\angle ABC\\) . Prove that \\(BC - BD = 2AB\\) if and only if \\(\\frac{1}{BD} - \\frac{1}{BC} = \\frac{1}{2AB}\\) .", "solution": ": (Kevin Shen) \n\nWlog let \\(AB = 1\\) and \\(BC = a\\) . Also let \\(BD = x\\) . We will try to find all the lengths in the diagram in terms of \\(a\\) . \n\n\n \n\nBy Pythagoras in \\(\\triangle ABC\\) we get \\(AC = \\sqrt{a^2 - 1}\\) . By the angle- bisector theorem we get \\(\\frac{AD}{DC} = \\frac{AB}{BC} = \\frac{1}{a}\\) , and so \\(DC = a \\times AD\\) . This can be substituted into \\(AD + DC = AC = \\sqrt{a^2 - 1}\\) to get \\(AD(1 + a) = \\sqrt{a^2 - 1}\\) . Therefore \n\n\\[AD = \\frac{AC}{1 + a} = \\frac{\\sqrt{a^2 - 1}}{a + 1}.\\] \n\nHence \\(AD^2 = \\frac{a^2 - 1}{(a + 1)^2} = \\frac{a - 1}{a + 1}\\) . Now consider Pythagoras in triangle \\(\\triangle BAD\\) . \n\n\\[x^{2} = 1^{2} + AD^{2} = 1 + \\frac{a - 1}{a + 1} = \\frac{2a}{a + 1}.\\] \n\nNow we do the two directions of the 'if and only if' separately: \n\nFirst, assuming \\(\\frac{1}{x} - \\frac{1}{a} = \\frac{1}{2}\\) we get \\(x = \\frac{2a}{a + 2}\\) . Hence \n\n\\[\\frac{2a}{a + 1} = x^2 = \\left(\\frac{2a}{a + 2}\\right)^2 = \\frac{4a^2}{a^2 + 4a + 4}\\] \n\n\\[a^{2} + 4a + 4 = 2a(a + 1)\\] \n\n\\[0 = a^{2} - 2a - 4\\] \n\nSo by the quadratic formula we get \\(a = 1 \\pm \\sqrt{5}\\) but since \\(a > 0\\) we must have \\(a = 1 + \\sqrt{5}\\) . \n\nTherefore \\(x = \\frac{2a}{a + 2} = \\frac{2(1 + \\sqrt{5})}{(1 + \\sqrt{5}) + 2} = \\sqrt{5} - 1\\) . \n\n\\[\\therefore a - x = (1 + \\sqrt{5}) - (\\sqrt{5} -1) = 2.\\] \n\nSo \\(\\frac{1}{x} - \\frac{1}{a} = \\frac{1}{2}\\) implies \\(a - x = 2\\) . \n\nSecond, assuming \\(a - x = 2\\) we get \\(x = a - 2\\) . Hence \n\n\\[\\frac{2a}{a + 1} = x^2 = (a - 2)^2 = a^2 -4a + 4\\] \n\n\\[2a = (a^2 -4a + 4)(a + 1)\\] \n\n\\[0 = a^3 -3a^2 -2a + 4\\] \n\n\\[0 = (a - 1)(a^2 -2a - 4)\\] \n\nSince \\(a - x = 2\\) we have \\(a > 2\\) so \\((a - 1) \\neq 0\\) and thus \\(a^2 - 2a - 4 = 0\\) . This gives us \\(a = 1 + \\sqrt{5}\\) and so again \\(x = a - 2 = \\sqrt{5} - 1\\) . \n\n\\[\\therefore \\frac{1}{x} - \\frac{1}{a} = \\frac{1}{\\sqrt{5} - 1} - \\frac{1}{\\sqrt{5} + 1} = \\frac{1}{2}.\\] \n\nSo \\(a - x = 2\\) implies \\(\\frac{1}{x} - \\frac{1}{a} = \\frac{1}{2}\\) . \n\nIn summary \\(\\frac{1}{x} - \\frac{1}{a} = \\frac{1}{2}\\) is equivalent to \\(a - x = 2\\) .\n\n\n\n**Solution B:** (Celine Yuan) \n\nLet \\(E\\) be the foot of the perpendicular from \\(D\\) to \\(BC\\) . Note that \\(E\\) is the reflection of \\(A\\) about \\(BD\\) (because \\(BD\\) is an angle bisector) and so we have congruent triangles \\(\\triangle ABD \\equiv \\triangle EBD\\) . Also construct point \\(N\\) on side \\(BC\\) to be the reflection of \\(B\\) about line \\(DE\\) . So we have three congruent triangles: \n\n\\[\\triangle ABD\\equiv \\triangle EBD\\equiv \\triangle END.\\] \n\nTherefore \\(AB = EB = EN\\) and hence \\(2AB = BE + EN = BN\\) . We also get \\(BD = DN\\) . Let \\(F\\) be the point on line \\(BC\\) such \\(BD = BF\\) . \n\n\n \n\nAssume \\(BC - BD = 2AB\\) . Since \\(BD = DN\\) and \\(2AB = BN\\) we can get \\(BC - DN = BN\\) . Therefore \n\n\\[DN = BC - BN = CN\\] \n\nand so triangle \\(DNC\\) is isosceles. Now let \\(\\theta = \\angle BCA\\) . Since \\(\\triangle DNC\\) is isosceles, we get \\(\\angle CDN = \\theta\\) and thus \\(\\angle END = 2\\theta\\) . Therefore \\(\\angle EBD = 2\\theta\\) and \\(\\angle DBA = 2\\theta\\) and thus \\(\\angle CBA = 4\\theta\\) . Thus \n\n\\[90^{\\circ} = \\angle BCA + \\angle CBA = \\theta +4\\theta .\\] \n\nHence \\(\\theta = 18^{\\circ}\\) . Therefore \n\n\\[\\frac{AB}{BD} = \\cos (\\angle ABD) = \\cos (2\\theta) = \\cos (36^{\\circ}) = \\frac{1 + \\sqrt{5}}{4}\\] \n\n\\[\\mathrm{and}\\quad \\frac{BC}{BD} = \\frac{\\sin(\\angle CDB)}{\\sin(\\angle BCD)} = \\frac{\\sin(126^{\\circ})}{\\sin(\\angle 18^{\\circ})} = \\frac{\\sqrt{5} + 1}{\\sqrt{5} - 1}\\] \n\nTherefore \\(1 - \\frac{BD}{BC} = 1 - \\frac{\\sqrt{5} - 1}{\\sqrt{5} + 1} = \\frac{2}{\\sqrt{5} + 1} = \\frac{BD}{2AB}\\) . Hence \n\n\\[\\frac{1}{BD} -\\frac{1}{BC} = \\frac{1}{2AB}\\] \n\nas required. \n\n- If \\(BC - BD > 2AB\\) then using similar logic we can get \\(\\angle BCA < 18^{\\circ}\\) which implies \n\n\\[\\frac{1}{BD} -\\frac{1}{BC} < \\frac{1}{2AB}\\] \n\n- Conversely if \\(BC - BD < 2AB\\) then we get \\(\\angle BCA > 18^{\\circ}\\) which implies \n\n\\[\\frac{1}{BD} -\\frac{1}{BC} >\\frac{1}{2AB}\\] \n\nTherefore we conclude that \\(BC - BD > 2AB\\) if and only if \\(\\frac{1}{BD} - \\frac{1}{BC} < \\frac{1}{2AB}\\) .", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2023_solutions.jsonl", "problem_match": "\n6. Problem:", "solution_match": "\nSolution"}}
+{"year": "2023", "tier": "T1", "problem_label": "6", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let triangle \\(ABC\\) be right-angled at \\(A\\) . Let \\(D\\) be the point on \\(AC\\) such that \\(BD\\) bisects angle \\(\\angle ABC\\) . Prove that \\(BC - BD = 2AB\\) if and only if \\(\\frac{1}{BD} - \\frac{1}{BC} = \\frac{1}{2AB}\\) .", "solution": ": (Kevin Shen) \n\nWlog let \\(AB = 1\\) and \\(BC = a\\) . Also let \\(BD = x\\) . We will try to find all the lengths in the diagram in terms of \\(a\\) . \n\n\n \n\nBy Pythagoras in \\(\\triangle ABC\\) we get \\(AC = \\sqrt{a^2 - 1}\\) . By the angle- bisector theorem we get \\(\\frac{AD}{DC} = \\frac{AB}{BC} = \\frac{1}{a}\\) , and so \\(DC = a \\times AD\\) . This can be substituted into \\(AD + DC = AC = \\sqrt{a^2 - 1}\\) to get \\(AD(1 + a) = \\sqrt{a^2 - 1}\\) . Therefore \n\n\\[AD = \\frac{AC}{1 + a} = \\frac{\\sqrt{a^2 - 1}}{a + 1}.\\] \n\nHence \\(AD^2 = \\frac{a^2 - 1}{(a + 1)^2} = \\frac{a - 1}{a + 1}\\) . Now consider Pythagoras in triangle \\(\\triangle BAD\\) . \n\n\\[x^{2} = 1^{2} + AD^{2} = 1 + \\frac{a - 1}{a + 1} = \\frac{2a}{a + 1}.\\] \n\nNow we do the two directions of the 'if and only if' separately: \n\nFirst, assuming \\(\\frac{1}{x} - \\frac{1}{a} = \\frac{1}{2}\\) we get \\(x = \\frac{2a}{a + 2}\\) . Hence \n\n\\[\\frac{2a}{a + 1} = x^2 = \\left(\\frac{2a}{a + 2}\\right)^2 = \\frac{4a^2}{a^2 + 4a + 4}\\] \n\n\\[a^{2} + 4a + 4 = 2a(a + 1)\\] \n\n\\[0 = a^{2} - 2a - 4\\] \n\nSo by the quadratic formula we get \\(a = 1 \\pm \\sqrt{5}\\) but since \\(a > 0\\) we must have \\(a = 1 + \\sqrt{5}\\) . \n\nTherefore \\(x = \\frac{2a}{a + 2} = \\frac{2(1 + \\sqrt{5})}{(1 + \\sqrt{5}) + 2} = \\sqrt{5} - 1\\) . \n\n\\[\\therefore a - x = (1 + \\sqrt{5}) - (\\sqrt{5} -1) = 2.\\] \n\nSo \\(\\frac{1}{x} - \\frac{1}{a} = \\frac{1}{2}\\) implies \\(a - x = 2\\) . \n\nSecond, assuming \\(a - x = 2\\) we get \\(x = a - 2\\) . Hence \n\n\\[\\frac{2a}{a + 1} = x^2 = (a - 2)^2 = a^2 -4a + 4\\] \n\n\\[2a = (a^2 -4a + 4)(a + 1)\\] \n\n\\[0 = a^3 -3a^2 -2a + 4\\] \n\n\\[0 = (a - 1)(a^2 -2a - 4)\\] \n\nSince \\(a - x = 2\\) we have \\(a > 2\\) so \\((a - 1) \\neq 0\\) and thus \\(a^2 - 2a - 4 = 0\\) . This gives us \\(a = 1 + \\sqrt{5}\\) and so again \\(x = a - 2 = \\sqrt{5} - 1\\) . \n\n\\[\\therefore \\frac{1}{x} - \\frac{1}{a} = \\frac{1}{\\sqrt{5} - 1} - \\frac{1}{\\sqrt{5} + 1} = \\frac{1}{2}.\\] \n\nSo \\(a - x = 2\\) implies \\(\\frac{1}{x} - \\frac{1}{a} = \\frac{1}{2}\\) . \n\nIn summary \\(\\frac{1}{x} - \\frac{1}{a} = \\frac{1}{2}\\) is equivalent to \\(a - x = 2\\) .\n\n\n\n**Solution B:** (Celine Yuan) \n\nLet \\(E\\) be the foot of the perpendicular from \\(D\\) to \\(BC\\) . Note that \\(E\\) is the reflection of \\(A\\) about \\(BD\\) (because \\(BD\\) is an angle bisector) and so we have congruent triangles \\(\\triangle ABD \\equiv \\triangle EBD\\) . Also construct point \\(N\\) on side \\(BC\\) to be the reflection of \\(B\\) about line \\(DE\\) . So we have three congruent triangles: \n\n\\[\\triangle ABD\\equiv \\triangle EBD\\equiv \\triangle END.\\] \n\nTherefore \\(AB = EB = EN\\) and hence \\(2AB = BE + EN = BN\\) . We also get \\(BD = DN\\) . Let \\(F\\) be the point on line \\(BC\\) such \\(BD = BF\\) . \n\n\n \n\nAssume \\(BC - BD = 2AB\\) . Since \\(BD = DN\\) and \\(2AB = BN\\) we can get \\(BC - DN = BN\\) . Therefore \n\n\\[DN = BC - BN = CN\\] \n\nand so triangle \\(DNC\\) is isosceles. Now let \\(\\theta = \\angle BCA\\) . Since \\(\\triangle DNC\\) is isosceles, we get \\(\\angle CDN = \\theta\\) and thus \\(\\angle END = 2\\theta\\) . Therefore \\(\\angle EBD = 2\\theta\\) and \\(\\angle DBA = 2\\theta\\) and thus \\(\\angle CBA = 4\\theta\\) . Thus \n\n\\[90^{\\circ} = \\angle BCA + \\angle CBA = \\theta +4\\theta .\\] \n\nHence \\(\\theta = 18^{\\circ}\\) . Therefore \n\n\\[\\frac{AB}{BD} = \\cos (\\angle ABD) = \\cos (2\\theta) = \\cos (36^{\\circ}) = \\frac{1 + \\sqrt{5}}{4}\\] \n\n\\[\\mathrm{and}\\quad \\frac{BC}{BD} = \\frac{\\sin(\\angle CDB)}{\\sin(\\angle BCD)} = \\frac{\\sin(126^{\\circ})}{\\sin(\\angle 18^{\\circ})} = \\frac{\\sqrt{5} + 1}{\\sqrt{5} - 1}\\] \n\nTherefore \\(1 - \\frac{BD}{BC} = 1 - \\frac{\\sqrt{5} - 1}{\\sqrt{5} + 1} = \\frac{2}{\\sqrt{5} + 1} = \\frac{BD}{2AB}\\) . Hence \n\n\\[\\frac{1}{BD} -\\frac{1}{BC} = \\frac{1}{2AB}\\] \n\nas required. \n\n- If \\(BC - BD > 2AB\\) then using similar logic we can get \\(\\angle BCA < 18^{\\circ}\\) which implies \n\n\\[\\frac{1}{BD} -\\frac{1}{BC} < \\frac{1}{2AB}\\] \n\n- Conversely if \\(BC - BD < 2AB\\) then we get \\(\\angle BCA > 18^{\\circ}\\) which implies \n\n\\[\\frac{1}{BD} -\\frac{1}{BC} >\\frac{1}{2AB}\\] \n\nTherefore we conclude that \\(BC - BD > 2AB\\) if and only if \\(\\frac{1}{BD} - \\frac{1}{BC} < \\frac{1}{2AB}\\) .", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2023_solutions.jsonl", "problem_match": "\n6. Problem:", "solution_match": "\nSolution"}}
{"year": "2023", "tier": "T1", "problem_label": "7", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(n,m\\) be positive integers. Let \\(A_{1},A_{2},A_{3},\\ldots ,A_{m}\\) be sets such that \\(A_{i}\\subseteq \\{1,2,3,\\ldots ,n\\}\\) and \\(|A_{i}| = 3\\) for all \\(i\\) (i.e. \\(A_{i}\\) consists of three different positive integers each at most \\(n\\) ). Suppose for all \\(i< j\\) we have \n\n\\[|A_{i}\\cap A_{j}|\\leqslant 1\\] \n\n(i.e. \\(A_{i}\\) and \\(A_{j}\\) have at most one element in common). \n\n(a) Prove that \\(m\\leqslant \\frac{n(n - 1)}{6}\\) . \n\n(b) Show that for all \\(n\\geq 3\\) it is possible to have \\(m\\geqslant \\frac{(n - 1)(n - 2)}{6}\\) .", "solution": ": (Ishan Nath) \n\nEach set \\(A_{i}\\) has exactly three pairs of elements. But each unordered pair chosen from \\(\\{1,2,\\ldots ,n\\}\\) can be in at most one such set. Therefore \n\n\\[{\\binom{n}{2}}\\geq3m.\\] \n\nThis establishes the required upper bound on the size of \\(m\\) . \n\nLet \\(T\\) be the set of all triples, i.e. \n\n\\[T = \\{(a,b,c)\\mid 1\\leq a< b< c\\leq n\\} .\\] \n\nWe now partition \\(T\\) into \\(n\\) parts, \\(T = T_{0}\\cup T_{1}\\cup T_{2}\\cup \\dots \\cup T_{n - 1}\\) based on the residue of \\(a + b + c\\) modulo \\(n\\) , i.e. \n\n\\[T_{i} = \\{(a,b,c)\\mid (a,b,c)\\in T\\mathrm{~and~}a + b + c\\equiv i\\mathrm{~(mod~}n)\\}\\] \n\nSo we have \\(\\binom{n}{3}\\) triples (our pigeons) and \\(n\\) pigeonholes (the parts \\(T_{0},T_{1},T_{2},\\ldots ,T_{n - 1}\\) ) therefore by the pigeonhole principle at least one part \\(T_{k}\\) must have at least \n\n\\[\\frac{1}{n}\\times \\binom{n}{3}\\] \n\ntriples in it. Now it suffices to show that \\(T_{k}\\) satisfies the problem. For the sake of contradiction, assume there exists two triples \\((a,b,c),(a^{\\prime},b^{\\prime},c^{\\prime})\\in T_{k}\\) such that \\(a = a^{\\prime}\\) and \\(b = b^{\\prime}\\) but \\(c\\neq c^{\\prime}\\) . This would imply \n\n\\[a + b + c\\equiv j\\equiv a^{\\prime} + b^{\\prime} + c^{\\prime}\\pmod {n}.\\] \n\nHence \\(c\\equiv c^{\\prime}\\) (mod \\(n\\) ) and thus \\(c = c^{\\prime}\\) and so we get \\((a,b,c) = (a^{\\prime},b^{\\prime},c^{\\prime})\\) , contradiction. Therefore no such pair \\((a,b,c),(a^{\\prime},b^{\\prime},c^{\\prime})\\in T_{k}\\) exists. Therefore, letting the elements of \\(T_{k}\\) be our sets \\(A_{i}\\) , we achieve \n\n\\[m\\geq \\frac{1}{n}\\times \\binom{n}{3} = \\frac{(n - 1)(n - 2)}{6}\\] \n\nas required.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2023_solutions.jsonl", "problem_match": "\n7. Problem:", "solution_match": "\nSolution"}}
{"year": "2023", "tier": "T1", "problem_label": "8", "problem_type": null, "exam": "NewZealand_MO", "problem": "Find all non-zero real numbers \\(a, b, c\\) such that the following polynomial has four (not necessarily distinct) positive real roots. \n\n\\[P(x) = ax^{4} - 8ax^{3} + bx^{2} - 32cx + 16c\\]", "solution": ": (Viet Hoang) \n\nAssume that \\(P(x)\\) has 4 positive real roots \\(x_{1}, x_{2}, x_{3}\\) and \\(x_{4}\\) . Using Viete's theorem, one can obtain the following equations \n\n\\[\\begin{array}{c}{x_{1} + x_{2} + x_{3} + x_{4} = \\frac{8a}{a} = 8}\\\\ {x_{1}x_{2} + x_{1}x_{3} + x_{1}x_{4} + x_{2}x_{3} + x_{2}x_{4} + x_{3}x_{4} = \\frac{b}{a}}\\\\ {x_{1}x_{2}x_{3} + x_{1}x_{2}x_{4} + x_{1}x_{3}x_{4} + x_{2}x_{3}x_{4} = \\frac{32c}{a}}\\\\ {x_{1}x_{2}x_{3}x_{4} = \\frac{16c}{a}} \\end{array} \\quad (5)\\] \n\nFrom 4 and 5, we have \n\n\\[\\frac{1}{x_{1}} +\\frac{1}{x_{2}} +\\frac{1}{x_{3}} +\\frac{1}{x_{4}} = \\frac{x_{1}x_{2}x_{3} + x_{1}x_{2}x_{4} + x_{1}x_{3}x_{4} + x_{2}x_{3}x_{4}}{x_{1}x_{2}x_{3}x_{4}i} = -\\frac{\\frac{32c}{a}}{\\frac{16c}{a}} = 2 \\quad (6)\\] \n\nUsing AM- GM (arithmetic mean geometric mean inequality) on the sums in equation (2) and (6) we get \n\n\\[\\frac{8}{4} = \\frac{x_{1} + x_{2} + x_{3} + x_{4}}{4}\\geqslant \\sqrt[4]{x_{1}\\cdot x_{2}\\cdot x_{3}\\cdot x_{4}}\\] \\[\\frac{2}{4} = \\frac{\\frac{1}{x_{1}} + \\frac{1}{x_{2}} + \\frac{1}{x_{3}} + \\frac{1}{x_{4}}}{4}\\geqslant \\sqrt[4]{\\frac{1}{x_{1}}\\cdot\\frac{1}{x_{2}}\\cdot\\frac{1}{x_{3}}\\cdot\\frac{1}{x_{4}}}\\] \n\nMultiplying these together yields \\(1\\geqslant 1\\) and so equality must occur in both AM- GM's. Therefore \\(x_{1} = x_{2} = x_{3} = x_{4}\\) . Furthermore for Equation (2) we get \\(x_{1} = x_{2} = x_{3} = x_{4} = 2\\) . Substituting this into equation (3) gives us \\(b = 24a\\) . Substituting this into equation (5) gives us \\(c = a\\) . So \\((a, b, c) = (a, 24a, a)\\) where \\(a\\) is any non- zero real number, and the polynomial \\(P(x)\\) is \n\n\\[P(x) = a(x - x_{1})(x - x_{2})(x - x_{3})(x - x_{4}) = a(x - 2)^{4}.\\]", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2023_solutions.jsonl", "problem_match": "\n8. Problem:", "solution_match": "\nSolution"}}
diff --git a/NewZealand_MO/segmented/en-nzmo1_2024_solutions.jsonl b/NewZealand_MO/segmented/en-nzmo1_2024_solutions.jsonl
index 6b3dc20fb474557af47df2673a1786bbd54510f4..1426c68f01174436fe4ec00adbe7cfa3c715ca50 100644
--- a/NewZealand_MO/segmented/en-nzmo1_2024_solutions.jsonl
+++ b/NewZealand_MO/segmented/en-nzmo1_2024_solutions.jsonl
@@ -1,9 +1,9 @@
{"year": "2024", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "NewZealand_MO", "problem": "Josie and Kevin are each thinking of a two digit positive integer. Josie's number is twice as big as Kevin's. One digit of Kevin's number is equal to the sum of digits of Josie's number. The other digit of Kevin's number is equal to the difference between the digits of Josie's number. What is the sum of Kevin and Josie's numbers?", "solution": ": (James Xu) \n\nWe'll use \\(\\overline{AB}\\) to denote a 2 digit number with \\(A\\) in the tens digit and \\(B\\) in the unit digit. \n\nLet Josie pick the number \\(\\overline{AB}\\) and Kevin \\(\\overline{CD}\\) . Then we have \n\n\\[\\overline{AB} = 2\\times \\overline{CD}\\] \n\n\\[10A + B = 20C + 2D\\] \n\nNow, \\(A \\geq 2C > C\\) so \\(C \\neq A + B\\) . Therefore, \\(C = |A - B|\\) and \\(D = A + B\\) . \n\n- Case 1: \\(A \\geq B\\) . So \\(C = |A - B| = A - B\\) . This yields \\(10A + B = 20C + 2D = 20(A - B) + 2(A + B)\\) . Which simplifies to give \n\n\\[19B = 12A.\\] \n\nThis can only happen when \\(A\\) is a multiple of 19 which is impossible since \\(A > 0\\) is a digit. \n\n- Case 2: \\(A < B\\) . \\(C = |A - B| = B - A\\) . This yields \\(10A + B = 20C + 2D = 20(B - A) + 2(A + B)\\) . Which simplifies to give \n\n\\[4A = 3B.\\] \n\nSince \\(A\\) and \\(B\\) are digits (and \\(A\\) is nonzero), this means either: \\((A, B) = (3, 4)\\) or \\((A, B) = (6, 8)\\) . \n\nTo verify: \\(\\overline{AB} = 34 \\Rightarrow \\overline{CD} = 17\\) which works, while \\(\\overline{AB} = 68 \\Rightarrow \\overline{CD} = 34\\) does not work. Therefore the final answer is \\(34 + 17 = 51\\) .", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2024_solutions.jsonl", "problem_match": "\n1. Problem:", "solution_match": "\nSolution"}}
{"year": "2024", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "NewZealand_MO", "problem": "Prove the following inequality \n\n\\[\\frac{6}{2024^{3}} < \\left(1 - \\frac{3}{4}\\right)\\left(1 - \\frac{3}{5}\\right)\\left(1 - \\frac{3}{6}\\right)\\left(1 - \\frac{3}{7}\\right)\\dots \\left(1 - \\frac{3}{2025}\\right).\\]", "solution": ": (Eric Liang) \n\n\\[\\frac{1}{4}\\times \\frac{2}{5}\\times \\frac{3}{6}\\times \\frac{4}{7}\\times \\frac{5}{8}\\times \\frac{6}{9}\\times \\frac{7}{10}\\times \\dots \\times \\frac{2022}{2025}\\] \\[= \\frac{1\\times 2\\times 3\\times 4\\times 5\\times 6\\times 7\\times \\dots \\times 2022}{4\\times 5\\times 6\\times 7\\times 8\\times 9\\times 10\\times \\dots \\times 2025}\\] \\[= \\frac{1\\times 2\\times 3}{2023\\times 2024\\times 2025}\\] \\[= \\frac{6}{2024(2024 - 1)(2024 + 1)}\\] \\[= \\frac{6}{2024(2024^{2} - 1)}\\] \\[> \\frac{6}{2024(2024^{2})}\\] \\[= \\frac{6}{2024^{3}}.\\]", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2024_solutions.jsonl", "problem_match": "\n2. Problem:", "solution_match": "\nSolution"}}
-{"year": "2024", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "NewZealand_MO", "problem": "A rectangular sheet of paper is folded so that one corner lies on top of the corner diagonally opposite. The resulting shape is a pentagon whose area is \\(20\\%\\) one-sheet-thick, and \\(80\\%\\) two-sheets-thick. Determine the ratio of the two sides of the original sheet of paper.", "solution": ": (Ross Atkins) \n\nLet the original rectangle be \\(ABCD\\) . Let the fold line be \\(XY\\) with point \\(X\\) lying on side \\(AB\\) and point \\(Y\\) lying on side \\(CD\\) . Consider the fold such that point \\(D\\) lands on point \\(B\\) . Let \\(P\\) be the location of corner \\(A\\) after the fold. \n\n\n \n\nThe two-sheets-thick area is \\(\\triangle BXY\\) which should be \\(80\\%\\) . By symmetry, triangles \\(\\triangle PXB\\) and \\(\\triangle CYB\\) are congruent with equal hypotenuses \n\n\\[B X = B Y.\\] \n\nTriangles \\(\\triangle PXB\\) and \\(\\triangle CYB\\) have equal area and their total area is \\(20\\%\\) of rectangle \\(ABCD\\) . Therefore \\(\\triangle BYC\\) has area \\(10\\%\\) . Hence the ratio of the areas \\(|BYC|:|BXY|\\) is \\(10\\% :80\\%\\) . Both triangles have height \\(BC\\) equal to the height of rectangle \\(ABCD\\) . Therefore their bases are also in the ratio \\(1:8\\) . i.e. \\(XB = 8YC\\) . \n\n\\[\\therefore BY = XB = 8YC.\\] \n\nNow we apply Pythagoras' to \\(\\triangle BCY\\) to get \n\n\\[BC^{2} = BY^{2} - CY^{2} = (8CY)^{2} - CY^{2} = 63CY^{2}.\\] \n\nTherefore \\(BC = \\sqrt{63} CY\\) . \n\nCombining this with \\(CD = CY + YD = CY + YB = 9CY\\) we get \n\n\\[BC:CD = \\sqrt{7}:3.\\]", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2024_solutions.jsonl", "problem_match": "\n3. Problem:", "solution_match": "\nSolution"}}
-{"year": "2024", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "NewZealand_MO", "problem": "A dot-trapezium consists of several rows of dots such that each row contains one more dot than the row immediately above (apart from the top row). For example here is a dot-trapezium consisting of 15 dots, having 3 rows and 4 dot in the top row. \n\n\n \n\nA positive integer \\(n\\) is called a trapezium- number if there exists a dot- trapezium consisting of exactly \\(n\\) dots, with at least two rows and at least two dots in the top row. How many trapezium- numbers are there less than 100?", "solution": "A: (Chris Tuffley) \n\nLet \\(n\\) be a trapezium number and suppose there are \\(a\\) dots in the first row and \\(b\\) dots in the last row. So the required conditions are \\(a \\geq 2\\) and \\(b \\geq a + 1\\) . Then the equation becomes: \n\n\\[2n = b(b + 1) - a(a - 1) = b^{2} - a^{2} + b + a = (a + b)(b - a + 1)\\] \n\nbecause it is the difference between two triangle numbers. Let the two factors on the RHS be \\(x = (a + b)\\) and \\(y = (b - a + 1)\\) . Rearranging gives us \n\n\\[a = \\frac{x - y + 1}{2}\\qquad \\mathrm{and}\\qquad b = \\frac{x + y - 1}{2}.\\] \n\nIn order for \\(a\\) and \\(b\\) to be integers we must have \\(x\\) and \\(y\\) being opposite parity. So we are looking for factorizations of the form \\(2n = xy\\) such that one of \\(x\\) and \\(y\\) is even while the other is odd. We also need \\(a < b\\) (so that the trapezium has at least two rows) which is equivalent to \\(y \\geq 2\\) . Finally we also need \\(a \\geq 2\\) (so there are at least two dots in the top row) which is equivalent to \\(x \\geq y + 3\\) . We need \n\nNow write \\(n = 2^{k}m\\) with \\(m\\) odd. \n\n- If \\(n = 2^{k}\\) is a power of 2, then the only factorization of \\(2n = 2^{k + 1} = xy\\) such that both \\(x\\) and \\(y\\) have opposite parity is \\(x = 2^{k + 1}\\) and \\(y = 1\\) . This doesn't work because we require \\(y > 1\\) . (when \\(y = 1\\) the \"trapezium\" would consist of only one row) \n\n- If \\(m > 2^{k + 1} + 1\\) or \\(1 < m < 2^{k + 1} - 1\\) then we can choose \n\n\\[x = \\max \\{2^{k + 1},m\\} \\quad \\mathrm{and}\\quad y = \\min \\{2^{k + 1},m\\} .\\] \n\nTo check that this works we simply check that \\(y > 1\\) and \\(x \\geq y + 3\\) . Note that this still works even when \\(k = 0\\) . \n\n- If \\(m = 2^{k + 1} \\pm 1\\) and \\(m\\) is prime then the only factorization of \\(2n = 2^{k + 1}m = xy\\) such that both \\(x\\) and \\(y\\) have opposite parity and \\(x > y > 1\\) is \n\n\\[x = \\max \\{2^{k + 1},m\\} \\quad \\mathrm{and}\\quad y = \\min \\{2^{k + 1},m\\} .\\] \n\nHowever this doesn't work because we require \\(x > x + 1\\) . (when \\(y = 1\\) the \"trapezium\" would consist have one dot in the top row)\n\n\n\nThe only remaining possibility is when \\(m = 2^{k + 1} \\pm 1\\) and \\(m\\) is composite. If \\(k > 2\\) then \\(m = 2^{k + 1} \\pm 1 \\geq 15\\) , so \\(n = 2^{k} m \\geq 8 \\times 15 > 100\\) and we don’t need to consider it. For \\(k \\leq 2\\) we get \\(m = 1, 3, 5, 7, 9\\) . We can’t have \\(m = 1\\) because then \\(n\\) would be a power of 2. We also can’t have \\(m = 3, 5, 7\\) because they are prime. Finally we consider \\(m = 9\\) and so \\(k = 2\\) . In this case we get \\(n = 2^{k} m = 36\\) , which is a trapezium number as seen here: \n\n\n \n\nTherefore all non- trapezium- numbers less than 100, are the powers of two and numbers of the form \\(n = 2^{k}(2^{k + 1} \\pm 1)\\) where \\((2^{k + 1} \\pm 1)\\) is prime (and \\(k \\leq 2\\) ). The powers of two are: \\(\\{1, 2, 4, 8, 16, 32, 64\\}\\) . The numbers of the form \\(2^{k}(2^{k + 1} \\pm 1)\\) with \\(k \\leq 2\\) and \\((2^{k + 1} \\pm 1)\\) being prime are \n\n\\[\\{2^{0}(2^{1} + 1) = 3, 2^{1}(2^{2} - 1) = 6, 2^{1}(2^{2} + 1) = 10, 2^{2}(2^{3} - 1) = 28\\} .\\] \n\nAll together the non- trapezium numbers are \\(\\{1, 2, 3, 4, 6, 8, 10, 16, 28, 32, 64\\}\\) . There are 99 positive integers less than 100 and exactly 11 of them are non- trapezium numbers. So the final answer is \\(99 - 11 = 88\\) .\n\n\n\n**Solution B:** (David Starshaw) \n\nFirst, notice that any odd integer can be written as a sum of consecutive integers: \\(2n + 1 = (n) + (n + 1)\\) , e.g. \\(23 = 11 + 12\\) . \n\nThen observe that any multiple of 3 can be written as a sum of consecutive integers: \\(3n = (n - 1) + (n) + (n + 1)\\) , for example, \\(18 = 5 + 6 + 7\\) . \n\nSimilarly, any multiple of 5 can be written as a sum of consecutive integers: \n\n\\[5n = (n - 2) + (n - 1) + (n) + (n + 1) + (n + 2).\\] \n\nIn general, any number with an odd factor can be written as a sum of consecutive integers. The only numbers that have no odd factors (other than 1) are the powers of two. There are 7 such numbers less than 100: \\(\\{1,2,4,8,16,32,64\\}\\) . \n\nIn general, any number with an odd factor (greater than one) can be written as a sum of consecutive integers. Note that if the odd factor is too large then some of these terms are negative. For example our above method for 26 would produce \n\n\\[26 = (-4) + (-3) + (-2) + (-1) + 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8.\\] \n\nBecause the centre of the consecutive run is always positive, there will always be more positive terms than negative terms. Therefore each negative term can be 'cancelled' with it's corresponding positive term. At a minimum, the centre of the run must be at least 1, so there will always be at least two positive terms remaining even after cancelling. \n\nBut we are also not allowed to have one dot in the top row, i.e. our sum (after cancelling) cannot include 1. This would seem to exclude the triangular numbers: \n\n\\[\\{1,3,6,10,15,21,28,36,45,55,66,78,91\\} .\\] \n\nExcept some can be rewritten as a different sum of consecutive integers in more than one way. For example \\(15 = 1 + 2 + 3 + 4 + 5\\) , but also \\(15 = 7 + 8\\) . In general, if the triangular number, \\(T_{n} = \\frac{1}{2} n(n + 1)\\) , is either odd, or has an odd factor less than \\(n\\) , it can be rewritten as a sum of consecutive integers that do not include 1. The following triangular numbers can be salvaged: \n\n- 15 is odd, so \\(15 = 7 + 8\\) - 21 is odd, so \\(21 = 10 + 11\\) - 36 is a multiple of 3, so \\(36 = 11 + 12 + 13\\) - 45 is odd, so \\(45 = 22 + 23\\) - 55 is odd, so \\(55 = 27 + 28\\) - 66 is a multiple of 3, so \\(66 = 21 + 22 + 23\\) - 78 is a multiple of 3, so \\(78 = 25 + 26 + 27\\) - 91 is odd, so \\(91 = 45 + 46\\) \n\nSo there are 5 triangular numbers that are not possible: \\(\\{1,3,6,10,28\\}\\) . Combined with the powers of two that are also impossible: \\(\\{1,2,4,8,16,32,64\\}\\) , there are 11 numbers \\((5 + 7 = 12\\) but the number 1 is included in both lists) that can not be written as trapezium numbers: \n\n\\[\\{1,2,3,4,6,8,10,16,28,32,64\\}\\] \n\nThere are 99 numbers less than 100, so there are \\(99 - 11 = 88\\) trapezium numbers.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2024_solutions.jsonl", "problem_match": "\n4. Problem:", "solution_match": "\nSolution"}}
+{"year": "2024", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "NewZealand_MO", "problem": "A rectangular sheet of paper is folded so that one corner lies on top of the corner diagonally opposite. The resulting shape is a pentagon whose area is \\(20\\%\\) one-sheet-thick, and \\(80\\%\\) two-sheets-thick. Determine the ratio of the two sides of the original sheet of paper.", "solution": ": (Ross Atkins) \n\nLet the original rectangle be \\(ABCD\\) . Let the fold line be \\(XY\\) with point \\(X\\) lying on side \\(AB\\) and point \\(Y\\) lying on side \\(CD\\) . Consider the fold such that point \\(D\\) lands on point \\(B\\) . Let \\(P\\) be the location of corner \\(A\\) after the fold. \n\n\n \n\nThe two-sheets-thick area is \\(\\triangle BXY\\) which should be \\(80\\%\\) . By symmetry, triangles \\(\\triangle PXB\\) and \\(\\triangle CYB\\) are congruent with equal hypotenuses \n\n\\[B X = B Y.\\] \n\nTriangles \\(\\triangle PXB\\) and \\(\\triangle CYB\\) have equal area and their total area is \\(20\\%\\) of rectangle \\(ABCD\\) . Therefore \\(\\triangle BYC\\) has area \\(10\\%\\) . Hence the ratio of the areas \\(|BYC|:|BXY|\\) is \\(10\\% :80\\%\\) . Both triangles have height \\(BC\\) equal to the height of rectangle \\(ABCD\\) . Therefore their bases are also in the ratio \\(1:8\\) . i.e. \\(XB = 8YC\\) . \n\n\\[\\therefore BY = XB = 8YC.\\] \n\nNow we apply Pythagoras' to \\(\\triangle BCY\\) to get \n\n\\[BC^{2} = BY^{2} - CY^{2} = (8CY)^{2} - CY^{2} = 63CY^{2}.\\] \n\nTherefore \\(BC = \\sqrt{63} CY\\) . \n\nCombining this with \\(CD = CY + YD = CY + YB = 9CY\\) we get \n\n\\[BC:CD = \\sqrt{7}:3.\\]", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2024_solutions.jsonl", "problem_match": "\n3. Problem:", "solution_match": "\nSolution"}}
+{"year": "2024", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "NewZealand_MO", "problem": "A dot-trapezium consists of several rows of dots such that each row contains one more dot than the row immediately above (apart from the top row). For example here is a dot-trapezium consisting of 15 dots, having 3 rows and 4 dot in the top row. \n\n\n \n\nA positive integer \\(n\\) is called a trapezium- number if there exists a dot- trapezium consisting of exactly \\(n\\) dots, with at least two rows and at least two dots in the top row. How many trapezium- numbers are there less than 100?", "solution": "A: (Chris Tuffley) \n\nLet \\(n\\) be a trapezium number and suppose there are \\(a\\) dots in the first row and \\(b\\) dots in the last row. So the required conditions are \\(a \\geq 2\\) and \\(b \\geq a + 1\\) . Then the equation becomes: \n\n\\[2n = b(b + 1) - a(a - 1) = b^{2} - a^{2} + b + a = (a + b)(b - a + 1)\\] \n\nbecause it is the difference between two triangle numbers. Let the two factors on the RHS be \\(x = (a + b)\\) and \\(y = (b - a + 1)\\) . Rearranging gives us \n\n\\[a = \\frac{x - y + 1}{2}\\qquad \\mathrm{and}\\qquad b = \\frac{x + y - 1}{2}.\\] \n\nIn order for \\(a\\) and \\(b\\) to be integers we must have \\(x\\) and \\(y\\) being opposite parity. So we are looking for factorizations of the form \\(2n = xy\\) such that one of \\(x\\) and \\(y\\) is even while the other is odd. We also need \\(a < b\\) (so that the trapezium has at least two rows) which is equivalent to \\(y \\geq 2\\) . Finally we also need \\(a \\geq 2\\) (so there are at least two dots in the top row) which is equivalent to \\(x \\geq y + 3\\) . We need \n\nNow write \\(n = 2^{k}m\\) with \\(m\\) odd. \n\n- If \\(n = 2^{k}\\) is a power of 2, then the only factorization of \\(2n = 2^{k + 1} = xy\\) such that both \\(x\\) and \\(y\\) have opposite parity is \\(x = 2^{k + 1}\\) and \\(y = 1\\) . This doesn't work because we require \\(y > 1\\) . (when \\(y = 1\\) the \"trapezium\" would consist of only one row) \n\n- If \\(m > 2^{k + 1} + 1\\) or \\(1 < m < 2^{k + 1} - 1\\) then we can choose \n\n\\[x = \\max \\{2^{k + 1},m\\} \\quad \\mathrm{and}\\quad y = \\min \\{2^{k + 1},m\\} .\\] \n\nTo check that this works we simply check that \\(y > 1\\) and \\(x \\geq y + 3\\) . Note that this still works even when \\(k = 0\\) . \n\n- If \\(m = 2^{k + 1} \\pm 1\\) and \\(m\\) is prime then the only factorization of \\(2n = 2^{k + 1}m = xy\\) such that both \\(x\\) and \\(y\\) have opposite parity and \\(x > y > 1\\) is \n\n\\[x = \\max \\{2^{k + 1},m\\} \\quad \\mathrm{and}\\quad y = \\min \\{2^{k + 1},m\\} .\\] \n\nHowever this doesn't work because we require \\(x > x + 1\\) . (when \\(y = 1\\) the \"trapezium\" would consist have one dot in the top row)\n\n\n\nThe only remaining possibility is when \\(m = 2^{k + 1} \\pm 1\\) and \\(m\\) is composite. If \\(k > 2\\) then \\(m = 2^{k + 1} \\pm 1 \\geq 15\\) , so \\(n = 2^{k} m \\geq 8 \\times 15 > 100\\) and we don’t need to consider it. For \\(k \\leq 2\\) we get \\(m = 1, 3, 5, 7, 9\\) . We can’t have \\(m = 1\\) because then \\(n\\) would be a power of 2. We also can’t have \\(m = 3, 5, 7\\) because they are prime. Finally we consider \\(m = 9\\) and so \\(k = 2\\) . In this case we get \\(n = 2^{k} m = 36\\) , which is a trapezium number as seen here: \n\n\n \n\nTherefore all non- trapezium- numbers less than 100, are the powers of two and numbers of the form \\(n = 2^{k}(2^{k + 1} \\pm 1)\\) where \\((2^{k + 1} \\pm 1)\\) is prime (and \\(k \\leq 2\\) ). The powers of two are: \\(\\{1, 2, 4, 8, 16, 32, 64\\}\\) . The numbers of the form \\(2^{k}(2^{k + 1} \\pm 1)\\) with \\(k \\leq 2\\) and \\((2^{k + 1} \\pm 1)\\) being prime are \n\n\\[\\{2^{0}(2^{1} + 1) = 3, 2^{1}(2^{2} - 1) = 6, 2^{1}(2^{2} + 1) = 10, 2^{2}(2^{3} - 1) = 28\\} .\\] \n\nAll together the non- trapezium numbers are \\(\\{1, 2, 3, 4, 6, 8, 10, 16, 28, 32, 64\\}\\) . There are 99 positive integers less than 100 and exactly 11 of them are non- trapezium numbers. So the final answer is \\(99 - 11 = 88\\) .\n\n\n\n**Solution B:** (David Starshaw) \n\nFirst, notice that any odd integer can be written as a sum of consecutive integers: \\(2n + 1 = (n) + (n + 1)\\) , e.g. \\(23 = 11 + 12\\) . \n\nThen observe that any multiple of 3 can be written as a sum of consecutive integers: \\(3n = (n - 1) + (n) + (n + 1)\\) , for example, \\(18 = 5 + 6 + 7\\) . \n\nSimilarly, any multiple of 5 can be written as a sum of consecutive integers: \n\n\\[5n = (n - 2) + (n - 1) + (n) + (n + 1) + (n + 2).\\] \n\nIn general, any number with an odd factor can be written as a sum of consecutive integers. The only numbers that have no odd factors (other than 1) are the powers of two. There are 7 such numbers less than 100: \\(\\{1,2,4,8,16,32,64\\}\\) . \n\nIn general, any number with an odd factor (greater than one) can be written as a sum of consecutive integers. Note that if the odd factor is too large then some of these terms are negative. For example our above method for 26 would produce \n\n\\[26 = (-4) + (-3) + (-2) + (-1) + 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8.\\] \n\nBecause the centre of the consecutive run is always positive, there will always be more positive terms than negative terms. Therefore each negative term can be 'cancelled' with it's corresponding positive term. At a minimum, the centre of the run must be at least 1, so there will always be at least two positive terms remaining even after cancelling. \n\nBut we are also not allowed to have one dot in the top row, i.e. our sum (after cancelling) cannot include 1. This would seem to exclude the triangular numbers: \n\n\\[\\{1,3,6,10,15,21,28,36,45,55,66,78,91\\} .\\] \n\nExcept some can be rewritten as a different sum of consecutive integers in more than one way. For example \\(15 = 1 + 2 + 3 + 4 + 5\\) , but also \\(15 = 7 + 8\\) . In general, if the triangular number, \\(T_{n} = \\frac{1}{2} n(n + 1)\\) , is either odd, or has an odd factor less than \\(n\\) , it can be rewritten as a sum of consecutive integers that do not include 1. The following triangular numbers can be salvaged: \n\n- 15 is odd, so \\(15 = 7 + 8\\) - 21 is odd, so \\(21 = 10 + 11\\) - 36 is a multiple of 3, so \\(36 = 11 + 12 + 13\\) - 45 is odd, so \\(45 = 22 + 23\\) - 55 is odd, so \\(55 = 27 + 28\\) - 66 is a multiple of 3, so \\(66 = 21 + 22 + 23\\) - 78 is a multiple of 3, so \\(78 = 25 + 26 + 27\\) - 91 is odd, so \\(91 = 45 + 46\\) \n\nSo there are 5 triangular numbers that are not possible: \\(\\{1,3,6,10,28\\}\\) . Combined with the powers of two that are also impossible: \\(\\{1,2,4,8,16,32,64\\}\\) , there are 11 numbers \\((5 + 7 = 12\\) but the number 1 is included in both lists) that can not be written as trapezium numbers: \n\n\\[\\{1,2,3,4,6,8,10,16,28,32,64\\}\\] \n\nThere are 99 numbers less than 100, so there are \\(99 - 11 = 88\\) trapezium numbers.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2024_solutions.jsonl", "problem_match": "\n4. Problem:", "solution_match": "\nSolution"}}
{"year": "2024", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "NewZealand_MO", "problem": "A shop sells golf balls, golf clubs and golf hats. Golf balls can be purchased at a rate of 25 cents for two balls. Golf hats cost \\(\\) 1\\$ each. Golf clubs cost \\(\\) 10\\$ each. At this shop, Ross purchased 100 items for a total cost of exactly \\(\\) 100\\((Ross purchased at least one of each type of item). How many golf hats did Ross purchase?", "solution": ": (James Xu) Let the number of pairs of balls, clubs, and hats purchased be \\(b, c, h\\) respectively. Then, we must have \n\n\\[\\mathrm{cost:}\\qquad \\qquad \\qquad \\frac{1}{4} b + h + 10c = 100\\] \\[\\mathrm{quantity:}\\qquad \\qquad \\qquad 2b + h + c = 100\\] \n\nSubtracting these yields \\(\\frac{7}{4} b - 9c = 0\\) and thus \n\n\\[7b = 36c.\\] \n\nSince \\(\\gcd (7,36) = 1\\) this implies \\(36|b\\) . Since \\(0< b< 51\\) , we must have \\(b = 36\\) . Therefore \\(c = 7\\) . Thus \\(h = 100 - 2\\times 36 - 7 = 21\\) . Hence the final answer is \\((b,h,c) = (36,21,7)\\) . Checking we get: \n\n\\[\\mathrm{cost:}\\qquad \\qquad \\frac{1}{4} b + h + 10c = 9 + 21 + 70 = 100\\] \\[\\mathrm{quantity:}\\qquad \\qquad 2b + h + c = 72 + 21 + 7 = 100.\\]", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2024_solutions.jsonl", "problem_match": "\n5. Problem:", "solution_match": "\nSolution"}}
-{"year": "2024", "tier": "T1", "problem_label": "6", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(\\omega\\) be the incircle of scalene triangle \\(ABC\\) . Let \\(\\omega\\) be tangent to \\(AB\\) and \\(AC\\) at points \\(X\\) and \\(Y\\) . Construct points \\(X'\\) and \\(Y'\\) on line segments \\(AB\\) and \\(AC\\) respectively such that \\(AX' = XB\\) and \\(AY' = YC\\) . Let line \\(CX'\\) intersects \\(\\omega\\) at points \\(P, Q\\) such that \\(P\\) is closer to \\(C\\) than \\(Q\\) . Also let \\(R\\) be the intersection of lines \\(CX'\\) and \\(BY'\\) . Prove that \\(CP = RX'\\) .", "solution": ": (Ross Atkins) \n\nLet \\(a, b, c\\) be the sidelengths \\(BC, AC, AB\\) respectively, and let \\(s\\) be the semiperimeter of triangle \\(ABC\\) (i.e. let \\(s = \\frac{a + b + c}{2}\\) ). Since \\(X\\) and \\(Y\\) are the points of contact of the incircle we get \\(AX = AY\\) . Similarly \\(BX = BZ\\) and \\(CY = CZ\\) where \\(Z\\) is the point of tangency between \\(\\omega\\) and side \\(BC\\) . Let \\(p = AX = AY\\) and \\(q = BX = BZ\\) and \\(r = CY + CZ\\) as in the diagram. \n\n\n \n\nWe have the system of equations: \n\n\\[p + q = c\\] \\[p + r = b\\] \\[q + r = a\\] \n\nAdding them together yields \\(2(p + q + r) = a + b + c\\) so therefore \\(p + q + r = s\\) . Then we simply get: \n\n\\[s - c = (p + q + r) - (p + q) = r.\\] \n\nSimilarly \\(p = (s - a)\\) and \\(q = (s - b)\\) . Hence \\(BX' = AX = AY = CY' = (s - a)\\) , and \\(AX' = (s - b)\\) .\n\n\n\nNow let \\(\\omega_{C}\\) be the excircle of triangle \\(A B C\\) opposite vertex \\(C\\) . The circle \\(\\omega_{C}\\) is tangent to lines \\(B C\\) , \\(A C\\) and \\(A B\\) at points \\(D\\) , \\(E\\) and \\(F\\) respectively. First we consider equal tangents from \\(C\\) to \\(\\omega_{C}\\) . \n\n\\[C D = B C + B D = a + B F\\] \\[C E = C A + A E = b + A F\\] \n\nAdding these together gives us \\(C D + C E = a + b + (A F + B F) = a + b + c\\) . But since \\(C D = C E\\) (equal tangents) this implies \n\n\\[C D = \\frac{a + b + c}{2} = s.\\] \n\nTherefore \\(C D = C E = s\\) . This means \\(B F = C D - a = (s - a)\\) and thus \\(X^{\\prime}\\) and \\(F\\) are the same point. \n\n\n \n\nNow consider the homothety (centred at \\(C\\) ) that carries \\(\\omega\\) to \\(\\omega_{C}\\) . This homothety sends point \\(P\\) to point \\(X^{\\prime}\\) (since \\(C, P, X^{\\prime}\\) are colinear). It also carries the tangency point \\(Y\\) to \\(E\\) . It follows that \n\n\\[\\frac{C P}{P X^{\\prime}} = \\frac{C Y}{Y E} = \\frac{C Y}{C E - C Y} = \\frac{s - c}{s - (s - c)} = \\frac{s - c}{c}.\\] \n\ni.e. Point \\(P\\) divides segment \\(C X^{\\prime}\\) into the ratio \\((s - c):c\\) . Also we can apply Menelaus' Theorem ( \\(B\\) , \\(R\\) , \\(Y^{\\prime}\\) colinear) to get \n\n\\[\\frac{C R}{R X^{\\prime}} \\times \\frac{X^{\\prime}B}{B A} \\times \\frac{A Y^{\\prime}}{Y^{\\prime}C} = 1\\] \n\n\\[\\frac{C R}{R X^{\\prime}} \\times \\frac{s - a}{c} \\times \\frac{s - c}{s - a} = 1\\] \n\nHence \\(\\frac{C R}{R X^{\\prime}} = \\frac{c}{s - c}\\) , so point \\(R\\) divides segment \\(C X^{\\prime}\\) into the ratio \\(c:(s - c)\\) . Therefore \\(C P = X^{\\prime}R\\) as required.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2024_solutions.jsonl", "problem_match": "\n6. Problem:", "solution_match": "\nSolution"}}
+{"year": "2024", "tier": "T1", "problem_label": "6", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(\\omega\\) be the incircle of scalene triangle \\(ABC\\) . Let \\(\\omega\\) be tangent to \\(AB\\) and \\(AC\\) at points \\(X\\) and \\(Y\\) . Construct points \\(X'\\) and \\(Y'\\) on line segments \\(AB\\) and \\(AC\\) respectively such that \\(AX' = XB\\) and \\(AY' = YC\\) . Let line \\(CX'\\) intersects \\(\\omega\\) at points \\(P, Q\\) such that \\(P\\) is closer to \\(C\\) than \\(Q\\) . Also let \\(R\\) be the intersection of lines \\(CX'\\) and \\(BY'\\) . Prove that \\(CP = RX'\\) .", "solution": ": (Ross Atkins) \n\nLet \\(a, b, c\\) be the sidelengths \\(BC, AC, AB\\) respectively, and let \\(s\\) be the semiperimeter of triangle \\(ABC\\) (i.e. let \\(s = \\frac{a + b + c}{2}\\) ). Since \\(X\\) and \\(Y\\) are the points of contact of the incircle we get \\(AX = AY\\) . Similarly \\(BX = BZ\\) and \\(CY = CZ\\) where \\(Z\\) is the point of tangency between \\(\\omega\\) and side \\(BC\\) . Let \\(p = AX = AY\\) and \\(q = BX = BZ\\) and \\(r = CY + CZ\\) as in the diagram. \n\n\n \n\nWe have the system of equations: \n\n\\[p + q = c\\] \\[p + r = b\\] \\[q + r = a\\] \n\nAdding them together yields \\(2(p + q + r) = a + b + c\\) so therefore \\(p + q + r = s\\) . Then we simply get: \n\n\\[s - c = (p + q + r) - (p + q) = r.\\] \n\nSimilarly \\(p = (s - a)\\) and \\(q = (s - b)\\) . Hence \\(BX' = AX = AY = CY' = (s - a)\\) , and \\(AX' = (s - b)\\) .\n\n\n\nNow let \\(\\omega_{C}\\) be the excircle of triangle \\(A B C\\) opposite vertex \\(C\\) . The circle \\(\\omega_{C}\\) is tangent to lines \\(B C\\) , \\(A C\\) and \\(A B\\) at points \\(D\\) , \\(E\\) and \\(F\\) respectively. First we consider equal tangents from \\(C\\) to \\(\\omega_{C}\\) . \n\n\\[C D = B C + B D = a + B F\\] \\[C E = C A + A E = b + A F\\] \n\nAdding these together gives us \\(C D + C E = a + b + (A F + B F) = a + b + c\\) . But since \\(C D = C E\\) (equal tangents) this implies \n\n\\[C D = \\frac{a + b + c}{2} = s.\\] \n\nTherefore \\(C D = C E = s\\) . This means \\(B F = C D - a = (s - a)\\) and thus \\(X^{\\prime}\\) and \\(F\\) are the same point. \n\n\n \n\nNow consider the homothety (centred at \\(C\\) ) that carries \\(\\omega\\) to \\(\\omega_{C}\\) . This homothety sends point \\(P\\) to point \\(X^{\\prime}\\) (since \\(C, P, X^{\\prime}\\) are colinear). It also carries the tangency point \\(Y\\) to \\(E\\) . It follows that \n\n\\[\\frac{C P}{P X^{\\prime}} = \\frac{C Y}{Y E} = \\frac{C Y}{C E - C Y} = \\frac{s - c}{s - (s - c)} = \\frac{s - c}{c}.\\] \n\ni.e. Point \\(P\\) divides segment \\(C X^{\\prime}\\) into the ratio \\((s - c):c\\) . Also we can apply Menelaus' Theorem ( \\(B\\) , \\(R\\) , \\(Y^{\\prime}\\) colinear) to get \n\n\\[\\frac{C R}{R X^{\\prime}} \\times \\frac{X^{\\prime}B}{B A} \\times \\frac{A Y^{\\prime}}{Y^{\\prime}C} = 1\\] \n\n\\[\\frac{C R}{R X^{\\prime}} \\times \\frac{s - a}{c} \\times \\frac{s - c}{s - a} = 1\\] \n\nHence \\(\\frac{C R}{R X^{\\prime}} = \\frac{c}{s - c}\\) , so point \\(R\\) divides segment \\(C X^{\\prime}\\) into the ratio \\(c:(s - c)\\) . Therefore \\(C P = X^{\\prime}R\\) as required.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2024_solutions.jsonl", "problem_match": "\n6. Problem:", "solution_match": "\nSolution"}}
{"year": "2024", "tier": "T1", "problem_label": "7", "problem_type": null, "exam": "NewZealand_MO", "problem": "Some of the 80960 lattice points in a \\(40 \\times 2024\\) lattice are coloured red. It is known that no four red lattice points are vertices of a rectangle with sides parallel to the axes of the lattice. What is the maximum possible number of red points in the lattice?", "solution": ": (Ross Atkins) \n\nLet \\(a_{1},a_{2},a_{3},\\ldots ,a_{2024}\\) be the number of red dots in rows \\(1,2,3,\\ldots ,2024\\) respectively. So \\(0\\leqslant a_{i}\\leqslant 40\\) for each \\(i\\) \n\nFor each of the \\(\\binom{40}{2}=780\\) pairs of columns, there can be at most one row with a red dot in both columns. Therefore we must have \n\n\\[\\binom{a_{1}}{2}+\\binom{a_{2}}{2}+\\binom{a_{3}}{2}+\\cdots+\\binom{a_{2024}}{2}\\leq\\binom{40}{2}=780.\\] \n\nAnd hence there is always guaranteed to be at least \\(2024 - 780\\) indices with \\(a_{i}\\leqslant 1\\) - i.e. At least \\(2024 - 780 = 1244\\) rows have at most one red dot in it. \n\nNow consider an arrangement in which the total number of red dots is maximised, and suppose for the sake of contradiction that \\(a_{j}\\geqslant 3\\) for some index \\(j\\) . Let \\(c_{1}\\) , \\(c_{2}\\) and \\(c_{3}\\) be three of the columns where row \\(j\\) has a red dot. Let \\(r_{1}\\) and \\(r_{2}\\) be any two rows which each currently contain one red dot. Consider the following operation: \n\n- Remove the dot in row \\(j\\) and column \\(c_{3}\\) . \n\n- Remove the dots in rows \\(r_{1}\\) and \\(r_{2}\\) (at most two dots removed). \n\n- Add dots in columns \\(c_{1}\\) and \\(c_{3}\\) in row \\(r_{1}\\) . \n\n- Add dots in columns \\(c_{2}\\) and \\(c_{3}\\) in row \\(r_{2}\\) . \n\nThis operation will increase the total number of dots which contradicts the assumption that a row with at least 3 dots exists in a maximal arrangement. \n\nHenceforth we assume \\(a_{i}\\leqslant 2\\) for all \\(i\\) . Now if there were more than 780 rows with two red dots in it (i.e. \\(a_{i} = 2\\) for more than 780 indices \\(i\\) ). Then by the pigeonhole principle, there would be two rows with dots in the same pair of columns and this would be a contradiction. Therefore there are at most 780 rows with two dots in it. Hence the total number of dots is at most \n\n\\[780\\times 2 + 1244\\times 1 = 2804.\\] \n\nTo achieve this let the first 780 rows each have a unique pair of columns in which their red dots lie. And the remaining 1244 rows each containing a single dot in the first column only.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2024_solutions.jsonl", "problem_match": "\n7. Problem:", "solution_match": "\nSolution"}}
{"year": "2024", "tier": "T1", "problem_label": "8", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(a\\) , \\(b\\) and \\(c\\) be any positive real numbers. Prove that \n\n\\[{\\frac{a^{2}+b^{2}}{2c}}+{\\frac{a^{2}+c^{2}}{2b}}+{\\frac{b^{2}+c^{2}}{2a}}\\geqslant a+b+c.\\]", "solution": "A: (Eric Liang) \n\nSince square numbers cannot be negative we trivially get \n\n\\[\\left(\\frac{y}{\\sqrt{x}} -\\sqrt{x}\\right)^{2}\\geqslant 0\\] \n\nfor any positive \\(x\\) and \\(y\\) . Expanding the brackets and rearranging gives us \\(\\frac{y^{2}}{x} +x\\geq 2y\\) . Since this holds for any \\(x\\) and \\(y\\) we get the following inequalities: \n\n\\[\\frac{a^{2}}{b} +b\\geq 2a\\] \\[\\frac{a^{2}}{c} +c\\geq 2a\\] \\[\\frac{b^{2}}{a} +a\\geq 2b\\] \\[\\frac{b^{2}}{c} +c\\geq 2b\\] \\[\\frac{c^{2}}{b} +b\\geq 2c\\] \\[\\frac{c^{2}}{a} +a\\geq 2c\\] \n\nSumming these together gives us \n\n\\[{\\frac{a^{2}+b^{2}}{c}}+{\\frac{a^{2}+c^{2}}{b}}+{\\frac{b^{2}+c^{2}}{a}}\\geqslant2a+2b+2c\\] \n\nand then dividing by 2 gives us the required result.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2024_solutions.jsonl", "problem_match": "\n8. Problem:", "solution_match": "\nSolution"}}
{"year": "2024", "tier": "T1", "problem_label": "8", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(a\\) , \\(b\\) and \\(c\\) be any positive real numbers. Prove that \n\n\\[{\\frac{a^{2}+b^{2}}{2c}}+{\\frac{a^{2}+c^{2}}{2b}}+{\\frac{b^{2}+c^{2}}{2a}}\\geqslant a+b+c.\\]", "solution": "B: (Ross Atkins) \n\nSince \\(a,b\\) are positive then \\((a + b) / ab > 0\\) . Furthermore, since square numbers are. nonnegative, we must have \\((a + b)(a - b)^{2}\\geq 0\\) . Expanding the brackets gives us, \n\n\\[(a + b)(a - b)^{2} = a^{3} - a^{2}b - ab^{2} + b^{3}\\geq 0\\] \n\nThus \\(a^{3} + b^{3}\\geq a^{2}b + ab^{2}\\) . Dividing both sides by \\(ab\\) gives us \n\n\\[\\frac{a^{2}}{b} +\\frac{b^{2}}{a}\\geq a + b.\\] \n\nSimilarly we have \\(\\frac{a^{2}}{c} +\\frac{c^{2}}{a}\\geq a + c\\) and \\(\\frac{b^{2}}{c} +\\frac{c^{2}}{b}\\geq b + c\\) . Adding these all together gives us \n\n\\[\\left(\\frac{a^{2}}{b} +\\frac{b^{2}}{a}\\right) + \\left(\\frac{a^{2}}{c} +\\frac{c^{2}}{a}\\right) + \\left(\\frac{b^{2}}{c} +\\frac{c^{2}}{b}\\right)\\geq (a + b) + (a + c) + (b + c)\\] \\[\\qquad \\Rightarrow \\frac{a^{2} + b^{2}}{c} +\\frac{a^{2} + c^{2}}{b} +\\frac{b^{2} + c^{2}}{a}\\geq 2(a + b + c)\\] \\[\\qquad \\Rightarrow \\frac{a^{2} + b^{2}}{2c} +\\frac{a^{2} + c^{2}}{2b} +\\frac{b^{2} + c^{2}}{2a}\\geq a + b + c.\\]", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2024_solutions.jsonl", "problem_match": "\n8. Problem:", "solution_match": "\nSolution"}}
diff --git a/NewZealand_MO/segmented/en-nzmo1_2025_solutions.jsonl b/NewZealand_MO/segmented/en-nzmo1_2025_solutions.jsonl
index 0e762904ac7f156f2d23799c3b7affc52ef71e55..9ddf368d37baf73afd4500e1efc1025b009e0e9c 100644
--- a/NewZealand_MO/segmented/en-nzmo1_2025_solutions.jsonl
+++ b/NewZealand_MO/segmented/en-nzmo1_2025_solutions.jsonl
@@ -1,9 +1,9 @@
{"year": "2025", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(a\\) and \\(b\\) be positive integers with no common factor greater than 1. What are the possible values for the greatest common divisor of \\((a + b)\\) and \\((a - b)\\) ?", "solution": ": (Kevin Shen) \n\nLet \\(d\\) be a common divisor of both \\((a + b)\\) and \\((a - b)\\) , and therefore divides linear combinations of \\((a + b)\\) and \\((a - b)\\) . In particular, \n\n\\[d\\mid [(a + b) + (a - b)] = 2a,\\quad d\\mid [(a + b) - (a - b)] = 2b.\\] \n\nAs \\(a\\) and \\(b\\) don't share any common factors, the only common factors \\(2a\\) and \\(2b\\) have is 2. Hence \\(d\\mid 2\\) , which means that \\(d = 1\\) or 2. This means that the greatest common divisor cannot be larger than 2. \n\nWe now show that both 1 and 2 can both occur. \n\nConsider \\(a = 2, b = 1\\) \n\n\\[\\gcd (2 + 1,2 - 1) = 1.\\] \n\nConsider \\(a = 5, b = 3\\) \n\n\\[\\gcd (5 + 3,5 - 3) = 2.\\] \n\nThus the only possible values for the greatest common divisor of \\((a + b)\\) and \\((a - b)\\) are 1 and 2.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2025_solutions.jsonl", "problem_match": "\n1. Problem:", "solution_match": "\nSolution"}}
-{"year": "2025", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(ABC\\) be a right-angled triangle with \\(\\angle BAC = 90^{\\circ}\\) , \\(\\angle ABC = 70^{\\circ}\\) , and \\(AB = 1\\) . Let \\(M\\) be the midpoint of \\(BC\\) . Let \\(D\\) be the point on the extension of \\(AM\\) beyond \\(M\\) such that \\(\\angle CDA = 110^{\\circ}\\) . Find the length of \\(CD\\) .", "solution": ": (Nico McKinlay) \n\nConstruct point \\(E\\) so that \\(ABEC\\) is a rectangle. The diagonals of any rectangle bisect each other, that is, they meet at each other's midpoints. Hence \\(AE\\) and \\(BC\\) meet at \\(M\\) , i.e. \\(E\\) lies on line \\(AM\\) . \n\n\n \n\nBy symmetry in rectangle \\(ABEC\\) , we have \n\n\\[\\angle CEA = \\angle ABC = 70^{\\circ}.\\] \n\nBy angles on a line, \n\n\\[\\angle CDE = 180^{\\circ} - \\angle CDA = 180^{\\circ} - 110^{\\circ} = 70^{\\circ}.\\] \n\nSo triangle \\(CDE\\) is isosceles with \\(CD = CE\\) , because \n\n\\[\\angle CED = \\angle CDE = 70^{\\circ}.\\] \n\nOpposite sides in a rectangle are equal so \\(CE = AB = 1\\) , hence \\(CD\\) has length 1.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2025_solutions.jsonl", "problem_match": "\n2. Problem:", "solution_match": "\nSolution"}}
+{"year": "2025", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(ABC\\) be a right-angled triangle with \\(\\angle BAC = 90^{\\circ}\\) , \\(\\angle ABC = 70^{\\circ}\\) , and \\(AB = 1\\) . Let \\(M\\) be the midpoint of \\(BC\\) . Let \\(D\\) be the point on the extension of \\(AM\\) beyond \\(M\\) such that \\(\\angle CDA = 110^{\\circ}\\) . Find the length of \\(CD\\) .", "solution": ": (Nico McKinlay) \n\nConstruct point \\(E\\) so that \\(ABEC\\) is a rectangle. The diagonals of any rectangle bisect each other, that is, they meet at each other's midpoints. Hence \\(AE\\) and \\(BC\\) meet at \\(M\\) , i.e. \\(E\\) lies on line \\(AM\\) . \n\n\n \n\nBy symmetry in rectangle \\(ABEC\\) , we have \n\n\\[\\angle CEA = \\angle ABC = 70^{\\circ}.\\] \n\nBy angles on a line, \n\n\\[\\angle CDE = 180^{\\circ} - \\angle CDA = 180^{\\circ} - 110^{\\circ} = 70^{\\circ}.\\] \n\nSo triangle \\(CDE\\) is isosceles with \\(CD = CE\\) , because \n\n\\[\\angle CED = \\angle CDE = 70^{\\circ}.\\] \n\nOpposite sides in a rectangle are equal so \\(CE = AB = 1\\) , hence \\(CD\\) has length 1.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2025_solutions.jsonl", "problem_match": "\n2. Problem:", "solution_match": "\nSolution"}}
{"year": "2025", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(P(x) = x^{3} + ax^{2} + bx - 8\\) be a polynomial with 3 real roots. Show that \\(a^{2} \\geq 2b + 12\\).", "solution": ": (Eric Liang) \n\nLet \\(\\alpha , \\beta , \\gamma\\) be the roots of the polynomial. By expanding \\(P(x) = (x - \\alpha)(x - \\beta)(x - \\gamma)\\) we get: \n\n\\[a = -\\alpha -\\beta -\\gamma\\] \\[b = \\alpha \\beta +\\beta \\gamma +\\gamma \\alpha\\] \\[-8 = -\\alpha \\beta \\gamma\\] \n\nUsing these results: \n\n\\[a^{2} = (-\\alpha -\\beta -\\gamma)^{2}\\] \\[\\quad = \\alpha^{2} + \\beta^{2} + \\gamma^{2} + 2\\alpha \\beta +2\\beta \\gamma +2\\gamma \\alpha\\] \\[\\quad = \\alpha^{2} + \\beta^{2} + \\gamma^{2} + 2b\\] \\[\\quad \\geq 3\\sqrt[3]{\\alpha^{2}\\beta^{2}\\gamma^{2} + 2b}\\] \\[\\quad = 3\\sqrt[3]{(-8)^{2} + 2b}\\] \\[\\quad = 2b + 12\\] \n\nAs required.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2025_solutions.jsonl", "problem_match": "\n3. Problem:", "solution_match": "\nSolution"}}
{"year": "2025", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "NewZealand_MO", "problem": "Find the largest integer \\(k\\) such that any string of 2025 letters consisting only of A's and B's contains a palindromic substring of length \\(k\\) or longer. A palindromic substring is a string of consecutive letters which reads the same backwards as forwards.", "solution": ": (Kevin Shen) \n\nWe claim that the largest integer is 4. We first prove that all strings \\(S\\) of 2025 letters contain a palindromic substring of length 4 or longer, which implies that \\(k \\geq 4\\) . Then we shall provide a construction to show that \\(k\\) cannot be 5 or more. \n\nWe first begin by proving \\(k \\geq 4\\) . Let \\(R\\) be the 2023 letter substring of \\(S\\) with the first and last letter removed. \n\nCase 1: \\(R\\) contains a letter that is repeated 4 or more times in a row. \n\nThis repeating letter (e.g. AAAA) is a palindromic substring or length 4 or more, so we are done. \n\nCase 2: \\(R\\) contains a letter that is repeated 3 times in a row, and no more. \n\nWithout loss of generality, assume this repeated letter is A. As it repeats only 3 times in a row, the letters immediately to the left and right of the substring AAA must be B, therefore \\(S\\) contains the substring BAAAB, which is palindromic with length \\(5 > 4\\) .\n\n\n\nCase 3: \\(R\\) contains a letter that is repeated 2 times in a row, and no more. \n\nWithout loss of generality, assume this repeated letter is A. As it repeats only 2 times in a row, the letters immediately to the left and right of the substring AA must be B, therefore \\(S\\) contains the substring BAAB, which is palindromic with length 4. \n\nCase 4: \\(R\\) does not contain a letter that is repeated in a row. \n\nAs no letter repeats, \\(R\\) must alternate between A and B. Therefore it contains the substring ABABA, which is palindromic with length \\(5 > 4\\) . \n\nThis proves that \\(k \\geq 4\\) . \n\nWe now provide a construction that does not contain palindromic substrings of length 5 or more. Consider the following sequence of length 2025 built from repeating the 6 letters \\(AABABB\\) . \n\n## AABABB AABABB AABABB... \n\nThere are only 6 possible five letter substrings as the sequence repeats every 6 letters, none of these substrings are palindromic. \n\nAABAB, ABABB, BABBA, ABBAA, BBAAB, BAABA \n\nSimilarly the six letter substrings are all not palindromic either. \n\nAABABB, ABABBA, BABBAA, ABBAAB, BBAABA, BAABAB \n\nAny longer palindromic substring of odd length must contain a palindromic substring with 5 letters, and any longer palindromic substring of even length must contain a palindromic substring with 6 letters. As we proved earlier there are no such substrings, there also cannot be palindromic substrings of length 7 or longer in our constructed sequence.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2025_solutions.jsonl", "problem_match": "\n4. Problem:", "solution_match": "\nSolution"}}
{"year": "2025", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "NewZealand_MO", "problem": "Alice plays a game with the Mad Hatter. The Mad Hatter will write two rows of numbers on a blackboard, each a distinct permutation of \\(\\{1,2,\\ldots ,n\\}\\) . On each move, Alice is allowed to swap the positions of the numbers \\(a\\) and \\(a + 1\\) in the first row, for some \\(1 \\leq a < n\\) . What is the minimum number of moves Alice needs in order to guarantee that she can turn the first row of numbers into the second, regardless of the permutations the Mad Hatter writes?", "solution": ": (Tony Wang) \n\nWe will show that the answer is \\(\\binom{n}{2} = \\frac{n(n - 1)}{2}\\) . To show that this is sufficient, we will use induction. \n\n- Base Case: Note that when \\(n = 1\\) , the two rows of numbers must be the same since there is only one permutation of \\(\\{1\\}\\) . Hence this case takes \\(0 = \\binom{1}{2}\\) turns.\n\n\n\n- Inductive Step: Suppose that we know that it takes \\(\\binom{n-1}{2}\\) moves for two permutations of \\(\\{1,2,\\ldots,n-1\\}\\) . Now consider two permutations of \\(\\{1,2,\\ldots,n\\}\\) . We will show that it takes at most \\(n-1\\) moves to put the number \\(n\\) into the correct position. \n\nSuppose that \\(n\\) is in the \\(i\\) - th position in the first row, and that the \\(i\\) - th number in the second row is occupied by \\(k\\) . We first swap the positions of \\(k\\) and \\(k + 1\\) , meaning that the \\(i\\) - th number in the second row is now \\(k + 1\\) . By repeating this argument with \\(k + 1\\) and \\(k + 2\\) , \\(k + 2\\) and \\(k + 3\\) , ..., \\(n - 1\\) and \\(n\\) , we will have \\(n\\) in the correct position. This takes \\(n - k\\) moves, and since \\(1 \\leq k \\leq n\\) , the whole process takes at most \\(n - 1\\) moves. \n\nOnce we have put \\(n\\) into the correct position, we can effectively remove the \\(n\\) from both rows, since we are able to arrange the rest of the numbers without touching \\(n\\) again, showing that the problem has now been reduced to the \\(n - 1\\) case. By the inductive hypothesis, the rest of the problem takes at most \\(\\binom{n-1}{2}\\) moves. \n\nHence, the maximum number of moves required is \\(\\binom{n-1}{2}+n-1=\\binom{n}{2}\\) . \n\nTo show that \\(\\binom{n}{2}\\) moves are necessary, we note that the Mad Hatter can write \n\n \n\nto force Alice to use \\(\\binom{n-2}{2}\\) moves. To show this, we define the warp of a permutation as follows: for each number in the permutation, its distance is the number of numbers to right of it which are smaller than itself. Then the warp is the sum of the distances over all the numbers in the permutation. \n\nNote that on each move, the warp of a permutation either increases by 1 or decreases by one. This is because, if we swap the positions of \\(a\\) and \\(a + 1\\) , then the only distance that can change is the distance of \\(a + 1\\) , and this may only increase by 1 (if \\(a + 1\\) started on the right) or decrease by 1 (if \\(a + 1\\) started on the left). \n\nNow, since the warp of the first row at the start is \\((n - 1) + (n - 2) + \\dots +1 = \\binom{n}{2}\\) , and the warp of the second row at the start is 0, the minimum number of moves required to turn the permutation \\(n\\) , \\(n - 1\\) , ..., 1 into the permutation 1, 2, ..., \\(n\\) is \\(\\binom{n}{2}\\) . \n\nWe have thus proven that the answer is \\(\\binom{n}{2}\\) .", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2025_solutions.jsonl", "problem_match": "\n5. Problem:", "solution_match": "\nSolution"}}
{"year": "2025", "tier": "T1", "problem_label": "6", "problem_type": null, "exam": "NewZealand_MO", "problem": "Determine the largest real number \\(M\\) such that for each infinite sequence \\(x_{0}, x_{1}, x_{2}, \\ldots\\) of real numbers satisfying \\(x_{0} = 1\\) , \\(x_{1} = 3\\) and \n\n\\[x_{0} + x_{1} + \\dots +x_{n - 1}\\geq 3x_{n} - x_{n + 1}\\quad \\mathrm{for~all~}n\\geq 1,\\] \n\nthe inequality \n\n\\[\\frac{x_{n + 1}}{x_{n}} >M\\] \n\nholds for all \\(n \\geq 0\\) .", "solution": ": (Eric Liang) \n\nWe claim that the largest real number \\(M\\) is 2. \n\nFirst we prove by induction on \\(n\\) that \\(\\frac{x_{n}}{x_{i}} > 2^{n - i}\\) for all \\(i \\leq n - 1\\) . \n\nBase case: For \\(n = 1\\) , \\(\\frac{x_{1}}{x_{0}} = 3 > 2\\) . \n\nInductive step: Suppose the inductive hypothesis holds true for \\(n = k\\) , i.e. \\(\\frac{x_{k}}{x_{i}} > 2^{k - i}\\) for all \\(i \\leq k - 1\\) . Then we have that, by the condition in the question, \n\n\\[x_{k + 1}\\geq 3x_{k} - x_{k - 1} - x_{k - 2} - \\dots -x_{0}\\] \\[\\qquad \\geq 3x_{k} - \\frac{1}{2} x_{k} - \\frac{1}{4} x_{k} - \\dots -\\frac{1}{2^{k}} x_{k}\\qquad \\mathrm{(Inductive~Hypothesis)}\\] \\[\\qquad >3x_{k} - x_{k}\\] \\[\\qquad = 2x_{k}\\] \\[\\qquad \\geq 2\\times 2^{k - i}x_{i} = 2^{k + 1 - i}x_{i}\\qquad \\mathrm{(Inductive~Hypothesis)}\\] \n\nThus we have proven that \\(\\frac{x_{k + 1}}{x_{i}} > 2^{k + 1 - i}\\) for all \\(i \\leq k\\) , i.e. the statement for \\(n = k + 1\\) . Thus by induction it is true for all integers \\(n \\geq 1\\) . \n\nHence we have shown that \\(M \\geq 2\\) . \n\nNow we claim that there is a sequence \\(x_{0}, x_{1}, \\ldots\\) satisfying the conditions of the question such that for any \\(\\epsilon > 0\\) which we can find an \\(n\\) such that \\(\\frac{x_{n + 1}}{x_{n}} < 2 + \\epsilon\\) . Let \\(s_{n} = x_{0} + x_{1} + \\dots + x_{n - 1}\\) . Then define the sequence \\(x_{i}\\) such that \\(x_{0} = 1\\) , \\(x_{1} = 3\\) , and \\(x_{n + 1} = 3x_{n} - s_{n}\\) . Note that \\(s_{n + 1} = s_{n} + x_{n}\\) . \n\nWe claim that \\(x_{n} = 2^{n} + n \\times 2^{n - 1}\\) and \\(s_{n} = n \\times 2^{n - 1}\\) . This can be shown by substituting into the recurrence. Hence \n\n\\[\\frac{x_{n + 1}}{x_{n}} = \\frac{2^{n + 1} + (n + 1)\\times 2^{n}}{2^{n} + n\\times 2^{n - 1}}\\] \\[\\qquad = \\frac{4 + 2n + 2}{2 + n}\\] \\[\\qquad = 2 + \\frac{2}{n + 2}\\] \n\nand we simply choose \\(n > \\frac{2}{\\epsilon}\\) to complete the proof.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2025_solutions.jsonl", "problem_match": "\n6. Problem:", "solution_match": "\nSolution"}}
-{"year": "2025", "tier": "T1", "problem_label": "7", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(ABC\\) be a triangle and let \\(D\\) be a point inside the triangle \\(ABC\\) such that \\(AD\\) bisects \\(\\angle BAC\\). Let line \\(BD\\) meet side \\(AC\\) at \\(E\\). Let line \\(CD\\) meet side \\(AB\\) at \\(F\\). Let \\(T\\) be the intersection of the (internal) angle bisectors of \\(\\angle AED\\) and \\(\\angle AFD\\). Prove that if \\(T\\) lies on segment \\(AD\\), then triangle \\(ABC\\) is isosceles.", "solution": "1: (Nico McKinlay) \n\n\n \n\nAssume \\(T\\) lies on \\(AD\\) . Then by the angle bisector theorem in triangles \\(AED\\) and \\(AFD\\) , we have \n\n\\[\\frac{AE}{ED} = \\frac{AT}{TD} = \\frac{AF}{FD}.\\] \n\nApplying the angle bisector theorem in triangles \\(AEB\\) and \\(AFC\\) , we also have: \n\n\\[\\frac{AE}{ED} = \\frac{AB}{BD\\] \\[\\frac{AF}{FD} = \\frac{AC}{CD\\] \n\nCombining all of the above, we get that \n\n\\[\\frac{AB}{BD} = \\frac{AC}{CD}, \\text { i.e. } \\frac{AB}{AC} = \\frac{BD}{CD}. \\quad (1)\\] \n\nNow extend \\(AD\\) beyond \\(D\\) to meet \\(BC\\) at \\(A'\\) . Applying the angle bisector theorem in triangle \\(ABC\\) , we have \n\n\\[\\frac{AB}{AC} = \\frac{BA'}{CA'}. \\quad (2)\\] \n\nCombining (1) and (2) gives \n\n\\[\\frac{BD}{CD} = \\frac{BA'}{CA'}\\] \n\nBy the converse of the angle bisector theorem, this implies \\(DA'\\) bisects \\(\\angle BDC\\) , i.e., \n\n\\[\\angle BDA' = \\angle CDA'.\\] \n\nTherefore, by angles on a line, we have \n\n\\[\\angle BDA = 180^{\\circ} - \\angle BDA' = 180^{\\circ} - \\angle CDA' = \\angle CDA.\\]\n\n\n\nSince \\(\\angle BAD = \\angle CAD\\) , \\(\\angle BDA = \\angle CDA\\) , and \\(AD = AD\\) , triangles \\(ABD\\) and \\(ACD\\) are congruent (ASA). Then \n\n\\[\\triangle ABD\\equiv \\triangle ACD\\Longrightarrow AB = AC\\] \n\nand we're done.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2025_solutions.jsonl", "problem_match": "\n7. Problem:", "solution_match": "\nSolution"}}
+{"year": "2025", "tier": "T1", "problem_label": "7", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(ABC\\) be a triangle and let \\(D\\) be a point inside the triangle \\(ABC\\) such that \\(AD\\) bisects \\(\\angle BAC\\). Let line \\(BD\\) meet side \\(AC\\) at \\(E\\). Let line \\(CD\\) meet side \\(AB\\) at \\(F\\). Let \\(T\\) be the intersection of the (internal) angle bisectors of \\(\\angle AED\\) and \\(\\angle AFD\\). Prove that if \\(T\\) lies on segment \\(AD\\), then triangle \\(ABC\\) is isosceles.", "solution": "1: (Nico McKinlay) \n\n\n \n\nAssume \\(T\\) lies on \\(AD\\) . Then by the angle bisector theorem in triangles \\(AED\\) and \\(AFD\\) , we have \n\n\\[\\frac{AE}{ED} = \\frac{AT}{TD} = \\frac{AF}{FD}.\\] \n\nApplying the angle bisector theorem in triangles \\(AEB\\) and \\(AFC\\) , we also have: \n\n\\[\\frac{AE}{ED} = \\frac{AB}{BD\\] \\[\\frac{AF}{FD} = \\frac{AC}{CD\\] \n\nCombining all of the above, we get that \n\n\\[\\frac{AB}{BD} = \\frac{AC}{CD}, \\text { i.e. } \\frac{AB}{AC} = \\frac{BD}{CD}. \\quad (1)\\] \n\nNow extend \\(AD\\) beyond \\(D\\) to meet \\(BC\\) at \\(A'\\) . Applying the angle bisector theorem in triangle \\(ABC\\) , we have \n\n\\[\\frac{AB}{AC} = \\frac{BA'}{CA'}. \\quad (2)\\] \n\nCombining (1) and (2) gives \n\n\\[\\frac{BD}{CD} = \\frac{BA'}{CA'}\\] \n\nBy the converse of the angle bisector theorem, this implies \\(DA'\\) bisects \\(\\angle BDC\\) , i.e., \n\n\\[\\angle BDA' = \\angle CDA'.\\] \n\nTherefore, by angles on a line, we have \n\n\\[\\angle BDA = 180^{\\circ} - \\angle BDA' = 180^{\\circ} - \\angle CDA' = \\angle CDA.\\]\n\n\n\nSince \\(\\angle BAD = \\angle CAD\\) , \\(\\angle BDA = \\angle CDA\\) , and \\(AD = AD\\) , triangles \\(ABD\\) and \\(ACD\\) are congruent (ASA). Then \n\n\\[\\triangle ABD\\equiv \\triangle ACD\\Longrightarrow AB = AC\\] \n\nand we're done.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2025_solutions.jsonl", "problem_match": "\n7. Problem:", "solution_match": "\nSolution"}}
{"year": "2025", "tier": "T1", "problem_label": "7", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(ABC\\) be a triangle and let \\(D\\) be a point inside the triangle \\(ABC\\) such that \\(AD\\) bisects \\(\\angle BAC\\). Let line \\(BD\\) meet side \\(AC\\) at \\(E\\). Let line \\(CD\\) meet side \\(AB\\) at \\(F\\). Let \\(T\\) be the intersection of the (internal) angle bisectors of \\(\\angle AED\\) and \\(\\angle AFD\\). Prove that if \\(T\\) lies on segment \\(AD\\), then triangle \\(ABC\\) is isosceles.", "solution": "2: (Nico McKinlay) \n\nLabel the following angles: \n\n\\[a = \\angle EAT = \\angle FAT\\qquad \\theta_{1} = \\angle EDT\\] \\[e = \\angle AET = \\angle DET\\qquad \\theta_{2} = \\angle FDT\\] \\[f = \\angle AFT = \\angle DFT\\] \n\nNow consider the product \n\n\\[\\frac{AT}{ET}\\times \\frac{ET}{DT}\\times \\frac{DT}{FT}\\times \\frac{FT}{AT} = 1.\\] \n\nBy the law of sines (in triangles \\(AET\\) , \\(EDT\\) , \\(DFT\\) and \\(FAT\\) ), this becomes \n\n\\[\\frac{\\sin e}{\\sin a}\\times \\frac{\\sin\\theta_{1}}{\\sin e}\\times \\frac{\\sin f}{\\sin\\theta_{2}}\\times \\frac{\\sin a}{\\sin f} = 1\\] \n\nwhich simplifies to \\(\\sin \\theta_{1} = \\sin \\theta_{2}\\) . Since \\(D\\) lies inside triangle \\(ABC\\) , \n\n\\[\\theta_{1} + \\theta_{2} = \\angle EDF = \\angle BDC = 180^{\\circ} - \\underbrace{\\angle BCD}_{>0} - \\underbrace{\\angle CBD}_{>0} < 180^{\\circ}.\\] \n\nTherefore we have \n\n\\[\\sin \\theta_{1} = \\sin \\theta_{2}\\Longrightarrow \\theta_{1} = \\theta_{2}.\\] \n\ni.e. \\(\\angle EDA = \\angle FDA\\) . We also have \\(\\angle BDF = \\angle CDE\\) (vertically opposite angles). Combining these, \n\n\\[\\angle BDA = \\angle BDF + \\angle FDA\\] \\[\\qquad = \\angle CDE + \\angle EDA\\] \\[\\qquad = \\angle CDA.\\] \n\nFrom here we conclude in the same manner as in Solution 1.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2025_solutions.jsonl", "problem_match": "\n7. Problem:", "solution_match": "\nSolution"}}
{"year": "2025", "tier": "T1", "problem_label": "8", "problem_type": null, "exam": "NewZealand_MO", "problem": "Show that there are infinitely many triples \\((a, b, c)\\) of positive integers such that \n\n\\[a^{2} + b^{2} + c^{2} + (a + b + c)^{2} = abc.\\]", "solution": ": (James Xu) \n\nNote that \\(a = b = c = 12\\) is a solution. Now, fix \\(c = 12\\) , the original equation becomes. \n\n\\[\\begin{array}{c}{{a^{2}+b^{2}+144+(a+b+12)^{2}=12a b}}\\\\ {{\\Rightarrow2a^{2}+(24-10b)a+(2b^{2}+24b+288)=0}}\\end{array} \\quad (1)\\] \n\nWe see that by Vieta's theorem taking (1) as a polynomial in \\(a\\) , there are 2 solutions of \\(a\\) , adding up to \\(5b - 12\\) . \n\nThus, if \\(a \\leq b\\) then we have a new set of solutions: \\((5b - 12 - a, b, 12)\\) , which by symmetry gives \\((b, 5b - 12 - a, 12)\\) as a bigger set of solutions in \\(a, b\\) (non- strict in \\(a\\) , strict in \\(b\\) as \\(5b - 12 - a \\geq 5b - b - b = 3b > b\\) ) when \\(a, b \\geq 12\\) . \n\nSince we can repeat this process infinitely, as \\(a, b\\) increases they always fulfills the requirement that \\(a, b \\geq 12\\) , thus we can generate infinitely many solutions.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo1_2025_solutions.jsonl", "problem_match": "\n8. Problem:", "solution_match": "\nSolution"}}
diff --git a/NewZealand_MO/segmented/en-nzmo2_2019_solutions.jsonl b/NewZealand_MO/segmented/en-nzmo2_2019_solutions.jsonl
index 5b5ae5900c151fc4a405dc13fbfc3ef0e8b42268..696dd7c4ec3b0f5784d5618c1f0a875e1dbaa443 100644
--- a/NewZealand_MO/segmented/en-nzmo2_2019_solutions.jsonl
+++ b/NewZealand_MO/segmented/en-nzmo2_2019_solutions.jsonl
@@ -3,4 +3,4 @@
{"year": "2019", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(X\\) be the intersection of the diagonals \\(AC\\) and \\(BD\\) of convex quadrilateral \\(ABCD\\) . Let \\(P\\) be the intersection of lines \\(AB\\) and \\(CD\\) , and let \\(Q\\) be the intersection of lines \\(PX\\) and \\(AD\\) . Suppose that \\(\\angle ABX = \\angle XCD = 90^{\\circ}\\) . Prove that \\(QP\\) is the angle bisector of \\(\\angle BQC\\) .", "solution": ": First note that quadrilateral \\(ABCD\\) is cyclic because \\(\\angle ABD = \\angle ACD = 90^{\\circ}\\) . Also, since \\(AP \\perp DX\\) and \\(DP \\perp AX\\) , we see that \\(X\\) is the orthocentre of triangle \\(APD\\) . Hence \\(PX \\perp AD\\) . Therefore quadrilaterals \\(ABXQ\\) and \\(QXCD\\) are cyclic (opposite angles are supplementary). Now we perform a simple angle chase \n\n\\[\\angle XQB = \\angle XAB = \\angle CAB = \\angle CDB = \\angle CDX = \\angle CQX.\\] \n\nSince \\(\\angle XQB = \\angle CQX\\) , it follows that \\(QX\\) is the angle bisector of \\(\\angle CQB\\) as required.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo2_2019_solutions.jsonl", "problem_match": "\n2.", "solution_match": "\nSolution"}}
{"year": "2019", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(a\\) , \\(b\\) and \\(c\\) be positive real numbers such that \\(a + b + c = 3\\) . Prove that \n\n\\[a^{a} + b^{b} + c^{c}\\geq 3.\\]", "solution": ": We start with a general fact about any positive real number \\(x\\) . There are two cases: either \\(x \\geq 1\\) or \\(x < 1\\) . \n\n- If \\(x \\geq 1\\) then \\(x^{p} \\geq x^{q}\\) for any \\(p > q\\) . Substituting \\(p = x\\) and \\(q = 1\\) gives us \\(x^{x} \\geq x^{1} = x\\) . \n\n- If \\(x < 1\\) then \\(x^{p} < x^{q}\\) for any \\(p > q\\) . Substituting \\(p = 1\\) and \\(q = x\\) gives us \\(x = x^{1} < x^{x}\\) . \n\nIn either case we get \\(x^{x} \\geq x\\) for all positive real numbers \\(x\\) . Applying this fact for \\(a\\) , \\(b\\) and \\(c\\) gives us \n\n\\[a^{a} + b^{b} + c^{c} \\geq a + b + c = 3\\] \n\nas required.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo2_2019_solutions.jsonl", "problem_match": "\n3.", "solution_match": "\nSolution"}}
{"year": "2019", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "NewZealand_MO", "problem": "Show that for all positive integers \\(k\\) , there exists a positive integer \\(n\\) such that \\(n2^{k} - 7\\) is a perfect square.", "solution": ": Proof by induction on \\(k\\) . For the base cases ( \\(k \\leq 3\\) ) we can simply choose \\(n = 2^{3 - k}\\) to get \\(n2^{k} - 7 = 2^{3} - 7 = 1^{2}\\) . For the inductive step let \\(k \\geq 3\\) and assume there exist integers \\(a\\) and \\(n\\) such that \n\n\\[a^{2} = n2^{k} - 7.\\] \n\nWe will now endeavour to find integers \\(b\\) and \\(m\\) such that \\(b^{2} = m2^{k + 1} - 7\\) . To do this we have two cases: \n\n- If \\(n\\) is even then choose \\(b = a\\) and \\(m = n / 2\\) . Thus \\(b^{2} = (2m)2^{k} - 7 = m2^{k + 1} - 7\\) , as required. \n\n- If \\(n\\) is odd, then note that \\(a\\) must also be odd. Let \\(n = 2x + 1\\) and let \\(a = 2y + 1\\) . Now consider \\((a + 2^{k - 1})^{2}\\) . \n\n\\[(a + 2^{k - 1})^{2} = a^{2} + 2^{k}a + 2^{2k - 2}\\] \\[\\qquad = \\left(n2^{k} - 7\\right) + 2^{k}a + 2^{2k - 2}\\] \\[\\qquad = {\\bigl (}(2x + 1)2^{k} - 7{\\bigr)} + 2^{k}\\left(2y + 1\\right) + 2^{2k - 2}\\] \\[\\qquad = \\left(x + y + 1 + 2^{k - 3}\\right)2^{k + 1} - 7.\\] \n\nSo in this case we can simply choose \\(b = a + 2^{k - 1}\\) and \\(m = x + y + 1 + 2^{k - 3}\\) . \n\nNote here that this inductive step only works when \\(k \\geq 3\\) (otherwise \\(m = x + y + 1 + 2^{k - 3}\\) is not an integer).", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo2_2019_solutions.jsonl", "problem_match": "\n4.", "solution_match": "\nSolution"}}
-{"year": "2019", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "NewZealand_MO", "problem": "An equilateral triangle is partitioned into smaller equilateral triangular pieces. Prove that two of the pieces are the same size.", "solution": ": For the purpose of this proof, we will consider a vertex to be any point which is a corner of at least one of the triangular pieces. Define an edge to be any line segment between two vertices, which is part of a side of a triangular piece but does not pass through any other vertex. Note that each vertex must be one of the following types: \n\n- Type \\(A\\) : incident with only 2 edges at \\(60^{\\circ}\\) .- Type \\(B\\) : incident with exactly 4 edges at angles \\(60^{\\circ}, 60^{\\circ}, 60^{\\circ}, 180^{\\circ}\\) .- Type \\(C\\) : incident with exactly 6 edges forming six \\(60^{\\circ}\\) angles. \n\nThere can be no other types of vertex, because all the angles must be \\(60^{\\circ}\\) or \\(180^{\\circ}\\) (or \\(300^{\\circ}\\) in the corners of the original large triangle). We will now colour each edge- end either green or blue, as shown in the diagram (the green edge- ends are also a bit thicker). \n\n\n \n\n\n \n\n\n\n\n\n\nAn edge- end is coloured green if it touches a Type \\(B\\) vertex along the \\(180^{\\circ}\\) angle, otherwise it is coloured blue. Now observe that there must be exactly 3 vertices of type \\(A\\) (the three corners of the original large triangle before it was partitioned). If there are \\(x\\) vertices of type \\(B\\) and \\(y\\) vertices of type \\(C\\) then this makes a total of \n\n\\((6 + 2x + 6y)\\) blue ends, but only \\((2x)\\) green ends. \n\nNote also that the 6 edges with an endpoint at a type \\(A\\) vertex must all have their other end being green. There are more blue edge- ends than green edge- ends, so it can't be the case that every blue edge- end is connected to a green edge- end. There must therefore be an edge such that both of its ends are blue. Since neither endpoint of such an edge is a type \\(A\\) vertex, we can conclude that all four of the angles at the endpoints of this double blue- ended edge must be \\(60^{\\circ}\\) . \n\n\n \n\nTherefore this edge is a shared side of two triangular pieces. These two triangular pieces must therefore be the same size. \\(\\square\\) \n\nSeptember 2019", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo2_2019_solutions.jsonl", "problem_match": "\n5.", "solution_match": "\nSolution"}}
+{"year": "2019", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "NewZealand_MO", "problem": "An equilateral triangle is partitioned into smaller equilateral triangular pieces. Prove that two of the pieces are the same size.", "solution": ": For the purpose of this proof, we will consider a vertex to be any point which is a corner of at least one of the triangular pieces. Define an edge to be any line segment between two vertices, which is part of a side of a triangular piece but does not pass through any other vertex. Note that each vertex must be one of the following types: \n\n- Type \\(A\\) : incident with only 2 edges at \\(60^{\\circ}\\) .- Type \\(B\\) : incident with exactly 4 edges at angles \\(60^{\\circ}, 60^{\\circ}, 60^{\\circ}, 180^{\\circ}\\) .- Type \\(C\\) : incident with exactly 6 edges forming six \\(60^{\\circ}\\) angles. \n\nThere can be no other types of vertex, because all the angles must be \\(60^{\\circ}\\) or \\(180^{\\circ}\\) (or \\(300^{\\circ}\\) in the corners of the original large triangle). We will now colour each edge- end either green or blue, as shown in the diagram (the green edge- ends are also a bit thicker). \n\n\n \n\n\n \n\n\n\n\n\n\nAn edge- end is coloured green if it touches a Type \\(B\\) vertex along the \\(180^{\\circ}\\) angle, otherwise it is coloured blue. Now observe that there must be exactly 3 vertices of type \\(A\\) (the three corners of the original large triangle before it was partitioned). If there are \\(x\\) vertices of type \\(B\\) and \\(y\\) vertices of type \\(C\\) then this makes a total of \n\n\\((6 + 2x + 6y)\\) blue ends, but only \\((2x)\\) green ends. \n\nNote also that the 6 edges with an endpoint at a type \\(A\\) vertex must all have their other end being green. There are more blue edge- ends than green edge- ends, so it can't be the case that every blue edge- end is connected to a green edge- end. There must therefore be an edge such that both of its ends are blue. Since neither endpoint of such an edge is a type \\(A\\) vertex, we can conclude that all four of the angles at the endpoints of this double blue- ended edge must be \\(60^{\\circ}\\) . \n\n\n \n\nTherefore this edge is a shared side of two triangular pieces. These two triangular pieces must therefore be the same size. \\(\\square\\) \n\nSeptember 2019", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo2_2019_solutions.jsonl", "problem_match": "\n5.", "solution_match": "\nSolution"}}
diff --git a/NewZealand_MO/segmented/en-nzmo2_2020_solutions.jsonl b/NewZealand_MO/segmented/en-nzmo2_2020_solutions.jsonl
index 5439c31e7ff6d944a17f8203c8b0c1b482e22fba..34213862c3c78d37d3c0f46dc60a4a16e0c217f6 100644
--- a/NewZealand_MO/segmented/en-nzmo2_2020_solutions.jsonl
+++ b/NewZealand_MO/segmented/en-nzmo2_2020_solutions.jsonl
@@ -1,8 +1,8 @@
{"year": "2020", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(P(x) = x^{3} - 2x + 1\\) and let \\(Q(x) = x^{3} - 4x^{2} + 4x - 1\\) . Show that \n\nif \\(P(r) = 0\\) then \\(Q(r^{2}) = 0\\)", "solution": ": Notice that \\(P(x) = x^{3} - 2x + 1 = (x - 1)(x^{2} + x - 1)\\) , and the roots of \\(x^{2} + x - 1\\) are \\(\\frac{- 1\\pm\\sqrt{5}}{2}\\) . Therefore the roots of \\(P(x)\\) are \n\n\\[1, \\frac{\\sqrt{5} - 1}{2} \\text{and} \\frac{-1 - \\sqrt{5}}{2}.\\] \n\nNotice also that \\(Q(x) = x^{3} - 4x^{2} + 4x - 1 = (x - 1)(x^{2} - 3x + 1)\\) , and the roots of \\(x^{2} - 3x + 1\\) are \\(\\frac{3\\pm\\sqrt{5}}{2}\\) . Therefore the roots of \\(Q(x)\\) are \n\n\\[1, \\frac{3 + \\sqrt{5}}{2} \\text{and} \\frac{3 - \\sqrt{5}}{2}.\\] \n\nSo it suffices to check that \\(1^{2} = 1\\) and that \\(\\left(\\frac{- 1\\pm\\sqrt{5}}{2}\\right)^{2} = \\frac{1\\pm 2\\sqrt{5} + 5}{4} = \\frac{3\\pm\\sqrt{5}}{2}\\) .", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo2_2020_solutions.jsonl", "problem_match": "\n1. Problem:", "solution_match": "\nSolution"}}
{"year": "2020", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(P(x) = x^{3} - 2x + 1\\) and let \\(Q(x) = x^{3} - 4x^{2} + 4x - 1\\) . Show that \n\nif \\(P(r) = 0\\) then \\(Q(r^{2}) = 0\\)", "solution": "First notice that \n\n\\[(x^{3} - 2x - 1)P(x) = (x^{3} - 2x - 1)(x^{3} - 2x + 1)\\] \\[\\qquad = x^{6} - 4x^{4} + 4x^{2} - 1\\] \\[\\qquad = Q(x^{2}).\\] \n\nIf \\(r\\) is any root of \\(P(x)\\) then \\(P(r) = 0\\) . This implies that \n\n\\[Q(r^{2}) = (r^{3} - 2r + 1)P(r) = (r^{3} - 2r + 1)\\times 0 = 0.\\] \n\nHence \\(r^{2}\\) is a root of \\(Q(x)\\) .", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo2_2020_solutions.jsonl", "problem_match": "\n1. Problem:", "solution_match": "# Alternative Solution:"}}
{"year": "2020", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "NewZealand_MO", "problem": "Find the smallest positive integer \\(N\\) satisfying the following three properties. \n\n- \\(N\\) leaves a remainder of 5 when divided by 7. \n\n- \\(N\\) leaves a remainder of 6 when divided by 8. \n\n- \\(N\\) leaves a remainder of 7 when divided by 9.", "solution": ": We notice that \\(\\{5,6,7\\}\\) are each 2 less than \\(\\{7,8,9\\}\\) respectively. Therefore \\(N + 2\\) must be a multiple of 7, 8 and 9. Since 7, 8 and 9 are pairwise coprime, this means that \n\n\\[(N + 2) \\text{is a multiple of} 7 \\times 8 \\times 9 = 504.\\] \n\nTherefore the smallest positive possibility for \\(N + 2\\) is 504. Thus \\(N = 502\\) .", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo2_2020_solutions.jsonl", "problem_match": "\n2. Problem:", "solution_match": "\nSolution"}}
-{"year": "2020", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "NewZealand_MO", "problem": "There are 13 marked points on the circumference of a circle with radius 13. Prove that we can choose three of the marked points which form a triangle with area less than 13.", "solution": ": Divide the circle into 6 equal ( \\(60^{\\circ}\\) ) arcs. By the pigeon-hole principle (since \\(13 > 2 \\times 6\\) ) there exists at least one arc which contains at least three of the marked points. Let this \\(60^{\\circ}\\) arc be \\(AB\\) , and let the three marked points be \\(X\\) , \\(Y\\) and \\(Z\\) in that order. \n\n\n \n\nLet \\(M\\) be the midpoint of arc \\(AB\\) . Points \\(X\\) and \\(Z\\) both lie within the minor arc \\(AB\\) so the base of triangle \\(\\triangle XYZ\\) is less than or equal to the base of triangle \\(\\triangle AMB\\) . Also the distance from \\(Y\\) to \\(XZ\\) is less than or equal to the distance from \\(Y\\) to \\(AB\\) , which is at most the distance from \\(M\\) to \\(AB\\) . Hence: the base and height of triangle \\(\\triangle XYZ\\) are less than or equal to the base and height of triangle \\(\\triangle AMB\\) respectively. Therefore \n\n\\[\\mathrm{area}(\\triangle XYZ)\\leq \\mathrm{area}(\\triangle AMB).\\] \n\nSo it suffices to show that \\(\\mathrm{area}(\\triangle AMB)< 13\\) . Let \\(O\\) be the centre of the circle and let \\(D\\) be the foot of the altitude from \\(M\\) to \\(AB\\) . Note that \\(\\triangle ABO\\) is equilateral and \\(D\\) is the midpoint of \\(AB\\) so \\(AD = \\frac{13}{2}\\) . Furthermore, by Pythagoras in \\(\\triangle ADO\\) we get \n\n\\[OD = \\sqrt{OA^{2} - AD^{2}} = \\sqrt{13^{2} - \\left(\\frac{13}{2}\\right)^{2}} = \\frac{13}{2}\\sqrt{3}.\\] \n\n\\[\\Rightarrow MD = MO - OD = 13 - \\frac{13}{2}\\sqrt{3} = 13\\left(1 - \\frac{\\sqrt{3}}{2}\\right).\\] \n\nNow we can calculate the area of triangle \\(\\triangle AMB\\) . The base is \\(AB = 13\\) because \\(ABO\\) is equilateral. The height is \\(MD = 13\\left(1 - \\frac{\\sqrt{3}}{2}\\right)\\) . Therefore \n\n\\[\\mathrm{area}(\\triangle AMB) = \\frac{13 \\times 13\\left(1 - \\frac{\\sqrt{3}}{2}\\right)}{2} = \\frac{169(2 - \\sqrt{3})}{4}.\\] \n\nAnd we can confirm that \\(\\frac{169(2 - \\sqrt{3})}{4} < 13\\) because \n\n\\[\\frac{169(2 - \\sqrt{3})}{4} < 13\\] \\[\\frac{13(2 - \\sqrt{3})< 4}{26 - 13\\sqrt{3} < 4\\] \\[\\frac{22< 13\\sqrt{3}}{22^{2}< 13^{2}\\times 3\\] \\[484< 507.\\] \n\nas required.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo2_2020_solutions.jsonl", "problem_match": "\n3. Problem:", "solution_match": "\nSolution"}}
-{"year": "2020", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(\\Gamma_{1}\\) and \\(\\Gamma_{2}\\) be circles internally tangent at point \\(A\\) , with \\(\\Gamma_{1}\\) inside \\(\\Gamma_{2}\\) . Let \\(BC\\) be a chord of \\(\\Gamma_{2}\\) which is tangent to \\(\\Gamma_{1}\\) at point \\(D\\) . Prove that line \\(AD\\) is the angle bisector of \\(\\angle BAC\\) .", "solution": ": Let \\(\\lambda\\) be the common tangent of \\(\\Gamma_{1}\\) and \\(\\Gamma_{2}\\) at point \\(A\\) . Let \\(P\\) be a point on \\(\\lambda\\) such that \\(P\\) and \\(C\\) are on opposite sides of line \\(AB\\) . Let \\(Q\\) and \\(R\\) be the points of intersection of \\(\\Gamma_{1}\\) with \\(AB\\) and \\(AC\\) respectively. \n\n\n \n\n\\[\\angle BCA = \\angle BAP\\] \\[\\qquad = \\angle QAP\\] \\[\\qquad = \\angle QRA\\] \n\nTherefore lines \\(BC\\) and \\(QR\\) are parallel. \n\nNow consider \\(\\angle BAD\\) . \n\n\\[\\angle BAD = \\angle QAD\\] \\[\\qquad = \\angle QRD\\] \\[\\qquad = \\angle CDR\\] \\[\\qquad = \\angle DAR\\] \\[\\qquad = \\angle DAC.\\] \n\nSince \\(\\angle BAD = \\angle DAC\\) , we are done.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo2_2020_solutions.jsonl", "problem_match": "\n4. Problem:", "solution_match": "\nSolution"}}
+{"year": "2020", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "NewZealand_MO", "problem": "There are 13 marked points on the circumference of a circle with radius 13. Prove that we can choose three of the marked points which form a triangle with area less than 13.", "solution": ": Divide the circle into 6 equal ( \\(60^{\\circ}\\) ) arcs. By the pigeon-hole principle (since \\(13 > 2 \\times 6\\) ) there exists at least one arc which contains at least three of the marked points. Let this \\(60^{\\circ}\\) arc be \\(AB\\) , and let the three marked points be \\(X\\) , \\(Y\\) and \\(Z\\) in that order. \n\n\n \n\nLet \\(M\\) be the midpoint of arc \\(AB\\) . Points \\(X\\) and \\(Z\\) both lie within the minor arc \\(AB\\) so the base of triangle \\(\\triangle XYZ\\) is less than or equal to the base of triangle \\(\\triangle AMB\\) . Also the distance from \\(Y\\) to \\(XZ\\) is less than or equal to the distance from \\(Y\\) to \\(AB\\) , which is at most the distance from \\(M\\) to \\(AB\\) . Hence: the base and height of triangle \\(\\triangle XYZ\\) are less than or equal to the base and height of triangle \\(\\triangle AMB\\) respectively. Therefore \n\n\\[\\mathrm{area}(\\triangle XYZ)\\leq \\mathrm{area}(\\triangle AMB).\\] \n\nSo it suffices to show that \\(\\mathrm{area}(\\triangle AMB)< 13\\) . Let \\(O\\) be the centre of the circle and let \\(D\\) be the foot of the altitude from \\(M\\) to \\(AB\\) . Note that \\(\\triangle ABO\\) is equilateral and \\(D\\) is the midpoint of \\(AB\\) so \\(AD = \\frac{13}{2}\\) . Furthermore, by Pythagoras in \\(\\triangle ADO\\) we get \n\n\\[OD = \\sqrt{OA^{2} - AD^{2}} = \\sqrt{13^{2} - \\left(\\frac{13}{2}\\right)^{2}} = \\frac{13}{2}\\sqrt{3}.\\] \n\n\\[\\Rightarrow MD = MO - OD = 13 - \\frac{13}{2}\\sqrt{3} = 13\\left(1 - \\frac{\\sqrt{3}}{2}\\right).\\] \n\nNow we can calculate the area of triangle \\(\\triangle AMB\\) . The base is \\(AB = 13\\) because \\(ABO\\) is equilateral. The height is \\(MD = 13\\left(1 - \\frac{\\sqrt{3}}{2}\\right)\\) . Therefore \n\n\\[\\mathrm{area}(\\triangle AMB) = \\frac{13 \\times 13\\left(1 - \\frac{\\sqrt{3}}{2}\\right)}{2} = \\frac{169(2 - \\sqrt{3})}{4}.\\] \n\nAnd we can confirm that \\(\\frac{169(2 - \\sqrt{3})}{4} < 13\\) because \n\n\\[\\frac{169(2 - \\sqrt{3})}{4} < 13\\] \\[\\frac{13(2 - \\sqrt{3})< 4}{26 - 13\\sqrt{3} < 4\\] \\[\\frac{22< 13\\sqrt{3}}{22^{2}< 13^{2}\\times 3\\] \\[484< 507.\\] \n\nas required.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo2_2020_solutions.jsonl", "problem_match": "\n3. Problem:", "solution_match": "\nSolution"}}
+{"year": "2020", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(\\Gamma_{1}\\) and \\(\\Gamma_{2}\\) be circles internally tangent at point \\(A\\) , with \\(\\Gamma_{1}\\) inside \\(\\Gamma_{2}\\) . Let \\(BC\\) be a chord of \\(\\Gamma_{2}\\) which is tangent to \\(\\Gamma_{1}\\) at point \\(D\\) . Prove that line \\(AD\\) is the angle bisector of \\(\\angle BAC\\) .", "solution": ": Let \\(\\lambda\\) be the common tangent of \\(\\Gamma_{1}\\) and \\(\\Gamma_{2}\\) at point \\(A\\) . Let \\(P\\) be a point on \\(\\lambda\\) such that \\(P\\) and \\(C\\) are on opposite sides of line \\(AB\\) . Let \\(Q\\) and \\(R\\) be the points of intersection of \\(\\Gamma_{1}\\) with \\(AB\\) and \\(AC\\) respectively. \n\n\n \n\n\\[\\angle BCA = \\angle BAP\\] \\[\\qquad = \\angle QAP\\] \\[\\qquad = \\angle QRA\\] \n\nTherefore lines \\(BC\\) and \\(QR\\) are parallel. \n\nNow consider \\(\\angle BAD\\) . \n\n\\[\\angle BAD = \\angle QAD\\] \\[\\qquad = \\angle QRD\\] \\[\\qquad = \\angle CDR\\] \\[\\qquad = \\angle DAR\\] \\[\\qquad = \\angle DAC.\\] \n\nSince \\(\\angle BAD = \\angle DAC\\) , we are done.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo2_2020_solutions.jsonl", "problem_match": "\n4. Problem:", "solution_match": "\nSolution"}}
{"year": "2020", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(\\Gamma_{1}\\) and \\(\\Gamma_{2}\\) be circles internally tangent at point \\(A\\) , with \\(\\Gamma_{1}\\) inside \\(\\Gamma_{2}\\) . Let \\(BC\\) be a chord of \\(\\Gamma_{2}\\) which is tangent to \\(\\Gamma_{1}\\) at point \\(D\\) . Prove that line \\(AD\\) is the angle bisector of \\(\\angle BAC\\) .", "solution": "Consider the dilation centered at \\(A\\) which sends \\(\\Gamma_{1}\\) to \\(\\Gamma_{2}\\) . This dilation sends point \\(D\\) to the point \\(D^{\\prime} \\in \\Gamma_{2}\\) such that points \\(A\\) , \\(D\\) and \\(D^{\\prime}\\) are colinear. This dilation also sends line \\(BC\\) to the line tangent to \\(\\Gamma_{2}\\) at point \\(D^{\\prime}\\) . Therefore the tangent to \\(\\Gamma_{2}\\) at point \\(D^{\\prime}\\) is parallel to chord \\(BC\\) . Hence \\(D^{\\prime}\\) is the midpoint of arc \\(BC\\) . I.e. \\(BD^{\\prime}\\) and \\(D^{\\prime}C\\) have the same arc- length. Since equal arcs subtend equal angles, we deduce that \\(\\angle BAD^{\\prime} = \\angle D^{\\prime}AC\\) as required.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo2_2020_solutions.jsonl", "problem_match": "\n4. Problem:", "solution_match": "# Alternative Solution:"}}
{"year": "2020", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "NewZealand_MO", "problem": "A sequence of \\(A s\\) and \\(B s\\) is called antipalindromic if writing it backwards, then turning all the \\(A s\\) into \\(B s\\) and vice versa, produces the original sequence. For example \\(A B B A A B\\) is antipalindromic. For any sequence of \\(A s\\) and \\(B s\\) we define the cost of the sequence to be the product of the positions of the \\(A s\\) . For example, the string \\(A B B A A B\\) has cost \\(1\\cdot 4\\cdot 5 = 20\\) . Find the sum of the costs of all antipalindromic sequences of length 2020.", "solution": ": For each integer \\(0\\leq k\\leq 1009\\) define a \\(k\\) - pal to be any sequence of 2020 \\(A s\\) and \\(B s\\) , where the first \\(k\\) terms are \\(B\\) , the last \\(k\\) terms are \\(B\\) , and the middle \\((2020 - 2k)\\) terms form an antipalindromic sequence. \n\nNow for any \\(k\\) , define \\(f(k)\\) to be sum of the costs of all \\(k\\) - pals. Note that any \\(k\\) - pal can be created from a \\((k + 1)\\) - pal by either \n\n- (A) replacing the \\(B\\) in position \\((k + 1)\\) with an \\(A\\) , or \n\n- (B) replacing the \\(B\\) in position \\((2021 - k)\\) with an \\(A\\) . \n\nTherefore the sum of the costs of all \\(k\\) - pals formed using operation (A) is \\((k + 1)\\times f(k + 1)\\) . Similarly the sum of the costs of all \\(k\\) - pals formed using operation (B) is \\((2021 - k)\\times f(k + 1)\\) . Hence \n\n\\[f(k) = (k + 1)f(k + 1) + (2020 - k)f(k + 1) = ((k + 1) + (2020 - k))f(k + 1) = 2021f(k + 1).\\] \n\nNow we note that there are two different 1009- pals, with costs equal to 1010 and 1011 respectively. So \n\n\\[f(1009) = 1010 + 1011 = 2021.\\] \n\nNow if we use the formula \\(f(k) = 2021f(k + 1)\\) iteratively, we get \\(f(1010 - i) = 2021^{i}\\) for each \\(i = 1,2,3,\\ldots\\) . Therefore \n\n\\[f(0) = 2021^{1010}\\] \n\nwhich is our final answer.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo2_2020_solutions.jsonl", "problem_match": "\n5. Problem:", "solution_match": "\nSolution"}}
{"year": "2020", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "NewZealand_MO", "problem": "A sequence of \\(A s\\) and \\(B s\\) is called antipalindromic if writing it backwards, then turning all the \\(A s\\) into \\(B s\\) and vice versa, produces the original sequence. For example \\(A B B A A B\\) is antipalindromic. For any sequence of \\(A s\\) and \\(B s\\) we define the cost of the sequence to be the product of the positions of the \\(A s\\) . For example, the string \\(A B B A A B\\) has cost \\(1\\cdot 4\\cdot 5 = 20\\) . Find the sum of the costs of all antipalindromic sequences of length 2020.", "solution": "Let \\(n\\) be a positive integer. We will find an expression (in terms of \\(n\\) ) for the sum of the costs of all antipalindromes of length \\(2n\\) . Note that a string of \\(A s\\) and \\(B s\\) of length \\(2n\\) is an antipalindrome if and only if for each \\(i\\) , exactly one of the \\(i^{\\mathrm{th}}\\) and \\((2n + 1 - i)^{\\mathrm{th}}\\) letters is an \\(A\\) (and the other is a \\(B\\) ). \n\nLet \\(x_{1},x_{2},\\ldots ,x_{2n}\\) be variables. For any \\(1\\leq a(1)< a(2)< \\dots < a(k)\\leq 2n\\) , consider the string of \\(A s\\) and \\(B s\\) of length \\(2n\\) , such that the \\(a(j)^{\\mathrm{th}}\\) letter is \\(A\\) for all \\(j\\) (and all the other letters are \\(B\\) ). Let this string correspond to the term \\(t = x_{a(1)}x_{a(2)}x_{a(3)}\\dots x_{a(k)}\\) . If \\(x_{i} = i\\) for all \\(i\\) then the value of \\(t\\) is equal to the cost of it's corresponding string. Now consider the expression \n\n\\[y = (x_{1} + x_{2n})(x_{2} + x_{2n - 1})\\cdot \\cdot \\cdot (x_{n} + x_{n + 1}) = \\prod_{j = 1}^{n}(x_{j} + x_{2n + 1 - j}).\\] \n\nIf we expand the brackets then we get \\(2^{n}\\) terms, each in the form \\(t = x_{a(1)}x_{a(2)}x_{a(3)}\\dots x_{a(n)}\\) such that for each \\(j = 1,2,\\ldots ,n\\) either \\(a(j) = j\\) or \\(a(j) = 2n + 1 - j\\) . Therefore \\(y\\) is the sum of all terms that correspond to antipalindromes. Hence if we substitute \\(x_{i} = i\\) for all \\(i\\) , then the value of \\(y\\) would be the sum of the costs of all antipalindromes. So the final answer is: \n\n\\[\\prod_{j = 1}^{n}(j + (2n + 1 - j)) = \\prod_{j = 1}^{n}(2n + 1) = (2n + 1)^{n}.\\]", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo2_2020_solutions.jsonl", "problem_match": "\n5. Problem:", "solution_match": "# Alternative Solution:"}}
diff --git a/NewZealand_MO/segmented/en-nzmo2_2021_solutions.jsonl b/NewZealand_MO/segmented/en-nzmo2_2021_solutions.jsonl
index fdb6c8de43fc4969c74b0b700e38357e2c726422..929dad9da5c398b94392d43fe110055ee136f0ca 100644
--- a/NewZealand_MO/segmented/en-nzmo2_2021_solutions.jsonl
+++ b/NewZealand_MO/segmented/en-nzmo2_2021_solutions.jsonl
@@ -1,8 +1,8 @@
-{"year": "2021", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(ABCD\\) be a convex quadrilateral such that \\(AB + BC = 2021\\) and \\(AD = CD\\) . We are also given that \n\n\\[\\angle ABC = \\angle CDA = 90^{\\circ}\\] \n\nDetermine the length of the diagonal \\(BD\\) .", "solution": ": Since \\(AD = DC\\) and \\(\\angle ADC = 90^{\\circ}\\) , we can fit four copies of quadrilateral \\(ABCD\\) around vertex \\(D\\) as shown in the diagram. \n\n\n \n\nThe outer shape is a quadrilateral because \\(\\angle DAB + \\angle BCD = 180^{\\circ}\\) . Moreover it is a rectangle because \\(\\angle ABC = 90^{\\circ}\\) . In fact it is a square with side- length 2021 because of rotational symmetry and \\(AB + BC = 2021\\) . Also \\(D\\) is the centre of the square because it is the centre of the rotational symmetry. So \\(BD\\) is the distance from a vertex to the centre of the square, which is half the length of the diagonal of the square. Thus \n\n\\[BD = \\frac{1}{2}\\left(2021\\sqrt{2}\\right) = \\frac{2021}{\\sqrt{2}}.\\]", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo2_2021_solutions.jsonl", "problem_match": "\n1. Problem:", "solution_match": "\nSolution"}}
+{"year": "2021", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(ABCD\\) be a convex quadrilateral such that \\(AB + BC = 2021\\) and \\(AD = CD\\) . We are also given that \n\n\\[\\angle ABC = \\angle CDA = 90^{\\circ}\\] \n\nDetermine the length of the diagonal \\(BD\\) .", "solution": ": Since \\(AD = DC\\) and \\(\\angle ADC = 90^{\\circ}\\) , we can fit four copies of quadrilateral \\(ABCD\\) around vertex \\(D\\) as shown in the diagram. \n\n\n \n\nThe outer shape is a quadrilateral because \\(\\angle DAB + \\angle BCD = 180^{\\circ}\\) . Moreover it is a rectangle because \\(\\angle ABC = 90^{\\circ}\\) . In fact it is a square with side- length 2021 because of rotational symmetry and \\(AB + BC = 2021\\) . Also \\(D\\) is the centre of the square because it is the centre of the rotational symmetry. So \\(BD\\) is the distance from a vertex to the centre of the square, which is half the length of the diagonal of the square. Thus \n\n\\[BD = \\frac{1}{2}\\left(2021\\sqrt{2}\\right) = \\frac{2021}{\\sqrt{2}}.\\]", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo2_2021_solutions.jsonl", "problem_match": "\n1. Problem:", "solution_match": "\nSolution"}}
{"year": "2021", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(ABCD\\) be a convex quadrilateral such that \\(AB + BC = 2021\\) and \\(AD = CD\\) . We are also given that \n\n\\[\\angle ABC = \\angle CDA = 90^{\\circ}\\] \n\nDetermine the length of the diagonal \\(BD\\) .", "solution": "First let \\(x = AD = DC\\) and \\(a = BD\\) and \\(y = AB\\) and \\(z = BC\\) . Now initially we can apply Pythagoras in triangles \\(CDA\\) and \\(ABC\\) to get \\(x^{2} + x^{2} = AC^{2}\\) and \\(y^{2} + z^{2} = AC^{2}\\) respectively. Putting this together gives us \n\n\\[x^{2} = \\frac{y^{2} + z^{2}}{2}.\\] \n\nNow note that the opposite angles \\(\\angle ABC\\) and \\(\\angle CDA\\) (in quad \\(ABCD\\) ) are supplementary. Therefore \\(ABCD\\) is a cyclic quadrilateral. Equal chords subtend equal arcs (and chords \\(AD = DC\\) are equal) so \\(\\angle ABD = \\angle DBC\\) . Furthermore, since \\(\\angle ABC\\) is a right angle, this means that \\(\\angle ABD = \\angle DBC = 45^{\\circ}\\) . \n\nFor any three points \\(P\\) , \\(Q\\) and \\(R\\) , let \\(|PQR|\\) denote the area of triangle \\(PQR\\) . Now consider the total area of quadrilateral \\(ABCD\\) calculated in two ways: \n\n\\[|ABC| + |CDA| = |ABD| + |DBC|.\\] \n\nWe calculate the areas of the right- angled triangles using the \\(\\triangle = \\frac{b h}{2}\\) formula, and we calculate the area of the \\(45^{\\circ}\\) - angled triangles using the \\(\\triangle = \\frac{1}{2} ab\\sin C\\) formula. \n\n\\[\\frac{yz}{2} +\\frac{x^{2}}{2} = \\frac{1}{2} ay\\sin (45^{\\circ}) + \\frac{1}{2} az\\sin (45^{\\circ})\\] \n\nAt this point we can substitute \\(x^{2} = \\frac{1}{2} (y^{2} + z^{2})\\) into this equation, and rearrange: \n\n\\[\\frac{yz}{2} +\\frac{x^{2}}{2} = \\frac{ay\\sin(45^{\\circ})}{2} +\\frac{az\\sin(45^{\\circ})}{2}\\] \\[\\frac{yz}{2} +\\frac{y^{2} + z^{2}}{4} = \\frac{ay}{2\\sqrt{2}} +\\frac{az}{2\\sqrt{2}}\\] \\[2yz + y^{2} + z^{2} = \\frac{ay + az}{2\\sqrt{2}}\\] \\[\\frac{(y + z)^{2}}{4} = \\frac{a(y + z)}{2\\sqrt{2}}\\] \\[\\frac{y + z}{\\sqrt{2}} = a.\\] \n\nFinally since \\(y + z = 2021\\) this gives our final answer of \\(BD = a = \\frac{2021}{\\sqrt{2}}\\) .", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo2_2021_solutions.jsonl", "problem_match": "\n1. Problem:", "solution_match": "# Alternative Solution:"}}
{"year": "2021", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "NewZealand_MO", "problem": "Prove that \n\n\\[x^{2} + \\frac{8}{xy} +y^{2}\\geq 8.\\] \n\nfor all positive real numbers \\(x\\) and \\(y\\) .", "solution": ": Since square numbers are always non- negative we have \n\n\\[(x - y)^{2}\\geq 0\\qquad \\mathrm{and}\\qquad (x y - 2)^{2}\\geq 0.\\] \n\nAlso since \\(x\\) and \\(y\\) are positive we have \\(\\frac{2}{x y} >0\\) . Combining this all together gives us: \n\n\\[(x - y)^{2} + \\frac{2}{xy} (x y - 2)^{2}\\geq 0.\\] \n\nFrom here we expand and simplify: \n\n\\[(x^{2} - 2x y + y^{2}) + \\frac{2}{x y} (x^{2}y^{2} - 4x y + 4)\\geq 0\\] \\[x^{2} - 2x y + y^{2} + 2x y - 8 + \\frac{8}{x y}\\geq 0\\] \\[x^{2} + \\frac{8}{x y} +y^{2}\\geq 8\\] \n\nas required.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo2_2021_solutions.jsonl", "problem_match": "\n2. Problem:", "solution_match": "\nSolution"}}
{"year": "2021", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "NewZealand_MO", "problem": "Prove that \n\n\\[x^{2} + \\frac{8}{xy} +y^{2}\\geq 8.\\] \n\nfor all positive real numbers \\(x\\) and \\(y\\) .", "solution": "Consider the AM- GM inequality applied to \\(\\left\\{x^{2},\\frac{4}{x y},\\frac{4}{x y},y^{2}\\right\\}\\) \n\n\\[\\frac{x^{2} + \\frac{4}{x y} + \\frac{4}{x y} + y^{2}}{4}\\geq \\sqrt[4]{x^{2}\\times\\frac{4}{x y}\\times\\frac{4}{x y}}\\times y^{2}\\] \\[\\frac{x^{2} + \\frac{8}{x y} + y^{2}}{4}\\geq 2\\] \\[x^{2} + \\frac{8}{x y} + y^{2}\\geq 8.\\]", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo2_2021_solutions.jsonl", "problem_match": "\n2. Problem:", "solution_match": "# Alternative Solution:"}}
{"year": "2021", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(\\{x_{1},x_{2},x_{3},\\ldots ,x_{n}\\}\\) be a set of \\(n\\) distinct positive integers, such that the sum of any 3 of them is a prime number. What is the maximum value of \\(n\\) ?", "solution": ": First we show that \\(n = 4\\) is possible with an example. The example \\(\\{x_{1},x_{2},x_{3},x_{4}\\} = \\{1,3,7,9\\}\\) satisfies the problem because: \n\n\\(\\cdot 1 + 3 + 7 = 11\\) is prime, \n\n\\(\\cdot 1 + 3 + 9 = 13\\) is prime, \n\n\\(\\cdot 1 + 7 + 9 = 17\\) is prime, and \n\n\\(\\cdot 3 + 7 + 9 = 19\\) is prime. \n\nWe still have to prove that \\(n\\geq 5\\) is impossible. \n\nConsider any set \\(\\{x_{1},x_{2},x_{3},\\ldots ,x_{n}\\}\\) such that the sum of any 3 of them is a prime number. Also consider the three \"pigeonholes\" modulo 3; the residue classes 0, 1 and 2. If all three pigeonholes were non- empty, then it would be possible to choose three numbers – one from each pigeonhole. This would result in a sum which is \\(0 + 1 + 2\\equiv 0\\) (mod 3), and since the numbers are distinct positive integers, this sum would be \\(>3\\) . Thus the sum would not be prime which is a contradiction. Hence at least one of the pigeonholes must be empty. i.e. \n\nThe numbers \\(\\{x_{1},\\ldots ,x_{n}\\}\\) are distributed amongst (at most) two different residue classes modulo 3. \n\nNow assume for the sake of contradiction that \\(n\\geq 5\\) . By the pigeonhole principle at least one residue class contains at least 3 of the numbers. The sum of any three numbers from the same residue class is always a multiple of 3 and so this is a contradiction. \n\nTherefore \\(n< 5\\) as required. \n\nComment: The example \\(\\{1,3,7,9\\}\\) is not the only example that satisfies the problem with \\(n = 4\\) . Here are many other examples: \\(\\{1,5,7,11\\}\\) , \\(\\{3,5,11,15\\}\\) , \\(\\{1,3,13,15\\}\\) , \\(\\{3,9,11,17\\}\\) , \\(\\{5,9,15,17\\}\\) , ... Searching for an example when \\(n = 4\\) is much easier if you conjecture that all the \\(x_{i}\\) must be odd.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo2_2021_solutions.jsonl", "problem_match": "\n3. Problem:", "solution_match": "\nSolution"}}
-{"year": "2021", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(AB\\) be a chord of circle \\(\\Gamma\\) . Let \\(O\\) be the centre of a circle which is tangent to \\(AB\\) at \\(C\\) and internally tangent to \\(\\Gamma\\) at \\(P\\) . Point \\(C\\) lies between \\(A\\) and \\(B\\) . Let the circumcircle of triangle \\(POC\\) intersect \\(\\Gamma\\) at distinct points \\(P\\) and \\(Q\\) . Prove that \\(\\angle AQP = \\angle CQB\\) .", "solution": ": Construct the tangent line to \\(\\Gamma\\) at \\(P\\) . Note that this line is also tangent to the circle through points \\(C\\) and \\(P\\) with centre \\(O\\) . Also construct point \\(E\\) on this tangent line to the right of \\(P\\) . Note that \\(\\angle EPO = 90^{\\circ}\\) and \\(\\angle OCA = 90^{\\circ}\\) because the radii and tangents are perpendicular. \n\n\n \n\nLet \\(x = \\angle PBQ\\) \n\n\\(\\angle EPQ = x\\) (by alternate segment theorem) \\(\\angle OPQ = x - 90^{\\circ}\\) (because \\(\\angle EPO = 90^{\\circ}\\) ) \\(\\angle QCO = 180^{\\circ} - \\angle OPQ\\) (because opposite angles in a cyclic quad \\(= 270^{\\circ} - x\\) are supplementary) \\(\\angle ACQ = 360^{\\circ} - \\angle QCO - \\angle OCA\\) (angles around point \\(C\\) are \\(360^{\\circ}\\) ) \\(= 360^{\\circ} - (270^{\\circ} - x) - 90^{\\circ}\\) \\(= x\\) . \n\nWe also have \\(\\angle QPB = \\angle QAB\\) (angles subtended by chord \\(QB\\) ) in cyclic quad \\(QAPB\\) . Therefore we have similar triangles \n\n\\[\\triangle QBP\\sim \\triangle QCA\\qquad (\\angle PBQ = \\angle ACQ\\mathrm{~and~}\\angle QPB = \\angle QAC)\\] \n\nHence \\(\\angle AQC = \\angle PQB\\) . Therefore \n\n\\[\\angle AQP = \\angle AQC + \\angle CQP = \\angle PQB + \\angle CQP = \\angle CQB.\\]", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo2_2021_solutions.jsonl", "problem_match": "\n4. Problem:", "solution_match": "\nSolution"}}
+{"year": "2021", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(AB\\) be a chord of circle \\(\\Gamma\\) . Let \\(O\\) be the centre of a circle which is tangent to \\(AB\\) at \\(C\\) and internally tangent to \\(\\Gamma\\) at \\(P\\) . Point \\(C\\) lies between \\(A\\) and \\(B\\) . Let the circumcircle of triangle \\(POC\\) intersect \\(\\Gamma\\) at distinct points \\(P\\) and \\(Q\\) . Prove that \\(\\angle AQP = \\angle CQB\\) .", "solution": ": Construct the tangent line to \\(\\Gamma\\) at \\(P\\) . Note that this line is also tangent to the circle through points \\(C\\) and \\(P\\) with centre \\(O\\) . Also construct point \\(E\\) on this tangent line to the right of \\(P\\) . Note that \\(\\angle EPO = 90^{\\circ}\\) and \\(\\angle OCA = 90^{\\circ}\\) because the radii and tangents are perpendicular. \n\n\n \n\nLet \\(x = \\angle PBQ\\) \n\n\\(\\angle EPQ = x\\) (by alternate segment theorem) \\(\\angle OPQ = x - 90^{\\circ}\\) (because \\(\\angle EPO = 90^{\\circ}\\) ) \\(\\angle QCO = 180^{\\circ} - \\angle OPQ\\) (because opposite angles in a cyclic quad \\(= 270^{\\circ} - x\\) are supplementary) \\(\\angle ACQ = 360^{\\circ} - \\angle QCO - \\angle OCA\\) (angles around point \\(C\\) are \\(360^{\\circ}\\) ) \\(= 360^{\\circ} - (270^{\\circ} - x) - 90^{\\circ}\\) \\(= x\\) . \n\nWe also have \\(\\angle QPB = \\angle QAB\\) (angles subtended by chord \\(QB\\) ) in cyclic quad \\(QAPB\\) . Therefore we have similar triangles \n\n\\[\\triangle QBP\\sim \\triangle QCA\\qquad (\\angle PBQ = \\angle ACQ\\mathrm{~and~}\\angle QPB = \\angle QAC)\\] \n\nHence \\(\\angle AQC = \\angle PQB\\) . Therefore \n\n\\[\\angle AQP = \\angle AQC + \\angle CQP = \\angle PQB + \\angle CQP = \\angle CQB.\\]", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo2_2021_solutions.jsonl", "problem_match": "\n4. Problem:", "solution_match": "\nSolution"}}
{"year": "2021", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "NewZealand_MO", "problem": "Find all pairs of integers \\(x, y\\) such that \n\n\\[y^{5} + 2x y = x^{2} + 2y^{4}.\\]", "solution": ": Rearrange and factorize to get \n\n\\[y^{2}(y - 1)(y^{2} - y - 1) = (x - y)^{2}.\\] \n\nNote that \\(y\\) and \\((y - 1)\\) are coprime (their greatest common divisor is 1) because they are consecutive integers. Note since \\(y(y - 1)\\) and \\((y^{2} - y - 1)\\) are consecutive integers, we see that \\((y^{2} - y - 1)\\) is coprime to both \\(y\\) and \\((y - 1)\\) . Therefore the three factors \n\n\\[y^{2},(y - 1)\\mathrm{~and~}(y^{2} - y - 1)\\mathrm{~are~pairwise~coprime}.\\] \n\nSince their product is a perfect square it follows that either: one of \\(y^{2}\\) , \\((y - 1)\\) and \\((y^{2} - y - 1)\\) is zero, or all three of them are perfect squares. So we have four cases: \n\nCase A: \\(y = 0\\) \n\nSubstituting this into the original equation yields \\((0)^{5} + 2x(0) = x^{2} + 2(0)^{4}\\) . Solving this quadratic yields \\(x = 1\\) and so \\((x,y) = (0,0)\\) is the only solution in this case. \n\nCase B: \\(y = 1\\) \n\nSubstituting this into the original equation yields \\((1)^{5} + 2x(1) = x^{2} + 2(1)^{4}\\) . Solving this quadratic yields \\(x = 1\\) and so \\((x,y) = (1,1)\\) is the only solution in this case. \n\nCase C: \\(y^{2} - y - 1< 0\\) \n\nThis rearranges to give us \\((2y - 1)^{2}< 5\\) . But the only odd square less than 5 is 1, and so we would have \\((2y - 1) = \\pm 1\\) . which leads to \\(y = 0,1\\) (but we have already covered this in Cases A and B). \n\nCase D: \\(y^{2} - y - 1 = k^{2}\\) \n\nIf \\(y > 2\\) then \\((y - 1)^{2} = y^{2} - 2y + 1< y^{2} - y - 1< y^{2}\\) . This would give us \\((y - 1)^{2}< k^{2}< y^{2}\\) which is a contradiction because \\((y - 1)^{2}\\) and \\(y^{2}\\) are consecutive squares. \n\nIf \\(y< - 1\\) then \\((- y)^{2}< y^{2} - y - 1< y^{2} - 2y + 1 = (- y + 1)^{2}\\) . This would give us \\(y^{2}< k^{2}< (y - 1)^{2}\\) which is a contradiction because \\((- y)^{2}\\) and \\((- y + 1)^{2}\\) are consecutive squares. \n\nTherefore we must have \\(- 1\\leq y\\leq 2\\) . We have already covered \\(y = 0\\) and \\(y = 1\\) in cases A and B respectively. So it suffices now only to consider \\(y = - 1\\) and \\(y = 2\\) . \n\nIf \\(y = - 1\\) then \\((- 1)^{5} + 2x(- 1) = x^{2} + 2(- 1)^{4}\\) . This rearranges into \\((x + 1)^{2} = - 2\\) which has no real solutions. \n\nIf \\(y = 2\\) then \\(2^{5} + 4x = x^{2} + 2\\times 2^{4}\\) . This rearranges to give \\(x^{2} - 4x = 0\\) which has solutions \\(x = 0\\) and \\(x = 4\\) . \n\nHence in this case we have solutions \\((x,y) = (0,2)\\) and \\((4,2)\\) . \n\nIn summary we have four distinct solutions for \\((x,y)\\) being: \n\n\\[(x,y) = (0,0),(1,1),(0,2)\\mathrm{~and~}(4,2).\\]", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo2_2021_solutions.jsonl", "problem_match": "\n5. Problem:", "solution_match": "\nSolution"}}
{"year": "2021", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "NewZealand_MO", "problem": "Find all pairs of integers \\(x, y\\) such that \n\n\\[y^{5} + 2x y = x^{2} + 2y^{4}.\\]", "solution": "In a similar manner to above get to Case D: \n\n\\[y^{2} - y - 1 = k^{2}.\\] \n\nThen rearrange it to give \\((2y - 1)^{2} - 4k^{2} = 5\\) . This then factorizes as a difference between two squares as \n\n\\[(2y - 1 + 2k)(2y - 1 - 2k) = 5\\] \n\nSince 5 is prime it can only be factored in two ways: \\(5 = 5\\times 1 = (- 5)\\times (- 1)\\) . The sum of these two factors is \\((2y - 1 + 2k) + (2y - 1 - 2k) = 4y - 2\\) . Therefore: \n\n\\[4y - 2 = 5 + 1\\qquad \\mathrm{or}\\qquad 4y - 2 = (-5) + (-1).\\] \n\nFrom \\((4y - 2) = 6\\) we get \\(y = 2\\) , and from \\((4y - 2) = - 6\\) we get \\(y = - 1\\) . \n\nThus we have ruled out all possibilities except for \\(y = - 1,0,1,2\\) . Checking these each individually yields the four answers. \n\n- \\(y = 2\\) yields \\((2)^{5} + 2x(2) = x^{2} + 2(2)^{4}\\) which simplifies to become \\(x^{2} - 4x = 0\\) , and has solutions \\(x = 0\\) and \\(x = 4\\) .- \\(y = 1\\) yields \\((1)^{5} + 2x(1) = x^{2} + 2(1)^{4}\\) which simplifies to become \\(x^{2} - 2x + 1 = 0\\) , and has solution \\(x = 1\\) only.- \\(y = 0\\) yields \\((0)^{5} + 2x(0) = x^{2} + 2(0)^{4}\\) which simplifies to become \\(x^{2} = 0\\) , and has solution \\(x = 0\\) only.- \\(y = -1\\) yields \\((-1)^{5} + 2x(-1) = x^{2} + 2(-1)^{4}\\) which simplifies to \\(x^{2} + 2x + 3 = 0\\) , but this has no real solutions. \n\nThus all solutions are \n\n\\[(x,y) = (0,2),(4,2),(1,1)\\mathrm{~and~}(0,0).\\]", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo2_2021_solutions.jsonl", "problem_match": "\n5. Problem:", "solution_match": "# Alternative Solution:"}}
diff --git a/NewZealand_MO/segmented/en-nzmo2_2022_solutions.jsonl b/NewZealand_MO/segmented/en-nzmo2_2022_solutions.jsonl
index b8861418cc085eb68cec597e3d4003849743db47..dcb6f9e9a046a15e78575b7dbd82eee6d0843c7d 100644
--- a/NewZealand_MO/segmented/en-nzmo2_2022_solutions.jsonl
+++ b/NewZealand_MO/segmented/en-nzmo2_2022_solutions.jsonl
@@ -1,6 +1,6 @@
{"year": "2022", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "NewZealand_MO", "problem": "Find all integers \\(a, b\\) such that \n\n\\[a^{2} + b = b^{2022}.\\]", "solution": ": (Ethan Ng) \n\nLet \\(g = \\gcd (b, b^{2021} - 1)\\) . Since we cannot have both \\(b\\) and \\((b^{2021} - 1)\\) being zero, \\(g\\) must be a positive integer. Since \\(g|b\\) we must have \\(g|b^{2021}\\) and therefore \n\n\\[g\\mid (b^{20221}) - (b^{2021} - 1) = 1.\\] \n\nHence \\(g = 1\\) and thus \\(b\\) and \\((b^{2021} - 1)\\) are coprime. Since the product \\(b\\times (b^{2021} - 1) =\\) \\(b^{2022} - b = a^{2}\\) is a perfect square, we get both factors \\(b\\) and \\((b^{2021} - 1)\\) must be perfect squares or the negatives of perfect squares, or one of them must be zero. \n\n- If both \\(b\\) and \\((b^{2021} - 1)\\) are positive perfect squares, then let \\(b = x^{2}\\) and \\((b^{2021} - 1) = y^{2}\\) . Therefore \\((x^{2})^{2021} - 1 = y^{2}\\) and thus \n\n\\[1 = x^{4042} - y^{2} = (x^{2021} - y)(x^{2021} + y).\\] \n\nHowever since 1 is prime and the only integer factorizations of 1 are \\(1\\times 1\\) and \\((- 1)\\times (- 1)\\) , we must have \n\n\\[(x^{2021} - y) = (x^{2021} + y)\\] \n\nHence \\(y = 0\\) which is a contradiction. \n\n- If both \\(b\\) and \\((b^{2021} - 1)\\) are negative perfect squares, then let \\(b = -x^{2}\\) and \\((b^{2021} - 1) = -y^{2}\\) . Therefore \\((- x^{2})^{2021} - 1 = -y^{2}\\) and thus \n\n\\[x^{4042} + y^{2} = 1.\\] \n\nHowever the minimum value of \\(x^{4042} + y^{2}\\) is \\(1 + 1 = 2\\) , so this is a contradiction too. \n\n- If \\(b = 0\\) then we get \\(a^{2} + 0 = 0^{2022}\\) . So \\(a = 0\\) and we get the solution \\((a, b) = (0, 0)\\) . \n\n- If \\((b^{2021} - 1) = 0\\) then we get \\(b = 1\\) . So \\(a^{2} + 1 = 1^{2022}\\) . So \\(a = 0\\) and we get the solution \\((a, b) = (0, 1)\\) . \n\nTherefore the only solutions for \\((a, b)\\) are: \\((0, 0)\\) and \\((0, 1)\\) .", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo2_2022_solutions.jsonl", "problem_match": "\n1. Problem:", "solution_match": "\nSolution"}}
{"year": "2022", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "NewZealand_MO", "problem": "Find all triples \\((a,b,c)\\) of real numbers such that \n\n\\[a^{2} + b^{2} + c^{2} = 1\\qquad \\mathrm{and}\\qquad a(2b - 2a - c)\\geq \\frac{1}{2}.\\]", "solution": ": (Viet Hoang) \n\nThe equations can be rewritten as \n\n\\[a^{2} + b^{2} + c^{2} = 1\\qquad \\mathrm{and}\\qquad 4ab - 4a^{2} - 2ac\\geq 1.\\] \n\nSubstituting the first equation into the second and rearranging yields \n\n\\[4a b - 4a^{2} - 2a c\\geq a^{2} + b^{2} + c^{2}\\] \\[5a^{2} + b^{2} + c^{2} + 2a c - 4a b\\leq 0\\] \\[(2a - b)^{2} + (a + c)^{2}\\leq 0.\\] \n\nHence \\(b = 2a\\) and \\(c = - a\\) . But \\(a^{2} + b^{2} + c^{2} = 1\\) . So \n\n\\[a^{2} + (2a)^{2} + (-a)^{2} = 1\\] \n\nand thus \\(a = \\pm \\frac{1}{\\sqrt{6}}\\) . Therefore the final answers are: \n\n\\[(a,b,c) = \\left(\\frac{1}{\\sqrt{6}},\\frac{2}{\\sqrt{6}},\\frac{-1}{\\sqrt{6}}\\right)\\mathrm{and}\\left(\\frac{-1}{\\sqrt{6}},\\frac{-2}{\\sqrt{6}},\\frac{1}{\\sqrt{6}}\\right).\\]", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo2_2022_solutions.jsonl", "problem_match": "\n2. Problem:", "solution_match": "\nSolution"}}
{"year": "2022", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(S\\) be a set of 10 positive integers. Prove that one can find two disjoint subsets \\(A = \\{a_{1}, \\ldots , a_{k}\\}\\) and \\(B = \\{b_{1}, \\ldots , b_{k}\\}\\) of \\(S\\) with \\(|A| = |B|\\) such that the sums \n\n\\[x = \\frac{1}{a_{1}} +\\dots +\\frac{1}{a_{k}}\\] \n\nand \n\n\\[y = \\frac{1}{b_{1}} +\\dots +\\frac{1}{b_{k}}\\] \n\ndiffer by less than 0.01; i.e., \\(|x - y|< 1 / 100\\)", "solution": ": (Ishan Nath) \n\nPartition the interval \\((0.00, 2.50]\\) into 250 intervals each of size 0.01. \n\n\\[(0.00, 2.50] = (0.00, 0.01] \\cup (0.01, 0.02] \\cup (0.02, 0.03] \\cup \\dots \\cup (2.49, 2.50].\\] \n\nNow consider all possible sets, \\(S\\) , we can choose from the given 10 positive integers with \\(|S| = 5\\) . Because each of the positive integers must be different, the smallest possible reciprocal sum of one of these sets is \n\n\\[1 + \\frac{1}{2} +\\frac{1}{3} +\\frac{1}{4} +\\frac{1}{5} < 2.50.\\] \n\nTherefore each of these different sets has a reciprocal sum lying somewhere in the interval \\((0.00, 2.50)\\) . The number of such sets \\(S\\) is \\(\\binom{10}{5} = 252\\) but the number of intervals in our partition is only 250. By the pigeonhole principle there exists at least one interval, \\((x, x + 0.01]\\) , and two distinct sets \\(S_{1}, S_{2}\\) such that both reciprocal sums, \n\n\\[\\sum_{s\\in S_{1}}\\frac{1}{s}\\quad \\mathrm{and}\\quad \\sum_{s\\in S_{2}}\\frac{1}{s}\\] \n\nlie in \\((x, x + 0.01]\\) . The reciprocal sums of \\(S_{1}\\) and \\(S_{2}\\) have difference less than \\(1 / 100\\) because they both lie in the interval \\((x, x + 0.01]\\) . If \\(S_{1}\\) and \\(S_{2}\\) are disjoint then we can simply choose \\(A = S_{1}\\) and \\(B = S_{2}\\) . Otherwise let \\(C = S_{1} \\cap S_{2}\\) be the intersection of \\(S_{1}\\) and \\(S_{2}\\) and then let \\(A = S_{1} \\setminus C\\) and let \\(B = S_{2} \\setminus C\\) . The sets \\(A\\) and \\(B\\) are disjoint and equisized because \\(|S_{1}| = |S_{2}|\\) and \\(S_{1} \\neq S_{2}\\) .", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo2_2022_solutions.jsonl", "problem_match": "\n3. Problem:", "solution_match": "\nSolution"}}
-{"year": "2022", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "NewZealand_MO", "problem": "Triangle \\(ABC\\) is right-angled at \\(B\\) and has incentre \\(I\\) . Points \\(D\\) , \\(E\\) and \\(F\\) are the points where the incircle of the triangle touches the sides \\(BC\\) , \\(AC\\) and \\(AB\\) respectively. Lines \\(CI\\) and \\(EF\\) intersect at point \\(P\\) . Lines \\(DP\\) and \\(AB\\) intersect at point \\(Q\\) . Prove that \\(AQ = BF\\) .", "solution": ": (Kevin Shen) \n\nFirst note that \\(ID = IE = IF\\) because they are all radii of the incircle, and \\(\\angle BFI = \\angle BDI = 90^{\\circ}\\) because tangents are perpendicular to radii. Since \\(\\angle ABC = 90^{\\circ}\\) we have \\(BFID\\) a square and so \\(BD = BF = ID\\) too. Thus \\(\\triangle EIF\\) is isosceles and so \\(\\angle IFE = \\angle FEI\\) . \n\n\n \n\nSince \\(CI\\) is the bisector of \\(\\angle DCE\\) , we see that \\(D\\) and \\(E\\) are reflections of each other over line \\(CIP\\) . Therefore \\(\\angle IDP = \\angle PEI\\) . Hence \n\n\\[\\angle IDP = \\angle FEI = \\angle IFP\\] \n\nand therefore quadrilateral \\(F PID\\) is cyclic. Therefore \\(\\angle FPD = \\angle FID = 90^{\\circ}\\) ( \\(BFID\\) is a square). Since \\(AI\\) is the bisector of \\(\\angle EAF\\) , we see that \\(E\\) and \\(F\\) are reflections of each other over line \\(AI\\) . Therefore \\(EF \\perp AI\\) . Hence \n\n\\[AI||DQ\\] \n\nbecause they are both perpendicular to \\(EF\\) . We also have \\(AQ||ID\\) (because \\(BFID\\) is a square) so \\(QAID\\) is a parallelogram. Therefore \n\n\\[AQ = ID = BF\\] \n\nas required.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo2_2022_solutions.jsonl", "problem_match": "\n4. Problem:", "solution_match": "\nSolution"}}
+{"year": "2022", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "NewZealand_MO", "problem": "Triangle \\(ABC\\) is right-angled at \\(B\\) and has incentre \\(I\\) . Points \\(D\\) , \\(E\\) and \\(F\\) are the points where the incircle of the triangle touches the sides \\(BC\\) , \\(AC\\) and \\(AB\\) respectively. Lines \\(CI\\) and \\(EF\\) intersect at point \\(P\\) . Lines \\(DP\\) and \\(AB\\) intersect at point \\(Q\\) . Prove that \\(AQ = BF\\) .", "solution": ": (Kevin Shen) \n\nFirst note that \\(ID = IE = IF\\) because they are all radii of the incircle, and \\(\\angle BFI = \\angle BDI = 90^{\\circ}\\) because tangents are perpendicular to radii. Since \\(\\angle ABC = 90^{\\circ}\\) we have \\(BFID\\) a square and so \\(BD = BF = ID\\) too. Thus \\(\\triangle EIF\\) is isosceles and so \\(\\angle IFE = \\angle FEI\\) . \n\n\n \n\nSince \\(CI\\) is the bisector of \\(\\angle DCE\\) , we see that \\(D\\) and \\(E\\) are reflections of each other over line \\(CIP\\) . Therefore \\(\\angle IDP = \\angle PEI\\) . Hence \n\n\\[\\angle IDP = \\angle FEI = \\angle IFP\\] \n\nand therefore quadrilateral \\(F PID\\) is cyclic. Therefore \\(\\angle FPD = \\angle FID = 90^{\\circ}\\) ( \\(BFID\\) is a square). Since \\(AI\\) is the bisector of \\(\\angle EAF\\) , we see that \\(E\\) and \\(F\\) are reflections of each other over line \\(AI\\) . Therefore \\(EF \\perp AI\\) . Hence \n\n\\[AI||DQ\\] \n\nbecause they are both perpendicular to \\(EF\\) . We also have \\(AQ||ID\\) (because \\(BFID\\) is a square) so \\(QAID\\) is a parallelogram. Therefore \n\n\\[AQ = ID = BF\\] \n\nas required.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo2_2022_solutions.jsonl", "problem_match": "\n4. Problem:", "solution_match": "\nSolution"}}
{"year": "2022", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "NewZealand_MO", "problem": "The sequence \\(x_{1},x_{2},x_{3},\\ldots\\) is defined by \\(x_{1} = 2022\\) and \\(x_{n + 1} = 7x_{n} + 5\\) for all positive integers \\(n\\) . Determine the maximum positive integer \\(m\\) such that \n\n\\[{\\frac{x_{n}(x_{n}-1)(x_{n}-2)\\ldots(x_{n}-m+1)}{m!}}\\] \n\nis never a multiple of 7 for any positive integer \\(n\\) .", "solution": "A: (Ishan Nath) \n\nWe claim the answer is 404. First, we notice that \\(m\\leq 2022\\) . Otherwise, \n\n\\[{\\frac{x_{1}(x_{1}-1)\\cdots(x_{1}-m+1)}{m!}}={\\frac{2022(2022-1)\\cdots(2022-m+1)}{m!}}=0,\\] \n\nwhich is a multiple of 7. Then, since \\(x_{n}\\geq x_{1} = 2022\\) for all \\(n\\) , we can write \n\n\\[{\\frac{x_{n}(x_{n}-1)\\cdots(x_{n}-m+1)}{m!}}={\\frac{x_{n}(x_{n}-1)\\cdots(x_{n}-m+ 1)(x_{n}-m)(x_{n}-m- 1)\\cdots 1}{m!\\times(x_{n}-m)(x_{n}-m- 1)\\cdots 1}}\\] \\[={\\frac{x_{n}!}{m!(x_{n}-m)!}}.\\] \n\nFor a positive integer \\(n\\) , we define \\(\\nu (n)\\) as the exponent of 7 in the prime factorization of \\(n\\) . For example, \\(\\nu (1) = 0\\) and \\(\\nu (98) = 2\\) . Note \\(\\nu (ab) = \\nu (a) + \\nu (b)\\) and \\(\\nu (a / b) = \\nu (a) - \\nu (b)\\) . We prove the following two Lemmas: \n\n- Lemma 1: \n\n\\[\\nu (n!) = \\sum_{i = 1}^{d}\\left\\lfloor {\\frac{n}{7^{i}}}\\right\\rfloor = \\left\\lfloor {\\frac{n}{7}}\\right\\rfloor +\\left\\lfloor {\\frac{n}{49}}\\right\\rfloor +\\dots +\\left\\lfloor {\\frac{n}{7^{d}}}\\right\\rfloor ,\\] \n\nwhere \\(d\\) is the largest integer such that \\(7^{d}\\leq n\\) , for all non- negative integers \\(n\\) . \n\nProof: Note that \n\n\\[\\nu (n!) = \\nu (1) + \\nu (2) + \\dots +\\nu (n).\\] \n\n\\(\\nu (n!)\\) has a contribution of \\(+1\\) for each multiple of seven (less than or equal to \\(n\\) ) and there are \\(\\lfloor n / 7\\rfloor\\) such numbers. Each multiple of 49 contributes another \\(+1\\) to \\(\\nu (n!)\\) , and there are \\(\\lfloor n / 49\\rfloor\\) such numbers. Generally, there is an additional \\(+1\\) contributed for each multiple of \\(7^{i}\\) , of which there are \\(\\lfloor n / 7^{i}\\rfloor\\) numbers. Adding up all these contributions, we get the desired result. \n\n- Lemma 2: \\(\\lfloor x\\rfloor -\\lfloor y\\rfloor -\\lfloor x-y\\rfloor \\geq 0\\) , for all reals \\(x\\) and \\(y\\) , with equality if and only if \\(\\{x\\} \\geq \\{y\\}\\) . \n\nProof: Let \\(x = \\lfloor x\\rfloor +\\{x\\}\\) and \\(y = \\lfloor y\\rfloor +\\{y\\}\\) . Then \\(x - y = \\lfloor x\\rfloor -\\lfloor y\\rfloor +\\{x\\} - \\{y\\}\\) . Since \\(0< \\{x\\} ,\\{y\\} < 1\\) , we have \\(- 1< \\{x\\} - \\{y\\} < 1\\) . \n\n(a) If \\(0\\leq \\{x\\} - \\{y\\} < 1\\) , i.e. \\(\\{x\\} \\geq \\{y\\}\\) , then \\(\\lfloor x - y\\rfloor = \\lfloor x\\rfloor - \\lfloor y\\rfloor\\) , so \\(\\lfloor x\\rfloor = \\lfloor y\\rfloor +\\lfloor x - y\\rfloor\\) . \n\n(b) Otherwise, \\(\\lfloor x - y\\rfloor = \\lfloor x\\rfloor - \\lfloor y\\rfloor - 1\\) , so \\(\\lfloor x\\rfloor >\\lfloor x\\rfloor - 1 = \\lfloor y\\rfloor +\\lfloor x - y\\rfloor\\) .\n\n\n\nNow we can use Lemma 1 to compute \n\n\\[\\nu \\left(\\frac{x_{n}!}{m!(x_{n} - m)!}\\right) = \\nu (x_{n}!) - \\nu (m!) - \\nu ((x_{n} - m)!)\\] \\[\\qquad = \\sum_{i = 1}^{d}\\left\\lfloor \\frac{x_{n}}{\\tau^{i}}\\right\\rfloor -\\sum_{i = 1}^{l_{1}}\\left\\lfloor \\frac{m}{\\tau^{i}}\\right\\rfloor -\\sum_{i = 1}^{l_{2}}\\left\\lfloor \\frac{x_{n} - m}{\\tau^{i}}\\right\\rfloor\\] \\[\\qquad = \\sum_{i = 1}^{d}\\left\\lfloor \\frac{x_{n}}{\\tau^{i}}\\right\\rfloor -\\sum_{i = 1}^{d}\\left\\lfloor \\frac{m}{\\tau^{i}}\\right\\rfloor -\\sum_{i = 1}^{d}\\left\\lceil \\frac{x_{n} - m}{\\tau^{i}}\\right\\rceil\\] \\[\\qquad = \\sum_{i = 1}^{d}\\left(\\left\\lfloor \\frac{x_{n}}{\\tau^{i}}\\right\\rfloor -\\left\\lfloor \\frac{m}{\\tau^{i}}\\right\\rfloor -\\left\\lfloor \\frac{x_{n} - m}{\\tau^{i}}\\right\\rfloor\\right).\\] \n\nHere \\(d\\) is the largest integer such that \\(7^{d} \\leq x_{n}\\) , \\(l_{1}\\) is the largest integer such that \\(7^{l_{1}} \\leq m\\) , and \\(l_{2}\\) is the largest integer such that \\(7^{l_{2}} \\leq x_{n} - m\\) . Increasing the range of the sums does not affect the result, as we are simply adding terms of the form \\(\\lfloor a / 7^{b}\\rfloor\\) , where \\(7^{b} > a\\) , which gives 0. \n\nIf we let \\(x = x_{n} / 7^{i}\\) and \\(y = m / 7^{i}\\) , then this final sum consists of terms of the form \\(\\lfloor x\\rfloor - \\lfloor y\\rfloor - \\lfloor x - y\\rfloor \\geq 0\\) . Therefore, we get that \n\n\\[7\\mathrm{doesn't~divide}\\frac{x_{n}(x_{n} - 1)\\cdot\\cdot\\cdot(x_{n} - m + 1)}{m!}\\] \\[\\mathrm{if~and~only~if~}\\nu \\left(\\frac{x_{n}!}{m!(x_{n} - m)!}\\right) = 0\\] \\[\\mathrm{if~and~only~if~}\\left\\lfloor \\frac{x_{n}}{\\tau^{i}}\\right\\rfloor = \\left\\lfloor \\frac{m}{\\tau^{i}}\\right\\rfloor +\\left\\lfloor \\frac{x_{n} - m}{\\tau^{i}}\\right\\rfloor \\mathrm{for~all~}0\\leq i\\leq d\\] \\[\\mathrm{if~and~only~if~}\\left\\{\\frac{x_{n}}{\\tau^{i}}\\right\\} \\geq \\left\\{\\frac{m}{\\tau^{i}}\\right\\}\\] \\[\\mathrm{for~all~}0\\leq i\\leq d.\\] \n\nby Lemma 2. This must hold for all \\(n\\) . Notice \n\n\\[\\left\\{\\frac{x_{n}}{\\tau^{i}}\\right\\} \\geq \\left\\{\\frac{m}{\\tau^{i}}\\right\\} \\mathrm{if~and~only~if~}\\tau^{i}\\left\\{\\frac{x_{n}}{\\tau^{i}}\\right\\} \\geq \\tau^{i}\\left\\{\\frac{m}{\\tau^{i}}\\right\\} ,\\] \n\nand \\(7^{b}\\{a / 7^{b}\\}\\) is simply the remainder of \\(a\\) modulo \\(7^{b}\\) . Hence we have \n\n\\[\\left\\{\\frac{x_{n}}{\\tau^{i}}\\right\\} \\geq \\left\\{\\frac{m}{\\tau^{i}}\\right\\} \\mathrm{if~and~only~if~}x_{n} \\pmod {7^{i}} \\geq m \\pmod {7^{i}}.\\] \n\nSince \\(x_{1} = 2022 = 5 \\cdot 7^{3} + 6 \\cdot 7^{2} + 1 \\cdot 7^{1} + 6 \\cdot 7^{0}\\) , we inductively get \n\n\\[x_{n} = 5 \\cdot 7^{n + 2} + 6 \\cdot 7^{n + 1} + 1 \\cdot 7^{n} + 6 \\cdot 7^{n - 1} + 5 \\cdot 7^{n - 2} + \\dots + 5 \\cdot 7^{0}.\\] \n\nUsing this, we can find the smallest value of \\(x_{n}\\) (mod \\(7^{i}\\) ): \n\n- For \\(i = 1\\) , the smallest value is \\(5 \\cdot 7^{0}\\) , when \\(n \\geq 2\\) . \n\n- For \\(i = 2\\) , the smallest value is \\(1 \\cdot 7^{1} + 6 \\cdot 7^{0}\\) , when \\(n = 1\\) . \n\n- For \\(i = 3\\) , the smallest value is \\(1 \\cdot 7^{2} + 6 \\cdot 7^{1} + 5 \\cdot 7^{0}\\) , when \\(n = 2\\) . \n\n- For \\(i = 4\\) , the smallest value is \\(1 \\cdot 7^{3} + 6 \\cdot 7^{2} + 5 \\cdot 7^{1} + 5 \\cdot 7^{0}\\) , when \\(n = 3\\) . \n\n- For \\(i \\geq 5\\) , the smallest value is \\(5 \\cdot 7^{3} + 6 \\cdot 7^{2} + 1 \\cdot 7^{1} + 6 \\cdot 7^{0}\\) , when \\(n = 1\\) .\n\n\n\nThus if we write \\(m\\) in the form \\(m = a_{3}\\cdot 7^{3} + a_{2}\\cdot 7^{2} + a_{1}\\cdot 7^{1} + a_{0}\\cdot 7^{0}\\) , where \\(0\\leq a_{i}\\leq 6\\) , we must have \\(a_{0}\\leq 5\\) , \\(a_{1}\\leq 1\\) , \\(a_{2}\\leq 1\\) , and \\(a_{3}\\leq 1\\) , which are necessary and sufficient. \n\nTherefore the maximum integer \\(m\\) is achieved when \\(a_{0} = 5\\) and \\(a_{1} = a_{2} = a_{3} = 1\\) . This gives \\(m = 7^{3} + 7^{2} + 7^{1} + 5 = 404\\) .", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo2_2022_solutions.jsonl", "problem_match": "\n5. Problem:", "solution_match": "\nSolution"}}
{"year": "2022", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "NewZealand_MO", "problem": "The sequence \\(x_{1},x_{2},x_{3},\\ldots\\) is defined by \\(x_{1} = 2022\\) and \\(x_{n + 1} = 7x_{n} + 5\\) for all positive integers \\(n\\) . Determine the maximum positive integer \\(m\\) such that \n\n\\[{\\frac{x_{n}(x_{n}-1)(x_{n}-2)\\ldots(x_{n}-m+1)}{m!}}\\] \n\nis never a multiple of 7 for any positive integer \\(n\\) .", "solution": "B: (Ishan Nath) \n\nAs in solution A, we make the observation that \n\n\\[\\frac{x_{n}(x_{n} - 1)\\cdot\\cdot\\cdot(x_{n} - m + 1)}{m!} = \\binom{x_{n}}{m}.\\] \n\n- Lemma: Let \\(a\\) and \\(b\\) be two positive integers with \\(\\overline{a_{k}a_{k - 1}\\cdot\\cdot\\cdot a_{0}}\\) and \\(\\overline{b_{k}b_{k - 1}\\cdot\\cdot\\cdot b_{0}}\\) the base 7 representations of \\(a\\) and \\(b\\) respectively, possibly with leading zeroes. \\(\\binom{a}{b}\\) is not a multiple of 7 if and only if \\(a_{i}\\geq b_{i}\\) for all \\(0\\leq i\\leq k\\) . \n\nProof: By Lucas' Theorem \n\n\\[\\binom{a}{b}\\equiv\\prod_{i=0}^{k}\\binom{a_{k}}{b_{k}}\\pmod{7}.\\] \n\nIn particular, we have \n\n\\[\\binom{a}{b}\\equiv0\\pmod{7}\\mathrm{if~and~only~if~}\\binom{a_{k}}{b_{k}}\\equiv0\\pmod{7}\\mathrm{for~some~}k\\] \\[\\mathrm{if~and~only~if~}a_{k}< b_{k}\\mathrm{~for~some~}k.\\] \n\nNow, note \\(x_{1} = 2022 = 5616\\) in base 7, and thus \\(x_{n} = 5616555\\dots 55\\) in base 7 (where the rightmost \\(n - 1\\) digits are '5's). Let \\(x_{n} = \\overline{a_{k}a_{k - 1}\\cdot\\cdot\\cdot a_{0}}\\) in base 7. Hence if \\(m =\\) \\(\\overline{m_{k}m_{k - 1}\\cdot\\cdot\\cdot m_{0}}\\) , we get that \n\n\\[7\\dagger \\binom{x_{n}}{m}\\mathrm{~for~all~}n\\mathrm{~if~and~only~if~}a_{k}\\geq m_{k}\\mathrm{~for~all~}n,k.\\] \n\nWhen \\(n = 1\\) we have \\(a_{k} = 0\\) for \\(k\\geq 4\\) , so this means \\(m_{k} = 0\\) for \\(k\\geq 4\\) . Also notice that \n\n- \\(a_{0}\\geq 5\\) with equality when \\(n\\geq 2\\) , \n\n- \\(a_{1}\\geq 1\\) with equality when \\(n = 1\\) , \n\n- \\(a_{2}\\geq 1\\) with equality when \\(n = 2\\) , and \n\n- \\(a_{3}\\geq 1\\) with equality when \\(n = 3\\) . \n\nThis implies \\(m_{0}\\leq 5\\) and \\(m_{1},m_{2},m_{3}\\leq 1\\) , which gives necessary and sufficient conditions for \\(m\\) . \n\nThe maximum value of \\(m\\) can now be found by taking the maximum values of all \\(m_{k}\\) . This gives \\(m = 1115\\) in base 7, or \\(m = 404\\) (in base 10).", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo2_2022_solutions.jsonl", "problem_match": "\n5. Problem:", "solution_match": "\nSolution"}}
diff --git a/NewZealand_MO/segmented/en-nzmo2_2023_solutions.jsonl b/NewZealand_MO/segmented/en-nzmo2_2023_solutions.jsonl
index 8a5b41de0ec40bdd316faff9c30f3317380e2db4..0ce67eb9d1a19076e3a726d29af116e38835f0fe 100644
--- a/NewZealand_MO/segmented/en-nzmo2_2023_solutions.jsonl
+++ b/NewZealand_MO/segmented/en-nzmo2_2023_solutions.jsonl
@@ -4,7 +4,7 @@
{"year": "2023", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(a\\) , \\(b\\) and \\(c\\) be positive real numbers such that \\(a + b + c = abc\\) . Prove that at least one of \\(a\\) , \\(b\\) or \\(c\\) is greater than \\(\\frac{17}{10}\\) .", "solution": "A: (Ross Atkins) \n\nWlog assume \\(a\\geq b\\geq c\\) . Therefore \\(a + b + c\\geq 3c\\) . Now for a proof by contradiction, assume \\(a\\leq \\frac{17}{10}\\) and \\(b\\leq \\frac{17}{10}\\) . Since \\(\\left(\\frac{17}{10}\\right)^2 = \\frac{289}{100} < 3\\) , it follows that \\(\\frac{17}{10} < \\sqrt{3}\\) and therefore \\(ab\\leq 3\\) . However this implies: \n\n\\[3c\\leq a + b + c = abc< 3c\\] \n\nwhich is a contradiction.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo2_2023_solutions.jsonl", "problem_match": "\n2. Problem:", "solution_match": "\nSolution"}}
{"year": "2023", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(a\\) , \\(b\\) and \\(c\\) be positive real numbers such that \\(a + b + c = abc\\) . Prove that at least one of \\(a\\) , \\(b\\) or \\(c\\) is greater than \\(\\frac{17}{10}\\) .", "solution": "B: (Michael Albert) \n\nFrom \\(abc = a + b + c\\) we get \n\n\\[1 = \\frac{a + b + c}{abc} = \\frac{1}{bc} +\\frac{1}{ca} +\\frac{1}{ab}\\] \n\nAs all three terms are positive, at least one must be less than or equal to \\(1 / 3\\) and, without loss of generality, we can assume \\(1 / bc\\leq 1 / 3\\) . But then \\(bc\\geq 3\\) and at least one of \\(b\\) or \\(c\\) must be greater than \\(\\sqrt{3}\\) . Since \\(\\sqrt{3} >17 / 10\\) we're done.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo2_2023_solutions.jsonl", "problem_match": "\n2. Problem:", "solution_match": "\nSolution"}}
{"year": "2023", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(a\\) , \\(b\\) and \\(c\\) be positive real numbers such that \\(a + b + c = abc\\) . Prove that at least one of \\(a\\) , \\(b\\) or \\(c\\) is greater than \\(\\frac{17}{10}\\) .", "solution": "C: (Ross Atkins) \n\nBy the AM- GM inequality we have \\(\\frac{a + b + c}{3} \\geq \\sqrt[3]{abc}\\) and thus \\(abc \\geq 3\\sqrt[3]{abc}\\) , which rearranges to give \\(\\sqrt[3]{abc} \\geq \\sqrt{3}\\) . Now wlog \\(a \\geq b \\geq c\\) and so \n\n\\[a = \\sqrt[3]{a^3} \\geq \\sqrt[3]{abc} \\geq \\sqrt{3} > \\frac{17}{10}.\\]", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo2_2023_solutions.jsonl", "problem_match": "\n2. Problem:", "solution_match": "\nSolution"}}
-{"year": "2023", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(ABCD\\) be a square (vertices labelled in clockwise order). Let \\(Z\\) be any point on diagonal \\(AC\\) between \\(A\\) and \\(C\\) such that \\(AZ > ZC\\) . Points \\(X\\) and \\(Y\\) exist such that \\(AXYZ\\) is a square (vertices labelled in clockwise order) and point \\(B\\) lies inside \\(AXYZ\\) . Let \\(M\\) be the point of intersection of lines \\(BX\\) and \\(DZ\\) (extended if necessary). Prove that \\(C\\) , \\(M\\) and \\(Y\\) are collinear.", "solution": ": (Kevin Shen) \n\nSince \\(Z\\) lies on diagonal \\(AC\\) , we have \\(\\angle DAZ = 45^{\\circ}\\) and \\(\\angle ZAB = 45^{\\circ}\\) . Therefore \\(B\\) lies on diagonal \\(AY\\) of square \\(AXYZ\\) and \\(\\angle BAX = 45^{\\circ}\\) . \n\n\n \n\nSince \\(AB = AD\\) and \\(AX = AZ\\) and \\(\\angle BAX = 45^{\\circ} = \\angle DAZ\\) , we have congruent triangles \n\n\\[\\triangle DAZ\\equiv \\triangle BAX\\qquad (SAS).\\] \n\nTherefore let \\(x = \\angle ZDA = \\angle XBA\\) and \\(y = \\angle AZD = \\angle AXB\\) . By the angle sum in triangle \\(DAZ\\) we have \\(\\angle DAZ + \\angle AZD + \\angle ZDA = 180^{\\circ}\\) . Therefore \\(x + y = 135^{\\circ}\\) . Now by the angle sum in quadrilateral \\(DAXM\\) we get \\(\\angle DAX + \\angle AXM + \\angle XMD + \\angle MDA = 360^{\\circ}\\) . Therefore \n\n\\[\\angle BMD = 90^{\\circ}.\\] \n\nHence \\(ABMCD\\) is cyclic (the circle with diameter \\(BD\\) ). Therefore \n\n\\[\\angle DMC = \\angle DAC = 45^{\\circ}.\\] \n\nAlso \\(AXYMZ\\) is cyclic (the circle with diameter \\(XZ\\) ). Therefore \n\n\\[\\angle YMX = \\angle YAX = 45^{\\circ}.\\] \n\nHence \\(\\angle YMC = \\angle YMX + \\angle YMD + \\angle DMC = 45^{\\circ} + 90^{\\circ} + 45^{\\circ} = 180^{\\circ}\\) as required.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo2_2023_solutions.jsonl", "problem_match": "\n3. Problem:", "solution_match": "\nSolution"}}
+{"year": "2023", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(ABCD\\) be a square (vertices labelled in clockwise order). Let \\(Z\\) be any point on diagonal \\(AC\\) between \\(A\\) and \\(C\\) such that \\(AZ > ZC\\) . Points \\(X\\) and \\(Y\\) exist such that \\(AXYZ\\) is a square (vertices labelled in clockwise order) and point \\(B\\) lies inside \\(AXYZ\\) . Let \\(M\\) be the point of intersection of lines \\(BX\\) and \\(DZ\\) (extended if necessary). Prove that \\(C\\) , \\(M\\) and \\(Y\\) are collinear.", "solution": ": (Kevin Shen) \n\nSince \\(Z\\) lies on diagonal \\(AC\\) , we have \\(\\angle DAZ = 45^{\\circ}\\) and \\(\\angle ZAB = 45^{\\circ}\\) . Therefore \\(B\\) lies on diagonal \\(AY\\) of square \\(AXYZ\\) and \\(\\angle BAX = 45^{\\circ}\\) . \n\n\n \n\nSince \\(AB = AD\\) and \\(AX = AZ\\) and \\(\\angle BAX = 45^{\\circ} = \\angle DAZ\\) , we have congruent triangles \n\n\\[\\triangle DAZ\\equiv \\triangle BAX\\qquad (SAS).\\] \n\nTherefore let \\(x = \\angle ZDA = \\angle XBA\\) and \\(y = \\angle AZD = \\angle AXB\\) . By the angle sum in triangle \\(DAZ\\) we have \\(\\angle DAZ + \\angle AZD + \\angle ZDA = 180^{\\circ}\\) . Therefore \\(x + y = 135^{\\circ}\\) . Now by the angle sum in quadrilateral \\(DAXM\\) we get \\(\\angle DAX + \\angle AXM + \\angle XMD + \\angle MDA = 360^{\\circ}\\) . Therefore \n\n\\[\\angle BMD = 90^{\\circ}.\\] \n\nHence \\(ABMCD\\) is cyclic (the circle with diameter \\(BD\\) ). Therefore \n\n\\[\\angle DMC = \\angle DAC = 45^{\\circ}.\\] \n\nAlso \\(AXYMZ\\) is cyclic (the circle with diameter \\(XZ\\) ). Therefore \n\n\\[\\angle YMX = \\angle YAX = 45^{\\circ}.\\] \n\nHence \\(\\angle YMC = \\angle YMX + \\angle YMD + \\angle DMC = 45^{\\circ} + 90^{\\circ} + 45^{\\circ} = 180^{\\circ}\\) as required.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo2_2023_solutions.jsonl", "problem_match": "\n3. Problem:", "solution_match": "\nSolution"}}
{"year": "2023", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "NewZealand_MO", "problem": "For any positive integer \\(n\\) , let \\(f(n)\\) be the number of subsets of \\(\\{1,2,\\ldots ,n\\}\\) whose sum is equal to \\(n\\) . Does there exist infinitely many positive integers \\(m\\) such that \\(f(m) = f(m + 1)\\) ? (Note that each element in a subset must be distinct.)", "solution": ": (Michael Albert) \n\nLet \\(S(n)\\) be the set of such subsets. Consider the map from \\(S(n)\\) to \\(S(n + 1)\\) that adds one to the largest element of each \\(A \\in S(n)\\) . This map is an injection (needs proof but easy) and not a surjection provided that \\(S(n + 1)\\) contains a set whose largest and second largest elements differ by one. For even \\(n = 2k \\geqslant 2\\) this is true since we can take \\(\\{k, k + 1\\} \\in S(n + 1)\\) and for odd \\(n = 2k + 1 \\geqslant 5\\) this is true since we can take \\(\\{1, k, k + 1\\}\\) . So for \\(n \\geqslant 5\\) , we must have \\(f(n) < f(n + 1)\\) and there do not exist infinitely many such pairs.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo2_2023_solutions.jsonl", "problem_match": "\n4. Problem:", "solution_match": "\nSolution"}}
{"year": "2023", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(x\\) , \\(y\\) and \\(z\\) be real numbers such that: \\(x^{2} = y + 2\\) , and \\(y^{2} = z + 2\\) , and \\(z^{2} = x + 2\\) . Prove that \\(x + y + z\\) is an integer.", "solution": "A: (Ross Atkins) \n\nFirst we exclude \\(- 1\\) and 2: \n\n\\(x = 2\\) implies \\(y = 2\\) implies \\(z = 2\\) implies \\(x = 2\\) \n\n\\(x = - 1\\) implies \\(y = - 1\\) implies \\(z = - 1\\) implies \\(x = - 1\\) \n\nIn both these cases we have \\(x + y + z\\) being an integer. So henceforth we assume none of \\(x,y,z\\) are 2 nor \\(- 1\\) . Now let \\(x,y,z\\) be the roots of the following cubic equation. \n\n\\[(\\lambda -x)(\\lambda -y)(\\lambda -z) = \\lambda^{3} - A\\lambda^{2} + B\\lambda -C.\\] \n\nApplying Viete's formula to this cubic gives us \\(A = x + y + z\\) , \\(B = xy + yz + zx\\) and \\(C = xyz\\) . This means that \\(x^{2} + y^{2} + z^{2} = (x + y + z)^{2} - 2(xy + yz + zx) = A^{2} - 2B\\) . Now sum the three given equations ( \\(x^{2} = y + 2\\) and \\(y^{2} = z + 2\\) and \\(z^{2} = x + 2\\) ) to get \n\n\\[A^{2} - 2B = x^{2} + y^{2} + z^{2} = (y + 2) + (z + 2) + (x + 2) = A + 6.\\] \n\n\\[A^{2} - A - 2B = 6. \\quad (1)\\] \n\nNext rearrange the equations to be \\(x^{2} - 1 = y + 1\\) and \\(y^{2} - 1 = z + 1\\) and \\(z^{2} - 1 = x + 1\\) . These can then be multiplied to get \n\n\\[(x^{2} - 1)(y^{2} - 1)(z^{2} - 1) = (y + 1)(z + 1)(x + 1)\\] \\[(x - 1)(x + 1)(y - 1)(y + 1)(z - 1)(z + 1) = (y + 1)(z + 1)(x + 1)\\] \\[(x - 1)(y - 1)(z - 1) = 1\\] \\[xyz - (xy + yz + zx) + (x + y + z) - 1 = 1\\] \n\n\\[B - A + 2 = C \\quad (2)\\] \n\nIn the above algebraic manipulation, we are allowed to cancel the \\((x + 1)(y + 1)(z + 1)\\) factor because none of \\(x,y,z\\) are equal to \\(- 1\\) . Finally rearrange the equations to be \\(x^{2} - 4 = y - 2\\) and \\(y^{2} - 4 = z - 2\\) and \\(z^{2} - 4 = x - 2\\) . These can then be multiplied to get \n\n\\[(x^{2} - 4)(y^{2} - 4)(z^{2} - 4) = (y - 2)(z - 2)(x - 2)\\] \\[(x - 2)(x + 2)(y - 2)(y + 2)(z - 2)(z + 2) = (y - 2)(z - 2)(x - 2)\\] \\[(x + 2)(y + 2)(z + 2) = 1\\] \\[xyz + 2(xy + yz + zx) + 4(x + y + z) + 8 = 1\\] \n\n\\[C + 2B + 4A = -7\\] \n\n\\[C = -4A - 2B - 7 \\quad (3)\\] \n\nIn the above algebraic manipulation, we are allowed to cancel the \\((x - 2)(y - 2)(z - 2)\\) factor because none of \\(x,y,z\\) are equal to 2. Combining equations (2) and (3) gives us \\(B - A + 2 = C = - 4A - 2B - 7\\) which rearranges to give us \\(B = - A - 3\\) . Substituting this into 1 gives us. \n\n\\[A^{2} - A - 2B = 6\\] \\[A^{2} - A - 2(-A - 3) = 6\\] \\[A(A + 1) = 0.\\] \n\nTherefore \\(A = 0\\) or \\(A = - 1\\) both of which are integers. Since \\(A = x + y + z\\) we are done.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo2_2023_solutions.jsonl", "problem_match": "\n5. Problem:", "solution_match": "\nSolution"}}
{"year": "2023", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(x\\) , \\(y\\) and \\(z\\) be real numbers such that: \\(x^{2} = y + 2\\) , and \\(y^{2} = z + 2\\) , and \\(z^{2} = x + 2\\) . Prove that \\(x + y + z\\) is an integer.", "solution": "B: (Ross Atkins) \n\nConsider the polynomial \\(P\\) defined by \n\n\\[P(\\lambda) = \\lambda^{8} - 8\\lambda^{6} + 20\\lambda^{4} - 16\\lambda^{2} - \\lambda +2\\] \\[\\qquad = (\\lambda +1)(\\lambda -2)(\\lambda^{3} - 3\\lambda +1)(\\lambda^{3} + \\lambda^{2} - 2\\lambda -1).\\] \n\nIf we substitute \\(z = y^{2} - 2\\) into \\(z^{2} = x + 2\\) gives us \\((y^{2} - 2)^{2} = x + 2\\) . Then substitute \\(y = x^{2} - 2\\) to get \\(\\left((x^{2} - 2)^{2} - 2\\right)^{2} = x + 2\\) . Expanding gives \n\n\\[x^{8} - 8x^{6} + 20x^{4} - 16x^{2} - x + 2 = 0.\\] \n\nTherefore \\(x\\) is a root of the polynomial \\(P\\) . By symmetry we must have all of \\(x,y,z\\) being roots of \\(P\\) . Now we consider cases: \n\nCase 1: at least one of \\(x,y,z\\) is equal to \\(- 1\\) . Wlog assume \\(x = - 1\\) . Using \\(y = x^{2} - 2\\) we get \\(y = - 1\\) . Then using \\(z = y^{2} - 2\\) we get \\(z = - 1\\) . In this case we get \\((x,y,z) = (- 1, - 1, - 1)\\) which has sum \\(- 3\\) which is an integer. \n\nCase 2: at least one of \\(x,y,z\\) is equal to 2. Wlog assume \\(x = 2\\) . Using \\(y = x^{2} - 2\\) we get \\(y = 2\\) . Then using \\(z = y^{2} - 2\\) we get \\(z = 2\\) . In this case we get \\((x,y,z) = (2,2,2)\\) which has sum 6 which is an integer. \n\nCase 3: at least one of \\(x,y,z\\) is a root of \\((\\lambda^{3} - 3\\lambda +1)\\) . Wlog assume \\(x^{3} - 3x + 1 = 0\\) . Note that \\(z = y^{2} - 2 = (x^{2} - 2)^{2} - 2 = x^{4} - 4x + 2\\) . Now consider the sum of \\(x\\) and \\(y = x^{2} - 2\\) and \\(z = x^{4} - 4x + 2\\) , \n\n\\[x + y + z = x + (x^{2} - 2) + (x^{4} - 4x^{2} + 2) = x^{4} - 3x^{2} + x = (x^{3} - 3x + 1)x.\\] \n\nBut since \\(x^{3} - 3x + 1 = 0\\) this means \\(x + y + z = 0\\) in this case. \n\nCase 4: some two of \\(x,y,z\\) are the same. Wlog assume \\(x = y\\) therefore \\(x = y = x^{2} - 2\\) . Hence \n\n\\[0 = x^{2} - x - 2 = (x - 2)(x + 1).\\] \n\nand so \\(x = - 1\\) or \\(x = 2\\) , and this was covered in cases 1 and 2. \n\nCase 5: Since \\(x,y,z\\) are all roots of \\(P\\) , the only remaining possibility is that \\(x,y\\) and \\(z\\) are distinct roots of \\(\\lambda^{3} + \\lambda^{2} - 2\\lambda - 1\\) . By Viete's formula this means that the sum of the roots is \\(- 1\\) in this case. \n\nIn all cases we conclude that \\(x + y + z\\) is an integer. \n\nComment: Each of the eight roots of \\(P\\) lead to genuine solutions. For example the roots of \\(\\lambda^{3} - 3\\lambda + 1\\) are approximately \\(0.3473\\ldots , - 1.8794\\ldots\\) and \\(1.5321\\ldots\\) , and so \n\n\\[(x,y,z) = (0.3473\\ldots , - 1.8794\\ldots ,1.5321\\ldots)\\] \n\nis a valid solution. Similarly, the roots of \\(\\lambda^{3} + \\lambda^{2} - 2\\lambda - 1\\) are approximately \\(1.2470\\ldots , - 0.44504\\ldots\\) and \\(- 1.8019\\ldots\\) , and so \n\n\\[(x,y,z) = (1.2470\\ldots , - 0.44504\\ldots , - 1.8019\\ldots)\\] \n\nis also a valid solution.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo2_2023_solutions.jsonl", "problem_match": "\n5. Problem:", "solution_match": "\nSolution"}}
diff --git a/NewZealand_MO/segmented/en-nzmo2_2024_solutions.jsonl b/NewZealand_MO/segmented/en-nzmo2_2024_solutions.jsonl
index 5a48688500013d7cc27aa9820738df0fea9a5d1a..a31ec9f5d7fa29ff835986ff8ae8da94ab446b9c 100644
--- a/NewZealand_MO/segmented/en-nzmo2_2024_solutions.jsonl
+++ b/NewZealand_MO/segmented/en-nzmo2_2024_solutions.jsonl
@@ -1,5 +1,5 @@
{"year": "2024", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "NewZealand_MO", "problem": "At each vertex of a regular 14-gon, lies a coin. Initially 7 coins are heads, and 7 coins are tails. Determine the minimum number \\(t\\) such that it's always possible to turn over at most \\(t\\) of the coins so that in the resulting 14-gon, no two adjacent coins are both heads and no two adjacent coins are both tails.", "solution": "A: (Ross Atkins) \n\nNumber the vertices from 1 to 14. We need to make sure that all the even- numbered vertices are tails and all the odd numbered vertices are heads or vice versa. \n\nLets look at the 7 odd numbered vertices. \n\nWithout loss of generality assume that there are initially more heads than tails on the odd numbered vertices. i.e. we assume that there are initially at least 4 heads on odd numbered vertices. This means there are at most 3 tails on odd numbered vertices and at most 3 heads on even numbered vertices. So we flip all the tails on odd numbered vertices and all the heads on even numbered vertices then the total number of flips required is at most 6. \n\nIf there are exactly 4 heads (and 3 tails) on odd numbered vertices, then we would require at least 3 flips to make all the odd numbered vertices the same, and similarly at least 3 flips to make the off numbered vertices the same. So we cannot guarantee to be able to achieve our goal in fewer than 6 flips.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo2_2024_solutions.jsonl", "problem_match": "\n1. Problem:", "solution_match": "# Solution"}}
{"year": "2024", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "NewZealand_MO", "problem": "Consider the sequence \\(a_{1}, a_{2}, a_{3}, \\ldots\\) defined by \\(a_{1} = 2024^{2024}\\) and for each positive integer \\(n\\) , \n\n\\[a_{n + 1} = \\left|a_{n} - \\sqrt{2}\\right|.\\] \n\nProve that there exists an integer \\(k\\) such that \\(a_{k + 2} = a_{k}\\) . \n\nHere \\(|x|\\) denotes the absolute value of \\(x\\) .", "solution": "A: (Ross Atkins) \n\nAs long as \\(a_{n} \\geqslant \\sqrt{2}\\) we have \\(a_{n + 1} = a_{n} - \\sqrt{2}\\) and so the sequence begins with an arithmetic progression with common difference \\(d = - \\sqrt{2}\\) . Let \\(k\\) be the first index such that \\(a_{k} < \\sqrt{2}\\) . So \\(a_{k - 1} \\geqslant \\sqrt{2}\\) . Therefore \n\n\\[a_{k} = a_{k - 1} - \\sqrt{2} \\geqslant \\sqrt{2} - \\sqrt{2} = 0.\\] \n\nThus we have \\(0 \\leq a_{k} < \\sqrt{2}\\) . From this we can calculate \\(a_{k} - \\sqrt{2} < 0\\) and so \\(\\left|a_{k} - \\sqrt{2}\\right| = \\sqrt{2} - a_{k}\\) . \n\n\\[\\Longrightarrow a_{k + 1} = \\left(\\sqrt{2} -a_{k}\\right)\\] \n\nSince \\(a_{k + 1} = \\left(\\sqrt{2} - a_{k}\\right) \\leqslant \\left(\\sqrt{2} - 0\\right) = \\sqrt{2}\\) , this means that \\(\\left(a_{k + 1} - \\sqrt{2}\\right)\\) is negative and thus \n\n\\[a_{k + 2} = \\left|a_{k + 1} - \\sqrt{2}\\right|\\] \\[= \\left(\\sqrt{2} -a_{k + 1}\\right)\\] \\[= \\left(\\sqrt{2} -\\left(\\sqrt{2} -a_{k}\\right)\\right)\\] \\[= a_{k}.\\] \n\nas required.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo2_2024_solutions.jsonl", "problem_match": "\n2. Problem:", "solution_match": "\nSolution"}}
-{"year": "2024", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(A, B, C, D, E\\) be five different points on the circumference of a circle in that (cyclic) order. Let \\(F\\) be the intersection of chords \\(BD\\) and \\(CE\\). Show that if \\(AB = AE = AF\\) then lines \\(AF\\) and \\(CD\\) are perpendicular.", "solution": "A: (Ross Atkins) \n\nLet \\(x = \\angle ABF\\) and let \\(y = \\angle BCA\\) . Since \\(\\triangle ABF\\) is isosceles, we get \\(\\angle BFA = x\\) . Since \\(AB = AE\\) , it follows that arcs \\(BA\\) and \\(AE\\) are equal. Since equal arcs subtend equal angles, every angle subtended by either arc \\(AB\\) or \\(AE\\) must be equal to \\(\\angle BCA = y\\) . \n\n\\[\\Rightarrow \\angle BCA = \\angle BDA = \\angle ACE = \\angle ADE = y.\\] \n\n\n \n\nSince opposite angles in a cyclic quadrilateral (ABDE) are supplementary, we get \\(\\angle ABD + \\angle DEA = 180^{\\circ}\\) . Therefore \\(\\angle DEA = 180^{\\circ} - x\\) . Now consider triangles \\(AED\\) and \\(AFD\\) . We have \n\n\\[\\angle ADE = y = \\angle ADF \\text{ and } \\angle AED = 180^{\\circ} - x = \\angle AFD\\] \n\nand side \\(AD\\) is shared. Therefore these triangles are congruent: \\(\\triangle AED \\equiv \\triangle AFD\\) . Hence \n\n\\[\\angle EAD = \\angle DAF = x - y. \\qquad (\\text{angle sum in} \\triangle AFD)\\] \n\nNow let \\(P\\) be the intersection of \\(AD\\) and \\(EF\\) . Also let \\(Q\\) be the intersection of \\(AF\\) and \\(CD\\) . Since \\(AP\\) is the angle bisector of isosceles triangle \\(AEF\\) , we have \n\n\\[\\angle APF = 90^{\\circ}.\\] \\[\\Rightarrow \\angle AFP = 90^{\\circ} + y - x \\qquad (\\text{angle sum in} \\triangle AFP)\\] \\[\\angle CFQ = 90^{\\circ} + y - x \\qquad (\\text{vertically opposite})\\] \n\nFinally we get \\(\\angle ECD = \\angle EAD = x - y\\) by the Bow Tie Theorem (ACDE cyclic). Therefore \\(\\angle FCQ = x - y\\) . Now consider the sum of the angles in triangle \\(FCQ\\) to get \n\n\\[\\angle CQF + (x - y) + (90^{\\circ} + y - x) = 180^{\\circ}\\] \n\n\\[\\Rightarrow \\angle CQF = 90^{\\circ}\\] \n\nas required.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo2_2024_solutions.jsonl", "problem_match": "\n3. Problem:", "solution_match": "\nSolution"}}
+{"year": "2024", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(A, B, C, D, E\\) be five different points on the circumference of a circle in that (cyclic) order. Let \\(F\\) be the intersection of chords \\(BD\\) and \\(CE\\). Show that if \\(AB = AE = AF\\) then lines \\(AF\\) and \\(CD\\) are perpendicular.", "solution": "A: (Ross Atkins) \n\nLet \\(x = \\angle ABF\\) and let \\(y = \\angle BCA\\) . Since \\(\\triangle ABF\\) is isosceles, we get \\(\\angle BFA = x\\) . Since \\(AB = AE\\) , it follows that arcs \\(BA\\) and \\(AE\\) are equal. Since equal arcs subtend equal angles, every angle subtended by either arc \\(AB\\) or \\(AE\\) must be equal to \\(\\angle BCA = y\\) . \n\n\\[\\Rightarrow \\angle BCA = \\angle BDA = \\angle ACE = \\angle ADE = y.\\] \n\n\n \n\nSince opposite angles in a cyclic quadrilateral (ABDE) are supplementary, we get \\(\\angle ABD + \\angle DEA = 180^{\\circ}\\) . Therefore \\(\\angle DEA = 180^{\\circ} - x\\) . Now consider triangles \\(AED\\) and \\(AFD\\) . We have \n\n\\[\\angle ADE = y = \\angle ADF \\text{ and } \\angle AED = 180^{\\circ} - x = \\angle AFD\\] \n\nand side \\(AD\\) is shared. Therefore these triangles are congruent: \\(\\triangle AED \\equiv \\triangle AFD\\) . Hence \n\n\\[\\angle EAD = \\angle DAF = x - y. \\qquad (\\text{angle sum in} \\triangle AFD)\\] \n\nNow let \\(P\\) be the intersection of \\(AD\\) and \\(EF\\) . Also let \\(Q\\) be the intersection of \\(AF\\) and \\(CD\\) . Since \\(AP\\) is the angle bisector of isosceles triangle \\(AEF\\) , we have \n\n\\[\\angle APF = 90^{\\circ}.\\] \\[\\Rightarrow \\angle AFP = 90^{\\circ} + y - x \\qquad (\\text{angle sum in} \\triangle AFP)\\] \\[\\angle CFQ = 90^{\\circ} + y - x \\qquad (\\text{vertically opposite})\\] \n\nFinally we get \\(\\angle ECD = \\angle EAD = x - y\\) by the Bow Tie Theorem (ACDE cyclic). Therefore \\(\\angle FCQ = x - y\\) . Now consider the sum of the angles in triangle \\(FCQ\\) to get \n\n\\[\\angle CQF + (x - y) + (90^{\\circ} + y - x) = 180^{\\circ}\\] \n\n\\[\\Rightarrow \\angle CQF = 90^{\\circ}\\] \n\nas required.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo2_2024_solutions.jsonl", "problem_match": "\n3. Problem:", "solution_match": "\nSolution"}}
{"year": "2024", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "NewZealand_MO", "problem": "Determine all positive integers \\(n\\) less than 2024 such that for all positive integers \\(x\\) , the greatest common divisor of \\(9x + 1\\) and \\(nx + 1\\) is 1.", "solution": "A: (Ross Atkins) \n\nLet \\(m = (n - 9)\\) and for the sake of contradiction assume \\(m\\) has a prime factor \\(p\\) with \\(p \\neq 3\\) . Express \\(p = 9q + r\\) where \\(0 \\leq r \\leq 8\\) ( \\(r\\) is the remainder when \\(p\\) is divided by 9 and \\(q\\) may be zero). Since \\(\\gcd (9, p) = 1\\) we must have \\(r = 1, 2, 4, 5, 7\\) or 8. For each of these possibilities we will use a different choice for \\(x\\) to get our contradiction. \n\nif \\(r = 1\\) then choose \\(x = q\\) so that \\((9x + 1) = 9q + 1 = p\\) \n\nif \\(r = 2\\) then choose \\(x = 5q + 1\\) so that \\((9x + 1) = 45q + 10 = 5p\\) \n\nif \\(r = 4\\) then choose \\(x = 7q + 3\\) so that \\((9x + 1) = 63q + 28 = 7p\\) \n\nif \\(r = 5\\) then choose \\(x = 2q + 1\\) so that \\((9x + 1) = 18q + 10 = 2p\\) \n\nif \\(r = 7\\) then choose \\(x = 4q + 3\\) so that \\((9x + 1) = 36q + 28 = 4p\\) \n\nif \\(r = 8\\) then choose \\(x = 8q + 7\\) so that \\((9x + 1) = 72q + 64 = 8p\\) \n\nIn any we can choose \\(x\\) so that \\((9x + 1)\\) is a multiple of \\(p\\) . For this particular value of \\(x\\) we have \n\n\\[(n x + 1) = m x + (9 x + 1),\\] \n\nand so \\((n x + 1)\\) is a multiple of \\(p\\) too (recall \\(p|m\\) ). Therefore \\(p\\) would be a common factor of \\((9x + 1)\\) and \\((n x + 1)\\) . This is a contradiction so no such \\(p\\) can exist. \n\nSo \\(m = (n - 9)\\) cannot have any prime factors other than 3. Hence \\(m = 3^{k}\\) or \\(m = - 3^{k}\\) for some integer \\(k \\geq 0\\) . i.e. \n\n\\[n = 9 + m = 9 + 3^{k} \\text{ or } 9 - 3^{k}\\] \n\nThe positive integers of this form, less than 2024 are: 6, 8, 10, 12, 18, 36, 90, 252, 738. \n\nTo show all these work, we now assume \\(n = 9 \\pm 3^{k}\\) for some integer \\(k \\geq 0\\) , and let \\(g = \\gcd (9x + 1, nx + 1)\\) . Since \\(9x + 1\\) is not a multiple of 3, we cannot have \\(g\\) being a multiple of 3. However \n\n\\[g \\mid (n x + 1) - (9 x + 1) = \\pm 3^{k}.\\] \n\nThe only divisors of \\(3^{k}\\) which are not a multiple of 3 are 1 and \\(- 1\\) . Therefore \\(g = 1\\) whenever \\(n = 9 \\pm 3^{k}\\) .", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo2_2024_solutions.jsonl", "problem_match": "\n4. Problem:", "solution_match": "\nSolution"}}
{"year": "2024", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "NewZealand_MO", "problem": "Determine the least real number \\(L\\) such that \n\n\\[\\frac{1}{a} +\\frac{a}{b} +\\frac{b}{c} +\\frac{c}{d}\\leqslant L\\] \n\nfor all quadruples \\((a,b,c,d)\\) of integers satisfying \\(1< a< b< c< d\\)", "solution": ": (Ross Atkins) \n\nAnswer: 3. To solve this problem, two parts are required. Part A shows that \\(L = 3\\) works. Part B shows that no \\(L^{\\prime}< 3\\) works. \n\nPart A We show that for all quadruples \\((a,b,c,d)\\) (with \\(1< a< b< c< d\\) ) we have \n\n\\[\\frac{1}{a} +\\frac{a}{b} +\\frac{b}{c} +\\frac{c}{d}\\leqslant 3.\\] \n\nWe do this by considering the two cases \\(b = a + 1\\) and \\(b\\geq a + 2\\) seperately. \n\nIf \\(b\\geq a + 2\\) then we can get \n\n\\[\\frac{1}{a} +\\frac{a}{b} +\\frac{b}{c} +\\frac{c}{d}\\leqslant \\frac{1}{a} +\\frac{a}{a + 2} +\\frac{b}{c} +\\frac{c}{d\\] \\[\\qquad < \\frac{1}{a} +\\frac{a}{a + 2} +1 + 1\\] \\[\\qquad = \\frac{2 - a}{a(a + 2)} +3\\quad \\leqslant 3\\] \n\nThe numerator \\((2 - a)\\) is non- positive because \\(1< a\\) and \\(a\\) is an integer. If \\(b = a + 1\\) then we can get \n\n\\[\\frac{1}{a} +\\frac{a}{b} +\\frac{b}{c} +\\frac{c}{d} = \\frac{1}{a} +\\frac{a}{a + 1} +\\frac{a + 1}{c} +\\frac{c}{d\\] \\[\\qquad < \\frac{1}{a} +\\frac{a}{a + 1} +\\frac{a + 1}{c} +1\\] \\[\\qquad \\leqslant \\frac{1}{a} +\\frac{a}{a + 1} +\\frac{a + 1}{a + 2} +1\\] \\[\\qquad = \\frac{2 - a^{2}}{a(a + 1)(a + 2)} +3\\quad < 3.\\] \n\nThe numerator \\((2 - a^{2})\\) is negative because \\(1< a\\) and \\(a\\) is an integer. In either case we get \\(\\frac{1}{a} +\\frac{a}{b} +\\frac{b}{c} +\\frac{c}{d}\\leqslant 3\\) for all quadruples \\((a,b,c,d)\\)\n\n\n\n- Part B For the sake of contradiction, suppose there existed some \\(L^{\\prime}< 3\\) which worked. We will show that there exists a quadruple \\((a,b,c,d)\\) such that \n\n\\[\\frac{1}{a} +\\frac{a}{b} +\\frac{b}{c} +\\frac{c}{d} >L^{\\prime}.\\] \n\nWe will do this explicitly by choosing \\(a,b,c,d\\) to be large consecutive integers. If \\(L^{\\prime} = 3 - \\epsilon\\) then we have \\(\\epsilon >0\\) , and so there exists some integer \\(n\\) such that \\(n > \\frac{3}{\\epsilon}\\) . This ensures \\(\\frac{1}{n + 1}\\) , \\(\\frac{1}{n + 2}\\) and \\(\\frac{1}{n + 3}\\) are each smaller than \\(\\frac{\\epsilon}{3}\\) . Now consider \\((a,b,c,d) = (n,n + 1,n + 2,n + 3)\\) . \n\n\\[{\\frac{1}{a}}+{\\frac{a}{b}}+{\\frac{b}{c}}+{\\frac{c}{d}}>{\\frac{a}{b}}+{\\frac{b}{c}}+{\\frac{c}{d}}\\] \\[\\qquad={\\frac{n}{n+1}}+{\\frac{n+1}{n+2}}+{\\frac{n+2}{n+3}}\\] \\[\\qquad=3-{\\left({\\frac{1}{n+1}}+{\\frac{1}{n+2}}+{\\frac{1}{n+3}}\\right)}\\] \\[\\qquad>3-{\\left({\\frac{\\epsilon}{3}}+{\\frac{\\epsilon}{3}}+{\\frac{\\epsilon}{3}}\\right)}\\] \\[\\qquad=L^{\\prime}\\] \n\nContradiction.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo2_2024_solutions.jsonl", "problem_match": "\n5. Problem:", "solution_match": "\nSolution"}}
diff --git a/NewZealand_MO/segmented/en-nzmo2_2025_solutions.jsonl b/NewZealand_MO/segmented/en-nzmo2_2025_solutions.jsonl
index 756b027fc1468406183f9915ea93da44c6572426..f28e3ad9721270bab83951102965ba36f2f0a579 100644
--- a/NewZealand_MO/segmented/en-nzmo2_2025_solutions.jsonl
+++ b/NewZealand_MO/segmented/en-nzmo2_2025_solutions.jsonl
@@ -1,5 +1,5 @@
{"year": "2025", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "NewZealand_MO", "problem": "Find all pairs of positive integers \\(m\\) and \\(n\\) such that the centres of the unit squares in a \\(m\\) by \\(n\\) grid of unit squares can be paired up so that the distance between the centres of each pair is exactly 2. \n\n(A unit square has side length 1. )", "solution": ": (Tony Wang) \n\nWe will show that the answer is all pairs \\((a,b)\\) where either \\(a\\) or \\(b\\) (or both) is a multiple of 4. First, partition the grid into four subgrids \\(A\\) , \\(B\\) , \\(C\\) , and \\(D\\) \n\n \n\nNote that, for any given square \\(s\\) , all the square centres that are exactly 2 away from the centre of \\(s\\) are in the same partition as \\(s\\) itself. This means that \\(s\\) must be paired with some square which is in the same partition as itself. Hence, each partition of squares must have an even number of squares. Meanwhile, if each partition of squares has an even number of squares, one of the dimensions of the subgrid must be even, and hence we can pair up the squares along that even dimension. \n\nHence, it suffices to find all values of \\(a\\) and \\(b\\) which create four subgrids which all have an even number of squares. Consider \\(a\\) modulo 4: if \\(a \\equiv 0\\) (mod 4), then all subgrids will have an even dimension along the axis of \\(a\\) . Otherwise, at least one of the subgrids will have an odd dimension along the axis of \\(a\\) . The same reasoning holds for \\(b\\) . Hence, if neither \\(a\\) nor \\(b\\) are multiples of 4, then one of the partitions will have an odd number of squares. However, if either \\(a\\) or \\(b\\) is a multiple of 4, then all partitions will have an even number of squares. Hence, we have shown that the answer is all pairs \\((a,b)\\) where either \\(a\\) or \\(b\\) (or both) is a multiple of 4.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo2_2025_solutions.jsonl", "problem_match": "\n1. Problem:", "solution_match": "\nSolution"}}
{"year": "2025", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "NewZealand_MO", "problem": "For which positive integers \\(n\\) , does there exist a sequence of real numbers \\((x_{1}, x_{2}, \\ldots , x_{n})\\) such that \n\n- \\(-2 < x_{i} < 2\\) for all \\(i\\) ,- \\(x_{1} + x_{2} + x_{3} + \\dots + x_{n} = 0\\) , and- \\(x_{1}^{4} + x_{2}^{4} + x_{3}^{4} + \\dots + x_{n}^{4} \\geqslant 32\\) .", "solution": ": (Eric Liang) \n\nNote that if \\(n = j\\) works then \\(n > j\\) also works for a positive integer \\(j\\) as we can just set \\(x_{i} = 0\\) for \\(n \\geq i > j\\) and have \\(x_{1}, \\ldots , x_{n}\\) be the sequence that worked for \\(n\\) . \n\nConsider \\(n = 4\\) . We take \\(x_{1}, x_{2} = \\sqrt[4]{8}\\) and \\(x_{3}, x_{4} = -\\sqrt[4]{8}\\) and all conditions are satisfied. Thus all \\(n \\geq 4\\) works.\n\n\n\nNow if \\(n = 2\\) , we note \\(|x_{1}|,|x_{2}|< 2\\) and thus \\(x_{1}^{4},x_{2}^{4}< 16\\) so \\(x_{1}^{4} + x_{2}^{4}< 32\\) . Contradiction. \n\nIf \\(n = 3\\) , then wlog \\(x_{1},x_{2}\\geq 0\\) and \\(x_{3}\\leq 0\\) (as flipping the signs won't affect any of the conditions). \n\nNow as \\(x_{1} + x_{2} = - x_{3}< 2\\) , we get that \\((x_{1} + x_{2})^{4}< 16\\) . But if we expand this out we get \\(16 > (x_{1} + x_{2})^{4} = x_{1}^{4} + 4x_{1}^{3}x_{2} + 6x_{1}^{2}x_{2}^{2} + 4x_{1}x_{2}^{3} + x_{2}^{4}\\geq x_{1}^{4} + x_{2}^{4}\\) (as \\(x_{1},x_{2}\\geq 0\\) ). \n\nThus \\(x_{1}^{4} + x_{2}^{4}< 16\\) but also note that \\(x_{3}^{4}< 16\\) , thus \\(x_{1}^{4} + x_{2}^{4} + x_{3}^{4}< 32\\) . Contradiction. \n\nThus, a sequence only exists for integers \\(n\\geq 4\\) .", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo2_2025_solutions.jsonl", "problem_match": "\n2. Problem:", "solution_match": "\nSolution"}}
-{"year": "2025", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(A B C\\) be an acute scalene triangle with \\(A C > B C > A B\\) . Let the orthocentre be \\(H\\) and circumcentre be \\(O\\) . Suppose that lines \\(B O\\) and \\(C H\\) intersect at a point \\(D\\) . Point \\(E\\) (where \\(E\\neq C\\) ) lies on side \\(A C\\) so that \\(O E C D\\) is cyclic. Point \\(F\\) (where \\(F\\neq C\\) ) lies on side \\(B C\\) such that \\(C E = F E\\) . Prove that \\(B H D F\\) is cyclic. \n\n(The orthocentre of a triangle is the point of intersection of its altitudes.)", "solution": ": (Nico McKinlay & George Zhu) \n\nLet \\(\\alpha = \\angle B A C\\) . Let \\(B B^{\\prime}\\) and \\(C C^{\\prime}\\) be altitudes in triangle \\(A B C\\) , as shown. \n\n\n \n\nClaim. Triangle \\(C D E\\) is isosceles with \\(C E = D E\\) . \n\nProof. \n\n\\[\\angle D E C = \\angle D O C\\] \\[\\qquad = \\angle B O C\\] \\[\\qquad = 2\\angle B A C\\] \\[\\qquad = 2\\alpha\\] \n\n\\[\\angle D C E = \\angle C^{\\prime}C A = 90^{\\circ} - \\angle C^{\\prime}A C\\] \\[\\qquad = 90^{\\circ} - \\alpha\\]\n\n\n\n\\[\\begin{array}{r l} & {\\angle C D E = 180^{\\circ} - \\angle D E C - \\angle D C E}\\\\ & {\\qquad = 180^{\\circ} - 2\\alpha -(90^{\\circ} - \\alpha)}\\\\ & {\\qquad = 90^{\\circ} - \\alpha}\\\\ & {\\qquad = \\angle D C E.} \\end{array} \\quad (angle sum in \\triangle C D E)\\] \n\nSince \\(C E = D E\\) and \\(C E = F E\\) , we have \\(C E = D E = F E\\) , therefore \\(E\\) is the circumcentre of triangle \\(C D F\\) . Consequently, \n\n\\[\\begin{array}{r l} & {\\angle H D F = 180^{\\circ} - \\angle C D F}\\\\ & {\\qquad = 180^{\\circ} - \\frac{1}{2}\\cdot \\angle C E F}\\\\ & {\\qquad = 180^{\\circ} - \\frac{1}{2}\\bullet (180^{\\circ} - 2\\angle E C F)}\\\\ & {\\qquad = 180^{\\circ} - \\frac{1}{2}\\bullet (180^{\\circ} - 2\\angle B^{\\prime}C B)}\\\\ & {\\qquad = 180^{\\circ} - (90^{\\circ} - \\angle B^{\\prime}C B)}\\\\ & {\\qquad = 180^{\\circ} - \\angle B^{\\prime}B C}\\\\ & {\\qquad = 180^{\\circ} - \\angle H B F} \\end{array} \\quad (angles in \\triangle B^{\\prime}B C)\\] \n\nso \\(B H D F\\) is cyclic.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo2_2025_solutions.jsonl", "problem_match": "# 3. Problem:", "solution_match": "\nSolution"}}
+{"year": "2025", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(A B C\\) be an acute scalene triangle with \\(A C > B C > A B\\) . Let the orthocentre be \\(H\\) and circumcentre be \\(O\\) . Suppose that lines \\(B O\\) and \\(C H\\) intersect at a point \\(D\\) . Point \\(E\\) (where \\(E\\neq C\\) ) lies on side \\(A C\\) so that \\(O E C D\\) is cyclic. Point \\(F\\) (where \\(F\\neq C\\) ) lies on side \\(B C\\) such that \\(C E = F E\\) . Prove that \\(B H D F\\) is cyclic. \n\n(The orthocentre of a triangle is the point of intersection of its altitudes.)", "solution": ": (Nico McKinlay & George Zhu) \n\nLet \\(\\alpha = \\angle B A C\\) . Let \\(B B^{\\prime}\\) and \\(C C^{\\prime}\\) be altitudes in triangle \\(A B C\\) , as shown. \n\n\n \n\nClaim. Triangle \\(C D E\\) is isosceles with \\(C E = D E\\) . \n\nProof. \n\n\\[\\angle D E C = \\angle D O C\\] \\[\\qquad = \\angle B O C\\] \\[\\qquad = 2\\angle B A C\\] \\[\\qquad = 2\\alpha\\] \n\n\\[\\angle D C E = \\angle C^{\\prime}C A = 90^{\\circ} - \\angle C^{\\prime}A C\\] \\[\\qquad = 90^{\\circ} - \\alpha\\]\n\n\n\n\\[\\begin{array}{r l} & {\\angle C D E = 180^{\\circ} - \\angle D E C - \\angle D C E}\\\\ & {\\qquad = 180^{\\circ} - 2\\alpha -(90^{\\circ} - \\alpha)}\\\\ & {\\qquad = 90^{\\circ} - \\alpha}\\\\ & {\\qquad = \\angle D C E.} \\end{array} \\quad (angle sum in \\triangle C D E)\\] \n\nSince \\(C E = D E\\) and \\(C E = F E\\) , we have \\(C E = D E = F E\\) , therefore \\(E\\) is the circumcentre of triangle \\(C D F\\) . Consequently, \n\n\\[\\begin{array}{r l} & {\\angle H D F = 180^{\\circ} - \\angle C D F}\\\\ & {\\qquad = 180^{\\circ} - \\frac{1}{2}\\cdot \\angle C E F}\\\\ & {\\qquad = 180^{\\circ} - \\frac{1}{2}\\bullet (180^{\\circ} - 2\\angle E C F)}\\\\ & {\\qquad = 180^{\\circ} - \\frac{1}{2}\\bullet (180^{\\circ} - 2\\angle B^{\\prime}C B)}\\\\ & {\\qquad = 180^{\\circ} - (90^{\\circ} - \\angle B^{\\prime}C B)}\\\\ & {\\qquad = 180^{\\circ} - \\angle B^{\\prime}B C}\\\\ & {\\qquad = 180^{\\circ} - \\angle H B F} \\end{array} \\quad (angles in \\triangle B^{\\prime}B C)\\] \n\nso \\(B H D F\\) is cyclic.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo2_2025_solutions.jsonl", "problem_match": "# 3. Problem:", "solution_match": "\nSolution"}}
{"year": "2025", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "NewZealand_MO", "problem": "The function \\(r_{n}(x)\\) is the remainder when \\(x\\) is divided by \\(n\\) , where \\(0 \\leq r_{n}(x) < n\\) . For which \\(n\\) does there exists some ordering \\(\\{a_{1}, \\ldots , a_{n - 1}\\}\\) of \\(\\{1, 2, \\ldots , n - 1\\}\\) such that \\(\\{r_{n}(a_{1}), r_{n}(2 \\times a_{2}), \\ldots , r_{n}((n - 1) \\times a_{n - 1})\\}\\) is an ordering of \\(\\{1, 2, \\ldots , n - 1\\}\\) ? \n\n(An ordering of \\(\\{1, 2, \\ldots , n - 1\\}\\) is the sequence of numbers 1 to \\(n - 1\\) in some order.)", "solution": ": (James Xu) \n\nNotice that \\(r_{n}(x)\\) is just \\(x\\) modulo \\(n\\) . Therefore \n\n\\[\\prod_{i}r_{n}(i a_{i})\\equiv \\prod_{i}i a_{i}\\pmod{n}\\] \n\nFor primes \\(p\\) , apply Wilson's theorem to see that we must have \n\n\\[\\prod_{i}i a_{i}\\equiv -1\\pmod{p}\\] \n\nHowever, \n\n\\[\\prod_{i}i a_{i} = \\prod_{i}i\\prod_{i}a_{i}\\equiv -1^{2}\\equiv 1\\pmod{p}\\] \n\nSo the only possibility in that case is \\(p = 2\\) . \n\nNow, if \\(n\\) is composite, then let \\(n\\) be the minimal solution. Let \\(n = pq\\) for prime \\(p\\) . Notice that we must have \\((p - 1)q\\) numbers in \\(\\{a_{1}, 2a_{2}, 3a_{3}, \\ldots , (n - 1)a_{n - 1}\\}\\) not divisible by \\(p\\) . The \\(q - 1\\) numbers of the form \\((kp)a_{kp}\\) are divisible by \\(p\\) and there are only \\(q - 1\\) multiples of \\(p\\) in \\(\\{1, 2, \\ldots , pq - 1\\}\\) . Therefore they must be the only numbers divisible by \\(p\\) , so \\(\\{a_{kp} | 1 \\leq k < q\\} = \\{kp | 1 \\leq k < q\\}\\) . Now, \\(p \\nmid q\\) as otherwise all of \\((kp)a_{kp}\\) are multiples of \\(p^{2}\\) , which is not true. \n\nLet \\(c \\equiv \\frac{1}{p} \\pmod {q}\\) . Consider \\(\\{\\frac{a_{p}}{p}, \\frac{a_{2p}}{p}, \\ldots , \\frac{a_{(q - 1)p}}{p}\\} = \\{1, 2, \\ldots , q - 1\\}\\) .\n\n\n\nThen, \n\n\\[\\{\\frac{a_{p}}{p},\\frac{2a_{2p}}{p},\\ldots ,\\frac{(q - 1)a_{(q - 1)p}}{p}\\} = \\{\\frac{p a_{p}}{p^{2}},\\frac{2p a_{2p}}{p^{2}},\\ldots ,\\frac{(q - 1)p a_{(q - 1)p}}{p^{2}}\\}\\] \n\n\\[\\equiv c\\{1,2,3,\\ldots ,q - 1\\} \\pmod {q}\\] \n\nAs \\((k p)a_{k p}\\equiv k^{\\prime}p\\) (mod \\(p q\\) ) for some \\(k^{\\prime}\\) , and \\(k a_{k p}\\equiv k^{\\prime}\\) (mod \\(q\\) ) \n\nSince \\(g c d(c,q) = 1\\) \n\n\\[c\\{1,2,3,\\ldots ,q - 1\\} \\equiv \\{1,2,3,\\ldots ,q - 1\\} \\pmod {q}\\] \n\nin some order. This is a solution for \\(n = q\\) , which contradicts the minimality of the solution for \\(n = p q\\) . Therefore no such solution \\(n = p q\\) can exist. \n\nHence, \\(n = 2\\) is the only solution.", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo2_2025_solutions.jsonl", "problem_match": "\n4. Problem:", "solution_match": "\nSolution"}}
{"year": "2025", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "NewZealand_MO", "problem": "Let \\(a,b,c\\) be positive real numbers satisfying \\(a b c = 1\\) . Determine the smallest possible value of \n\n\\[\\frac{a^{2} + 2025}{a^{3}(b + c)} +\\frac{b^{2} + 2025}{b^{3}(c + a)} +\\frac{c^{2} + 2025}{c^{3}(a + b)}\\]", "solution": ": (Eric Liang) \n\nNote, \n\n\\[\\begin{array}{r l r}{{\\frac{1}{a^{3}(b+c)}+\\frac{1}{b^{3}(a+c)}+\\frac{1}{c^{3}(a+b)}=\\frac{(\\frac{1}{a})^{2}}{a(b+c)}+\\frac{(\\frac{1}{b})^{2}}{b(a+c)}+\\frac{(\\frac{1}{c})^{2}}{c(a+b)}}}\\\\ &{}&{\\geq\\frac{(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c})^{2}}{2(a b+a c+b c)}}\\\\ &{}&{=\\frac{(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c})^{2}}{2(\\frac{1}{c}+\\frac{1}{b}+\\frac{1}{a})}}\\\\ &{}&{=\\frac{(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c})}{2}}\\\\ &{}&{\\geq\\frac{3}{2\\sqrt[3]{a b c}}}\\\\ &{}&{=\\frac{3}{2}}\\end{array} \\quad (AM-GM)\\] \n\nNow, \\((a + b + c)^{2} \\geq 3(a b + a c + b c)\\) for positive reals \\(a, b, c\\) . (†) \n\nSo we also get, \n\n\\[\\begin{array}{r l r}{{\\frac{1}{a^{2}(b+c)}+\\frac{1}{b^{2}(a+c)}+\\frac{1}{c^{2}(a+b)}\\geq\\frac{(\\frac{1}{a})^{2}}{(b+c)}+\\frac{(\\frac{1}{b})^{2}}{(a+c)}+\\frac{(\\frac{1}{c})^{2}}{(a+b)}}}\\\\ &{}&{\\geq\\frac{(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c})^{2}}{2(a+b+c)}}\\\\ &{}&{=\\frac{(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c})^{2}}{2(\\frac{1}{b}\\cdot\\frac{1}{c}+\\frac{1}{a}\\cdot\\frac{1}{c}+\\frac{1}{a}\\cdot\\frac{1}{b})}}\\\\ &{}&{=\\frac{3}{2}}\\end{array} \\quad (†)\\]\n\n\n\nFinally, as \\(a^{2} + 1 \\geq 2a\\) , we get \n\n\\[\\frac{a^{2} + 2025}{a^{3}(b + c)} +\\frac{b^{2} + 2025}{b^{3}(a + c)} +\\frac{c^{2} + 2025}{c^{3}(a + b)}\\geq \\frac{2a + 2024}{a^{3}(b + c)} +\\frac{2b + 2024}{b^{3}(a + c)} +\\frac{2c + 2024}{c^{3}(a + b)}\\] \\[\\qquad \\geq 2\\cdot \\frac{3}{2} +2024\\cdot \\frac{3}{2}\\] \\[\\qquad =3039\\] \n\nEquality can be achieved when \\(a = b = c = 1\\) .", "metadata": {"resource_path": "NewZealand_MO/segmented/en-nzmo2_2025_solutions.jsonl", "problem_match": "\n5. Problem:", "solution_match": "\nSolution"}}
diff --git a/SovietUnion/md/en-ASU-1961-1991.md b/SovietUnion/md/en-ASU-1961-1991.md
index a98ed43287b3a9e8dcca8b04a14aa6646f9d4671..09a0d22e71249ce4ee751d49c380214194db2246 100644
--- a/SovietUnion/md/en-ASU-1961-1991.md
+++ b/SovietUnion/md/en-ASU-1961-1991.md
@@ -7,7 +7,7 @@
Given 12 vertices and 16 edges arranged as follows:
-
+
Draw any curve which does not pass through any vertex. Prove that the curve cannot intersect each edge just once. Intersection means that the curve crosses the edge from one side to the other. For example, a circle which had one of the edges as tangent would not intersect that edge.
@@ -573,7 +573,7 @@ Given a lattice of regular hexagons. A bug crawls from vertex A to vertex B alon
## Solution
-
+
Suppose vertex A is that marked \* at the bottom left. Without loss of generality, B is in a 60 degree sector as shown. Assume the edges have unit length. The vertices can be partitioned into two sets (marked ° and . in the diagram). Each set forms a skewed lattice with axes at 60 degrees. Any path must alternate between the two lattices.
@@ -1295,7 +1295,7 @@ Given point O inside the acute- angled triangle ABC, and point O' inside the acu
## Solution
-
+
Let \(\Gamma\) be the circumcircle of DEF. Let OD, OE, OF meet it again at A", B", C" respectively. Then the figure O'A'B'C' must be similar to OA"B"C". So to prove that OD is parallel to O'A', we have to prove that AO is perpendicular to B"C".
@@ -2464,13 +2464,13 @@ Some unit squares in an infinite sheet of squared paper are colored red so that
## Solution
-
+
There cannot be two red squares with a common side. For consider as the diagram shows we can immediately conclude that the squares with a \(\ast\) are not red, but now the bold rectangle has at most 1 red square. Contradiction.
-
+
Consider a red square. One of the two diagonally adjacent squares marked \* must be red. But it is now easy to show that all red squares on that diagonal are red and that the other red squares are those on every third parallel diagonal line. Any \(9 \times 11\) rectangle must have just three such diagonals on a 9 cell border row, and hence just 3 red cells in that border row. But the remaining \(9 \times 10\) rectangle can easily be partitioned into fifteen \(3 \times 2\) rectangles, each with 2 red squares.
@@ -2511,7 +2511,7 @@ ABC is equilateral. A line parallel to AC meets AB at M and BC at P. D is the ce
## Solution
-
+
Let K be the midpoint of BP and L the midpoint of AC. EL is parallel to BC, so \(\angle ELC = 120^{\circ}\) . EK is parallel to AB, so \(\angle EKC = 60^{\circ}\) , so ELCK is cyclic. But \(\angle DKC = \angle DLC = 90^{\circ}\) , so DLCK is cyclic. Hence D, K, C, L, E all lie on a circle. Hence \(\angle DEC = \angle DLC = 90^{\circ}\) , and \(\angle EDC = \angle EKC = 60^{\circ}\) .
@@ -3522,7 +3522,7 @@ Answer 6
## Solution
-
+
Suppose one endpoint of a segment length 3 is at A. Evidently the other end could be at the edge midpoints B, C, D. It could also be on the circular arc connecting B and C (with center O and radius \(\sqrt{5}\) ). Similarly, it could be on arcs connecting C and D, or B and D. We claim that if X is a point of one of these arcs other than its endpoints, then the only possible segment length 3 with an endpoint at X (and the other endpoint on the surface of the cube) is AX. wlog we can consider X to be on the arc BC. Take axes with origin O, so that A is (0,0,2). Suppose X is (a,b,0) and that the other endpoint of the segment is Y (x,y,z). Then \(\mathrm{XY}^{2} = (\mathrm{x} - \mathrm{a})^{2} + (\mathrm{y} - \mathrm{b})^{2} + \mathrm{z}^{2} = \mathrm{a}^{2} + \mathrm{b}^{2} + \mathrm{z}^{2} - \mathrm{x}(2\mathrm{a} - \mathrm{x}) - \mathrm{y}(2\mathrm{b} - \mathrm{y}) = 5 + \mathrm{z}^{2} - \mathrm{x}(2\mathrm{a} - \mathrm{x}) - \mathrm{y}(2\mathrm{b} - \mathrm{y})\) . But a, \(\mathrm{b} > 1\) since X is not an endpoint of the arc, so (2a- x) and (2b- y) are both positive. Hence - x(2a- x) - y(2b- y) \(\leq 0\) with equality iff \(\mathrm{x} = \mathrm{y} = 0\) . Similarly, \(\mathrm{z}^{2} \leq 4\) with equality iff \(\mathrm{z} = 2\) . Hence \(\mathrm{XY}^{2} \leq 9\) with equality iff \(\mathrm{Y} = \mathrm{A}\) , which proves the claim.
@@ -3689,7 +3689,7 @@ The line joining the midpoints of two opposite sides of a convex quadrilateral m
## Solution
-
+
Let L, M, N be the midpoints of BC, CD, DA. Assume that NL makes equal angles with AC and BD, so \(\angle \mathrm{NLM} = \angle \mathrm{BEL} = \angle \mathrm{AFN} = \angle \mathrm{LNM}\) , so \(\mathrm{LM} = \mathrm{MN}\) and hence \(\mathrm{BD} = \mathrm{AC}\) .
@@ -3718,7 +3718,7 @@ yes, yes
## Solution
-
+
By stretching vertically we get a square.
@@ -3729,7 +3729,7 @@ The point P lies inside the triangle ABC. A line is drawn through P parallel to
## Solution
-
+
The three small triangles are similar, so a/a" = c'/c = b"/b' and a/a' = c'/c" = b"/b. Hence (a/a")(b/b") = (c'/c)(c"/c') = c"/c, so abc = a"b"c". Similarly, (a/a')(c/c') = (b"/b)(b"/b") = b/b, so abc = a'b'c'.
@@ -3762,7 +3762,7 @@ An equilateral triangle of side n is divided into \(\mathrm{n}^2\) equilateral t
## Solution
-
+
The diagram has \(1 + 2 + \ldots + \mathrm{n} = \mathrm{n}(\mathrm{n} + 1) / 2\) upright triangles and \(1 + 2 + \ldots + \mathrm{n} - 1 = \mathrm{n}(\mathrm{n} - 1) / 2\) upside down triangles. It has \(1 + 2 + \ldots + \mathrm{n} + 1 = (\mathrm{n} + 1)(\mathrm{n} + 2) / 2\) vertices. So the path must be \((\mathrm{n} + 1)(\mathrm{n} + 2) / 2 - 1 = (\mathrm{n}^2 +3\mathrm{n}) / 2\) units long. Each unit length of the path is in just one upright triangle. The path cannot contain all three sides of a small triangle, or it would pass through a vertex more than once. So it must contain two sides of \((\mathrm{n}^2 +3\mathrm{n}) / 2 - \mathrm{n}(\mathrm{n} + 1) / 2 = \mathrm{n}\) triangles. But if it contains two sides of a triangle, then it must make an acute angle at the vertex where they meet.
@@ -3776,7 +3776,7 @@ Answer no
Solution
-
+
Suppose it can be done. Divide the board into 4 quadrants. Suppose there are b black and \(995^{2}\) - b white squares in the top left quadrant. Then there are \(995^{2}\) - b black and b white squares in the bottom right quadrant (by the symmetry property).
@@ -3891,7 +3891,7 @@ Answer no, yes
## Solution
-
+
If a triangle has integer sides, one of which is 1, then it must be isosceles. So the only candidates for the segment length 1 are AB and AE. wlog \(\mathrm{AB} = 1\) , so \(\mathrm{BF} = \mathrm{AF}\) . Hence \(\cos\)
@@ -3900,7 +3900,7 @@ If a triangle has integer sides, one of which is 1, then it must be isosceles. S
DFE = 1 - 1/(2 AF²). Hence ED² = DF² + EF² + 2DF. EF(1 - 1/2AF²) = DF² + EF² + 2DF. EF - DF. EF/AF². But the first three terms are integers and the last term is < 1. Contradiction. (Careful, looking at the figure one is tempted to conclude that ED < AB, but a more realistic figure shows that is false.).
-
+
Building on the 3,4,5 triangle we get the figure above.
@@ -3941,7 +3941,7 @@ ABCD is a rectangle. Points K, L, M, N are chosen on AB, BC, CD, DA respectively
## Solution
-
+
Let LN and KM meet at O. \(\angle \mathrm{NOM} = \angle \mathrm{NDM} = 90^{\circ}\) , so OMDN is cyclic. Hence \(\angle \mathrm{NOD} = \angle \mathrm{NMD}\) . Similarly, BLOK is cyclic and \(\angle \mathrm{LOB} = \angle \mathrm{LKB}\) . But NM is parallel to LK and AB is parallel to CD, so \(\angle \mathrm{LKB} = \angle \mathrm{NMD}\) . Hence \(\angle \mathrm{NOD} = \angle \mathrm{LOB}\) , so DOB is a straight line.
@@ -3966,12 +3966,12 @@ Answer only the central square
## Solution
-
+
We take the 5x5 square, the two yellow 3x3 squares, which overlap at the center, and the two blue 2x2 squares. Then every square except the center square is changed an even number of times. So this works if the central square was selected.
-
+
It is easy to check that any 2x2, 3x3, 4x4 or 5x5 square has an even number of green squares, so if the selected square was green, and we change it an odd number of times, then some other green square must also be changed an odd number of times and hence end up with a minus. So if all the squares end up plus, then the selected square was not green, so it must belong to the central white column. Similarly, it must belong to the central row and hence must be the center square.
@@ -4024,7 +4024,7 @@ numbers 1, 2, ..., \(2^{\mathrm{a}}\) - b- 1 can either be converted to powers o
The figure below is cut along the lines into polygons (which need not be convex). No polygon contains a 2 x 2 square. What is the smallest possible number of polygons?
-
+
Answer 12
@@ -4113,7 +4113,7 @@ E is a point on the diagonal BD of the square ABCD. Show that the points A, E an
## Solution
-
+
Let O, O' be the circumcenters of ABE, ADE respectively. Then OA = OE and \(\angle AOB = 2\) \(\angle ABE = 90^{\circ}\) . Similarly, \(\mathrm{O'A = O'E}\) and \(\angle AOE = 2\angle ADE = 90^{\circ}\) . Hence AOE \(\mathrm{O'}\) is a square.
@@ -4182,7 +4182,7 @@ A and B lie on a circle. P lies on the minor arc AB. Q and R (distinct from P) a
## Solution
-
+
Let AR and BQ meet at X. Since arcs QA and AP are equal, we have \(\angle \mathrm{ABX} = \angle \mathrm{ABP}\) . Similarly, \(\angle \mathrm{BAX} = \angle \mathrm{BAP}\) . Side AB is common, so triangles ABX and ABP are congruent. Hence X is the reflection of P.
@@ -4255,7 +4255,7 @@ Circles C and C' intersect at O and X. A circle center O meets C at Q and R and
We show first that YRSX is cyclic.
-
+
It is sufficient to show that \(\angle \mathrm{RYS} = \angle \mathrm{RXS}\) . We have \(\angle \mathrm{RYS} = \angle \mathrm{PRQ} - \angle \mathrm{RQY} = \angle \mathrm{PSQ} - \angle \mathrm{RQY} = \angle \mathrm{OSQ} - \angle \mathrm{OSP} - \angle \mathrm{RQY} = \angle \mathrm{OQS} - \angle \mathrm{RQY} - \angle \mathrm{OSP} = \angle \mathrm{OQR} - \angle \mathrm{OSP}\) . But \(\angle \mathrm{RXS}\)
@@ -4280,7 +4280,7 @@ ABCD is a parallelogram. The excircle of ABC opposite A has center E and touches
## Solution
-
+
We have the familiar result that AY is perimeter ADC (chase round using the fact that the two tangents from the same point have the same length). Similarly, \(\mathrm{AX} =\) perimeter ABC = perimeter ADC. So \(\mathrm{AX} = \mathrm{AY} (*)\)
@@ -4333,7 +4333,7 @@ A plane intersects a sphere in a circle C. The points A and B lie on the sphere
All points of the circle C are equidistant from A. The plane p also meets the sphere in a circle C'. Let C" be the circle center A radius AP. Provided that AB is not a diameter of C' one of the lines BP, BQ will meet C" again at some point R (see diagram). Now since arcs AP, AQ are equal, so are the angles ABP, ABQ. Hence triangles ABP, ABQ are congruent and so BP \(= \mathrm{BR}\) . Hence BP. \(\mathrm{BQ} = \mathrm{BR}\) . BQ. But the square of the tangent from B to C" is \(\mathrm{AB}^2 - \mathrm{AP}^2\) , so \(\mathrm{BR}\cdot \mathrm{BQ} = \mathrm{AB}^2 - \mathrm{AP}^2\) , which is independent of the position of p.
-
+
## Problem 19
@@ -4372,7 +4372,7 @@ If \(\mathrm{k} = 6\) or 9, then \(\mathrm{kb - a - 1}\geq 5\) , so the first di
An equilateral triangle side 10 is divided into 100 equilateral triangles of side 1 by lines parallel to its sides. There are m equilateral tiles of 4 unit triangles and 25 - m straight tiles of 4 unit triangles (as shown below). For which values of m can they be used to tile the original triangle. [The straight tiles may be turned over.]
-
+
Answer \(\mathrm{m} = 5,7,9,\ldots\) or 25.
@@ -4380,7 +4380,7 @@ Answer \(\mathrm{m} = 5,7,9,\ldots\) or 25.
Solution
-
+
There are 45 upside down triangles (yellow in the diagram) and 55 right way up. A straight tile always covers two of each. A triangular tile covers 3 of one and 1 of the other. So suppose there are t1 triangular tiles placed the right way up and t2 triangular tiles placed upside down. Then we have \(3\mathrm{t}_1 + \mathrm{t}_2 + 2(25 - \mathrm{t}_1 - \mathrm{t}_2) = 55\) , so \(\mathrm{t}_1 - \mathrm{t}_2 = 5\) . So \(\mathrm{t}_1\) and \(\mathrm{t}_2\) have opposite parity and hence \(\mathrm{m} = \mathrm{t}_1 + \mathrm{t}_2\) must be odd. Also \(\mathrm{m}\geq \mathrm{t}_1\geq 5\) . Now each rhombus shown may be filled with two triangular tiles or two straight tiles. So any tiling with m odd and \(\geq 5\) is possible.
diff --git a/SovietUnion/segmented/en-ASU-1961-1991.jsonl b/SovietUnion/segmented/en-ASU-1961-1991.jsonl
index 99365ef3fa8d5943e38c405e05f3a2a8bd38de90..91147464b68c75cb0bfa548f2d62ff0f984c8c55 100644
--- a/SovietUnion/segmented/en-ASU-1961-1991.jsonl
+++ b/SovietUnion/segmented/en-ASU-1961-1991.jsonl
@@ -1,4 +1,4 @@
-{"year": "1961", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "AllSovietUnion", "problem": "Given 12 vertices and 16 edges arranged as follows: \n\n\n \n\nDraw any curve which does not pass through any vertex. Prove that the curve cannot intersect each edge just once. Intersection means that the curve crosses the edge from one side to the other. For example, a circle which had one of the edges as tangent would not intersect that edge.", "solution": "A- - - - B- - - - C | | | D- - E- - F- - G- - H | | | I- - J- - - - K- - L \n\nIf a curve intersects the boundary of a region R (such as ABFED), then it moves from inside R to outside or vice versa. Hence if R has an odd number of edges (like ABFED) then a curve intersecting all of them just once must have one endpoint inside R. But there are four such regions (ABFED, BCHGF, EFGKJ and the outside of ABCHLKJID) and only two endpoints. Note that we can easily intersect all edges but one. For example, start above AB, then cross successively AB, AD, DI, DE, EF, EJ, IJ, JK, GK, KL, HL, GH, CH, BC, FG.", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 1", "solution_match": "# Solution"}}
+{"year": "1961", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "AllSovietUnion", "problem": "Given 12 vertices and 16 edges arranged as follows: \n\n\n \n\nDraw any curve which does not pass through any vertex. Prove that the curve cannot intersect each edge just once. Intersection means that the curve crosses the edge from one side to the other. For example, a circle which had one of the edges as tangent would not intersect that edge.", "solution": "A- - - - B- - - - C | | | D- - E- - F- - G- - H | | | I- - J- - - - K- - L \n\nIf a curve intersects the boundary of a region R (such as ABFED), then it moves from inside R to outside or vice versa. Hence if R has an odd number of edges (like ABFED) then a curve intersecting all of them just once must have one endpoint inside R. But there are four such regions (ABFED, BCHGF, EFGKJ and the outside of ABCHLKJID) and only two endpoints. Note that we can easily intersect all edges but one. For example, start above AB, then cross successively AB, AD, DI, DE, EF, EJ, IJ, JK, GK, KL, HL, GH, CH, BC, FG.", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 1", "solution_match": "# Solution"}}
{"year": "1961", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "AllSovietUnion", "problem": "Given a rectangle ABCD with AC length e and four circles centers A, B, C, D and radii a, b, c, d respectively, satisfying \\(\\mathrm{a + c = b + d< e}\\) . Prove you can inscribe a circle inside the quadrilateral whose sides are the two outer common tangents to the circles center A and C, and the two outer common tangents to the circles center B and D.", "solution": "Let O be the center of the rectangle. Let \\(\\mathrm{r} = (\\mathrm{a} + \\mathrm{c}) / 2 = (\\mathrm{b} + \\mathrm{d}) / 2\\) . The required circle has center O, radius r. Let an outer common tangent touch the circle center A at W, and the circle center C at X. Let P be the midpoint of WX, then OP is parallel to AW and CX and has length r, hence the circle center O touches AW at P. Similarly for the other common tangents.", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 2", "solution_match": "# Solution"}}
{"year": "1961", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "AllSovietUnion", "problem": "Prove that any 39 successive natural numbers include at least one whose digit sum is divisible by 11.", "solution": "Let n be the smallest number in the sequence and m the smallest with last digit 0. m and m+10 have different digit sums unless (possibly) the penultimate digit of m is 9, but in that case m+10 and m+20 have different digit sums. So two of m, m+10, m+20 are sure to have different digit sums. Hence at least one has a digit sum not congruent to 1 mod 11. Adding the appropriate final digit gives a number whose digit sum is divisible by 11. This number lies in the range m to m+29 and m<=n+9. Hence the result. n=999981 shows it is best possible.", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 3", "solution_match": "# Solution"}}
{"year": "1961", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "AllSovietUnion", "problem": "(a) Arrange 7 stars in the 16 places of a 4 x 4 array, so that no 2 rows and 2 columns contain all the stars. \n(b) Prove this is not possible for <7 stars.", "solution": "(a) \\(* * .\\) \\(* .* .\\) \\(* * .\\) \\(* .* .\\) \n\nPick any two rows. The unpicked stars lie in different columns. \n\n(b) If there is a row with at least 3 stars, pick it. That leaves at most 3 stars, pick the row for one and the columns for the others. Now assume no row has more than 2 stars. 6 stars in <6 rows, so we can pick a row with 2 stars. That leaves 4 stars in 3 rows, so we can pick another row with 2 stars. That leaves 2 stars. Pick their columns. [This glosses over the case of <6 stars. In this case we can add extra stars to make the number up to 6. Now the procedure above deals with the original stars and the extra stars, and in particular with the original stars.]", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 4", "solution_match": "# Solution"}}
@@ -47,7 +47,7 @@
{"year": "1964", "tier": "T1", "problem_label": "6", "problem_type": null, "exam": "AllSovietUnion", "problem": "Find all possible integer solutions for \\(\\mathrm{sqrt(x + sqrt(x\\ldots(x + sqrt(x)) \\ldots)) = y}\\) , where there are 1998 square roots.", "solution": "Let \\(s_{1} = \\mathrm{sqrt(x)}\\) , \\(s_{2} = \\mathrm{sqrt(x + s_{1})}\\) , \\(s_{3} = \\mathrm{sqrt(x + s_{2})}\\) and so on. So the equation given is \\(y = s_{1998}\\) . We show first that all \\(s_{\\mathrm{n}}\\) must be integral for \\(1 \\ll n \\ll 1998\\) . \\(y\\) is integral, so \\(s_{1998}\\) is integral. Now suppose \\(s_{\\mathrm{n}}\\) is integral. Then \\(s_{\\mathrm{n - 1}} = s_{\\mathrm{n}}^{2} - x\\) is integral, proving the claim. \n\nSo in particular \\(s_{1}\\) and \\(s_{2}\\) are integers and \\(s_{2}^{2} = s_{1}^{2} + s_{1}\\) . But if \\(s_{1} > 0\\) , then \\(s_{1}^{2} < s_{1}^{2} + s_{1} < (s_{1} + 1)^{2}\\) , which is impossible. Similarly \\(s_{1} < 0\\) is impossible. So the only possible solution is \\(s_{1} = 0\\) and hence \\(x = 0\\) and \\(y = 0\\) .", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 6", "solution_match": "# Solution"}}
{"year": "1964", "tier": "T1", "problem_label": "7", "problem_type": null, "exam": "AllSovietUnion", "problem": "ABCD is a convex quadrilateral. A' is the foot of the perpendicular from A to the diagonal BD, B' is the foot of the perpendicular from B to the diagonal AC, and so on. Prove that A'B'C'D' is similar to ABCD.", "solution": "Let the diagonals meet at O. Then \\(\\mathrm{CC^{\\prime}O}\\) is similar to AA'O (because \\(\\mathrm{CC^{\\prime}O = AA^{\\prime}O = 90}\\) , and COC', AOA' are opposite angles), so \\(\\mathrm{A^{\\prime}O / C^{\\prime}O = AO / CO}\\) . Similarly, \\(\\mathrm{B^{\\prime}O / D^{\\prime}O = BO / DO}\\) . AA'O is also similar to BB'O, so \\(\\mathrm{A^{\\prime}O / B^{\\prime}O = AO / BO}\\) . Thus \\(\\mathrm{OA^{\\prime}:OB^{\\prime}:OC^{\\prime}:OD^{\\prime} =}\\) OA:OB:OC:OD. Hence triangles OA'B' and OAB are similar. Likewise OB'C' and OBC, OC'D' and OCD, and OD'A' and ODA. Hence result.", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 7", "solution_match": "# Solution"}}
{"year": "1964", "tier": "T1", "problem_label": "8", "problem_type": null, "exam": "AllSovietUnion", "problem": "Find all natural numbers n such that \\(\\mathrm{n}^2\\) does not divide n!.", "solution": "Answer: \\(\\mathrm{n} = 4\\) or prime. \n\nIf \\(\\mathrm{n} = \\mathrm{rs}\\) , with \\(1< \\mathrm{r}< \\mathrm{s}\\) , then \\(\\mathrm{r}< \\mathrm{s}< \\mathrm{n}\\) , and hence \\(\\mathrm{rsn} = \\mathrm{n}^2\\) divides n!. Similarly, if \\(\\mathrm{n} = \\mathrm{r}^2\\) with \\(\\mathrm{r} > 2\\) , then \\(\\mathrm{r}< 2\\mathrm{r}< \\mathrm{n}\\) , and hence \\(\\mathrm{n}^2\\) divides n!. This covers all possibilities except \\(\\mathrm{n} = 4\\) or \\(\\mathrm{n} =\\) prime, and it is easy to see that in these cases \\(\\mathrm{n}^2\\) does not divide n!.", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 8", "solution_match": "# Solution"}}
-{"year": "1964", "tier": "T1", "problem_label": "9", "problem_type": null, "exam": "AllSovietUnion", "problem": "Given a lattice of regular hexagons. A bug crawls from vertex A to vertex B along edges of the hexagons, taking the shortest possible path (or one of them). Prove that it travels a distance at least AB/2 in one direction. If it travels exactly AB/2 in one direction, how many edges does it traverse?", "solution": "\n \n\nSuppose vertex A is that marked \\* at the bottom left. Without loss of generality, B is in a 60 degree sector as shown. Assume the edges have unit length. The vertices can be partitioned into two sets (marked ° and . in the diagram). Each set forms a skewed lattice with axes at 60 degrees. Any path must alternate between the two lattices. \n\nIf B is on the same lattice as A, then we can give B coordinates (m,n) relative to A and the shortest path from A to B must move m units east and n units east of north. The shortest path between a lattice point and the next lattice point east is evidently one edge in direction 3 followed by one edge in direction 2. Similarly, the shortest path between a lattice point and the next lattice point east of north is one edge in direction 1, followed by one edge in direction 2. So a shortest path from A to B must have \\(\\mathrm{m + n}\\) edges in direction 2. \n\nB is a distance \\(\\sqrt[3]{(m + n / 2)}\\) east of A and a distance \\(3\\mathrm{n} / 2\\) north of A, so \\(\\mathrm{AB}^2 = (3\\mathrm{m}^2 + 3\\mathrm{mn} + 3\\mathrm{n}^2)\\) \\(< (4\\mathrm{m}^2 + 8\\mathrm{mn} + 4\\mathrm{n}^2) = 4(\\mathrm{m} + \\mathrm{n})^2\\) . So in this case the bug must travel more than AB/2 in direction 2.\n\n\n\nNow suppose B is on the other lattice. Let C be the lattice point immediately north of A and D the lattice point in direction 3 from A. Then a shortest path from A to B must either be A to C and then a shortest path from C to B, or A to D and then a shortest path from D to B. Take B to have coordinates (m, n) relative to C or D. \n\nIn the first case, \\(\\mathrm{AB}^2 = (\\sqrt{3} (\\mathrm{m} + \\mathrm{n} / 2))^2 + (3\\mathrm{n} / 2 + 1)^2 = (3\\mathrm{m}^2 + 3\\mathrm{mn} + 3\\mathrm{n}^2) + 3\\mathrm{n} + 1\\) and a shortest path has \\(\\mathrm{m} + \\mathrm{n}\\) units in direction 2. But \\(4(\\mathrm{m} + \\mathrm{n})^2 > (3\\mathrm{m}^2 + 3\\mathrm{mn} + 3\\mathrm{n}^2) + 3\\mathrm{n} + 1\\) , if \\(\\mathrm{m}^2 + \\mathrm{n}^2 + 5\\mathrm{mn} > 3\\mathrm{n} + 1\\) , which is true for m, n at least 1. If \\(\\mathrm{m} = 0\\) and \\(\\mathrm{n} = 1\\) , then a shortest path has 2 units in direction 1 and \\(\\mathrm{AB} = \\sqrt{7} < 4\\) . If \\(\\mathrm{m} = 1\\) and \\(\\mathrm{n} = 0\\) , then \\(\\mathrm{AB} = 2\\) and a shortest path has 1 unit in each direction. So in this case (the only one so far) we have equality. \n\nIt remains to consider the case where the path starts out towards D. In this case \\(\\mathrm{AB}^2 = (\\sqrt{3} (\\mathrm{m} + \\mathrm{n} / 2) + \\sqrt{3} /2)^2 + (3\\mathrm{n} / 2 - 1 / 2)^2 = (3\\mathrm{m}^2 + 3\\mathrm{mn} + 3\\mathrm{n}^2) + 3\\mathrm{m} + 1\\) and a path has \\(\\mathrm{m} + \\mathrm{n}\\) units in direction 2. But \\(4(\\mathrm{m} + \\mathrm{n})^2 > (3\\mathrm{m}^2 + 3\\mathrm{mn} + 3\\mathrm{n}^2) + 3\\mathrm{m} + 1\\) for \\(\\mathrm{m}^2 + \\mathrm{n}^2 + 5\\mathrm{mn} > 3\\mathrm{m} + 1\\) , which is true for m, n at least 1. If \\(\\mathrm{m} = 1\\) , \\(\\mathrm{n} = 0\\) , then a shortest path has 2 units in direction 3 and \\(\\mathrm{AB} = \\sqrt{7} < 4\\) . Finally, if \\(\\mathrm{m} = 0\\) and \\(\\mathrm{n} = 1\\) , then a shortest path has 1 unit in each direction and \\(\\mathrm{AB} = 2\\) . \n\nThus the answer to the final question is 3, because the only cases where the bug travels exactly \\(\\mathrm{AB} / 2\\) in one direction are where it goes to the opposite vertex of a hexagon it is on.", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 9", "solution_match": "# Solution"}}
+{"year": "1964", "tier": "T1", "problem_label": "9", "problem_type": null, "exam": "AllSovietUnion", "problem": "Given a lattice of regular hexagons. A bug crawls from vertex A to vertex B along edges of the hexagons, taking the shortest possible path (or one of them). Prove that it travels a distance at least AB/2 in one direction. If it travels exactly AB/2 in one direction, how many edges does it traverse?", "solution": "\n \n\nSuppose vertex A is that marked \\* at the bottom left. Without loss of generality, B is in a 60 degree sector as shown. Assume the edges have unit length. The vertices can be partitioned into two sets (marked ° and . in the diagram). Each set forms a skewed lattice with axes at 60 degrees. Any path must alternate between the two lattices. \n\nIf B is on the same lattice as A, then we can give B coordinates (m,n) relative to A and the shortest path from A to B must move m units east and n units east of north. The shortest path between a lattice point and the next lattice point east is evidently one edge in direction 3 followed by one edge in direction 2. Similarly, the shortest path between a lattice point and the next lattice point east of north is one edge in direction 1, followed by one edge in direction 2. So a shortest path from A to B must have \\(\\mathrm{m + n}\\) edges in direction 2. \n\nB is a distance \\(\\sqrt[3]{(m + n / 2)}\\) east of A and a distance \\(3\\mathrm{n} / 2\\) north of A, so \\(\\mathrm{AB}^2 = (3\\mathrm{m}^2 + 3\\mathrm{mn} + 3\\mathrm{n}^2)\\) \\(< (4\\mathrm{m}^2 + 8\\mathrm{mn} + 4\\mathrm{n}^2) = 4(\\mathrm{m} + \\mathrm{n})^2\\) . So in this case the bug must travel more than AB/2 in direction 2.\n\n\n\nNow suppose B is on the other lattice. Let C be the lattice point immediately north of A and D the lattice point in direction 3 from A. Then a shortest path from A to B must either be A to C and then a shortest path from C to B, or A to D and then a shortest path from D to B. Take B to have coordinates (m, n) relative to C or D. \n\nIn the first case, \\(\\mathrm{AB}^2 = (\\sqrt{3} (\\mathrm{m} + \\mathrm{n} / 2))^2 + (3\\mathrm{n} / 2 + 1)^2 = (3\\mathrm{m}^2 + 3\\mathrm{mn} + 3\\mathrm{n}^2) + 3\\mathrm{n} + 1\\) and a shortest path has \\(\\mathrm{m} + \\mathrm{n}\\) units in direction 2. But \\(4(\\mathrm{m} + \\mathrm{n})^2 > (3\\mathrm{m}^2 + 3\\mathrm{mn} + 3\\mathrm{n}^2) + 3\\mathrm{n} + 1\\) , if \\(\\mathrm{m}^2 + \\mathrm{n}^2 + 5\\mathrm{mn} > 3\\mathrm{n} + 1\\) , which is true for m, n at least 1. If \\(\\mathrm{m} = 0\\) and \\(\\mathrm{n} = 1\\) , then a shortest path has 2 units in direction 1 and \\(\\mathrm{AB} = \\sqrt{7} < 4\\) . If \\(\\mathrm{m} = 1\\) and \\(\\mathrm{n} = 0\\) , then \\(\\mathrm{AB} = 2\\) and a shortest path has 1 unit in each direction. So in this case (the only one so far) we have equality. \n\nIt remains to consider the case where the path starts out towards D. In this case \\(\\mathrm{AB}^2 = (\\sqrt{3} (\\mathrm{m} + \\mathrm{n} / 2) + \\sqrt{3} /2)^2 + (3\\mathrm{n} / 2 - 1 / 2)^2 = (3\\mathrm{m}^2 + 3\\mathrm{mn} + 3\\mathrm{n}^2) + 3\\mathrm{m} + 1\\) and a path has \\(\\mathrm{m} + \\mathrm{n}\\) units in direction 2. But \\(4(\\mathrm{m} + \\mathrm{n})^2 > (3\\mathrm{m}^2 + 3\\mathrm{mn} + 3\\mathrm{n}^2) + 3\\mathrm{m} + 1\\) for \\(\\mathrm{m}^2 + \\mathrm{n}^2 + 5\\mathrm{mn} > 3\\mathrm{m} + 1\\) , which is true for m, n at least 1. If \\(\\mathrm{m} = 1\\) , \\(\\mathrm{n} = 0\\) , then a shortest path has 2 units in direction 3 and \\(\\mathrm{AB} = \\sqrt{7} < 4\\) . Finally, if \\(\\mathrm{m} = 0\\) and \\(\\mathrm{n} = 1\\) , then a shortest path has 1 unit in each direction and \\(\\mathrm{AB} = 2\\) . \n\nThus the answer to the final question is 3, because the only cases where the bug travels exactly \\(\\mathrm{AB} / 2\\) in one direction are where it goes to the opposite vertex of a hexagon it is on.", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 9", "solution_match": "# Solution"}}
{"year": "1964", "tier": "T1", "problem_label": "10", "problem_type": null, "exam": "AllSovietUnion", "problem": "A circle center O is inscribed in ABCD (touching every side). Prove that angle AOB + angle COD equals 180 degrees.", "solution": "Let AB touch the circle at W, BC at X, CD at Y, and DA at Z. Then AO bisects angle ZOW and BO bisects angle XOW. So angle AOB is half angle ZOX. Similarly angle COD is half angle XOZ and hence angle AOB + angle COD equals 180.", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 10", "solution_match": "# Solution"}}
{"year": "1964", "tier": "T1", "problem_label": "11", "problem_type": null, "exam": "AllSovietUnion", "problem": "The natural numbers a, b, n are such that for every natural number k not equal to b, b - k divides a - \\(\\mathrm{k^n}\\) . Prove that \\(\\mathrm{a} = \\mathrm{b^n}\\) .", "solution": "We have \\(\\mathrm{k^n - a = b^n - a}\\) (mod b - k). Hence \\(\\mathrm{b^n - a = 0}\\) (mod b - k) for every k not equal to b. But if \\(\\mathrm{b^n}\\) does not equal a, then by taking \\(\\mathrm{k - b > b^n - a}\\) we could render the equation false.", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 11", "solution_match": "# Solution"}}
{"year": "1964", "tier": "T1", "problem_label": "12", "problem_type": null, "exam": "AllSovietUnion", "problem": "How many (algebraically) different expressions can we obtain by placing parentheses in \\(\\mathrm{a_1 / a_2 / ... / a_n?}\\)", "solution": "Answer \\(2^{\\mathrm{n} - 2}\\) . \\(\\mathrm{a_1}\\) must be in the numerator, and \\(\\mathrm{a_2}\\) must be in the denominator, but the other symbols can be in either. This is easily proved by induction.", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 12", "solution_match": "# Solution"}}
@@ -99,7 +99,7 @@
{"year": "1968", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "AllSovietUnion", "problem": "Prove that: \n\n\\[2 / (\\mathrm{x}^{2} - 1) + 4 / (\\mathrm{x}^{2} - 4) + 6 / (\\mathrm{x}^{2} - 9) + \\ldots +20 / (\\mathrm{x}^{2} - 100) =\\] \\[11 / ((\\mathrm{x} - 1)(\\mathrm{x} + 10)) + 11 / ((\\mathrm{x} - 2)(\\mathrm{x} + 9)) + \\ldots +11 / ((\\mathrm{x} - 10)(\\mathrm{x} + 1)).\\]", "solution": "\\[\\mathrm{lhs} = 1 / (\\mathrm{x} - 1) - 1 / (\\mathrm{x} + 1) + 1 / (\\mathrm{x} - 2) - 1 / (\\mathrm{x} + 2) + \\ldots +1 / (\\mathrm{x} + 10) - 1 / (\\mathrm{x} - 10) =\\] \\[1 / (\\mathrm{x} - 1) - 1 / (\\mathrm{x} + 10) + 1 / (\\mathrm{x} - 2) - 1 / (\\mathrm{x} + 9) + \\ldots +1 / (\\mathrm{x} - 10) - 1 / (\\mathrm{x} + 1) = \\mathrm{rhs}.\\]", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 5", "solution_match": "# Solution"}}
{"year": "1968", "tier": "T1", "problem_label": "6", "problem_type": null, "exam": "AllSovietUnion", "problem": "The difference between the longest and shortest diagonals of the regular n- gon equals its side. Find all possible n.", "solution": "Answer: \\(\\mathrm{n} = 9\\) \n\nFor \\(\\mathrm{n}< 6\\) , there is at most one length of diagonal. For \\(\\mathrm{n} = 6\\) , 7 the longest and shortest, and a side of the n- gon form a triangle, so the difference between the longest and shortest is less than the side. \n\nFor \\(\\mathrm{n} > 7\\) the side has length 2R sin \\(\\pi /\\mathrm{n}\\) , the shortest diagonal has length 2R sin \\(2\\pi /\\mathrm{n}\\) , and the longest diagonal has length 2R for n even and 2R cos \\(\\pi /2\\mathrm{n}\\) for n odd (where R is the radius of the circumcircle). Thus we require: \n\n\\(\\sin 2\\pi /\\mathrm{n} + \\sin \\pi /\\mathrm{n} = 1\\) and n even, or\n\n\n\n\\(\\sin 2\\pi /\\mathrm{n} + \\sin \\pi /\\mathrm{n} = \\cos \\pi /2\\mathrm{n}\\) and n odd. \n\nEvidently the lhs is a strictly decreasing function of n and the rhs is an increasing function of n, so there can be at most one solution of each equation. The second equation is satisfied by n \\(= 9\\) , although it is easier to see that there is a quadrilateral with the longest diagonal and shortest diagonals as one pair of opposite sides, and 9- gon sides as the other pair of opposite sides. The angle between the longest side and an adjacent side is 60, so that its length is the length of the shortest diagonal plus \\(2 \\times 9\\) - gon side x cos 60. Hence that is the only solution for n odd. \n\nFor \\(\\mathrm{n} = 8\\) we have the same quadrilateral as for the 9- gon except that the angle is 67.5 and hence the difference is less than 1. For \\(\\mathrm{n} = 10\\) , \\(\\sin 2\\pi /10 + \\sin \\pi /10 = \\sin \\pi /10\\) (2 cos \\(\\pi /10 + 1\\) ) \\(< 3\\) sin \\(\\pi /10< 3\\pi /10< 1\\) . So there are no solutions for n even \\(\\geq 10\\) , and hence no solutions for n even.", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 6", "solution_match": "# Solution"}}
{"year": "1968", "tier": "T1", "problem_label": "7", "problem_type": null, "exam": "AllSovietUnion", "problem": "The sequence \\(\\mathrm{a}_{\\mathrm{n}}\\) is defined as follows: \\(\\mathrm{a}_{1} = 1\\) , \\(\\mathrm{a}_{\\mathrm{n} + 1} = \\mathrm{a}_{\\mathrm{n}} + 1 / \\mathrm{a}_{\\mathrm{n}}\\) for \\(\\mathrm{n} \\geq 1\\) . Prove that \\(\\mathrm{a}_{100} > 14\\) .", "solution": "First we must notice that for \\(1 \\leq \\mathrm{a}\\) , b we have \\(\\mathrm{a} < \\mathrm{b}\\) , then \\(\\mathrm{a} + 1 / \\mathrm{a} < \\mathrm{b} + 1 / \\mathrm{b}\\) . This is basic to any estimation. \n\nThe obvious approach is to notice that if \\(\\mathrm{a}_{\\mathrm{i}} \\leq \\mathrm{n}\\) , then \\(\\mathrm{a}_{\\mathrm{i} + 1} \\geq \\mathrm{a}_{\\mathrm{i}} + 1 / \\mathrm{n}\\) . Hence it takes at most n steps to get from \\(\\mathrm{n} - 1\\) to \\(\\mathrm{n}\\) . Unfortunately, this does not quite work: we need \\(2 + 3 + \\ldots + 14 = 104\\) steps to get from 1 to 14. \n\nThe trick is to notice that \\(\\mathrm{a}_{\\mathrm{n} + 1}^{2} > \\mathrm{a}_{\\mathrm{n}}^{2} + 2\\) . But \\(\\mathrm{a}_{2} = 2\\) , so \\(\\mathrm{a}_{\\mathrm{n}}^{2} > 2\\mathrm{n}\\) . That gives \\(\\mathrm{a}_{100}^{2} > 200 > 14^{2}\\) .", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 7", "solution_match": "# Solution"}}
-{"year": "1968", "tier": "T1", "problem_label": "8", "problem_type": null, "exam": "AllSovietUnion", "problem": "Given point O inside the acute- angled triangle ABC, and point O' inside the acute- angled triangle A'B'C'. D, E, F are the feet of the perpendiculars from O to BC, CA, AB respectively, and D', E', F' are the feet of the perpendiculars from O' to B'C', C'A', A'B' respectively. OD is parallel to O'A', OE is parallel to O'B' and OF is parallel to O'C'. Also OD- O'A' = OE- O'B' = OF- O'C'. Prove that O'D' is parallel to OA, O'E' to OB and O'F' to OC, and that O'D' OA = O'E' OB = O'F' OC.", "solution": "\n \n\nLet \\(\\Gamma\\) be the circumcircle of DEF. Let OD, OE, OF meet it again at A\", B\", C\" respectively. Then the figure O'A'B'C' must be similar to OA\"B\"C\". So to prove that OD is parallel to O'A', we have to prove that AO is perpendicular to B\"C\". \n\nSo AO meets B\"C\" at D\". Now since \\(\\square \\mathrm{AFC}'' = 90^{0}\\) and \\(\\square \\mathrm{AD}''\\mathrm{C}'' = 90^{0}\\) , both F and D' lie on the circle diameter AC\". Hence AO- OD\" = OF- OC\". Similarly, BO meets C\"A\" at E\", and CO meets A\"B\" at F\", and BO- OE\" = OD- OA\" and CO- OF\" = OE- OB\". Hence OD\". OA = OE\". OB = OF\". OC. So using the similarity, OD'- OA = OE'- OB = OF'. OC.", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 8", "solution_match": "# Solution"}}
+{"year": "1968", "tier": "T1", "problem_label": "8", "problem_type": null, "exam": "AllSovietUnion", "problem": "Given point O inside the acute- angled triangle ABC, and point O' inside the acute- angled triangle A'B'C'. D, E, F are the feet of the perpendiculars from O to BC, CA, AB respectively, and D', E', F' are the feet of the perpendiculars from O' to B'C', C'A', A'B' respectively. OD is parallel to O'A', OE is parallel to O'B' and OF is parallel to O'C'. Also OD- O'A' = OE- O'B' = OF- O'C'. Prove that O'D' is parallel to OA, O'E' to OB and O'F' to OC, and that O'D' OA = O'E' OB = O'F' OC.", "solution": "\n \n\nLet \\(\\Gamma\\) be the circumcircle of DEF. Let OD, OE, OF meet it again at A\", B\", C\" respectively. Then the figure O'A'B'C' must be similar to OA\"B\"C\". So to prove that OD is parallel to O'A', we have to prove that AO is perpendicular to B\"C\". \n\nSo AO meets B\"C\" at D\". Now since \\(\\square \\mathrm{AFC}'' = 90^{0}\\) and \\(\\square \\mathrm{AD}''\\mathrm{C}'' = 90^{0}\\) , both F and D' lie on the circle diameter AC\". Hence AO- OD\" = OF- OC\". Similarly, BO meets C\"A\" at E\", and CO meets A\"B\" at F\", and BO- OE\" = OD- OA\" and CO- OF\" = OE- OB\". Hence OD\". OA = OE\". OB = OF\". OC. So using the similarity, OD'- OA = OE'- OB = OF'. OC.", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 8", "solution_match": "# Solution"}}
{"year": "1968", "tier": "T1", "problem_label": "9", "problem_type": null, "exam": "AllSovietUnion", "problem": "Prove that any positive integer not exceeding n! can be written as a sum of at most n distinct factors of n!.", "solution": "Given \\(\\mathrm{m}\\leq \\mathrm{n}!\\) write \\(\\mathrm{m} = \\mathrm{nq} + \\mathrm{r}\\) (\\*), with \\(0\\leq \\mathrm{q},0\\leq \\mathrm{r}< \\mathrm{m}\\) . Then \\(\\mathrm{q}\\leq (\\mathrm{n} - 1)!\\) , so \\(\\mathrm{q}\\) is a sum of at most n- 1 distinct factors of (n- 1)!. r is itself a factor of n! and is not divisible by n, so (\\*) expresses m as a sum of at most n distinct factors of n!.", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 9", "solution_match": "# Solution"}}
{"year": "1968", "tier": "T1", "problem_label": "10", "problem_type": null, "exam": "AllSovietUnion", "problem": "Given a triangle ABC, and D on the segment AB, E on the segment AC, such that AD = DE = AC, BD = AE, and DE is parallel to BC. Prove that BD equals the side of a regular 10- gon inscribed in a circle with radius AC.", "solution": "\\(\\mathrm{DA} = \\mathrm{DE}\\) , so DAE is isosceles. DE is parallel to BC, so ABC is isosceles, so \\(\\mathrm{BA} = \\mathrm{AC} / (2 \\cos \\mathrm{A})\\) . Hence \\(\\mathrm{BD} = \\mathrm{AC} / (2 \\cos \\mathrm{A}) - \\mathrm{AC}\\) . But \\(\\mathrm{AE} = 2 \\mathrm{AC} \\cos \\mathrm{A}\\) , so we have an equation for \\(\\mathrm{c} = \\mathrm{cos} \\mathrm{A}: 4 \\mathrm{c}^{2} + 2 \\mathrm{c} - 1 = 0\\) . \n\n\\(2\\pi /5,4\\pi /5,6\\pi /5,8\\pi /5\\) and \\(10\\pi /5\\) are the roots of: real part of \\((\\cos \\theta + \\mathrm{i}\\sin \\theta)^{5} = 1\\) . Expanding this gives that \\(\\cos 2\\pi /5\\) , \\(\\cos 4\\pi /5\\) , \\(\\cos 6\\pi /5\\) , \\(\\cos 8\\pi /5\\) and 1 are the roots of \\(16\\mathrm{c}^{5} - 20\\mathrm{c}^{3} + 5\\mathrm{c} - 1 = 0\\) . Dividing by \\((\\mathrm{c} - 1)\\) gives \\(16\\mathrm{c}^{4} + 16\\mathrm{c}^{3} - 4\\mathrm{c}^{2} - 4\\mathrm{c} + 1 = (4\\mathrm{c}^{2} + 2\\mathrm{c} - 1)^{2}\\) . So \\(\\cos 2\\pi /5\\) ( \\(=\\) cos \\(8\\pi /5\\) ) and \\(\\cos 4\\pi /5\\) ( \\(=\\) cos \\(6\\pi /5\\) ) are the roots of \\(4\\mathrm{c}^{2} + 2\\mathrm{c} - 1 = 0\\) .\n\n\n\nWe know that \\(\\mathrm{A}< 90^{\\circ}\\) (since \\(\\mathrm{A} = \\mathrm{C}\\) and their sum is less than \\(180^{\\circ}\\) ). Hence \\(\\mathrm{A} = 2\\pi /5\\) . So BD \\(= 2\\) AC cos \\(2\\pi /5 = 2\\) AC sin \\(\\pi /10\\) , which is the side length for a regular 10- gon inscribed in a circle radius AC.", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 10", "solution_match": "# Solution"}}
{"year": "1968", "tier": "T1", "problem_label": "11", "problem_type": null, "exam": "AllSovietUnion", "problem": "Given a regular tetrahedron ABCD, prove that it is contained in the three spheres on diameters AB, BC and AD. Is this true for any tetrahedron?", "solution": "Let the tetrahedron have side 1. Then the center O is a distance \\(1 / \\sqrt{8}\\) from the center of each of the spheres, so it is contained in each of the spheres. We now use convexity. \n\nTwo circles with diameters two of the sides of a triangle cover the triangle (consider the foot of the altitude to the third side), so faces ABC and ABD are certainly contained in the spheres. Consider face ACD. The sphere on BC passes through the midpoints of AC and CD, and through C, so it contains the triangle formed by these three points (by convexity). But the rest of ACD is contained in the sphere on AD. Similarly for the face BCD. Hence all the faces are contained in the spheres. But now take any point P inside the tetrahedron. Extend OP to meet a face at X. X lies in one of the spheres, but O also lies in the sphere and hence all points on OX, including P (by convexity). \n\nFalse in general. Take ABCD to be a plane square, then no points on CD are in the spheres except C and D (and we can obviously distort this slightly to make it less degenerate).", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 11", "solution_match": "# Solution"}}
@@ -299,10 +299,10 @@
{"year": "1980", "tier": "T1", "problem_label": "9", "problem_type": null, "exam": "AllSovietUnion", "problem": "Find all real solutions to: \\(\\sin x + 2\\sin (x + y + z) = 0\\) \\(\\sin y + 3\\sin (x + y + z) = 0\\) \\(\\sin z + 4\\sin (x + y + z) = 0\\)", "solution": "", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 9", "solution_match": ""}}
{"year": "1980", "tier": "T1", "problem_label": "10", "problem_type": null, "exam": "AllSovietUnion", "problem": "Given 1980 vectors in the plane. The sum of every 1979 vectors is a multiple of the other vector. Not all the vectors are multiples of each other. Show that the sum of all the vectors is zero.", "solution": "Let the vectors be \\(\\mathbf{x}_{\\mathrm{i}}\\) and their sum \\(\\mathbf{s}\\) . Then we have \\(\\mathbf{s} - \\mathbf{x}_{\\mathrm{i}} = \\mathbf{n}_{\\mathrm{i}}\\mathbf{x}_{\\mathrm{i}}\\) for some scalar \\(\\mathbf{n}_{\\mathrm{i}}\\) . Hence \\((\\mathbf{n}_{\\mathrm{i}} + 1)\\mathbf{x}_{\\mathrm{i}} = \\mathbf{s}\\) . If \\(\\mathbf{s}\\) is non- zero, then it follows that every vector is a multiple of \\(\\mathbf{s}\\) and hence all the vectors are multiples of each other. But we are told that is not true. Hence \\(\\mathbf{s}\\) is zero.", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 10", "solution_match": "# Solution"}}
{"year": "1980", "tier": "T1", "problem_label": "11", "problem_type": null, "exam": "AllSovietUnion", "problem": "Let \\(\\mathbf{f}(\\mathbf{n})\\) be the sum of \\(\\mathbf{n}\\) and its digits. For example, \\(\\mathbf{f}(34) = 41\\) . Is there an integer such that \\(\\mathbf{f}(\\mathbf{n}) = 1980\\) ? Show that given any positive integer \\(\\mathbf{m}\\) we can find \\(\\mathbf{n}\\) such that \\(\\mathbf{f}(\\mathbf{n}) = \\mathbf{m}\\) or \\(\\mathbf{m} + 1\\) . \n\n## Answer \n\n\\[\\mathrm{f}(1962) = 1962 + 18 = 1980\\]", "solution": "If the last digit of \\(\\mathbf{n}\\) is not 9, then \\(\\mathbf{f}(\\mathbf{n} + 1) = \\mathbf{f}(\\mathbf{n}) + 2\\) . If the last digit of \\(\\mathbf{n}\\) is 9, then \\(\\mathbf{f}(\\mathbf{n} + 1)< \\mathbf{f}(\\mathbf{n})\\) . On the other hand \\(\\mathbf{f}\\) clearly achieves arbitrarily large values. Also \\(\\mathbf{f}(1) = 1\\) . Now consider any \\(\\mathbf{m} > 1\\) . Let \\(\\mathbf{M}\\) be the smallest integer such that \\(\\mathbf{f}(\\mathbf{M}) > \\mathbf{m}\\) . Then \\(\\mathbf{f}(\\mathbf{M} - 1)\\leq \\mathbf{m}\\) . Since \\(\\mathbf{f}(\\mathbf{M}) > \\mathbf{f}(\\mathbf{M} - 1)\\) we must have \\(\\mathbf{f}(\\mathbf{M}) = \\mathbf{f}(\\mathbf{M} - 1) + 2\\) . Hence either \\(\\mathbf{f}(\\mathbf{M}) = \\mathbf{m} + 1\\) or \\(\\mathbf{f}(\\mathbf{M} - 1) = \\mathbf{m}\\) .", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 11", "solution_match": "# Solution"}}
-{"year": "1980", "tier": "T1", "problem_label": "12", "problem_type": null, "exam": "AllSovietUnion", "problem": "Some unit squares in an infinite sheet of squared paper are colored red so that every \\(2 \\times 3\\) and \\(3 \\times 2\\) rectangle contains exactly two red squares. How many red squares are there in a \\(9 \\times 11\\) rectangle? \n\n## Answer \n\n33", "solution": "\n \n\nThere cannot be two red squares with a common side. For consider as the diagram shows we can immediately conclude that the squares with a \\(\\ast\\) are not red, but now the bold rectangle has at most 1 red square. Contradiction.\n\n\n\n \n\nConsider a red square. One of the two diagonally adjacent squares marked \\* must be red. But it is now easy to show that all red squares on that diagonal are red and that the other red squares are those on every third parallel diagonal line. Any \\(9 \\times 11\\) rectangle must have just three such diagonals on a 9 cell border row, and hence just 3 red cells in that border row. But the remaining \\(9 \\times 10\\) rectangle can easily be partitioned into fifteen \\(3 \\times 2\\) rectangles, each with 2 red squares.", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 12", "solution_match": "# Solution"}}
+{"year": "1980", "tier": "T1", "problem_label": "12", "problem_type": null, "exam": "AllSovietUnion", "problem": "Some unit squares in an infinite sheet of squared paper are colored red so that every \\(2 \\times 3\\) and \\(3 \\times 2\\) rectangle contains exactly two red squares. How many red squares are there in a \\(9 \\times 11\\) rectangle? \n\n## Answer \n\n33", "solution": "\n \n\nThere cannot be two red squares with a common side. For consider as the diagram shows we can immediately conclude that the squares with a \\(\\ast\\) are not red, but now the bold rectangle has at most 1 red square. Contradiction.\n\n\n\n \n\nConsider a red square. One of the two diagonally adjacent squares marked \\* must be red. But it is now easy to show that all red squares on that diagonal are red and that the other red squares are those on every third parallel diagonal line. Any \\(9 \\times 11\\) rectangle must have just three such diagonals on a 9 cell border row, and hence just 3 red cells in that border row. But the remaining \\(9 \\times 10\\) rectangle can easily be partitioned into fifteen \\(3 \\times 2\\) rectangles, each with 2 red squares.", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 12", "solution_match": "# Solution"}}
{"year": "1980", "tier": "T1", "problem_label": "13", "problem_type": null, "exam": "AllSovietUnion", "problem": "There is a flu epidemic in elf city. The course of the disease is always the same. An elf is infected one day, he is sick the next, recovered and immune the third, recovered but not immune thereafter. Every day every elf who is not sick visits all his sick friends. If he is not immune he is sure to catch flu if he visits a sick elf. On day 1 no one is immune and one or more elves are infected from some external source. Thereafter there is no further external infection and the epidemic spreads as described above. Show that it is sure to die out (irrespective of the number of elves, the number of friends each has, and the number infected on day 1). Show that if one or more elves is immune on day 1, then it is possible for the epidemic to continue indefinitely.", "solution": "This is curiously easy. Write S for sick, N for not sick and not immune, and I for immune. Suppose group A are S on day 1 and group B are N on day 1. Then on day 2, B are S, and A are I. So on day 3 no one is sick, A are N and B are I. Thereafter no one can get sick, so the epidemic has died out. \n\nSuppose on day 1, there is also group C who are I. Then on day 2, B are S, A are I and C are N. So on day 3, C are S, A are N and B are I. On day 4, A are S, B are N and C are I, the same as day 1, so the epidemic continues indefinitely.", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 13", "solution_match": "# Solution"}}
{"year": "1980", "tier": "T1", "problem_label": "14", "problem_type": null, "exam": "AllSovietUnion", "problem": "Define the sequence \\(\\mathbf{a}_{\\mathrm{n}}\\) of positive integers as follows. \\(\\mathbf{a}_{1} = \\mathbf{m}\\) . \\(\\mathbf{a}_{\\mathrm{n} + 1} = \\mathbf{a}_{\\mathrm{n}}\\) plus the product of the digits of \\(\\mathbf{a}_{\\mathrm{n}}\\) . For example, if \\(\\mathrm{m} = 5\\) , we have 5, 10, 10, ... . Is there an m for which the sequence is unbounded? \n\n## Answer \n\nNo.", "solution": "Put \\(\\mathrm{p(n)}\\) for the product of the digits of n. We show that, for sufficiently large n, a sequence starting below it cannot get past the \"gap\" from \\(10_{\\mathrm{n}}\\) to \\(10_{\\mathrm{n}} + 10_{\\mathrm{n} - 1}\\) . For suppose N is the last member of the sequence below the gap. Then N has at most n digits, so \\(\\mathrm{p(N)} \\leq 9^{\\mathrm{n}}\\) . But for sufficiently large n (in fact for \\(\\mathrm{n} \\geq 21\\) ) we have \\(9^{\\mathrm{n}} < 10^{\\mathrm{n} - 1}\\) . So \\(\\mathrm{N} + \\mathrm{p(N)} < 10^{\\mathrm{n}} + 10^{\\mathrm{n} - 1}\\) . But \\(\\mathrm{N} + \\mathrm{p(N)} > 10^{\\mathrm{n}}\\) by assumption. Hence \\(\\mathrm{N} + \\mathrm{p(N)}\\) is sure to have second digit (from the left) zero.\n\n\n\nSo all further terms of the sequence are the same. But for any m there is certainly a gap above m, and, as shown, the sequence will not be able to get beyond it. So it is bounded.", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 14", "solution_match": "# Solution"}}
-{"year": "1980", "tier": "T1", "problem_label": "15", "problem_type": null, "exam": "AllSovietUnion", "problem": "ABC is equilateral. A line parallel to AC meets AB at M and BC at P. D is the center of the equilateral triangle BMP. E is the midpoint of AP. Find the angles of DEC. \n\n## Answer \n\n\\[\\mathrm{D} = 60^{\\circ},\\mathrm{E} = 90^{\\circ}\\]", "solution": "\n \n\nLet K be the midpoint of BP and L the midpoint of AC. EL is parallel to BC, so \\(\\angle ELC = 120^{\\circ}\\) . EK is parallel to AB, so \\(\\angle EKC = 60^{\\circ}\\) , so ELCK is cyclic. But \\(\\angle DKC = \\angle DLC = 90^{\\circ}\\) , so DLCK is cyclic. Hence D, K, C, L, E all lie on a circle. Hence \\(\\angle DEC = \\angle DLC = 90^{\\circ}\\) , and \\(\\angle EDC = \\angle EKC = 60^{\\circ}\\) .", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 15", "solution_match": "# Solution"}}
+{"year": "1980", "tier": "T1", "problem_label": "15", "problem_type": null, "exam": "AllSovietUnion", "problem": "ABC is equilateral. A line parallel to AC meets AB at M and BC at P. D is the center of the equilateral triangle BMP. E is the midpoint of AP. Find the angles of DEC. \n\n## Answer \n\n\\[\\mathrm{D} = 60^{\\circ},\\mathrm{E} = 90^{\\circ}\\]", "solution": "\n \n\nLet K be the midpoint of BP and L the midpoint of AC. EL is parallel to BC, so \\(\\angle ELC = 120^{\\circ}\\) . EK is parallel to AB, so \\(\\angle EKC = 60^{\\circ}\\) , so ELCK is cyclic. But \\(\\angle DKC = \\angle DLC = 90^{\\circ}\\) , so DLCK is cyclic. Hence D, K, C, L, E all lie on a circle. Hence \\(\\angle DEC = \\angle DLC = 90^{\\circ}\\) , and \\(\\angle EDC = \\angle EKC = 60^{\\circ}\\) .", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 15", "solution_match": "# Solution"}}
{"year": "1980", "tier": "T1", "problem_label": "16", "problem_type": null, "exam": "AllSovietUnion", "problem": "A rectangular box has sides \\(\\mathbf{x}< \\mathbf{y}< \\mathbf{z}\\) . Its perimeter is \\(\\mathrm{p} = 4(\\mathrm{x} + \\mathrm{y} + \\mathrm{z})\\) , its surface area is \\(\\mathrm{s} =\\) \\(2(\\mathrm{xy} + \\mathrm{yz} + \\mathrm{zx})\\) and its main diagonal has length \\(\\mathrm{d} = \\sqrt{(\\mathrm{x}^2 + \\mathrm{y}^2 + \\mathrm{z}^2)}\\) . Show that \\(3\\mathrm{x}< (\\mathrm{p} / 4-\\) \\(\\sqrt{(\\mathrm{d}^2 - \\mathrm{s} / 2)}\\) and \\(3\\mathrm{z} > (\\mathrm{p} / 4 + \\sqrt{(\\mathrm{d}^2 - \\mathrm{s} / 2)}}\\)", "solution": "We have \\(3(\\mathrm{y - x})(\\mathrm{z - x}) > 0\\) , so \\(3\\mathrm{x}^2 +3\\mathrm{yz} > 3\\mathrm{xy} + 3\\mathrm{xz}\\) . Hence \\(\\mathrm{y}^2 +\\mathrm{z}^2 +4\\mathrm{x}^2 +2\\mathrm{yz} - 4\\mathrm{xy} - 4\\mathrm{xz} > \\mathrm{x}^2 + \\mathrm{y}^2 +\\mathrm{z}^2 - \\mathrm{xy} - \\mathrm{yz} - \\mathrm{xz}\\) or \\((\\mathrm{y} + \\mathrm{z} - 2\\mathrm{x})^2 >(\\mathrm{d}^2 - \\mathrm{s} / 2)\\) . Hence \\((\\mathrm{x} + \\mathrm{y} + \\mathrm{z}) > 3\\mathrm{x} + \\sqrt{(\\mathrm{d}^2 - \\mathrm{s} / 2)}\\) . So \\(3\\mathrm{x}< \\mathrm{p} / 4 - \\sqrt{(\\mathrm{d}^2 - \\mathrm{s} / 2)}\\) . \n\nSimilarly, \\(3(\\mathrm{z - x})(\\mathrm{z - y}) > 0\\) , so \\(\\mathrm{x}^2 +\\mathrm{y}^2 +4\\mathrm{z}^2 >\\mathrm{x}^2 +\\mathrm{y}^2 +\\mathrm{z}^2 +3\\mathrm{zx} + 3\\mathrm{zy} - 3\\mathrm{xy}\\) , so \\((2\\mathrm{z} - \\mathrm{x} - \\mathrm{y})^2 >\\mathrm{x}^2 +\\mathrm{y}^2 +\\mathrm{z}^2 - \\mathrm{xy} - \\mathrm{yz} - \\mathrm{zx}\\) or \\((3\\mathrm{z} - \\mathrm{p} / 4)^2 >(\\mathrm{d}^2 - \\mathrm{s} / 2)\\) . Hence \\(3\\mathrm{z} > \\mathrm{p} / 4 + \\sqrt{(\\mathrm{d}^2 - \\mathrm{s} / 2)}\\) .", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 16", "solution_match": "# Solution"}}
{"year": "1980", "tier": "T1", "problem_label": "17", "problem_type": null, "exam": "AllSovietUnion", "problem": "S is a set of integers. Its smallest element is 1 and its largest element is 100. Every element of S except 1 is the sum of two distinct members of the set or double a member of the set. What is the smallest possible number of integers in S? \n\n## Answer", "solution": "Let \\(\\mathrm{n} = \\mathrm{M}(\\mathrm{n}) + \\mathrm{m}(\\mathrm{n})\\) , where \\(\\mathrm{M}(\\mathrm{n})\\geq \\mathrm{m}(\\mathrm{n})\\) . Put \\(\\mathrm{M}^{1}(\\mathrm{n}) = \\mathrm{M}(\\mathrm{n})\\) , \\(\\mathrm{M}^{2}(\\mathrm{n}) = \\mathrm{M}(\\mathrm{M}(\\mathrm{n}))\\) etc. Then \\(\\mathrm{M}(100)\\geq 50\\) , \\(\\mathrm{M}^{2}(100)\\geq 25\\) , \\(\\mathrm{M}^{3}(100)\\geq 13\\) , \\(\\mathrm{M}^{4}(100)\\geq 7\\) , \\(\\mathrm{M}^{5}(100)\\geq 4\\) , \\(\\mathrm{M}^{6}(100)\\geq 2\\) (and obviously \\(\\mathrm{n} > \\mathrm{M}(\\mathrm{n})\\) ), so we need at least 8 numbers. There are several ways of using 9 numbers. For example, {1, 2, 4, 8, 16, 32, 36, 64, 100}, where \\(36 = 4 + 32\\) , \\(100 = 36 + 64\\) and the others are double another number. \n\nDoubling every time does not work: 1, 2, 4, 8, 16, 32, 64, 128. But if we do not double every time, then we cannot get a number larger than 96 with 8 numbers: the best we can do is \\(1\\cdot 2^{6}\\) . \\((3 / 2) = 96\\) (on the occasion when we do not double the best we can do is to the largest plus the next largest, or \\(3 / 2\\) x the largest). Hence we need at least 9 numbers. [To be more formal, write the elements as \\(1 = \\mathrm{a}_{1}< \\mathrm{a}_{2}< \\ldots < \\mathrm{a}_{\\mathrm{n}}\\) , then each \\(\\mathrm{a}_{\\mathrm{i}}\\) must be a sum of preceding elements. The largest possible \\(\\mathrm{a}_{\\mathrm{i}}\\) is \\(2\\mathrm{a}_{\\mathrm{i - 1}}\\) and the next largest \\(\\mathrm{a}_{\\mathrm{i - 1}} + \\mathrm{a}_{\\mathrm{i - 2}}\\) and so on.]", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 17", "solution_match": "# Solution"}}
{"year": "1980", "tier": "T1", "problem_label": "18", "problem_type": null, "exam": "AllSovietUnion", "problem": "Show that there are infinitely many positive integers n such that \\([\\mathrm{a}^{3 / 2}] + [\\mathrm{b}^{3 / 2}] = \\mathrm{n}\\) has at least 1980 integer solutions.", "solution": "Consider all a, b in the range 1, 2, 3, ..., \\(\\mathrm{N}^{2}\\) . There are \\(\\mathrm{N}^{4}\\) possible pairs of values. But \\([\\mathrm{a}^{3 / 2}]\\) and \\([\\mathrm{b}^{3 / 2}]\\) are in the range 1, 2, ..., \\(\\mathrm{N}^{3}\\) , so their sum is in the range 1, 2, ..., \\(2\\mathrm{N}^{3}\\) . Hence one of these values has at least \\(\\mathrm{N} / 2\\) solutions. By taking \\(\\mathrm{N}\\) sufficiently large we can get \\(\\mathrm{a}_{1} > 1980\\) solutions for some \\(\\mathrm{N}_{1}\\leq 2\\mathrm{N}^{3}\\) . But now by taking \\(\\mathrm{N}\\) sufficiently large we can get \\(\\mathrm{a}_{2} > \\mathrm{a}_{1}\\) solutions for some \\(\\mathrm{N}_{2}\\) . Since \\(\\mathrm{a}_{2}\\neq \\mathrm{a}_{1}\\) , we must have \\(\\mathrm{N}_{2}\\neq \\mathrm{N}_{1}\\) . In other words, we have a different n, also with \\(>1980\\) solutions. Continuing, we get an infinite sequence of distinct n each with at least 1980 solutions.", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 18", "solution_match": "# Solution"}}
@@ -485,7 +485,7 @@
{"year": "1988", "tier": "T1", "problem_label": "16", "problem_type": null, "exam": "AllSovietUnion", "problem": "\\(\\mathrm{n}^{2}\\) real numbers are written in a square \\(\\mathrm{n}\\times \\mathrm{n}\\) table so that the sum of the numbers in each row and column equals zero. A move is to add a row to one column and subtract it from another (so if the entries are \\(\\mathrm{a_{ij}}\\) and we select row i, column h and column k, then column h becomes \\(\\mathrm{a_{1h} + a_{i1}}\\) , \\(\\mathrm{a_{2h} + a_{i2}}\\) , ..., \\(\\mathrm{a_{nh} + a_{in}}\\) , column k becomes \\(\\mathrm{a_{1k} - a_{i1}}\\) , \\(\\mathrm{a_{2k} - a_{i2}}\\) , ..., \\(\\mathrm{a_{nk} - a_{in}}\\) , and the other entries are unchanged). Show that we can make all the entries zero by a series of moves.", "solution": "", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 16", "solution_match": ""}}
{"year": "1988", "tier": "T1", "problem_label": "17", "problem_type": null, "exam": "AllSovietUnion", "problem": "In the acute- angled triangle ABC, the altitudes BD and CE are drawn. Let F and G be the points of the line ED such that BF and CG are perpendicular to ED. Prove that \\(\\mathrm{EF} = \\mathrm{DG}\\) .", "solution": "\\(\\angle \\mathrm{BDC} = \\angle \\mathrm{BEC} = 90^{\\circ}\\) , so BCDE is cyclic, so \\(\\angle \\mathrm{BDE} = \\angle \\mathrm{BCE} = 90^{\\circ} - \\angle \\mathrm{B}\\) . Hence \\(\\angle \\mathrm{DGC} = 90^{\\circ} - \\angle \\mathrm{CDG} = \\angle \\mathrm{BDE} = 90^{\\circ} - \\mathrm{B}\\) . So \\(\\mathrm{DG} = \\mathrm{CD}\\) sin \\(\\mathrm{DGC} = \\mathrm{BC}\\) sin CBD sin \\(\\mathrm{DGC} = \\mathrm{BC}\\) cos C cos B. Similarly EF.", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 17", "solution_match": "# Solution"}}
{"year": "1988", "tier": "T1", "problem_label": "18", "problem_type": null, "exam": "AllSovietUnion", "problem": "Find the minimum value of \\(\\mathrm{xy / z + yz / x + zx / y}\\) for positive reals \\(\\mathrm{x}\\) , \\(\\mathrm{y}\\) , \\(\\mathrm{z}\\) with \\(\\mathrm{x}^{2} + \\mathrm{y}^{2} + \\mathrm{z}^{2} = 1\\) .", "solution": "Answer: min \\(\\sqrt{3}\\) when all equal. \n\nLet us consider \\(\\mathrm{z}\\) to be fixed and focus on \\(\\mathrm{x}\\) and \\(\\mathrm{y}\\) . Put \\(\\mathrm{f(x,y,z) = xy / z + yz / x + zx / y}\\) . We have \\(\\mathrm{f(x,y,z) = p / z + z(1 - z^{2}) / p = (p + k^{2} / p) / z}\\) , where \\(\\mathrm{p} = \\mathrm{xy}\\) , and \\(\\mathrm{k} = z\\sqrt{(1 - z^{2})}\\) . Now \\(\\mathrm{p}\\) can take any value in the range \\(0< \\mathrm{p}\\leq (1 - z^{2}) / 2\\) . The upper limit is achieved when \\(\\mathrm{x} = \\mathrm{y}\\) . \n\nWe have \\(\\mathrm{p} + \\mathrm{k}^{2} / \\mathrm{p} = (\\mathrm{p} - \\mathrm{k})^{2} / \\mathrm{p}\\) . For \\(\\mathrm{p}\\leq \\mathrm{k}\\) , \\((\\mathrm{p} - \\mathrm{k})\\) and \\(1 / \\mathrm{p}\\) are both decreasing functions of \\(\\mathrm{p}\\) , so \\(\\mathrm{p} + \\mathrm{k}^{2} / \\mathrm{p}\\) is a decreasing function of \\(\\mathrm{p}\\) . Thus if \\(\\mathrm{p}\\) is restricted to the interval \\((0,\\mathrm{h}]\\) , then for \\(\\mathrm{k}\\leq\\) \\(\\mathrm{h}\\) the minimum value of \\(\\mathrm{p} + \\mathrm{k}^{2} / \\mathrm{p}\\) is \\(2\\mathrm{k}\\) and occurs at \\(\\mathrm{p} = \\mathrm{k}\\) . For \\(\\mathrm{k}\\geq \\mathrm{h}\\) the minimum is \\(\\mathrm{h} + \\mathrm{k}^{2} / \\mathrm{h}\\) and occurs at \\(\\mathrm{p} = \\mathrm{h}\\) . \n\nWe have \\(\\mathrm{h} = (1 - \\mathrm{z}^{2}) / 2\\) , \\(\\mathrm{k} = z\\sqrt{(1 - \\mathrm{z}^{2})}\\) . So \\(\\mathrm{k}\\leq \\mathrm{h}\\) iff \\(\\mathrm{z}\\leq = 1 / \\sqrt{5}\\) . So if \\(\\mathrm{z}\\leq 1 / \\sqrt{5}\\) , then \\(\\mathrm{f(x,y,z)}\\geq 2\\mathrm{k} / \\mathrm{z}\\) \\(= 2\\sqrt{(1 - \\mathrm{z}^{2})}\\geq 27\\sqrt{(1 - 1 / 5)} = 4 / \\sqrt{5} >\\sqrt{3}\\) .\n\n\n\nIf \\(z > 1 / \\sqrt{5}\\) , then the minimum of \\(\\mathrm{f}(\\mathrm{x}, \\mathrm{y}, \\mathrm{z})\\) occurs at \\(\\mathrm{x} = \\mathrm{y}\\) and is \\(\\mathrm{x}^{2} / \\mathrm{z} + \\mathrm{z} + \\mathrm{z} = (1 - \\mathrm{z}^{2}) / (2\\mathrm{z}) + 2\\mathrm{z} = 3\\mathrm{z} / 2 + 1 / (2\\mathrm{z}) = (\\sqrt{3}) / 2 (\\mathrm{z} \\sqrt{3} + 1 / (\\mathrm{z} \\sqrt{3}) \\geq \\sqrt{3}\\) with equality at \\(\\mathrm{z} = 1 / \\sqrt{3}\\) (and hence \\(\\mathrm{x} = \\mathrm{y} = 1 / \\sqrt{3}\\) also).", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 18", "solution_match": "# Solution"}}
-{"year": "1988", "tier": "T1", "problem_label": "19", "problem_type": null, "exam": "AllSovietUnion", "problem": "A polygonal line connects two opposite vertices of a cube with side 2. Each segment of the line has length 3 and each vertex lies on the faces (or edges) of the cube. What is the smallest number of segments the line can have? \n\nAnswer 6", "solution": "\n \n\nSuppose one endpoint of a segment length 3 is at A. Evidently the other end could be at the edge midpoints B, C, D. It could also be on the circular arc connecting B and C (with center O and radius \\(\\sqrt{5}\\) ). Similarly, it could be on arcs connecting C and D, or B and D. We claim that if X is a point of one of these arcs other than its endpoints, then the only possible segment length 3 with an endpoint at X (and the other endpoint on the surface of the cube) is AX. wlog we can consider X to be on the arc BC. Take axes with origin O, so that A is (0,0,2). Suppose X is (a,b,0) and that the other endpoint of the segment is Y (x,y,z). Then \\(\\mathrm{XY}^{2} = (\\mathrm{x} - \\mathrm{a})^{2} + (\\mathrm{y} - \\mathrm{b})^{2} + \\mathrm{z}^{2} = \\mathrm{a}^{2} + \\mathrm{b}^{2} + \\mathrm{z}^{2} - \\mathrm{x}(2\\mathrm{a} - \\mathrm{x}) - \\mathrm{y}(2\\mathrm{b} - \\mathrm{y}) = 5 + \\mathrm{z}^{2} - \\mathrm{x}(2\\mathrm{a} - \\mathrm{x}) - \\mathrm{y}(2\\mathrm{b} - \\mathrm{y})\\) . But a, \\(\\mathrm{b} > 1\\) since X is not an endpoint of the arc, so (2a- x) and (2b- y) are both positive. Hence - x(2a- x) - y(2b- y) \\(\\leq 0\\) with equality iff \\(\\mathrm{x} = \\mathrm{y} = 0\\) . Similarly, \\(\\mathrm{z}^{2} \\leq 4\\) with equality iff \\(\\mathrm{z} = 2\\) . Hence \\(\\mathrm{XY}^{2} \\leq 9\\) with equality iff \\(\\mathrm{Y} = \\mathrm{A}\\) , which proves the claim. \n\nThus if the next link of the polygonal line goes from A to anywhere except B, C, D, then it has to go back to A. So a minimal line must go to B, C, or D. \n\nNow from D the line can only go to A or O. For if it goes to Z (x,y,z), then we have \\(\\mathrm{DZ}^{2} = (\\mathrm{x} - 2)^{2} + (\\mathrm{y} - 2)^{2} + (\\mathrm{z} - 1)^{2} \\leq 2^{2} + 2^{2} + 1^{2} = 3^{2}\\) with equality iff \\(\\mathrm{x} = 0\\) , \\(\\mathrm{y} = 0\\) and \\(\\mathrm{z} = 0\\) or 2. \n\nSo let us take A as the starting point of the polygonal line. wlog the first segment is AD. Then the second segment must be DO (for a minimal line). Thus the best we can do with 2 segments is to move along an edge. It takes three such moves to get to the opposite corner, and hence at least 6 segments. But it is obvious that it can be done with 6 segments.", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 19", "solution_match": "# Solution"}}
+{"year": "1988", "tier": "T1", "problem_label": "19", "problem_type": null, "exam": "AllSovietUnion", "problem": "A polygonal line connects two opposite vertices of a cube with side 2. Each segment of the line has length 3 and each vertex lies on the faces (or edges) of the cube. What is the smallest number of segments the line can have? \n\nAnswer 6", "solution": "\n \n\nSuppose one endpoint of a segment length 3 is at A. Evidently the other end could be at the edge midpoints B, C, D. It could also be on the circular arc connecting B and C (with center O and radius \\(\\sqrt{5}\\) ). Similarly, it could be on arcs connecting C and D, or B and D. We claim that if X is a point of one of these arcs other than its endpoints, then the only possible segment length 3 with an endpoint at X (and the other endpoint on the surface of the cube) is AX. wlog we can consider X to be on the arc BC. Take axes with origin O, so that A is (0,0,2). Suppose X is (a,b,0) and that the other endpoint of the segment is Y (x,y,z). Then \\(\\mathrm{XY}^{2} = (\\mathrm{x} - \\mathrm{a})^{2} + (\\mathrm{y} - \\mathrm{b})^{2} + \\mathrm{z}^{2} = \\mathrm{a}^{2} + \\mathrm{b}^{2} + \\mathrm{z}^{2} - \\mathrm{x}(2\\mathrm{a} - \\mathrm{x}) - \\mathrm{y}(2\\mathrm{b} - \\mathrm{y}) = 5 + \\mathrm{z}^{2} - \\mathrm{x}(2\\mathrm{a} - \\mathrm{x}) - \\mathrm{y}(2\\mathrm{b} - \\mathrm{y})\\) . But a, \\(\\mathrm{b} > 1\\) since X is not an endpoint of the arc, so (2a- x) and (2b- y) are both positive. Hence - x(2a- x) - y(2b- y) \\(\\leq 0\\) with equality iff \\(\\mathrm{x} = \\mathrm{y} = 0\\) . Similarly, \\(\\mathrm{z}^{2} \\leq 4\\) with equality iff \\(\\mathrm{z} = 2\\) . Hence \\(\\mathrm{XY}^{2} \\leq 9\\) with equality iff \\(\\mathrm{Y} = \\mathrm{A}\\) , which proves the claim. \n\nThus if the next link of the polygonal line goes from A to anywhere except B, C, D, then it has to go back to A. So a minimal line must go to B, C, or D. \n\nNow from D the line can only go to A or O. For if it goes to Z (x,y,z), then we have \\(\\mathrm{DZ}^{2} = (\\mathrm{x} - 2)^{2} + (\\mathrm{y} - 2)^{2} + (\\mathrm{z} - 1)^{2} \\leq 2^{2} + 2^{2} + 1^{2} = 3^{2}\\) with equality iff \\(\\mathrm{x} = 0\\) , \\(\\mathrm{y} = 0\\) and \\(\\mathrm{z} = 0\\) or 2. \n\nSo let us take A as the starting point of the polygonal line. wlog the first segment is AD. Then the second segment must be DO (for a minimal line). Thus the best we can do with 2 segments is to move along an edge. It takes three such moves to get to the opposite corner, and hence at least 6 segments. But it is obvious that it can be done with 6 segments.", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 19", "solution_match": "# Solution"}}
{"year": "1988", "tier": "T1", "problem_label": "20", "problem_type": null, "exam": "AllSovietUnion", "problem": "Let m, n, k be positive integers with \\(\\mathrm{m} \\geq \\mathrm{n}\\) and \\(1 + 2 + \\ldots + \\mathrm{n} = \\mathrm{mk}\\) . Prove that the numbers 1, 2, ..., n can be divided into k groups in such a way that the sum of the numbers in each group equals m.", "solution": "Induction on n, then m. For \\(\\mathrm{n} = 1\\) , 2 there is nothing to prove. Assume the result is proved for \\(< \\mathrm{n}\\) and consider the case n. If n is odd, we have \\(\\mathrm{n} = \\mathrm{n} - 1 + 1 = \\mathrm{n} - 2 + 2 = \\ldots = (\\mathrm{n} + 1) / 2 + (\\mathrm{n} - 1) / 2\\) , so the result is true for \\(\\mathrm{m} = \\mathrm{n}\\) , \\(\\mathrm{k} = (\\mathrm{n} + 1) / 2\\) . If n is even, we have \\(\\mathrm{n} + 1 = \\mathrm{n} - 1 + 2 = \\ldots = (\\mathrm{n} / 2 + 1) + (\\mathrm{n} / 2 - 1)\\) , so the result is true for \\(\\mathrm{m} = \\mathrm{n} + 1\\) and \\(\\mathrm{k} = \\mathrm{n} / 2\\) . Now suppose it is true for \\(< \\mathrm{m}\\) . \n\nIf \\(2\\mathrm{n} > \\mathrm{m} > \\mathrm{n} + 1\\) , then for m odd we can take the sums \\(\\mathrm{m} = \\mathrm{n} + \\mathrm{m} - \\mathrm{n} = \\mathrm{n} - 1 + \\mathrm{m} - \\mathrm{n} + 1 = \\ldots = (\\mathrm{m} + 1) / 2 + (\\mathrm{m} - 1) / 2\\) . These use up the numbers \\(\\mathrm{m} - \\mathrm{n}\\) , \\(\\mathrm{m} - \\mathrm{n} + 1\\) , ..., n and give some sums of m. By induction the remaining numbers 1, 2, ..., \\(\\mathrm{m} - \\mathrm{n} - 1\\) will give the remaining sums of m (obviously \\(\\mathrm{m} > \\mathrm{m} - \\mathrm{n} - 1\\) ). If m is even, we can take the sums \\(\\mathrm{m} = \\mathrm{n} + \\mathrm{m} - \\mathrm{n} = \\mathrm{n} - 1 + \\mathrm{m} - \\mathrm{n} + 1 = \\ldots = (\\mathrm{m} / 2 + 1) + (\\mathrm{m} / 2 - 1)\\) . That gives some sums of m and leaves us with the integers 1, 2, ..., \\(\\mathrm{m} - \\mathrm{n} - 1\\) and \\(\\mathrm{m} / 2\\) . But since \\(\\mathrm{m} < 2(\\mathrm{n} + 1)\\) , \\(\\mathrm{m} / 2 > \\mathrm{m} - \\mathrm{n} - 1\\) and hence we can use the integers 1, 2, ..., \\(\\mathrm{m} - \\mathrm{n} - 1\\) to form sums of \\(\\mathrm{m} / 2\\) . With the integer \\(\\mathrm{m} / 2\\) that gives us sums of m (we know that the parity must come out right because we know that the sum of all the remaining numbers is divisible by m). \n\nFinally, consider \\(\\mathrm{m} \\geq 2\\mathrm{n}\\) . In that case we can form k sums of \\(2\\mathrm{n} - 2\\mathrm{k} + 1\\) : \\(\\mathrm{n} + (\\mathrm{n} - 2\\mathrm{k} + 1)\\) , \\((\\mathrm{n} - 1) + (\\mathrm{n} - 2\\mathrm{k} + 2)\\) , ..., \\((\\mathrm{n} - \\mathrm{k} + 1) + (\\mathrm{n} - \\mathrm{k})\\) . So we are home provided the remaining integers 1, 2, ..., \\(\\mathrm{n} - 2\\mathrm{k}\\) can be used to form k sums of \\(\\mathrm{m} - (2\\mathrm{n} - 2\\mathrm{k} + 1)\\) . That follows by induction provided that \\(\\mathrm{m} - (2\\mathrm{n} - 2\\mathrm{k} + 1) \\geq \\mathrm{n} - 2\\mathrm{k}\\) , or \\(\\mathrm{m} + 4\\mathrm{k} - 1 \\geq 3\\mathrm{n}\\) , or \\(\\mathrm{m} + 2\\mathrm{n}(\\mathrm{n} + 1) / \\mathrm{m} - 1 \\geq 3\\mathrm{n}\\) or \\((\\mathrm{m} - 2\\mathrm{n})(\\mathrm{m} - \\mathrm{n} - 1) \\geq 0\\) , which is true.", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 20", "solution_match": "# Solution"}}
{"year": "1988", "tier": "T1", "problem_label": "21", "problem_type": null, "exam": "AllSovietUnion", "problem": "A polygonal line with a finite number of segments has all its vertices on a parabola. Any two adjacent segments make equal angles with the tangent to the parabola at their point of intersection. One end of the polygonal line is also on the axis of the parabola. Show that the other vertices of the polygonal line are all on the same side of the axis.", "solution": "", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 21", "solution_match": ""}}
{"year": "1988", "tier": "T1", "problem_label": "22", "problem_type": null, "exam": "AllSovietUnion", "problem": "What is the smallest n for which there is a solution to \\(\\sin x_{1} + \\sin x_{2} + \\ldots +\\sin x_{n} = 0\\) , \\(\\sin x_{1} + 2 \\sin x_{2} + \\ldots + n \\sin x_{n} = 100\\) ?", "solution": "Put \\(x_{1} = x_{2} = \\ldots = x_{10} = 3\\pi /2\\) , \\(x_{11} = x_{12} = \\ldots = x_{20} = \\pi /2\\) . Then \\(\\sin x_{1} + \\sin x_{2} + \\ldots +\\sin x_{20} = (- 1 - 1 - 1 - \\ldots - 1) + (1 + 1 + \\ldots + 1) = 0\\) , and \\(\\sin x_{1} + 2 \\sin x_{2} + \\ldots + 20 \\sin x_{20} = - (1 + 2 + \\ldots + 10) + (11 + 12 + \\ldots + 20) = 100\\) . So there is a solution with \\(\\mathrm{n} = 20\\) . If there is a solution with \\(\\mathrm{n} < 20\\) , then there must be a solution for \\(\\mathrm{n} = 19\\) (put any extra \\(x_{i} = 0\\) ). But then \\(100 = (\\sin x_{1} + 2 \\sin x_{2} + \\ldots + 19 \\sin x_{19}) - 10 (\\sin x_{1} + \\sin x_{2} + \\ldots + \\sin x_{19}) = - 9 \\sin x_{1} - 8 \\sin x_{2} - 7 \\sin x_{3} - \\ldots - \\sin x_{9} + \\sin x_{11} + 2 \\sin x_{12} + \\ldots + 9 \\sin x_{19}\\) . But \\(\\mathrm{l} \\mathrm{rhs} \\mathrm{l} \\leq (9 + 8 + \\ldots + 1) + (1 + 2 + \\ldots + 9) = 90\\) . Contradiction. So there is no solution for \\(\\mathrm{n} < 20\\) .", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 22", "solution_match": "# Solution"}}
@@ -516,14 +516,14 @@
{"year": "1989", "tier": "T1", "problem_label": "23", "problem_type": null, "exam": "AllSovietUnion", "problem": "N is the set of positive integers. Does there exist a function f: \\(\\mathrm{N}\\to \\mathrm{N}\\) such that \\(\\mathrm{f(n + 1)} = \\mathrm{f}(\\mathrm{f(n)}\\) ) + f( \\(\\mathrm{f(n + 2)}\\) ) for all n.", "solution": "", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 23", "solution_match": ""}}
{"year": "1989", "tier": "T1", "problem_label": "24", "problem_type": null, "exam": "AllSovietUnion", "problem": "A convex polygon is such that any segment dividing the polygon into two parts of equal area which has at least one end at a vertex has length \\(< 1\\) . Show that the area of the polygon is \\(< \\pi /4\\) .", "solution": "", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 24", "solution_match": ""}}
{"year": "1990", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "AllSovietUnion", "problem": "Show that \\(\\mathrm{x}^4 > \\mathrm{x} - 1 / 2\\) for all real x.", "solution": "\\(\\mathrm{x}^4 - \\mathrm{x} + 1 / 2 = (\\mathrm{x}^2 - 1 / 2)^2 +(\\mathrm{x} - 1 / 2)^2\\geq 0\\) . We could only have equality if \\(\\mathrm{x}^2 = \\mathrm{x} = 1 / 2\\) , which is impossible, so the inequality is strict.", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 1", "solution_match": "# Solution"}}
-{"year": "1990", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "AllSovietUnion", "problem": "The line joining the midpoints of two opposite sides of a convex quadrilateral makes equal angles with the diagonals. Show that the diagonals are equal.", "solution": "\n \n\nLet L, M, N be the midpoints of BC, CD, DA. Assume that NL makes equal angles with AC and BD, so \\(\\angle \\mathrm{NLM} = \\angle \\mathrm{BEL} = \\angle \\mathrm{AFN} = \\angle \\mathrm{LNM}\\) , so \\(\\mathrm{LM} = \\mathrm{MN}\\) and hence \\(\\mathrm{BD} = \\mathrm{AC}\\) .", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 2", "solution_match": "# Solution"}}
+{"year": "1990", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "AllSovietUnion", "problem": "The line joining the midpoints of two opposite sides of a convex quadrilateral makes equal angles with the diagonals. Show that the diagonals are equal.", "solution": "\n \n\nLet L, M, N be the midpoints of BC, CD, DA. Assume that NL makes equal angles with AC and BD, so \\(\\angle \\mathrm{NLM} = \\angle \\mathrm{BEL} = \\angle \\mathrm{AFN} = \\angle \\mathrm{LNM}\\) , so \\(\\mathrm{LM} = \\mathrm{MN}\\) and hence \\(\\mathrm{BD} = \\mathrm{AC}\\) .", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 2", "solution_match": "# Solution"}}
{"year": "1990", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "AllSovietUnion", "problem": "A graph has 30 points and each point has 6 edges. Find the total number of triples such that each pair of points is joined or each pair of points is not joined. \n\n## Answer \n\n1990", "solution": "There are 30. 29. \\(28 / 6 = 4060\\) triples in all. Let m be the number of triples with 0 or 3 edges, and let n be the number of triples with 1 or 2 edges. So \\(\\mathrm{m} + \\mathrm{n} = 4060\\) . Each point is joined to 6 others, so it is in 6. \\(5 / 2 = 15\\) triples where it is joined to both the other points, and it is in 23. \\(22 / 2 = 253\\) triples where it is not joined to either of the other points. So the total number of triples (a,b,c), where a is joined to b and c or not joined to b or c is \\(30(15 + 253) = 8040\\) . This counts the m triples 3 times each, and the n triples once each, so \\(3\\mathrm{m} + \\mathrm{n} = 8040\\) . Hence \\(\\mathrm{m} = 1990\\) .", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 3", "solution_match": "# Solution"}}
-{"year": "1990", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "AllSovietUnion", "problem": "Does there exist a rectangle which can be dissected into 15 congruent polygons which are not rectangles? Can a square be dissected into 15 congruent polygons which are not rectangles? \n\n## Answer \n\nyes, yes", "solution": "\n \n\nBy stretching vertically we get a square.", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 4", "solution_match": "# Solution"}}
-{"year": "1990", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "AllSovietUnion", "problem": "The point P lies inside the triangle ABC. A line is drawn through P parallel to each side of the triangle. The lines divide AB into three parts length c, c', c\" (in that order), and BC into three parts length a, a', a\" (in that order), and CA into three parts length b, b', b\" (in that order). Show that abc = a'b'c' = a\"b\"c\".", "solution": "\n \n\nThe three small triangles are similar, so a/a\" = c'/c = b\"/b' and a/a' = c'/c\" = b\"/b. Hence (a/a\")(b/b\") = (c'/c)(c\"/c') = c\"/c, so abc = a\"b\"c\". Similarly, (a/a')(c/c') = (b\"/b)(b\"/b\") = b/b, so abc = a'b'c'.", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 5", "solution_match": "# Solution"}}
+{"year": "1990", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "AllSovietUnion", "problem": "Does there exist a rectangle which can be dissected into 15 congruent polygons which are not rectangles? Can a square be dissected into 15 congruent polygons which are not rectangles? \n\n## Answer \n\nyes, yes", "solution": "\n \n\nBy stretching vertically we get a square.", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 4", "solution_match": "# Solution"}}
+{"year": "1990", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "AllSovietUnion", "problem": "The point P lies inside the triangle ABC. A line is drawn through P parallel to each side of the triangle. The lines divide AB into three parts length c, c', c\" (in that order), and BC into three parts length a, a', a\" (in that order), and CA into three parts length b, b', b\" (in that order). Show that abc = a'b'c' = a\"b\"c\".", "solution": "\n \n\nThe three small triangles are similar, so a/a\" = c'/c = b\"/b' and a/a' = c'/c\" = b\"/b. Hence (a/a\")(b/b\") = (c'/c)(c\"/c') = c\"/c, so abc = a\"b\"c\". Similarly, (a/a')(c/c') = (b\"/b)(b\"/b\") = b/b, so abc = a'b'c'.", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 5", "solution_match": "# Solution"}}
{"year": "1990", "tier": "T1", "problem_label": "6", "problem_type": null, "exam": "AllSovietUnion", "problem": "Find three non- zero reals such that all quadratics with those numbers as coefficients have two distinct rational roots. \n\nAnswer 1,2,- 3", "solution": "If a + b + c = 0, then 1 is a root of ax² + bx + c, and so the other root is - b/a - 1, which is rational.", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 6", "solution_match": "# Solution"}}
{"year": "1990", "tier": "T1", "problem_label": "7", "problem_type": null, "exam": "AllSovietUnion", "problem": "What is the largest possible value of \\(|\\ldots |\\mathrm{la}_1 - \\mathrm{a}_2| - \\mathrm{a}_3| - \\ldots - \\mathrm{a}_{1990}|\\) , where \\(a_1, a_2, \\ldots , a_{1990}\\) is a permutation of 1, 2, 3, ..., 1990? \n\nAnswer 1989", "solution": "Since \\(\\mathrm{la - bl} \\leq \\max (\\mathrm{a,b})\\) , a trivial induction shows that the expression does not exceed \\(\\max (\\mathrm{a}_1, \\mathrm{a}_2, \\ldots , \\mathrm{a}_{1990}) = 1990\\) . But for integers \\(\\mathrm{la - bl}\\) has the same parity as \\(\\mathrm{a + b}\\) , so a trivial induction shows that the expression has the same parity as \\(\\mathrm{a}_1 + \\mathrm{a}_2 + \\ldots + \\mathrm{a}_{1990} = 1990\\) . 1991/2, which is odd. So it cannot exceed 1989. That can be attained by the permutation 2, 4, 5, 3, 6, 8, 9, 7, ... , \\(4\\mathrm{k} + 2\\) , \\(4\\mathrm{k} + 4\\) , \\(4\\mathrm{k} + 5\\) , \\(4\\mathrm{k} + 3\\) , ..., \\(1984 + 2\\) , \\(1984 + 4\\) , \\(1984 + 5\\) , \\(1984 + 3\\) , 1990, 1. Because we get successively 2, 3, 0; 6, 2, 7, 0; 10, 2, 11, 0; ... ; \\(4\\mathrm{k} + 2\\) , 2, \\(4\\mathrm{k} + 3\\) , 0; ... ; 1986, 2, 1987, 0; 1990, 1989.", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 7", "solution_match": "# Solution"}}
-{"year": "1990", "tier": "T1", "problem_label": "8", "problem_type": null, "exam": "AllSovietUnion", "problem": "An equilateral triangle of side n is divided into \\(\\mathrm{n}^2\\) equilateral triangles of side 1. A path is drawn along the sides of the triangles which passes through each vertex just once. Prove that the path makes an acute angle at at least n vertices.", "solution": "\n \n\nThe diagram has \\(1 + 2 + \\ldots + \\mathrm{n} = \\mathrm{n}(\\mathrm{n} + 1) / 2\\) upright triangles and \\(1 + 2 + \\ldots + \\mathrm{n} - 1 = \\mathrm{n}(\\mathrm{n} - 1) / 2\\) upside down triangles. It has \\(1 + 2 + \\ldots + \\mathrm{n} + 1 = (\\mathrm{n} + 1)(\\mathrm{n} + 2) / 2\\) vertices. So the path must be \\((\\mathrm{n} + 1)(\\mathrm{n} + 2) / 2 - 1 = (\\mathrm{n}^2 +3\\mathrm{n}) / 2\\) units long. Each unit length of the path is in just one upright triangle. The path cannot contain all three sides of a small triangle, or it would pass through a vertex more than once. So it must contain two sides of \\((\\mathrm{n}^2 +3\\mathrm{n}) / 2 - \\mathrm{n}(\\mathrm{n} + 1) / 2 = \\mathrm{n}\\) triangles. But if it contains two sides of a triangle, then it must make an acute angle at the vertex where they meet.", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 8", "solution_match": "# Solution"}}
-{"year": "1990", "tier": "T1", "problem_label": "9", "problem_type": null, "exam": "AllSovietUnion", "problem": "Can the squares of a 1990 x 1990 chessboard be colored black or white so that half the squares in each row and column are black and cells symmetric with respect to the center are of opposite color? \n\nAnswer no", "solution": "\n \n\nSuppose it can be done. Divide the board into 4 quadrants. Suppose there are b black and \\(995^{2}\\) - b white squares in the top left quadrant. Then there are \\(995^{2}\\) - b black and b white squares in the bottom right quadrant (by the symmetry property). \n\nIf half the squares in each row are black, then half the squares in the first 995 rows are black, so the number of black squares in the top right quadrant is 995. \\(1990 / 2 - \\mathrm{b} = 995^{2} - \\mathrm{b}\\) . So if half the squares in each column are black, then half the squares in the right- hand half of the board are black, so \\((995^{2} - \\mathrm{b}) + (995^{2} - \\mathrm{b}) = 995^{2}\\) , in other words, \\(\\mathrm{b} = 995^{2} / 2\\) , which is impossible.", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 9", "solution_match": "\nSolution"}}
+{"year": "1990", "tier": "T1", "problem_label": "8", "problem_type": null, "exam": "AllSovietUnion", "problem": "An equilateral triangle of side n is divided into \\(\\mathrm{n}^2\\) equilateral triangles of side 1. A path is drawn along the sides of the triangles which passes through each vertex just once. Prove that the path makes an acute angle at at least n vertices.", "solution": "\n \n\nThe diagram has \\(1 + 2 + \\ldots + \\mathrm{n} = \\mathrm{n}(\\mathrm{n} + 1) / 2\\) upright triangles and \\(1 + 2 + \\ldots + \\mathrm{n} - 1 = \\mathrm{n}(\\mathrm{n} - 1) / 2\\) upside down triangles. It has \\(1 + 2 + \\ldots + \\mathrm{n} + 1 = (\\mathrm{n} + 1)(\\mathrm{n} + 2) / 2\\) vertices. So the path must be \\((\\mathrm{n} + 1)(\\mathrm{n} + 2) / 2 - 1 = (\\mathrm{n}^2 +3\\mathrm{n}) / 2\\) units long. Each unit length of the path is in just one upright triangle. The path cannot contain all three sides of a small triangle, or it would pass through a vertex more than once. So it must contain two sides of \\((\\mathrm{n}^2 +3\\mathrm{n}) / 2 - \\mathrm{n}(\\mathrm{n} + 1) / 2 = \\mathrm{n}\\) triangles. But if it contains two sides of a triangle, then it must make an acute angle at the vertex where they meet.", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 8", "solution_match": "# Solution"}}
+{"year": "1990", "tier": "T1", "problem_label": "9", "problem_type": null, "exam": "AllSovietUnion", "problem": "Can the squares of a 1990 x 1990 chessboard be colored black or white so that half the squares in each row and column are black and cells symmetric with respect to the center are of opposite color? \n\nAnswer no", "solution": "\n \n\nSuppose it can be done. Divide the board into 4 quadrants. Suppose there are b black and \\(995^{2}\\) - b white squares in the top left quadrant. Then there are \\(995^{2}\\) - b black and b white squares in the bottom right quadrant (by the symmetry property). \n\nIf half the squares in each row are black, then half the squares in the first 995 rows are black, so the number of black squares in the top right quadrant is 995. \\(1990 / 2 - \\mathrm{b} = 995^{2} - \\mathrm{b}\\) . So if half the squares in each column are black, then half the squares in the right- hand half of the board are black, so \\((995^{2} - \\mathrm{b}) + (995^{2} - \\mathrm{b}) = 995^{2}\\) , in other words, \\(\\mathrm{b} = 995^{2} / 2\\) , which is impossible.", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 9", "solution_match": "\nSolution"}}
{"year": "1990", "tier": "T1", "problem_label": "10", "problem_type": null, "exam": "AllSovietUnion", "problem": "Let \\(\\mathbf{x}_1\\) , \\(\\mathbf{x}_2\\) , ..., \\(\\mathbf{x}_n\\) be positive reals with sum 1. Show that \\(\\mathbf{x}_1^{2} / (\\mathbf{x}_1 + \\mathbf{x}_2) + \\mathbf{x}_2^{2} / (\\mathbf{x}_2 + \\mathbf{x}_3) + \\ldots + \\mathbf{x}_n\\) . \\(\\frac{1}{2} /(\\mathbf{x}_{\\mathrm{n - 1}} + \\mathbf{x}_{\\mathrm{n}}) + \\mathbf{x}_{\\mathrm{n}}^{2} / (\\mathbf{x}_{\\mathrm{n}} + \\mathbf{x}_{\\mathrm{1}}) \\geq 1 / 2\\) .", "solution": "\\[\\sum \\mathbf{x}_1^{2} / (\\mathbf{x}_1 + \\mathbf{x}_{i + 1}) - \\sum \\mathbf{x}_{i + 1}^{2} / (\\mathbf{x}_i + \\mathbf{x}_{i + 1}) = \\sum (\\mathbf{x}_i - \\mathbf{x}_{i + 1}) = 0.\\] \\[\\mathbf{x}_i^{2} + \\mathbf{x}_{i + 1}^{2} / (\\mathbf{x}_i + \\mathbf{x}_{i + 1}) \\geq \\frac{1}{4} \\sum (\\mathbf{x}_i + \\mathbf{x}_{i + 1}) = \\frac{1}{2}.\\]", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 10", "solution_match": "# Solution"}}
{"year": "1990", "tier": "T1", "problem_label": "11", "problem_type": null, "exam": "AllSovietUnion", "problem": "ABCD is a convex quadrilateral. X is a point on the side AB. AC and DE intersect at Y. Show that the circumcircles of ABC, CDY and BDX have a common point.", "solution": "", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 11", "solution_match": ""}}
{"year": "1990", "tier": "T1", "problem_label": "12", "problem_type": null, "exam": "AllSovietUnion", "problem": "Two grasshoppers sit at opposite ends of the interval [0, 1]. A finite number of points (greater than zero) in the interval are marked. A move is for a grasshopper to select a marked point and jump over it to the equidistant point the other side. This point must lie in the interval for the move to be allowed, but it does not have to be marked. What is the smallest n such that if each grasshopper makes n moves or less, then they end up with no marked points between them?", "solution": "", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 12", "solution_match": ""}}
@@ -541,16 +541,16 @@
{"year": "1990", "tier": "T1", "problem_label": "24", "problem_type": null, "exam": "AllSovietUnion", "problem": "Given 2n genuine coins and 2n fake coins. The fake coins look the same as genuine coins but weigh less (but all fake coins have the same weight). Show how to identify each coin as genuine or fake using a balance at most 3n times.", "solution": "", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 24", "solution_match": ""}}
{"year": "1991", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "AllSovietUnion", "problem": "Find all integers a, b, c, d such that ab - 2cd = 3, ac + bd = 1. \n\n## Answer \n\n\\[(\\mathrm{a},\\mathrm{b},\\mathrm{c},\\mathrm{d}) = (1,3,1,0),(-1, - 3, - 1,0), (3,1,0,1),(-3, - 1,0, - 1)\\]", "solution": "\\(11 = (\\mathrm{ab} - 2\\mathrm{cd})^{2} + 2(\\mathrm{ac} + \\mathrm{bd})^{2} = (\\mathrm{a}^{2} + 2\\mathrm{d}^{2})(\\mathrm{b}^{2} + 2\\mathrm{c}^{2})\\) , so we must have either (1) \\(\\mathrm{a}^{2} + 2\\mathrm{d}^{2} = 1\\) , \\(\\mathrm{b}^{2} + 2\\mathrm{c}^{2} = 11\\) , or (2) \\(\\mathrm{a}^{2} + 2\\mathrm{d}^{2} = 11\\) , \\(\\mathrm{b}^{2} + 2\\mathrm{c}^{2} = 1\\) . \n\n(1) gives \\(\\mathrm{a} = \\pm 1\\) , \\(\\mathrm{d} = 0\\) , \\(\\mathrm{b} = \\pm 3\\) , \\(\\mathrm{c} = \\pm 1\\) . If \\(\\mathrm{a} = 1\\) and \\(\\mathrm{d} = 0\\) , then \\(\\mathrm{ac} + \\mathrm{bd} = 1\\) implies \\(\\mathrm{c} = 1\\) , and \\(\\mathrm{ab} - 2\\mathrm{cd} = 3\\) implies \\(\\mathrm{b} = 3\\) . Similarly, if \\(\\mathrm{a} = -1\\) , then \\(\\mathrm{c} = -1\\) , and \\(\\mathrm{b} = -3\\) . Similarly, (2) gives \\((\\mathrm{a},\\mathrm{b},\\mathrm{c},\\mathrm{d}) = (3,1,0,1),(-3, - 1,0, - 1)\\) .", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 1", "solution_match": "# Solution"}}
{"year": "1991", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "AllSovietUnion", "problem": "n numbers are written on a blackboard. Someone then repeatedly erases two numbers and writes half their arithmetic mean instead, until only a single number remains. If all the original numbers were 1, show that the final number is not less than \\(1 / \\mathrm{n}\\) .", "solution": "Put \\(\\mathrm{c} = (\\mathrm{a} + \\mathrm{b}) / 4\\) . We have \\(1 / \\mathrm{c} = 4 / (\\mathrm{a} + \\mathrm{b}) \\leq 1 / \\mathrm{a} + 1 / \\mathrm{b}\\) , so each move does not increase the sum of the reciprocals of the numbers. If the final number is \\(\\mathrm{k}\\) , then the final sum of reciprocals is \\(1 / \\mathrm{k}\\) . The initial sum is \\(\\mathrm{n}\\) , so \\(1 / \\mathrm{k} \\leq \\mathrm{n}\\) , or \\(\\mathrm{k} \\geq 1 / \\mathrm{n}\\) .", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 2", "solution_match": "# Solution"}}
-{"year": "1991", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "AllSovietUnion", "problem": "Four lines in the plane intersect in six points. Each line is thus divided into two segments and two rays. Is it possible for the eight segments to have lengths 1, 2, 3, ..., 8? Can the lengths of the eight segments be eight distinct integers? \n\nAnswer no, yes", "solution": "\n \n\nIf a triangle has integer sides, one of which is 1, then it must be isosceles. So the only candidates for the segment length 1 are AB and AE. wlog \\(\\mathrm{AB} = 1\\) , so \\(\\mathrm{BF} = \\mathrm{AF}\\) . Hence \\(\\cos\\)\n\n\n\nDFE = 1 - 1/(2 AF²). Hence ED² = DF² + EF² + 2DF. EF(1 - 1/2AF²) = DF² + EF² + 2DF. EF - DF. EF/AF². But the first three terms are integers and the last term is < 1. Contradiction. (Careful, looking at the figure one is tempted to conclude that ED < AB, but a more realistic figure shows that is false.). \n\n\n \n\nBuilding on the 3,4,5 triangle we get the figure above.", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 3", "solution_match": "# Solution"}}
+{"year": "1991", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "AllSovietUnion", "problem": "Four lines in the plane intersect in six points. Each line is thus divided into two segments and two rays. Is it possible for the eight segments to have lengths 1, 2, 3, ..., 8? Can the lengths of the eight segments be eight distinct integers? \n\nAnswer no, yes", "solution": "\n \n\nIf a triangle has integer sides, one of which is 1, then it must be isosceles. So the only candidates for the segment length 1 are AB and AE. wlog \\(\\mathrm{AB} = 1\\) , so \\(\\mathrm{BF} = \\mathrm{AF}\\) . Hence \\(\\cos\\)\n\n\n\nDFE = 1 - 1/(2 AF²). Hence ED² = DF² + EF² + 2DF. EF(1 - 1/2AF²) = DF² + EF² + 2DF. EF - DF. EF/AF². But the first three terms are integers and the last term is < 1. Contradiction. (Careful, looking at the figure one is tempted to conclude that ED < AB, but a more realistic figure shows that is false.). \n\n\n \n\nBuilding on the 3,4,5 triangle we get the figure above.", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 3", "solution_match": "# Solution"}}
{"year": "1991", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "AllSovietUnion", "problem": "A lottery ticket has 50 cells into which one must put a permutation of 1, 2, 3, ..., 50. Any ticket with at least one cell matching the winning permutation wins a prize. How many tickets are needed to be sure of winning a prize? \n\nAnswer 26", "solution": "Take the tickets: \n\n1 2 3 ... 25 26 27 ... 50 2 3 4 ... 26 1 27 ... 50 3 4 5 ... 1 2 27 ... 50 \n\n26 1 2 ... 24 25 27 ... 50 Each of the numbers 1, 2, ..., 26 occurs in each of the places 1, 2, ..., 26, but the winning ticket cannot have all these numbers in the last 24 places. So there must be at least one match. So 26 tickets suffice. \n\nNow given any 25 tickets we show that they could all fail to match the winning permutation. In other words, we construct a permutation which fails to match any of the 25 tickets in any cell. We place the numbers 1, 2, 3, ..., 50 in turn. We start by placing 1. Clearly at most 25 places are ruled out, so we can place the 1. Now suppose we have placed 1, 2, ..., a. There must be at least 25 places where a+1 is not ruled out. If any of them are still unoccupied, then we are done. If not, they must be occupied by numbers \\(\\mathbf{x}_1\\) , \\(\\mathbf{x}_2\\) , ..., \\(\\mathbf{x}_{25}\\) already placed. Take any empty place. 26 numbers cannot be ruled out for it, and we know that a+1 is ruled out, so at least one of the \\(\\mathbf{x}_i\\) is not ruled out. So we can move that \\(\\mathbf{x}_i\\) to it and then place a+1 where the \\(\\mathbf{x}_i\\) came from.", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 4", "solution_match": "# Solution"}}
{"year": "1991", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "AllSovietUnion", "problem": "Find unequal integers m, n such that mn + n and mn + m are both squares. Can you find such integers between 988 and 1991? \n\nAnswer no", "solution": "For example, \\(49 = 7^{2}\\) , \\(50 = 2\\) . \\(5^{2}\\) , \\(8 = 2\\) . \\(2^{2}\\) , \\(9 = 3^{2}\\) , so \\(49\\) . \\(8 + 8 = 20^{2}\\) , \\(49\\) . \\(8 + 49 = 21^{2}\\) . \n\n\\(\\mathrm{wlog m}< \\mathrm{n}\\) . Then \\(\\mathrm{mn} + \\mathrm{m} = (\\mathrm{m} + \\mathrm{h})^{2}\\) , \\(\\mathrm{mn} + \\mathrm{n} = (\\mathrm{m} + \\mathrm{k})^{2}\\) , with \\(\\mathrm{k} > \\mathrm{h}\\) . So \\(\\mathrm{n} - \\mathrm{m} = (\\mathrm{m} + \\mathrm{k})^{2} - (\\mathrm{m} + \\mathrm{h})^{2} = (\\mathrm{k} - \\mathrm{h})(2\\mathrm{m} + \\mathrm{k} + \\mathrm{h}) > 2\\mathrm{m}\\) , so \\(\\mathrm{n} > 3\\mathrm{m}\\) . Hence we cannot have \\(\\mathrm{m}\\) and \\(\\mathrm{n}\\) between 988 and 1991.", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 5", "solution_match": "# Solution"}}
-{"year": "1991", "tier": "T1", "problem_label": "6", "problem_type": null, "exam": "AllSovietUnion", "problem": "ABCD is a rectangle. Points K, L, M, N are chosen on AB, BC, CD, DA respectively so that KL is parallel to MN, and KM is perpendicular to LN. Show that the intersection of KM and LN lies on BD.", "solution": "\n \n\nLet LN and KM meet at O. \\(\\angle \\mathrm{NOM} = \\angle \\mathrm{NDM} = 90^{\\circ}\\) , so OMDN is cyclic. Hence \\(\\angle \\mathrm{NOD} = \\angle \\mathrm{NMD}\\) . Similarly, BLOK is cyclic and \\(\\angle \\mathrm{LOB} = \\angle \\mathrm{LKB}\\) . But NM is parallel to LK and AB is parallel to CD, so \\(\\angle \\mathrm{LKB} = \\angle \\mathrm{NMD}\\) . Hence \\(\\angle \\mathrm{NOD} = \\angle \\mathrm{LOB}\\) , so DOB is a straight line.", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 6", "solution_match": "# Solution"}}
+{"year": "1991", "tier": "T1", "problem_label": "6", "problem_type": null, "exam": "AllSovietUnion", "problem": "ABCD is a rectangle. Points K, L, M, N are chosen on AB, BC, CD, DA respectively so that KL is parallel to MN, and KM is perpendicular to LN. Show that the intersection of KM and LN lies on BD.", "solution": "\n \n\nLet LN and KM meet at O. \\(\\angle \\mathrm{NOM} = \\angle \\mathrm{NDM} = 90^{\\circ}\\) , so OMDN is cyclic. Hence \\(\\angle \\mathrm{NOD} = \\angle \\mathrm{NMD}\\) . Similarly, BLOK is cyclic and \\(\\angle \\mathrm{LOB} = \\angle \\mathrm{LKB}\\) . But NM is parallel to LK and AB is parallel to CD, so \\(\\angle \\mathrm{LKB} = \\angle \\mathrm{NMD}\\) . Hence \\(\\angle \\mathrm{NOD} = \\angle \\mathrm{LOB}\\) , so DOB is a straight line.", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 6", "solution_match": "# Solution"}}
{"year": "1991", "tier": "T1", "problem_label": "7", "problem_type": null, "exam": "AllSovietUnion", "problem": "An investigator works out that he needs to ask at most 91 questions on the basis that all the answers will be yes or no and all will be true. The questions may depend upon the earlier answers. Show that he can make do with 105 questions if at most one answer could be a lie.", "solution": "Suppose he asks n questions as usual, and then asks \"did you lie to any of the last n questions?\" If the reply is a truthful no, then the n answers were correct. If the reply is a lying no, then the n answers were still correct. On the other hand if the answer is yes, then the n answers might have been correct and might not. However, a lie has certainly been told, so all future answers must be truthful and so he could ask the n questions again. \n\n\\(91 = 7\\) . 13, so the obvious candidates for n are 7 and 13. If we take \\(\\mathrm{n} = 7\\) , then the worst case is 13 check questions and 7 repeat questions. That does not work because he needs 20 extra questions and only has 14. A little thought suggests reducing n each time. So the first batch of questions is 13, followed by a check question. If the check answer is yes, then he knows a lie has been told and asks the 13 questions again. No further check questions are needed, and he has used exactly 14 extra questions. If the check answer is no, then the lie may not have been told, so the next batch of questions is 12, followed by a check question, and so on. That allows him to ask \\(13 + 12 + \\ldots +1 = 91\\) questions. If he gets a yes to the check question after the batch of i, then he ignores the answers to that batch and asks them again, thus asking a total of 14 extra questions, but thereafter asks no check questions.", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 7", "solution_match": "# Solution"}}
-{"year": "1991", "tier": "T1", "problem_label": "8", "problem_type": null, "exam": "AllSovietUnion", "problem": "A minus sign is placed on one square of a \\(5 \\mathrm{x} 5\\) board and plus signs are placed on the remaining squares. A move is to select a \\(2 \\mathrm{x} 2\\) , \\(3 \\mathrm{x} 3\\) , \\(4 \\mathrm{x} 4\\) or \\(5 \\mathrm{x} 5\\) square and change all the signs in it. Which initial positions allow a series of moves to change all the signs to plus? \n\nAnswer only the central square", "solution": "\n \n\nWe take the 5x5 square, the two yellow 3x3 squares, which overlap at the center, and the two blue 2x2 squares. Then every square except the center square is changed an even number of times. So this works if the central square was selected. \n\n\n \n\nIt is easy to check that any 2x2, 3x3, 4x4 or 5x5 square has an even number of green squares, so if the selected square was green, and we change it an odd number of times, then some other green square must also be changed an odd number of times and hence end up with a minus. So if all the squares end up plus, then the selected square was not green, so it must belong to the central white column. Similarly, it must belong to the central row and hence must be the center square.", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 8", "solution_match": "# Solution"}}
+{"year": "1991", "tier": "T1", "problem_label": "8", "problem_type": null, "exam": "AllSovietUnion", "problem": "A minus sign is placed on one square of a \\(5 \\mathrm{x} 5\\) board and plus signs are placed on the remaining squares. A move is to select a \\(2 \\mathrm{x} 2\\) , \\(3 \\mathrm{x} 3\\) , \\(4 \\mathrm{x} 4\\) or \\(5 \\mathrm{x} 5\\) square and change all the signs in it. Which initial positions allow a series of moves to change all the signs to plus? \n\nAnswer only the central square", "solution": "\n \n\nWe take the 5x5 square, the two yellow 3x3 squares, which overlap at the center, and the two blue 2x2 squares. Then every square except the center square is changed an even number of times. So this works if the central square was selected. \n\n\n \n\nIt is easy to check that any 2x2, 3x3, 4x4 or 5x5 square has an even number of green squares, so if the selected square was green, and we change it an odd number of times, then some other green square must also be changed an odd number of times and hence end up with a minus. So if all the squares end up plus, then the selected square was not green, so it must belong to the central white column. Similarly, it must belong to the central row and hence must be the center square.", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 8", "solution_match": "# Solution"}}
{"year": "1991", "tier": "T1", "problem_label": "9", "problem_type": null, "exam": "AllSovietUnion", "problem": "Show that \\((\\mathrm{x} + \\mathrm{y} + \\mathrm{z})^{2} / 3 \\geq \\mathrm{x}\\sqrt{(\\mathrm{y}\\mathrm{z})} + \\mathrm{y}\\sqrt{(\\mathrm{z}\\mathrm{x})} + \\mathrm{z}\\sqrt{(\\mathrm{x}\\mathrm{y})}\\) for all non- negative reals \\(\\mathrm{x}, \\mathrm{y}, \\mathrm{z}\\) .", "solution": "By AM/GM xy + yz \\(\\geq 2\\mathrm{x}\\sqrt{(\\mathrm{y}\\mathrm{z})}\\) . Adding the similar results gives \\(2(\\mathrm{xy} + \\mathrm{yz} + \\mathrm{zx})\\geq 2(\\mathrm{x}\\sqrt{(\\mathrm{y}\\mathrm{z})} +\\) \\(\\mathrm{y}\\sqrt{(\\mathrm{z}\\mathrm{x})} + \\mathrm{z}\\sqrt{(\\mathrm{x}\\mathrm{y})}\\) ). \n\nBy AM/GM \\(\\mathrm{x}^{2} + \\mathrm{x}^{2} + \\mathrm{y}^{2} + \\mathrm{z}^{2}\\geq 4\\mathrm{x}\\sqrt{(\\mathrm{y}\\mathrm{z})}\\) . Adding the similar results gives \\(\\mathrm{x}^{2} + \\mathrm{y}^{2} + \\mathrm{z}^{2}\\geq \\mathrm{x}\\sqrt{(\\mathrm{y}\\mathrm{z})}\\) \\(+\\mathrm{y}\\sqrt{(\\mathrm{z}\\mathrm{x})} +\\mathrm{z}\\sqrt{(\\mathrm{x}\\mathrm{y})}\\) . Adding the first result gives \\((\\mathrm{x} + \\mathrm{y} + \\mathrm{z})^{2} / 3\\geq \\mathrm{x}\\sqrt{(\\mathrm{y}\\mathrm{z})} +\\mathrm{y}\\sqrt{(\\mathrm{z}\\mathrm{x})} +\\mathrm{z}\\sqrt{(\\mathrm{x}\\mathrm{y})}\\)", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 9", "solution_match": "# Solution"}}
{"year": "1991", "tier": "T1", "problem_label": "10", "problem_type": null, "exam": "AllSovietUnion", "problem": "Does there exist a triangle in which two sides are integer multiples of the median to that side? Does there exist a triangle in which every side is an integer multiple of the median to that side? \n\nAnswer yes, no", "solution": "The obvious approach is to make the triangle isosceles. So suppose the sides are a, b, b. Then the length m of a median to one of the sides length b satisfies: \\(\\mathrm{a}^2 +\\mathrm{b}^2 = 2\\mathrm{m}^2 +\\mathrm{b}^2 /2\\) . The simplest possibility is to take \\(\\mathrm{m} = \\mathrm{b}\\) , so \\(\\mathrm{a}^2 = 3\\mathrm{b}^2 /2\\) . Thus if \\(\\mathrm{b} = 2\\) , \\(\\mathrm{a} = \\sqrt{6}\\) . \n\nSuppose we have a triangle ABC, with medians AD, BE, CF, and BC/AD, CA/BE, AB/CF all integers. If \\(\\mathrm{AD} = \\mathrm{BC} / 2\\) , then \\(\\angle A = 90^{\\circ}\\) . If \\(\\mathrm{AD}< \\mathrm{BC} / 2\\) , then \\(\\angle A\\) is obtuse, so at least two of the medians must be equal to the corresponding sides. So wlog we have \\(\\mathrm{b}^2 +\\mathrm{c}^2 = 5\\mathrm{a}^2 /2\\) , \\(\\mathrm{c}^2 +\\mathrm{a}^2 = 5\\mathrm{b}^2 /2\\) . Subtracting, \\(\\mathrm{b}^2 -\\mathrm{a}^2 = (5 / 2)(\\mathrm{a}^2 -\\mathrm{b}^2)\\) , so \\(\\mathrm{a} = \\mathrm{b}\\) . Hence \\(\\mathrm{c} / \\mathrm{a} = \\sqrt{(3 / 2)}\\) . So the third median has length m where \\(\\mathrm{a}^2 +\\mathrm{a}^2 = (3 / 4)\\mathrm{a}^2 +2\\mathrm{m}^2\\) , so \\(\\mathrm{a} / \\mathrm{m} = \\sqrt{(8 / 5)}\\) , which is not integral. Contradiction.", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 10", "solution_match": "# Solution"}}
{"year": "1991", "tier": "T1", "problem_label": "11", "problem_type": null, "exam": "AllSovietUnion", "problem": "The numbers 1, 2, 3, ..., n are written on a blackboard (where \\(\\mathrm{n} \\geq 3\\) ). A move is to replace two numbers by their sum and non- negative difference. A series of moves makes all the numbers equal k. Find all possible k. \n\nAnswer all powers of \\(2 \\geq \\mathrm{n}\\)", "solution": "If a prime p divides \\(\\mathrm{a} + \\mathrm{b}\\) and a- b, then it divides 2a and 2b, so if p is odd, it divides a and b. Thus if an odd prime p divides k, then it must divide all the original numbers including 1. So k must be a power of 2. Note that \\(\\mathrm{k}, \\mathrm{k} \\rightarrow 0, 2\\mathrm{k} \\rightarrow 2\\mathrm{k}, 2\\mathrm{k}\\) and \\(\\mathrm{k}, \\mathrm{k}, \\mathrm{k} \\rightarrow 0, \\mathrm{k}, 2\\mathrm{k} \\rightarrow \\mathrm{k}, \\mathrm{k}, 2\\mathrm{k} \\rightarrow 0, 2\\mathrm{k}, 2\\mathrm{k} \\rightarrow 2\\mathrm{k}, 2\\mathrm{k}, 2\\mathrm{k}\\) . So (by a trivial induction) if we get all the numbers equal to k, then we can get them all to equal 2k. Finally, note that we can never decrease the largest number on the board, so the answer must be all powers of 2 greater than some minimum, which must be at least n. \n\nWe use induction to show that if \\(2^{\\mathrm{m}}\\) is the smallest power of 2 which is \\(\\geq \\mathrm{n}\\) , then we can get all numbers equal to \\(2^{\\mathrm{m}}\\) . Note that \\(0, \\mathrm{k} \\rightarrow \\mathrm{k}, \\mathrm{k} \\rightarrow 0, 2\\mathrm{k}\\) , so with a zero we can double each member of any set of numbers as often as we wish and finally convert the zero. For example, we could convert 0, 2, 4 to 8, 8, 8. It is convenient to take the induction hypothesis as \\(\\mathrm{S}_{\\mathrm{n}}\\) : we can convert 1, 2, ..., n to 0, \\(2^{\\mathrm{k}}, 2^{\\mathrm{k}}, \\ldots , 2^{\\mathrm{k}}\\) , where \\(2^{\\mathrm{k}}\\) is the smallest power of 2 which is \\(\\geq \\mathrm{n}\\) . \n\nWe show first that \\(\\mathrm{S}_{\\mathrm{n}}\\) is true for \\(\\mathrm{n} \\leq 8\\) . For \\(\\mathrm{n} = 3\\) , we take \\(1,3 \\rightarrow 2,4\\) , then \\(2,2 \\rightarrow 0,4\\) . For \\(\\mathrm{n} = 4\\) , we ignore the 4 and use the case \\(\\mathrm{n} = 3\\) . For \\(\\mathrm{n} = 5\\) , we take \\(3,5 \\rightarrow 2,8\\) . Then \\(2,2 \\rightarrow 0,4\\) . Then we use the 0 to convert the remaining powers of 2 (1,4,4) to 8. For \\(\\mathrm{n} = 6\\) , we take \\(2,6 \\rightarrow 4,8\\) and \\(3,5 \\rightarrow 2,8\\) , then \\(4,4 \\rightarrow 0,8\\) . Finally, we use the 0 to convert 1 and 2 to 8. For \\(\\mathrm{n} = 7\\) , we take \\(1,7 \\rightarrow 6,8\\) , then \\(2,6 \\rightarrow 4,8\\) , then \\(3,5 \\rightarrow 2,8\\) , then \\(4,4 \\rightarrow 0,8\\) , then \\(2,6 \\rightarrow 4,8\\) and finally use the 0 to convert the remaining 4 to 8. \n\nLet \\(\\mathrm{n} = 2^{\\mathrm{a}} + \\mathrm{b}\\) , where \\(0 < \\mathrm{b} \\leq 2^{\\mathrm{a}}\\) and assume \\(\\mathrm{S}_{\\mathrm{m}}\\) is true for all \\(\\mathrm{m} < \\mathrm{n}\\) . If \\(\\mathrm{b} = 1\\) , we convert the pair \\(2^{\\mathrm{a}} - 1\\) , \\(2^{\\mathrm{a}} + 1\\) to 2, \\(2^{\\mathrm{a} + 1}\\) . We have \\(2^{\\mathrm{a}} - 2 > 2\\) , so by induction we can convert 1, 2, ..., \\(2^{\\mathrm{a}} - 2\\) to 0, \\(2^{\\mathrm{a}}\\) , ..., \\(2^{\\mathrm{a}}\\) . Now all the numbers except 0 are powers of 2 and we can use the 0 to convert them each to \\(2^{\\mathrm{a} + 1}\\) . Similarly, if \\(\\mathrm{b} = 2\\) , we convert \\(2^{\\mathrm{a}} - 1\\) , \\(2^{\\mathrm{a}} + 1\\) to 2, \\(2^{\\mathrm{a} + 1}\\) and \\(2^{\\mathrm{a}} - 2\\) , \\(2^{\\mathrm{a}} + 2\\) to 4, \\(2^{\\mathrm{a} + 1}\\) and then proceed as in the previous case. If \\(3 \\leq \\mathrm{b} < 2^{\\mathrm{a}}\\) , then we start by converting the pairs \\((2^{\\mathrm{a}} + \\mathrm{b}, 2^{\\mathrm{a}} - \\mathrm{b})\\) , \\((2^{\\mathrm{a}} + \\mathrm{b} - 1, 2^{\\mathrm{a}} - \\mathrm{b} + 1)\\) , \\((2^{\\mathrm{a}} + \\mathrm{b} - 2, 2^{\\mathrm{a}} - \\mathrm{b} + 2)\\) , ..., \\((2^{\\mathrm{a}} + 1, 2^{\\mathrm{a}} - 1)\\) . That gives some \\(2^{\\mathrm{a} + 1}\\) s and 2, 4, ..., 2b. Now by \\(\\mathrm{S}_{\\mathrm{b}}\\) we can convert 2, 4, ..., 2b to 0, \\(2^{\\mathrm{a} + 1}\\) , ..., \\(2^{\\mathrm{a} + 1}\\) . The remaining\n\n\n\nnumbers 1, 2, ..., \\(2^{\\mathrm{a}}\\) - b- 1 can either be converted to powers of 2 by \\(S^{2\\mathrm{a - b - 1}}\\) (if \\(2^{\\mathrm{a}}\\) - b- 1 \\(\\geq 3\\) ) or are already powers of 2. Finally we use the 0 to bring all powers of 2 up to \\(2^{\\mathrm{a + 1}}\\) . In the case \\(\\mathrm{b} = 2^{\\mathrm{a}}\\) , we ignore \\(2^{\\mathrm{a}} + \\mathrm{b} (= 2^{\\mathrm{a + 1}})\\) and use the case b- 1 to convert the others.", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 11", "solution_match": "# Solution"}}
-{"year": "1991", "tier": "T1", "problem_label": "12", "problem_type": null, "exam": "AllSovietUnion", "problem": "The figure below is cut along the lines into polygons (which need not be convex). No polygon contains a 2 x 2 square. What is the smallest possible number of polygons? \n\n\n \n\nAnswer 12", "solution": "We can clearly cut the polygon into 12 strips width 1, so the smallest number is \\(\\leq 12\\) . \n\nThere are 84 unit squares in the figure. Each cut along the edge of a unit square not already cut (and not on the boundary) increases the number of pieces by at most 1. So it is sufficient to show that at most 72 edges remain uncut (after cutting into polygons). Because then cutting the remaining edges would increase the total number of pieces by at most 72. But the final number of pieces is 84, so we would have to start with at least 12. \n\nInitially, there are 144 edges, so we have to show that at least 72 of them are cut to make the polygons. An interior vertex has 4 edges. At least two of them must be cut, or the vertex would be the center of an uncut 2 x 2 square. If we take alternate interior vertices (36 in total, as shown below), then each has at least two cut edges, so in total at least 72 edges are cut to make the polygons.", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 12", "solution_match": "# Solution"}}
+{"year": "1991", "tier": "T1", "problem_label": "12", "problem_type": null, "exam": "AllSovietUnion", "problem": "The figure below is cut along the lines into polygons (which need not be convex). No polygon contains a 2 x 2 square. What is the smallest possible number of polygons? \n\n\n \n\nAnswer 12", "solution": "We can clearly cut the polygon into 12 strips width 1, so the smallest number is \\(\\leq 12\\) . \n\nThere are 84 unit squares in the figure. Each cut along the edge of a unit square not already cut (and not on the boundary) increases the number of pieces by at most 1. So it is sufficient to show that at most 72 edges remain uncut (after cutting into polygons). Because then cutting the remaining edges would increase the total number of pieces by at most 72. But the final number of pieces is 84, so we would have to start with at least 12. \n\nInitially, there are 144 edges, so we have to show that at least 72 of them are cut to make the polygons. An interior vertex has 4 edges. At least two of them must be cut, or the vertex would be the center of an uncut 2 x 2 square. If we take alternate interior vertices (36 in total, as shown below), then each has at least two cut edges, so in total at least 72 edges are cut to make the polygons.", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 12", "solution_match": "# Solution"}}
{"year": "1991", "tier": "T1", "problem_label": "13", "problem_type": null, "exam": "AllSovietUnion", "problem": "ABC is an acute- angled triangle with circumcenter O. The circumcircle of ABO intersects AC and BC at M and N. Show that the circumradii of ABO and MNC are the same.", "solution": "It is sufficient to show that \\(\\angle \\mathrm{MBN} = \\angle \\mathrm{C}\\) . But \\(\\angle \\mathrm{MBN} = \\angle \\mathrm{MBO} + \\angle \\mathrm{OBN} = \\angle \\mathrm{MAO} + \\angle \\mathrm{OBN} = \\angle \\mathrm{MCO} + \\angle \\mathrm{OCN} = \\angle \\mathrm{C}\\) .", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 13", "solution_match": "# Solution"}}
{"year": "1991", "tier": "T1", "problem_label": "14", "problem_type": null, "exam": "AllSovietUnion", "problem": "A polygon can be transformed into a new polygon by making a straight cut, which creates two new pieces each with a new edge. One piece is then turned over and the two new edges are reattached. Can repeated transformations of this type turn a square into a triangle?", "solution": "", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 14", "solution_match": ""}}
{"year": "1991", "tier": "T1", "problem_label": "15", "problem_type": null, "exam": "AllSovietUnion", "problem": "An \\(h \\times k\\) minor of an n x n table is the hk cells which lie in h rows and k columns. The semiperimeter of the minor is \\(\\mathrm{h} + \\mathrm{k}\\) . A number of minors each with semiperimeter at least n together include all the cells on the main diagonal. Show that they include at least half the cells in the table.", "solution": "", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 15", "solution_match": ""}}
diff --git a/USAJMO/md/en-JMO-2025-notes.md b/USAJMO/md/en-JMO-2025-notes.md
index 0c3b3a8912c420ff5bbed7622a1914fd4c277162..3aae49ccc835339b30761c9da8f94635e3569f6e 100644
--- a/USAJMO/md/en-JMO-2025-notes.md
+++ b/USAJMO/md/en-JMO-2025-notes.md
@@ -152,7 +152,7 @@ Case 4 If some two consecutive rows of the staircase differ by more than two cel
Needless to say, each induction step preserves that the number of black squares equals the number of white squares. Hence the remaining cells can then be tiled by the induction hypothesis. See the image below for an illustration of all four cases.
-
+
I claim that at least one of these four cases must apply unless the staircase is empty. This could only occur if the heights of the staircase are \(1,2,\ldots ,k\) in that order. However, in that case the number of black cells and white cells are obviously not equal (unless \(k = 0\) ), so this can never occur. This completes the induction. \(\square\)
@@ -235,7 +235,7 @@ Claim — \(AHQC\) is a parallelogram, and \(APCQ\) is an isosceles trapezoid.
Proof. As \(\overline{AH} \perp \overline{BC} \perp \overline{CQ}\) and \(\overline{CF} \perp \overline{AB} \perp \overline{AQ}\) .
-
+
Let \(M\) be the midpoint of \(\overline{QC}\) .
diff --git a/USAJMO/segmented/en-JMO-2025-notes.jsonl b/USAJMO/segmented/en-JMO-2025-notes.jsonl
index ffa67bfdc2a8090bdb6c5a3ebc9fae2a1927e8d9..6a62de66118a872f41c7710a2b95e9ac0e00177c 100644
--- a/USAJMO/segmented/en-JMO-2025-notes.jsonl
+++ b/USAJMO/segmented/en-JMO-2025-notes.jsonl
@@ -1,6 +1,6 @@
{"year": "2025", "tier": "T3", "problem_label": "1", "problem_type": null, "exam": "USAJMO", "problem": "Prove that if \\(f\\colon \\mathbb{Z}\\to \\mathbb{Z}\\) is any function, then there are infinitely many integers \\(c\\) such that the function \\(g(x) = f(x) + cx\\) is not a bijection.", "solution": "Assume for contradiction that there exists a finite \"bad\" set \\(S\\) such that \\(f(x) + cx\\) is bijective for all \\(c\\notin S\\) \n\nThe first observation is basically that given just \\(f(0)\\) and \\(f(1)\\) , or any two consecutive \\(f\\) - values, we can already find bad values of \\(c\\) . \n\nClaim — The numbers \\(f(0) - f(1)\\) , \\(f(1) - f(2)\\) , \\(f(2) - f(3)\\) , ... are all in \\(S\\) . That is, consecutive differences of \\(f\\) only take finitely many values. \n\nProof. To see that \\(f(0) - f(1)\\in S\\) , just note that \n\n\\[x\\mapsto f(x) + (f(0) - f(1))\\cdot x\\] \n\nis obviously not a bijective function, since it takes the same values at \\(x = 0\\) and \\(x = 1\\) , namely \\(f(0)\\) . The same is true for other elements. \\(\\square\\) \n\nIn particular, suppose \\(M > \\max_{s\\in S}|s|\\) . Then the function \n\n\\[g(x) = f(x) + (M + 100)x\\] \n\nis supposed to be a bijection (because \\(M + 100\\notin S\\) ), and yet \\(g(x + 1) - g(x) > 100\\) for all \\(x\\) , which is a contradiction.", "metadata": {"resource_path": "USAJMO/segmented/en-JMO-2025-notes.jsonl", "problem_match": "1. ", "solution_match": "## \\(\\S 1.1\\) JMO 2025/1, proposed by John Berman \n"}}
{"year": "2025", "tier": "T3", "problem_label": "2", "problem_type": null, "exam": "USAJMO", "problem": "Fix positive integers \\(k\\) and \\(d\\) . Prove that for all sufficiently large odd positive integers \\(n\\) , the digits of the base- \\(2n\\) representation of \\(n^k\\) are all greater than \\(d\\) .", "solution": "The problem actually doesn't have much to do with digits: the idea is to pick any length \\(\\ell \\leq k\\) , and look at the rightmost \\(\\ell\\) digits of \\(n^k\\) ; that is, the remainder upon division by \\((2n)^\\ell\\) . We compute it exactly: \n\nClaim — Let \\(n \\geq 1\\) be an odd integer, and \\(k \\geq \\ell \\geq 1\\) integers. Then \n\n\\[n^k \\bmod (2n)^i = c(k, \\ell) \\cdot n^i\\] \n\nfor some odd integer \\(1 \\leq c(k, \\ell) \\leq 2^\\ell - 1\\) . \n\nProof. This follows directly by the Chinese remainder theorem, with \\(c(k, \\ell)\\) being the residue class of \\(n^{k-i} \\pmod {2^\\ell}\\) (which makes sense because \\(n\\) was odd). \\(\\square\\) \n\nWe can now stake the required threshold: \n\nClaim — The problem statement holds once \\(n \\geq (d + 1) \\cdot 2^{k- 1}\\) . \n\nProof. Suppose \\(n\\) is that large. Then \\(n^k\\) has \\(k\\) digits in base- \\(2n\\) . Moreover, for each \\(1 \\leq \\ell \\leq k\\) we have \n\n\\[c(k, \\ell) \\cdot n^\\ell \\geq (d + 1) \\cdot (2n)^{\\ell - 1}\\] \n\nbecause \\(n\\) is large enough; that implies the \\(\\ell^{\\mathrm{th}}\\) digit from the right is at least \\(d + 1\\) . Hence the problem is solved. \\(\\square\\) \n\nRemark. Note it doesn't really matter that \\(c(k, i)\\) is odd per se; we only need that \\(c(k, i) \\geq 1\\) .", "metadata": {"resource_path": "USAJMO/segmented/en-JMO-2025-notes.jsonl", "problem_match": "2. ", "solution_match": "## \\(\\S 1.2\\) JMO 2025/2, proposed by John Berman \n"}}
-{"year": "2025", "tier": "T3", "problem_label": "3", "problem_type": null, "exam": "USAJMO", "problem": "Let \\(m\\) and \\(n\\) be positive integers, and let \\(\\mathcal{R}\\) be a \\(2m\\times 2n\\) grid of unit squares. A domino is a \\(1\\times 2\\) or \\(2\\times 1\\) rectangle. An up-right path is a path from the lower-left corner of \\(\\mathcal{R}\\) to the upper-right corner of \\(\\mathcal{R}\\) formed by exactly \\(2m + 2n\\) edges of the grid squares. \n\nIn terms of \\(m\\) and \\(n\\) , find the number of up-right paths that divide \\(\\mathcal{R}\\) into two subsets (possibly empty), each of which can be tiled with dominoes.", "solution": "Let. \n\nApply the usual black/white checkerboard coloring. We define a staircase to be the below above an up- right path (equivalently, a Young diagram rotated \\(180^{\\circ}\\) ). \n\nThe proof is composed of two steps. One is rewriting the \"domino- tileable\" condition to one that is actually usable; the other is the counting part. \n\n\\(\\P\\) Determining when a staircase can be domino- tileable. It turns out the obvious guess is right. \n\n## Lemma \n\nA staircase can be tiled with dominoes if and only if the number of black and white cells in it is equal. \n\nProof. It's obvious equal black and white cells is necessary; we prove it's sufficient. \n\nAssume the number of black and white cells is equal. The proof is by induction with empty staircases being vacuous. We consider the following four cases (which could overlap, pick one arbitrarily if so). Note that the latter two cases are just mirrors of the first two. \n\nCase 1 If the rightmost two columns of the staircase have the same height, tile those two columns. \n\nCase 2 If some two consecutive columns of the staircase differ by more than two cells in height, say the \\((i - 1)^{\\mathrm{st}}\\) and the \\(i^{\\mathrm{th}}\\) , then tile the top two cells in the \\(i^{\\mathrm{th}}\\) column and rightwards. \n\nCase 3 If the bottom two rows of the staircase have the same length, tile those two rows. \n\nCase 4 If some two consecutive rows of the staircase differ by more than two cells in length, say the \\((j - 1)^{\\mathrm{st}}\\) and the \\(j^{\\mathrm{th}}\\) , then tile the leftmost two cells in the \\(j^{\\mathrm{th}}\\) column and below. \n\nNeedless to say, each induction step preserves that the number of black squares equals the number of white squares. Hence the remaining cells can then be tiled by the induction hypothesis. See the image below for an illustration of all four cases.\n\n\n\n \n\nI claim that at least one of these four cases must apply unless the staircase is empty. This could only occur if the heights of the staircase are \\(1,2,\\ldots ,k\\) in that order. However, in that case the number of black cells and white cells are obviously not equal (unless \\(k = 0\\) ), so this can never occur. This completes the induction. \\(\\square\\) \n\nCounting. Back to the main problem. The point is that we need to find the number of up- right paths for which the two resulting regions have equal numbers of black and white squares; call such paths balanced. Since the overall \\(2m\\times 2n\\) grid has \\(2mn\\) black and \\(2mn\\) white squares, it's sufficient for just the bottom- right staircase to have equal black and white cells. \n\nThen in general, an up- right path is characterized by integer sequences \n\n\\[0\\leq h_{1}\\leq h_{2}\\leq \\dots \\leq h_{2n}\\leq 2m\\] \n\ncorresponding to the heights of the staircase. \n\nClaim — The \\((h_{i})\\) correspond to a balanced up- right path if and only if \\(\\{i:h_{i}\\equiv 1\\) (mod 2)} has an equal number of even and odd indices. \n\nProof. WLOG let's fix our coloring so that the bottom- left square is black. Then in the \\(i^{\\mathrm{th}}\\) column, for \\(i = 1,\\ldots ,2n\\) , has (i) one more black square than white if \\(h_{i}\\) is odd and \\(i\\equiv 1\\) (mod 2); (ii) one more white square than black if \\(h_{i}\\) is odd and \\(i\\equiv 0\\) (mod 2); (iii) equal black and white squares if \\(h_{i}\\) is even. The conclusion follows immediately. \\(\\square\\) \n\nThe trick is to instead define \n\n\\[t_{i}:= h_{i} + i.\\] \n\nThat is, we will instead describe the up- right paths by integer sequences \\(t_{i}\\) with \n\n\\[1\\leq t_{1}< t_{2}< \\dots < t_{2n}\\leq 2n + 2m.\\] \n\nClaim — The \\((t_{i})\\) correspond to a balanced up- right path if and only if the set \\(\\{t_{1},\\ldots ,t_{2n}\\}\\) has an equal number of even and odd elements.\n\n\n\nProof. Translating the four cases between the two notations gives the following table: \n\n | i even | i odd | | i even | i odd |
| hi≡1 (mod 2) | a | b | ⇔ | ti≡1 (mod 2) | a 2n-b |
| hi≡0 (mod 2) | 2n-a | 2n-b | | ti≡0 (mod 2) | 2n-a b. |
\n\nThen we have \n\nbalanced \\(\\iff a = b\\iff a + (2n - b) = (2n - a) + b\\iff \\# \\{\\mathrm{odd} t_i\\} = \\# \\{\\mathrm{even} t_i\\}\\) as desired. \n\nHence we must count the number of \\(2m\\) - element subsets of \\(\\{1,2,\\ldots ,2n + 2m\\}\\) with \\(m\\) even and \\(m\\) odd terms. Since the even and odd terms can be chosen separately, this gives an answer of \\(\\binom{n+m}{m}^2\\) .\n\n\n\n## \\(\\S 2\\) Solutions to Day 2", "metadata": {"resource_path": "USAJMO/segmented/en-JMO-2025-notes.jsonl", "problem_match": "3. ", "solution_match": "## \\(\\S 1.3\\) JMO 2025/3, proposed by Wilbert Chu \n"}}
+{"year": "2025", "tier": "T3", "problem_label": "3", "problem_type": null, "exam": "USAJMO", "problem": "Let \\(m\\) and \\(n\\) be positive integers, and let \\(\\mathcal{R}\\) be a \\(2m\\times 2n\\) grid of unit squares. A domino is a \\(1\\times 2\\) or \\(2\\times 1\\) rectangle. An up-right path is a path from the lower-left corner of \\(\\mathcal{R}\\) to the upper-right corner of \\(\\mathcal{R}\\) formed by exactly \\(2m + 2n\\) edges of the grid squares. \n\nIn terms of \\(m\\) and \\(n\\) , find the number of up-right paths that divide \\(\\mathcal{R}\\) into two subsets (possibly empty), each of which can be tiled with dominoes.", "solution": "Let. \n\nApply the usual black/white checkerboard coloring. We define a staircase to be the below above an up- right path (equivalently, a Young diagram rotated \\(180^{\\circ}\\) ). \n\nThe proof is composed of two steps. One is rewriting the \"domino- tileable\" condition to one that is actually usable; the other is the counting part. \n\n\\(\\P\\) Determining when a staircase can be domino- tileable. It turns out the obvious guess is right. \n\n## Lemma \n\nA staircase can be tiled with dominoes if and only if the number of black and white cells in it is equal. \n\nProof. It's obvious equal black and white cells is necessary; we prove it's sufficient. \n\nAssume the number of black and white cells is equal. The proof is by induction with empty staircases being vacuous. We consider the following four cases (which could overlap, pick one arbitrarily if so). Note that the latter two cases are just mirrors of the first two. \n\nCase 1 If the rightmost two columns of the staircase have the same height, tile those two columns. \n\nCase 2 If some two consecutive columns of the staircase differ by more than two cells in height, say the \\((i - 1)^{\\mathrm{st}}\\) and the \\(i^{\\mathrm{th}}\\) , then tile the top two cells in the \\(i^{\\mathrm{th}}\\) column and rightwards. \n\nCase 3 If the bottom two rows of the staircase have the same length, tile those two rows. \n\nCase 4 If some two consecutive rows of the staircase differ by more than two cells in length, say the \\((j - 1)^{\\mathrm{st}}\\) and the \\(j^{\\mathrm{th}}\\) , then tile the leftmost two cells in the \\(j^{\\mathrm{th}}\\) column and below. \n\nNeedless to say, each induction step preserves that the number of black squares equals the number of white squares. Hence the remaining cells can then be tiled by the induction hypothesis. See the image below for an illustration of all four cases.\n\n\n\n \n\nI claim that at least one of these four cases must apply unless the staircase is empty. This could only occur if the heights of the staircase are \\(1,2,\\ldots ,k\\) in that order. However, in that case the number of black cells and white cells are obviously not equal (unless \\(k = 0\\) ), so this can never occur. This completes the induction. \\(\\square\\) \n\nCounting. Back to the main problem. The point is that we need to find the number of up- right paths for which the two resulting regions have equal numbers of black and white squares; call such paths balanced. Since the overall \\(2m\\times 2n\\) grid has \\(2mn\\) black and \\(2mn\\) white squares, it's sufficient for just the bottom- right staircase to have equal black and white cells. \n\nThen in general, an up- right path is characterized by integer sequences \n\n\\[0\\leq h_{1}\\leq h_{2}\\leq \\dots \\leq h_{2n}\\leq 2m\\] \n\ncorresponding to the heights of the staircase. \n\nClaim — The \\((h_{i})\\) correspond to a balanced up- right path if and only if \\(\\{i:h_{i}\\equiv 1\\) (mod 2)} has an equal number of even and odd indices. \n\nProof. WLOG let's fix our coloring so that the bottom- left square is black. Then in the \\(i^{\\mathrm{th}}\\) column, for \\(i = 1,\\ldots ,2n\\) , has (i) one more black square than white if \\(h_{i}\\) is odd and \\(i\\equiv 1\\) (mod 2); (ii) one more white square than black if \\(h_{i}\\) is odd and \\(i\\equiv 0\\) (mod 2); (iii) equal black and white squares if \\(h_{i}\\) is even. The conclusion follows immediately. \\(\\square\\) \n\nThe trick is to instead define \n\n\\[t_{i}:= h_{i} + i.\\] \n\nThat is, we will instead describe the up- right paths by integer sequences \\(t_{i}\\) with \n\n\\[1\\leq t_{1}< t_{2}< \\dots < t_{2n}\\leq 2n + 2m.\\] \n\nClaim — The \\((t_{i})\\) correspond to a balanced up- right path if and only if the set \\(\\{t_{1},\\ldots ,t_{2n}\\}\\) has an equal number of even and odd elements.\n\n\n\nProof. Translating the four cases between the two notations gives the following table: \n\n | i even | i odd | | i even | i odd |
| hi≡1 (mod 2) | a | b | ⇔ | ti≡1 (mod 2) | a 2n-b |
| hi≡0 (mod 2) | 2n-a | 2n-b | | ti≡0 (mod 2) | 2n-a b. |
\n\nThen we have \n\nbalanced \\(\\iff a = b\\iff a + (2n - b) = (2n - a) + b\\iff \\# \\{\\mathrm{odd} t_i\\} = \\# \\{\\mathrm{even} t_i\\}\\) as desired. \n\nHence we must count the number of \\(2m\\) - element subsets of \\(\\{1,2,\\ldots ,2n + 2m\\}\\) with \\(m\\) even and \\(m\\) odd terms. Since the even and odd terms can be chosen separately, this gives an answer of \\(\\binom{n+m}{m}^2\\) .\n\n\n\n## \\(\\S 2\\) Solutions to Day 2", "metadata": {"resource_path": "USAJMO/segmented/en-JMO-2025-notes.jsonl", "problem_match": "3. ", "solution_match": "## \\(\\S 1.3\\) JMO 2025/3, proposed by Wilbert Chu \n"}}
{"year": "2025", "tier": "T3", "problem_label": "4", "problem_type": null, "exam": "USAJMO", "problem": "Let \\(n\\) be a positive integer, and let \\(a_0 \\geq a_1 \\geq \\dots \\geq a_n \\geq 0\\) be integers. Prove that \n\n\\[\\sum_{i = 0}^{n}i\\binom{a_{i}}{2}\\leq \\frac{1}{2}\\binom{a_{0} + a_{1} + \\cdots + a_{n}}{2}.\\]", "solution": "Le] \n\nFor \\(n = 0\\) (which we permit) there is nothing to prove. Hence to prove by induction on \\(n\\) , it would be sufficient to verify \n\n\\[2n\\binom{a_{n}}{2}\\leq \\binom{a_{0} + a_{1} + \\cdots + a_{n}}{2} -\\binom{a_{0} + a_{1} + \\cdots + a_{n - 1}}{2}.\\] \n\nRearranging the terms around, that's equivalent to proving \n\n\\[\\iff 2n(a_{n}^{2} - a_{n})\\leq a_{n}^{2} + a_{n}\\cdot (2(a_{0} + \\cdot \\cdot \\cdot +a_{n - 1}) - 1)\\] \\[\\iff 0\\leq 2a_{n}(a_{0} + \\cdot \\cdot \\cdot +a_{n - 1} - na_{n}) + a_{n}(a_{n} + 2n - 1).\\] \n\nHowever, the last line is obvious because \\(\\min (a_{0},\\ldots ,a_{n - 1})\\geq a_{n}\\) , and \\(a_{n}\\geq 0\\) \n\nRemark. The only equality case is when \\(a_{0}\\in \\{0,1\\}\\) and \\(a_{i} = 0\\) for \\(i\\geq 1\\) \n\nThe bound in the problem is extremely loose and pretty much anything will work.", "metadata": {"resource_path": "USAJMO/segmented/en-JMO-2025-notes.jsonl", "problem_match": "4. ", "solution_match": "## \\(\\S 2.1\\) JMO 2025/4, proposed by Hung- Hsun Yu \n"}}
-{"year": "2025", "tier": "T3", "problem_label": "5", "problem_type": null, "exam": "USAJMO", "problem": "Let \\(H\\) be the orthocenter of an acute triangle \\(ABC\\) , let \\(F\\) be the foot of the altitude from \\(C\\) to \\(AB\\) , and let \\(P\\) be the reflection of \\(H\\) across \\(BC\\) . Suppose that the circumcircle of triangle \\(AFP\\) intersects line \\(BC\\) at two distinct points \\(X\\) and \\(Y\\) . Prove that \\(CX = CY\\) .", "solution": "Let \\(Q\\) be the antipode of \\(B\\) . \n\nClaim — \\(AHQC\\) is a parallelogram, and \\(APCQ\\) is an isosceles trapezoid. \n\nProof. As \\(\\overline{AH} \\perp \\overline{BC} \\perp \\overline{CQ}\\) and \\(\\overline{CF} \\perp \\overline{AB} \\perp \\overline{AQ}\\) . \n\n\n \n\nLet \\(M\\) be the midpoint of \\(\\overline{QC}\\) . \n\nClaim — Point \\(M\\) is the circumcenter of \\(\\triangle AFP\\) . \n\nProof. It's clear that \\(MA = MP\\) from the isosceles trapezoid. As for \\(MA = MF\\) , let \\(N\\) denote the midpoint of \\(\\overline{AF}\\) ; then \\(\\overline{MN}\\) is a midline of the parallelogram, so \\(\\overline{MN} \\perp \\overline{AF}\\) . \\(\\square\\) \n\nSince \\(\\overline{CM} \\perp \\overline{BC}\\) and \\(M\\) is the center of \\((AFP)\\) , it follows \\(CX = CY\\) .", "metadata": {"resource_path": "USAJMO/segmented/en-JMO-2025-notes.jsonl", "problem_match": "5. ", "solution_match": "## \\(\\S 2.2\\) JMO 2025/5, proposed by Carl Schildkraut \n"}}
+{"year": "2025", "tier": "T3", "problem_label": "5", "problem_type": null, "exam": "USAJMO", "problem": "Let \\(H\\) be the orthocenter of an acute triangle \\(ABC\\) , let \\(F\\) be the foot of the altitude from \\(C\\) to \\(AB\\) , and let \\(P\\) be the reflection of \\(H\\) across \\(BC\\) . Suppose that the circumcircle of triangle \\(AFP\\) intersects line \\(BC\\) at two distinct points \\(X\\) and \\(Y\\) . Prove that \\(CX = CY\\) .", "solution": "Let \\(Q\\) be the antipode of \\(B\\) . \n\nClaim — \\(AHQC\\) is a parallelogram, and \\(APCQ\\) is an isosceles trapezoid. \n\nProof. As \\(\\overline{AH} \\perp \\overline{BC} \\perp \\overline{CQ}\\) and \\(\\overline{CF} \\perp \\overline{AB} \\perp \\overline{AQ}\\) . \n\n\n \n\nLet \\(M\\) be the midpoint of \\(\\overline{QC}\\) . \n\nClaim — Point \\(M\\) is the circumcenter of \\(\\triangle AFP\\) . \n\nProof. It's clear that \\(MA = MP\\) from the isosceles trapezoid. As for \\(MA = MF\\) , let \\(N\\) denote the midpoint of \\(\\overline{AF}\\) ; then \\(\\overline{MN}\\) is a midline of the parallelogram, so \\(\\overline{MN} \\perp \\overline{AF}\\) . \\(\\square\\) \n\nSince \\(\\overline{CM} \\perp \\overline{BC}\\) and \\(M\\) is the center of \\((AFP)\\) , it follows \\(CX = CY\\) .", "metadata": {"resource_path": "USAJMO/segmented/en-JMO-2025-notes.jsonl", "problem_match": "5. ", "solution_match": "## \\(\\S 2.2\\) JMO 2025/5, proposed by Carl Schildkraut \n"}}
{"year": "2025", "tier": "T3", "problem_label": "6", "problem_type": null, "exam": "USAJMO", "problem": "Let \\(S\\) be a set of integers with the following properties: \n\n\\(\\{1,2,\\ldots ,2025\\} \\subseteq S.\\) If \\(a,b\\in S\\) and \\(\\gcd (a,b) = 1\\) , then \\(a b\\in S\\) If \\(s + 1\\) is composite for some \\(s\\in S\\) , then all positive divisors of \\(s + 1\\) are in \\(S\\) Prove that \\(S\\) contains all positive integers.", "solution": "Let \\rs. \n\nWe prove by induction on \\(N\\) that \\(S\\) contains \\(\\{1,\\ldots ,N\\}\\) with the base cases being \\(N = 1,\\ldots ,2025\\) already given. \n\nFor the inductive step, to show \\(N + 1\\in S\\) \n\nIf \\(N + 1\\) is composite we're already done from the third bullet. \n\nOtherwise, assume \\(N + 1 = p\\geq 2025\\) is an (odd) prime number. We say a number is good if the prime powers in its prime factorization are all less than \\(p\\) . Hence by the second bullet (repeatedly), good numbers are in \\(S\\) . Now our proof is split into three cases: \n\nCase 1. Suppose neither \\(p - 1\\) nor \\(p + 1\\) is a power of 2 (but both are still even). We claim that the number \n\n\\[s:= p^{2} - 1 = (p - 1)(p + 1)\\] \n\nis good. Indeed, one of the numbers has only a single factor of 2, and the other by hypothesis is not a power of 2 (but still even). So the largest power of 2 dividing \\(p^{2} - 1\\) is certainly less than \\(p\\) . And every other prime power divides at most one of \\(p - 1\\) and \\(p + 1\\) . \n\nHence \\(s:= p^{2} - 1\\) is good. As \\(s + 1 = p^{2}\\) , Case 1 is done. \n\nCase 2. Suppose \\(p + 1\\) is a power of 2; that is \\(p = 2^{q} - 1\\) . Since \\(p > 2025\\) , we assume \\(q\\geq 11\\) is odd. First we contend that the number \n\n\\[s^{\\prime}:= 2^{q + 1} - 1 = \\left(2^{(q + 1) / 2} - 1\\right)\\left(2^{(q + 1) / 2} + 1\\right)\\] \n\nis good. Indeed, this follows from the two factors being coprime and both less than \\(p\\) . Hence \\(s^{\\prime} + 1 = 2^{q + 1}\\) is in \\(S\\) . \n\nThus, we again have \n\n\\[s:= p^{2} - 1 = (p - 1)(p + 1)\\in S\\] \n\nas we did in the previous case, because the largest power of 2 dividing \\(p^{2} - 1\\) will be exactly \\(2^{q + 1}\\) which is known to be in \\(S\\) . And since \\(s + 1 = p^{2}\\) , Case 2 is done.\n\n\n\nCase 3. Finally suppose \\(p - 1\\) is a power of 2; that is \\(p = 2^{2^{e}} + 1\\) is a Fermat prime. Then in particular, \\(p \\equiv 2\\) (mod 3). Now observe that \n\n\\[s:= 2p - 1\\equiv 0\\pmod {3}\\] \n\nand moreover \\(2p - 1\\) is not a power of 3 (it would imply \\(2^{2^{e} + 1} + 1 = 3^{k}\\) , which is impossible for \\(k \\geq 3\\) by Zsigmondy/Mihailescu/etc.). So \\(s\\) is good, and since \\(s + 1 = 2p\\) , Case 3 is done. \n\nHaving finished all the cases, we conclude \\(p \\in S\\) and the induction is done. \n\nRemark. In fact just \\(2025 \\in S\\) is sufficient as a base case; however this requires a bit more work to check. Here is how: \n\n- From \\(2025 \\in S\\) we get \\(2026 = 2 \\cdot 1013\\) , so \\(2,1013 \\in S\\) . \n\n- From \\(1013 + 1 = 1014 = 2 \\cdot 3 \\cdot 13^{2}\\) we get \\(3,13 \\in S\\) . \n\n- From \\(3 + 1 = 4\\) we get \\(4 \\in S\\) . \n\n- \\(3 \\cdot 13 + 1 = 40 = 2^{3} \\cdot 5\\) we get \\(5 \\in S\\) . \n\n- Once \\(\\{1,2,\\ldots ,5\\} \\subseteq S\\) , the induction above actually works fine; that is, \\(N \\leq 5\\) are sufficient as base cases for the earlier cases to finish the rest of the problem. (Case 2 works once \\(q \\geq 3\\) , and Case 3 works once \\(e \\geq 2\\) .) \n\nHowever, \\(\\{1,2,3,4\\} \\subseteq S\\) is not sufficient; for example \\(S = \\{1,2,3,4,6,12\\}\\) satisfies all the problem conditions.", "metadata": {"resource_path": "USAJMO/segmented/en-JMO-2025-notes.jsonl", "problem_match": "6. ", "solution_match": "## \\(\\S 2.3\\) JMO 2025/6, proposed by Enrique Treviño \n"}}
diff --git a/USAMO/md/en-USAMO-2025-notes.md b/USAMO/md/en-USAMO-2025-notes.md
index 534dc78d6172641d7d2e086c22bd4ce073f5def6..41ffcf7a2f87581e2164a6317a9bc3c533c55ede 100644
--- a/USAMO/md/en-USAMO-2025-notes.md
+++ b/USAMO/md/en-USAMO-2025-notes.md
@@ -178,7 +178,7 @@ Proof that every Bob- graph is connected. Assume for contradiction the graph is
Hence, \(r\) is in a different connected component from at least one of \(p\) or \(q\) — let's say point \(p\) . Then we repeat the same argument on the disk with diameter \(\overline{pr}\) to find a new point \(s\) , non- adjacent to either \(p\) or \(r\) . See the figure below, where the \(X\) 'ed out dashed edges indicate points which are not only non- adjacent but in different connected components.
-
+
In this way we generate an infinite sequence of distances \(\delta_{1}\) , \(\delta_{2}\) , \(\delta_{3}\) , ... among the non- edges in the picture above. By the "Pythagorean theorem" (or really the inequality for it), we have
@@ -206,7 +206,7 @@ However for infinite Bob- sets the descending condition is insufficient, and con
A cartoon of the graph is shown below.
-
+
In that case, \(\{A_{n}\}\) and \(\{B_{n}\}\) will be disconnected from each other: none of the edges \(A_{n}B_{n}\) or \(B_{n}A_{n + 1}\) are formed. In this case the relative neighbor graph consists of the edges \(A_{1}A_{2}A_{3}A_{4}\dots\) and \(B_{1}B_{2}B_{3}B_{4}\dots\) . That's why for the present problem, the inequality
@@ -233,7 +233,7 @@ Claim — \(AHQC\) is a parallelogram, and \(APCQ\) is an isosceles trapezoid.
Proof. As \(\overline{AH} \perp \overline{BC} \perp \overline{CQ}\) and \(\overline{CF} \perp \overline{AB} \perp \overline{AQ}\) .
-
+
Let \(M\) be the midpoint of \(\overline{QC}\) .
@@ -341,7 +341,7 @@ We now consider the following algorithm, which takes several steps.
A cartoon of this picture is shown below.
-
+
We commit to assigning each of person in \(\mathcal{M}\) their matched arc (in particular if there are no bad sets at all, the problem is already solved). Now we finish the problem by induction on \(n\) (for the remaining people) by simply deleting the arcs used up by \(\mathcal{M}\) .
@@ -352,7 +352,7 @@ To see why this deletion- induction works, consider any particular person Quinn
meaning the induction is OK. See below for a cartoon of the deletion, where Pip's arcs are drawn in blue while Quinn's arcs and scores are drawn in red (in this example \(n = 3\) ).
-
+
Remark. This deletion argument can be thought of in some special cases even before the realization of Hall, in the case where \(\mathcal{M}\) has only one person (Pip). This amounts to saying that if one of Pip's arcs isn't liked by anybody, then that arc can be deleted and the induction carries through.
diff --git a/USAMO/segmented/en-USAMO-2025-notes.jsonl b/USAMO/segmented/en-USAMO-2025-notes.jsonl
index cbb761b7228e4d9abb6452e0cd872d01c1f19e9d..426d32b77634297dadb17555ac98aa0fe1e85b26 100644
--- a/USAMO/segmented/en-USAMO-2025-notes.jsonl
+++ b/USAMO/segmented/en-USAMO-2025-notes.jsonl
@@ -1,6 +1,6 @@
{"year": "2025", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "USAMO", "problem": "Fix positive integers \\(k\\) and \\(d\\) . Prove that for all sufficiently large odd positive integers \\(n\\) , the digits of the base-2 \\(n\\) representation of \\(n^{k}\\) are all greater than \\(d\\) .", "solution": "The problem actually doesn't have much to do with digits: the idea is to pick any length \\(\\ell \\leq k\\) , and look at the rightmost \\(\\ell\\) digits of \\(n^{k}\\) ; that is, the remainder upon division by \\((2n)^{\\ell}\\) . We compute it exactly: \n\nClaim — Let \\(n \\geq 1\\) be an odd integer, and \\(k \\geq \\ell \\geq 1\\) integers. Then \n\n\\[n^{k} \\bmod (2n)^{i} = c(k, \\ell) \\cdot n^{i}\\] \n\nfor some odd integer \\(1 \\leq c(k, \\ell) \\leq 2^{\\ell} - 1\\) . \n\nProof. This follows directly by the Chinese remainder theorem, with \\(c(k, \\ell)\\) being the residue class of \\(n^{k - i} \\pmod{2^{\\ell}}\\) (which makes sense because \\(n\\) was odd). \\(\\square\\) \n\nWe can now stake the required threshold: \n\nClaim — The problem statement holds once \\(n \\geq (d + 1) \\cdot 2^{k - 1}\\) . \n\nProof. Suppose \\(n\\) is that large. Then \\(n^{k}\\) has \\(k\\) digits in base- \\(2n\\) . Moreover, for each \\(1 \\leq \\ell \\leq k\\) we have \n\n\\[c(k, \\ell) \\cdot n^{\\ell} \\geq (d + 1) \\cdot (2n)^{\\ell - 1}\\] \n\nbecause \\(n\\) is large enough; that implies the \\(\\ell^{\\mathrm{th}}\\) digit from the right is at least \\(d + 1\\) . Hence the problem is solved. \\(\\square\\) \n\nRemark. Note it doesn't really matter that \\(c(k, i)\\) is odd per se; we only need that \\(c(k, i) \\geq 1\\) .", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-2025-notes.jsonl", "problem_match": "1. ", "solution_match": "## \\(\\S 1.1\\) USAMO 2025/1, proposed by John Berman \n"}}
{"year": "2025", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "USAMO", "problem": "Let \\(n > k \\geq 1\\) be integers. Let \\(P(x) \\in \\mathbb{R}[x]\\) be a polynomial of degree \\(n\\) with no repeated roots and \\(P(0) \\neq 0\\) . Suppose that for any real numbers \\(a_0, \\ldots , a_k\\) such that the polynomial \\(a_k x^k + \\dots + a_1 x + a_0\\) divides \\(P(x)\\) , the product \\(a_0 a_1 \\ldots a_k\\) is zero. Prove that \\(P(x)\\) has a nonreal root.", "solution": "By considering any \\(k + 1\\) of the roots of \\(P\\) , we may as well assume WLOG that \\(n = k + 1\\) . Suppose that \\(P(x) = (x + r_{1})\\ldots (x + r_{n})\\in \\mathbb{R}[x]\\) has \\(P(0)\\neq 0\\) . Then the problem hypothesis is that each of the \\(n\\) polynomials (of degree \\(n - 1\\) ) given by \n\n\\[P_{1}(x) = (x + r_{2})(x + r_{3})(x + r_{4})\\ldots (x + r_{n})\\] \\[P_{2}(x) = (x + r_{1})(x + r_{3})(x + r_{4})\\ldots (x + r_{n})\\] \\[P_{3}(x) = (x + r_{1})(x + r_{2})(x + r_{4})\\ldots (x + r_{n})\\] \\[\\vdots\\] \\[P_{n}(x) = (x + r_{1})(x + r_{2})(x + r_{3})\\ldots (x + r_{n - 1})\\] \n\nhas at least one coefficient equal to zero. (Explicitly, \\(P_{i}(x) = \\frac{P(x)}{x + r_{i}}\\) .) We'll prove that at least one \\(r_{i}\\) is not real. \n\nObviously the leading and constant coefficients of each \\(P_{i}\\) are nonzero, and there are \\(n - 2\\) other coefficients to choose between. So by pigeonhole principle, we may assume, say, that \\(P_{1}\\) and \\(P_{2}\\) share the position of a zero coefficient, say the \\(x^{k}\\) one, for some \\(1\\leq k< n - 1\\) \n\nClaim — If \\(P_{1}\\) and \\(P_{2}\\) both have \\(x^{k}\\) coefficient equal to zero, then the polynomial \n\n\\[Q(x) = (x + r_{3})(x + r_{4})\\ldots (x + r_{n})\\] \n\nhas two consecutive zero coefficients, namely \\(b_{k} = b_{k - 1} = 0\\) \n\nProof. Invoking Vieta formulas, suppose that \n\n\\[Q(x) = x^{n - 2} + b_{n - 3}x^{n - 3} + \\dots +b_{0}.\\] \n\n(And let \\(b_{n - 2} = 1\\) .) Then the fact that the \\(x^{k}\\) coefficient of \\(P_{1}\\) and \\(P_{2}\\) are both zero means \n\n\\[r_{1}b_{k} + b_{k - 1} = r_{2}b_{k} + b_{k - 1} = 0\\] \n\nand hence that \\(b_{k} = b_{k - 1} = 0\\) (since the \\(r_{i}\\) are nonzero). \n\nTo solve the problem, we use: \n\n## Lemma \n\nIf \\(F(x)\\in \\mathbb{R}[x]\\) is a polynomial with two consecutive zero coefficients, it cannot have all distinct real roots.\n\n\n\nHere are two possible proofs of the lemma I know (there are more). \n\nFirst proof using Rolle's theorem. Say \\(x^{t}\\) and \\(x^{t + 1}\\) coefficients of \\(F\\) are both zero. \n\nAssume for contradiction all the roots of \\(F\\) are real and distinct. Then by Rolle's theorem, every higher- order derivative of \\(F\\) should have this property too. However, the \\(t\\) th order derivative of \\(F\\) has a double root of 0, contradiction. \\(\\square\\) \n\nSecond proof using Descartes rule of signs. The number of (nonzero) roots of \\(F\\) is bounded above by the number of sign changes of \\(F(x)\\) (for the positive roots) and the number of sign changes of \\(F(- x)\\) (for the negative roots). Now consider each pair of consecutive nonzero coefficients in \\(F\\) , say \\(x x^{i}\\) and \\(x x^{j}\\) for \\(i > j\\) . \n\n- If \\(i - j = 1\\) , then this sign change will only count for one of \\(F(x)\\) or \\(F(-x)\\) \n\n- If \\(i - j \\geq 2\\) , then the sign change could count towards both \\(F(x)\\) or \\(F(-x)\\) (i.e. counted twice), but also there is at least one zero coefficient between them. \n\nHence if \\(b\\) is the number of nonzero coefficients of \\(F\\) , and \\(z\\) is the number of consecutive runs of zero coefficients of \\(F\\) , then the number of real roots is bounded above by \n\n\\[1\\cdot (b - 1 - z) + 2\\cdot z = b - 1 + z\\leq \\deg F.\\] \n\nHowever, if \\(F\\) has two consecutive zero coefficients, then the inequality is strict. \\(\\square\\) \n\nRemark. The final claim has appeared before apparently in the HUST Team Selection Test for the Vietnamese Math Society's undergraduate olympiad; see https://aops.com/ community/p33893374 for citation.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-2025-notes.jsonl", "problem_match": "2. ", "solution_match": "## \\(\\S 1.2\\) USAMO 2025/2, proposed by Carl Schildkraut \n"}}
-{"year": "2025", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "USAMO", "problem": "Alice the architect and Bob the builder play a game. First, Alice chooses two points \\(P\\) and \\(Q\\) in the plane and a subset \\(S\\) of the plane, which are announced to Bob. Next, Bob marks infinitely many points in the plane, designating each a city. He may not place two cities within distance at most one unit of each other, and no three cities he places may be collinear. Finally, roads are constructed between the cities as follows: each pair \\(A\\) , \\(B\\) of cities is connected with a road along the line segment \\(AB\\) if and only if the following condition holds: \n\nFor every city \\(C\\) distinct from \\(A\\) and \\(B\\) , there exists \\(R \\in S\\) such that \n\n\\(\\triangle PQR\\) is directly similar to either \\(\\triangle ABC\\) or \\(\\triangle BAC\\) . \n\nAlice wins the game if (i) the resulting roads allow for travel between any pair of cities via a finite sequence of roads and (ii) no two roads cross. Otherwise, Bob wins. Determine, with proof, which player has a winning strategy.", "solution": "The answer is that Alice wins. Let's define a Bob- set \\(V\\) to be a set of points in the plane with no three collinear and with all distances at least 1. The point of the problem is to prove the following fact. \n\nClaim — Given a Bob- set \\(V\\subseteq \\mathbb{R}^{2}\\) , consider the Bob- graph with vertex set \\(V\\) defined as follows: draw edge \\(ab\\) if and only if the disk with diameter \\(\\overline{ab}\\) contains no other points of \\(V\\) on or inside it. Then the Bob- graph is (i) connected, and (ii) planar. \n\nProving this claim shows that Alice wins since Alice can specify \\(S\\) to be the set of points outside the disk of diameter \\(PQ\\) . \n\nProof that every Bob- graph is connected. Assume for contradiction the graph is disconnected. Let \\(p\\) and \\(q\\) be two points in different connected components. Since \\(pq\\) is not an edge, there exists a third point \\(r\\) inside the disk with diameter \\(\\overline{pq}\\) . \n\nHence, \\(r\\) is in a different connected component from at least one of \\(p\\) or \\(q\\) — let's say point \\(p\\) . Then we repeat the same argument on the disk with diameter \\(\\overline{pr}\\) to find a new point \\(s\\) , non- adjacent to either \\(p\\) or \\(r\\) . See the figure below, where the \\(X\\) 'ed out dashed edges indicate points which are not only non- adjacent but in different connected components.\n\n\n\n \n\nIn this way we generate an infinite sequence of distances \\(\\delta_{1}\\) , \\(\\delta_{2}\\) , \\(\\delta_{3}\\) , ... among the non- edges in the picture above. By the \"Pythagorean theorem\" (or really the inequality for it), we have \n\n\\[\\delta_{i}^{2}\\leq \\delta_{i - 1}^{2} - 1\\] \n\nand this eventually generates a contradiction for large \\(i\\) , since we get \\(0 \\leq \\delta_{i}^{2} \\leq \\delta_{1}^{2} - (i - 1)\\) . \\(\\square\\) \n\nProof that every Bob- graph is planar. Assume for contradiction edges \\(ac\\) and \\(bd\\) meet, meaning \\(abcd\\) is a convex quadrilateral. WLOG assume \\(\\angle bad \\geq 90^{\\circ}\\) (each quadrilateral has an angle at least \\(90^{\\circ}\\) ). Then the disk with diameter \\(\\overline{bd}\\) contains \\(a\\) , contradiction. \\(\\square\\) \n\nRemark. In real life, the Bob- graph is actually called the Gabriel graph. Note that we never require the Bob- set to be infinite; the solution works unchanged for finite Bob- sets. \n\nHowever, there are approaches that work for finite Bob- sets that don't work for infinite sets, such as the relative neighbor graph, in which one joins \\(a\\) and \\(b\\) iff there is no \\(c\\) such that \\(d(a,b) \\leq \\max \\{d(a,c),d(b,c)\\}\\) . In other words, edges are blocked by triangles where \\(ab\\) is the longest edge (rather than by triangles where \\(ab\\) is the longest edge of a right or obtuse triangle as in the Gabriel graph). \n\nThe relative neighbor graph has fewer edges than the Gabriel graph, so it is planar too. When the Bob- set is finite, the relative distance graph is still connected. The same argument above works where the distances now satisfy \n\n\\[\\delta_{1} > \\delta_{2} > \\ldots\\] \n\ninstead, and since there are finitely many distances one arrives at a contradiction. \n\nHowever for infinite Bob- sets the descending condition is insufficient, and connectedness actually fails altogether. A counterexample (communicated to me by Carl Schildkraut) is to start by taking \\(A_{n} \\approx (2n,0)\\) and \\(B_{n} \\approx (2n + 1,\\sqrt{3})\\) for all \\(n \\geq 1\\) , then perturb all the points slightly so that \n\n\\[B_{1}A_{1} > A_{1}A_{2} > A_{2}B_{1} > B_{1}B_{2} > B_{2}A_{2\\] \\[> A_{2}A_{3} > A_{3}B_{2} > B_{2}B_{3} > B_{3}A_{3\\] \\[> \\dots\\] \n\nA cartoon of the graph is shown below.\n\n\n\n \n\nIn that case, \\(\\{A_{n}\\}\\) and \\(\\{B_{n}\\}\\) will be disconnected from each other: none of the edges \\(A_{n}B_{n}\\) or \\(B_{n}A_{n + 1}\\) are formed. In this case the relative neighbor graph consists of the edges \\(A_{1}A_{2}A_{3}A_{4}\\dots\\) and \\(B_{1}B_{2}B_{3}B_{4}\\dots\\) . That's why for the present problem, the inequality \n\n\\[\\delta_{i}^{2}\\leq \\delta_{i - 1}^{2} - 1\\] \n\nplays such an important role, because it causes the (squared) distances to decrease appreciably enough to give the final contradiction.\n\n\n\n## \\(\\S 2\\) Solutions to Day 2", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-2025-notes.jsonl", "problem_match": "3. ", "solution_match": "## \\(\\S 1.3\\) USAMO 2025/3, proposed by Carl Schildkraut \n"}}
-{"year": "2025", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "USAMO", "problem": "Let \\(H\\) be the orthocenter of an acute triangle \\(ABC\\) , let \\(F\\) be the foot of the altitude from \\(C\\) to \\(AB\\) , and let \\(P\\) be the reflection of \\(H\\) across \\(BC\\) . Suppose that the circumcircle of triangle \\(AFP\\) intersects line \\(BC\\) at two distinct points \\(X\\) and \\(Y\\) . Prove that \\(CX = CY\\) .", "solution": "Let \\(Q\\) be the antipode of \\(B\\) . \n\nClaim — \\(AHQC\\) is a parallelogram, and \\(APCQ\\) is an isosceles trapezoid. \n\nProof. As \\(\\overline{AH} \\perp \\overline{BC} \\perp \\overline{CQ}\\) and \\(\\overline{CF} \\perp \\overline{AB} \\perp \\overline{AQ}\\) . \n\n\n \n\nLet \\(M\\) be the midpoint of \\(\\overline{QC}\\) . \n\nClaim — Point \\(M\\) is the circumcenter of \\(\\triangle AFP\\) . \n\nProof. It's clear that \\(MA = MP\\) from the isosceles trapezoid. As for \\(MA = MF\\) , let \\(N\\) denote the midpoint of \\(\\overline{AF}\\) ; then \\(\\overline{MN}\\) is a midline of the parallelogram, so \\(\\overline{MN} \\perp \\overline{AF}\\) . \\(\\square\\) \n\nSince \\(\\overline{CM} \\perp \\overline{BC}\\) and \\(M\\) is the center of \\((AFP)\\) , it follows \\(CX = CY\\) .", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-2025-notes.jsonl", "problem_match": "4. ", "solution_match": "## \\(\\S 2.1\\) USAMO 2025/4, proposed by Carl Schildkraut \n"}}
+{"year": "2025", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "USAMO", "problem": "Alice the architect and Bob the builder play a game. First, Alice chooses two points \\(P\\) and \\(Q\\) in the plane and a subset \\(S\\) of the plane, which are announced to Bob. Next, Bob marks infinitely many points in the plane, designating each a city. He may not place two cities within distance at most one unit of each other, and no three cities he places may be collinear. Finally, roads are constructed between the cities as follows: each pair \\(A\\) , \\(B\\) of cities is connected with a road along the line segment \\(AB\\) if and only if the following condition holds: \n\nFor every city \\(C\\) distinct from \\(A\\) and \\(B\\) , there exists \\(R \\in S\\) such that \n\n\\(\\triangle PQR\\) is directly similar to either \\(\\triangle ABC\\) or \\(\\triangle BAC\\) . \n\nAlice wins the game if (i) the resulting roads allow for travel between any pair of cities via a finite sequence of roads and (ii) no two roads cross. Otherwise, Bob wins. Determine, with proof, which player has a winning strategy.", "solution": "The answer is that Alice wins. Let's define a Bob- set \\(V\\) to be a set of points in the plane with no three collinear and with all distances at least 1. The point of the problem is to prove the following fact. \n\nClaim — Given a Bob- set \\(V\\subseteq \\mathbb{R}^{2}\\) , consider the Bob- graph with vertex set \\(V\\) defined as follows: draw edge \\(ab\\) if and only if the disk with diameter \\(\\overline{ab}\\) contains no other points of \\(V\\) on or inside it. Then the Bob- graph is (i) connected, and (ii) planar. \n\nProving this claim shows that Alice wins since Alice can specify \\(S\\) to be the set of points outside the disk of diameter \\(PQ\\) . \n\nProof that every Bob- graph is connected. Assume for contradiction the graph is disconnected. Let \\(p\\) and \\(q\\) be two points in different connected components. Since \\(pq\\) is not an edge, there exists a third point \\(r\\) inside the disk with diameter \\(\\overline{pq}\\) . \n\nHence, \\(r\\) is in a different connected component from at least one of \\(p\\) or \\(q\\) — let's say point \\(p\\) . Then we repeat the same argument on the disk with diameter \\(\\overline{pr}\\) to find a new point \\(s\\) , non- adjacent to either \\(p\\) or \\(r\\) . See the figure below, where the \\(X\\) 'ed out dashed edges indicate points which are not only non- adjacent but in different connected components.\n\n\n\n \n\nIn this way we generate an infinite sequence of distances \\(\\delta_{1}\\) , \\(\\delta_{2}\\) , \\(\\delta_{3}\\) , ... among the non- edges in the picture above. By the \"Pythagorean theorem\" (or really the inequality for it), we have \n\n\\[\\delta_{i}^{2}\\leq \\delta_{i - 1}^{2} - 1\\] \n\nand this eventually generates a contradiction for large \\(i\\) , since we get \\(0 \\leq \\delta_{i}^{2} \\leq \\delta_{1}^{2} - (i - 1)\\) . \\(\\square\\) \n\nProof that every Bob- graph is planar. Assume for contradiction edges \\(ac\\) and \\(bd\\) meet, meaning \\(abcd\\) is a convex quadrilateral. WLOG assume \\(\\angle bad \\geq 90^{\\circ}\\) (each quadrilateral has an angle at least \\(90^{\\circ}\\) ). Then the disk with diameter \\(\\overline{bd}\\) contains \\(a\\) , contradiction. \\(\\square\\) \n\nRemark. In real life, the Bob- graph is actually called the Gabriel graph. Note that we never require the Bob- set to be infinite; the solution works unchanged for finite Bob- sets. \n\nHowever, there are approaches that work for finite Bob- sets that don't work for infinite sets, such as the relative neighbor graph, in which one joins \\(a\\) and \\(b\\) iff there is no \\(c\\) such that \\(d(a,b) \\leq \\max \\{d(a,c),d(b,c)\\}\\) . In other words, edges are blocked by triangles where \\(ab\\) is the longest edge (rather than by triangles where \\(ab\\) is the longest edge of a right or obtuse triangle as in the Gabriel graph). \n\nThe relative neighbor graph has fewer edges than the Gabriel graph, so it is planar too. When the Bob- set is finite, the relative distance graph is still connected. The same argument above works where the distances now satisfy \n\n\\[\\delta_{1} > \\delta_{2} > \\ldots\\] \n\ninstead, and since there are finitely many distances one arrives at a contradiction. \n\nHowever for infinite Bob- sets the descending condition is insufficient, and connectedness actually fails altogether. A counterexample (communicated to me by Carl Schildkraut) is to start by taking \\(A_{n} \\approx (2n,0)\\) and \\(B_{n} \\approx (2n + 1,\\sqrt{3})\\) for all \\(n \\geq 1\\) , then perturb all the points slightly so that \n\n\\[B_{1}A_{1} > A_{1}A_{2} > A_{2}B_{1} > B_{1}B_{2} > B_{2}A_{2\\] \\[> A_{2}A_{3} > A_{3}B_{2} > B_{2}B_{3} > B_{3}A_{3\\] \\[> \\dots\\] \n\nA cartoon of the graph is shown below.\n\n\n\n \n\nIn that case, \\(\\{A_{n}\\}\\) and \\(\\{B_{n}\\}\\) will be disconnected from each other: none of the edges \\(A_{n}B_{n}\\) or \\(B_{n}A_{n + 1}\\) are formed. In this case the relative neighbor graph consists of the edges \\(A_{1}A_{2}A_{3}A_{4}\\dots\\) and \\(B_{1}B_{2}B_{3}B_{4}\\dots\\) . That's why for the present problem, the inequality \n\n\\[\\delta_{i}^{2}\\leq \\delta_{i - 1}^{2} - 1\\] \n\nplays such an important role, because it causes the (squared) distances to decrease appreciably enough to give the final contradiction.\n\n\n\n## \\(\\S 2\\) Solutions to Day 2", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-2025-notes.jsonl", "problem_match": "3. ", "solution_match": "## \\(\\S 1.3\\) USAMO 2025/3, proposed by Carl Schildkraut \n"}}
+{"year": "2025", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "USAMO", "problem": "Let \\(H\\) be the orthocenter of an acute triangle \\(ABC\\) , let \\(F\\) be the foot of the altitude from \\(C\\) to \\(AB\\) , and let \\(P\\) be the reflection of \\(H\\) across \\(BC\\) . Suppose that the circumcircle of triangle \\(AFP\\) intersects line \\(BC\\) at two distinct points \\(X\\) and \\(Y\\) . Prove that \\(CX = CY\\) .", "solution": "Let \\(Q\\) be the antipode of \\(B\\) . \n\nClaim — \\(AHQC\\) is a parallelogram, and \\(APCQ\\) is an isosceles trapezoid. \n\nProof. As \\(\\overline{AH} \\perp \\overline{BC} \\perp \\overline{CQ}\\) and \\(\\overline{CF} \\perp \\overline{AB} \\perp \\overline{AQ}\\) . \n\n\n \n\nLet \\(M\\) be the midpoint of \\(\\overline{QC}\\) . \n\nClaim — Point \\(M\\) is the circumcenter of \\(\\triangle AFP\\) . \n\nProof. It's clear that \\(MA = MP\\) from the isosceles trapezoid. As for \\(MA = MF\\) , let \\(N\\) denote the midpoint of \\(\\overline{AF}\\) ; then \\(\\overline{MN}\\) is a midline of the parallelogram, so \\(\\overline{MN} \\perp \\overline{AF}\\) . \\(\\square\\) \n\nSince \\(\\overline{CM} \\perp \\overline{BC}\\) and \\(M\\) is the center of \\((AFP)\\) , it follows \\(CX = CY\\) .", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-2025-notes.jsonl", "problem_match": "4. ", "solution_match": "## \\(\\S 2.1\\) USAMO 2025/4, proposed by Carl Schildkraut \n"}}
{"year": "2025", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "USAMO", "problem": "Find all positive integers \\(k\\) such that: for every positive integer \\(n\\) , the sum \n\n\\[\\binom{n}{0}^{k} + \\binom{n}{1}^{k} + \\dots +\\binom{n}{n}^{k}\\] \n\nis divisible by \\(n + 1\\) .", "solution": "The answer is all even \\(k\\) . \n\nLet's abbreviate \\(S(n) := \\binom{n}{0}^{k} + \\dots + \\binom{n}{n}^{k}\\) for the sum in the problem. \n\n\\(\\P\\) Proof that even \\(k\\) is necessary. Choose \\(n = 2\\) . We need \\(3 \\mid S(2) = 2 + 2^{k}\\) , which requires \\(k\\) to be even. \n\nRemark. It's actually not much more difficult to just use \\(n = p - 1\\) for prime \\(p\\) , since \\(\\binom{p- 1}{i} \\equiv (- 1)^{i} \\pmod{p}\\) . Hence \\(S(p- 1) \\equiv 1 + (- 1)^{k} + 1 + (- 1)^{k} + \\dots + 1 \\pmod{p}\\) , and this also requires \\(k\\) to be even. This special case is instructive in figuring out the proof to follow. \n\n\\(\\P\\) Proof that \\(k\\) is sufficient. From now on we treat \\(k\\) as fixed, and we let \\(p^{e}\\) be a prime fully dividing \\(n + 1\\) . The basic idea is to reduce from \\(n + 1\\) to \\((n + 1) / p\\) by an induction. \n\nRemark. Here is a concrete illustration that makes it clear what's going on. Let \\(p = 5\\) . When \\(n = p - 1 = 4\\) , we have \n\n\\[S(4) = 1^{k} + 4^{k} + 6^{k} + 4^{k} + 1^{k} \\equiv 1 + 1 + 1 + 1 + 1 \\equiv 0 \\pmod{5}.\\] \n\nWhen \\(n = p^{2} - 1 = 24\\) , the 25 terms of \\(S(24)\\) in order are, modulo 25, \n\n\\[S(24)\\equiv 1^{k}+1^{k}+1^{k}+1^{k}+1^{k}\\] \\[\\qquad+4^{k}+4^{k}+4^{k}+4^{k}+4^{k}\\] \\[\\qquad+6^{k}+6^{k}+6^{k}+6^{k}+6^{k}\\] \\[\\qquad+4^{k}+4^{k}+4^{k}+4^{k}+4^{k}\\] \\[\\qquad+1^{k}+1^{k}+1^{k}+1^{k}+1^{k}\\] \\[\\qquad=5(1^{k}+4^{k}+6^{k}+4^{k}+1^{k}).\\] \n\nThe point is that \\(S(24)\\) has five copies of \\(S(4)\\) , modulo 25. \n\nTo make the pattern in the remark explicit, we prove the following lemma on each individual binomial coefficient.\n\n\n\n## Lemma 2.1 \n\nSuppose \\(p^{e}\\) is a prime power which fully divides \\(n + 1\\) . Then \n\n\\[\\binom{n}{i}\\equiv\\pm\\binom{\\frac{n+1}{p}-1}{\\lfloor i/p\\rfloor}\\pmod{p^{e}}.\\] \n\nProof of lemma. It's easiest to understand the proof by looking at the cases \\(\\lfloor i / p\\rfloor \\in\\) \\(\\{0,1,2\\}\\) first. \n\nFor \\(0\\leq i< p\\) , since \\(n\\equiv - 1\\) mod \\(p^{e}\\) , we have \n\n\\[\\binom{n}{i}=\\frac{n(n-1)\\ldots(n-i+1)}{1\\cdot2\\cdot\\cdots\\cdot i}\\equiv\\frac{(-1)(-2)\\ldots(-i)}{1\\cdot2\\cdot\\cdots\\cdot i}\\equiv\\pm 1\\pmod{p^{e}}.\\] \n\nFor \\(p\\leq i< 2p\\) we have \n\n\\[\\binom{n}{i}\\equiv\\pm 1\\cdot\\frac{n-p+1}{p}\\cdot\\frac{(n-p)(n-p-1)\\ldots(n-i+1)}{(p+1)(p+2)\\ldots i}\\] \\[\\qquad\\equiv\\pm 1\\cdot\\frac{\\frac{n-p+1}{p}\\cdot\\pm 1}{1}\\] \\[\\qquad\\equiv\\pm\\binom{\\frac{n+1}{p}-1}{1}\\pmod{p^{e}}.\\] \n\nFor \\(2p\\leq i< 3p\\) the analogous reasoning gives \n\n\\[\\binom{n}{i}\\equiv\\pm 1\\cdot\\frac{n-p+1}{p}\\cdot\\pm 1\\cdot\\frac{n-2p+1}{2p}\\cdot\\pm 1\\] \\[\\qquad\\equiv\\pm\\frac{\\binom{n+1}{p}-1}{1\\cdot2}\\binom{\\frac{n+1}{p}-2}{2}\\] \\[\\qquad\\equiv\\pm\\binom{\\frac{n+1}{p}-1}{2}\\pmod{p^{e}}.\\] \n\nAnd so on. The point is that in general, if we write \n\n\\[\\binom{n}{i}=\\prod_{1\\leq j\\leq i}\\frac{n-(j-1)}{j}\\] \n\nthen the fractions for \\(p\\nmid j\\) are all \\(\\pm 1\\) (mod \\(p^{e}\\) ). So only considers those \\(j\\) with \\(p\\mid j\\) ; in that case one obtains the claimed \\(\\binom{\\frac{n+1}{p}-1}{\\lfloor i/p\\rfloor}\\) exactly (even without having to take modulo \\(p^{e}\\) ). \\(\\square\\) \n\nFrom the lemma, it follows if \\(p^{e}\\) is a prime power which fully divides \\(n + 1\\) , then \n\n\\[S(n)\\equiv p\\cdot S\\left(\\frac{n + 1}{p} -1\\right)\\pmod{p^{e}}\\] \n\nby grouping the \\(n + 1\\) terms (for \\(0\\leq i\\leq n\\) ) into consecutive ranges of length \\(p\\) (by the value of \\(\\lfloor i / p\\rfloor\\) ).\n\n\n\nRemark. Actually, with the exact same proof (with better \\(\\pm\\) bookkeeping) one may show that \n\n\\[n + 1\\mid \\sum_{i = 0}^{n}\\left((-1)^{i}\\binom{n}{i}\\right)^{k}\\] \n\nholds for all nonnegative integers \\(k\\) , not just \\(k\\) even. So in some sense this result is more natural than the one in the problem statement.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-2025-notes.jsonl", "problem_match": "5. ", "solution_match": "## \\(\\S 2.2\\) USAMO 2025/5, proposed by John Berman \n"}}
-{"year": "2025", "tier": "T1", "problem_label": "6", "problem_type": null, "exam": "USAMO", "problem": "Let \\(m\\) and \\(n\\) be positive integers with \\(m \\geq n\\) . There are \\(m\\) cupcakes of different flavors arranged around a circle and \\(n\\) people who like cupcakes. Each person assigns a nonnegative real number score to each cupcake, depending on how much they like the cupcake. Suppose that for each person \\(P\\) , it is possible to partition the circle of \\(m\\) cupcakes into \\(n\\) groups of consecutive cupcakes so that the sum of \\(P\\) 's scores of the cupcakes in each group is at least 1. Prove that it is possible to distribute the \\(m\\) cupcakes to the \\(n\\) people so that each person \\(P\\) receives cupcakes of total score at least 1 with respect to \\(P\\) .", "solution": "Arbitrarily pick any one person — call her Pip — and her \\(n\\) arcs. The initial idea is to try to apply Hall's marriage lemma to match the \\(n\\) people with Pip's arcs (such that each such person is happy with their matched arc). To that end, construct the obvious bipartite graph \\(\\mathfrak{G}\\) between the people and the arcs for Pip. \n\nWe now consider the following algorithm, which takes several steps. \n\n- If a perfect matching of \\(\\mathfrak{G}\\) exists, we're done!- We're probably not that lucky. Per Hall's condition, this means there is a bad set \\(B_{1}\\) of people, who are compatible with fewer than \\(|B_{1}|\\) of the arcs. Then delete \\(B_{1}\\) and the neighbors of \\(B_{1}\\) , then try to find a matching on the remaining graph.- If a matching exists now, terminate the algorithm. Otherwise, that means there's another bad set \\(B_{2}\\) for the remaining graph. We again delete \\(B_{2}\\) and the fewer than \\(B_{2}\\) neighbors.- Repeat until some perfect matching \\(\\mathcal{M}\\) is possible in the remaining graph, i.e. there are no more bad sets (and then terminate once that occurs). Since Pip is a universal vertex, it's impossible to delete Pip, so the algorithm does indeed terminate with nonempty \\(\\mathcal{M}\\) . \n\nA cartoon of this picture is shown below.\n\n\n\n \n\nWe commit to assigning each of person in \\(\\mathcal{M}\\) their matched arc (in particular if there are no bad sets at all, the problem is already solved). Now we finish the problem by induction on \\(n\\) (for the remaining people) by simply deleting the arcs used up by \\(\\mathcal{M}\\) . \n\nTo see why this deletion- induction works, consider any particular person Quinn not in \\(\\mathcal{M}\\) . By definition, Quinn is not happy with any of the arcs in \\(\\mathcal{M}\\) . So when an arc \\(\\mathcal{A}\\) of \\(\\mathcal{M}\\) is deleted, it had value less than 1 for Quinn so in particular it couldn't contain entirely any of Quinn's arcs. Hence at most one endpoint among Quinn's arcs was in the deleted arc \\(\\mathcal{A}\\) . When this happens, this causes two arcs of Quinn to merge, and the merged value is \n\n\\[(\\geq 1) + (\\geq 1) - (\\leq 1)\\qquad \\geq \\qquad 1\\] \n\nmeaning the induction is OK. See below for a cartoon of the deletion, where Pip's arcs are drawn in blue while Quinn's arcs and scores are drawn in red (in this example \\(n = 3\\) ). \n\n\n \n\nRemark. This deletion argument can be thought of in some special cases even before the realization of Hall, in the case where \\(\\mathcal{M}\\) has only one person (Pip). This amounts to saying that if one of Pip's arcs isn't liked by anybody, then that arc can be deleted and the induction carries through.\n\n\n\nRemark. Conversely, it should be reasonable to expect Hall's theorem to be helpful even before finding the deletion argument. While working on this problem, one of the first things I said was: \n\n\"We should let Hall do the heavy lifting for us: find a way to make \\(n\\) groups that satisfy Hall's condition, rather than an assignment of \\(n\\) groups to \\(n\\) people.\" \n\nAs a general heuristic, for any type of \"compatible matching\" problem, Hall's condition is usually the go- to tool. (It is much easier to verify Hall's condition than actually find the matching yourself.) Actually in most competition problems, if one realizes one is in a Hall setting, one is usually close to finishing the problem. This is a relatively rare example in which one needs an additional idea to go alongside Hall's theorem.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-2025-notes.jsonl", "problem_match": "6. ", "solution_match": "## \\(\\S 2.3\\) USAMO 2025/6, proposed by Cheng-Yin Chang and Hung-Hsun Yu \n"}}
+{"year": "2025", "tier": "T1", "problem_label": "6", "problem_type": null, "exam": "USAMO", "problem": "Let \\(m\\) and \\(n\\) be positive integers with \\(m \\geq n\\) . There are \\(m\\) cupcakes of different flavors arranged around a circle and \\(n\\) people who like cupcakes. Each person assigns a nonnegative real number score to each cupcake, depending on how much they like the cupcake. Suppose that for each person \\(P\\) , it is possible to partition the circle of \\(m\\) cupcakes into \\(n\\) groups of consecutive cupcakes so that the sum of \\(P\\) 's scores of the cupcakes in each group is at least 1. Prove that it is possible to distribute the \\(m\\) cupcakes to the \\(n\\) people so that each person \\(P\\) receives cupcakes of total score at least 1 with respect to \\(P\\) .", "solution": "Arbitrarily pick any one person — call her Pip — and her \\(n\\) arcs. The initial idea is to try to apply Hall's marriage lemma to match the \\(n\\) people with Pip's arcs (such that each such person is happy with their matched arc). To that end, construct the obvious bipartite graph \\(\\mathfrak{G}\\) between the people and the arcs for Pip. \n\nWe now consider the following algorithm, which takes several steps. \n\n- If a perfect matching of \\(\\mathfrak{G}\\) exists, we're done!- We're probably not that lucky. Per Hall's condition, this means there is a bad set \\(B_{1}\\) of people, who are compatible with fewer than \\(|B_{1}|\\) of the arcs. Then delete \\(B_{1}\\) and the neighbors of \\(B_{1}\\) , then try to find a matching on the remaining graph.- If a matching exists now, terminate the algorithm. Otherwise, that means there's another bad set \\(B_{2}\\) for the remaining graph. We again delete \\(B_{2}\\) and the fewer than \\(B_{2}\\) neighbors.- Repeat until some perfect matching \\(\\mathcal{M}\\) is possible in the remaining graph, i.e. there are no more bad sets (and then terminate once that occurs). Since Pip is a universal vertex, it's impossible to delete Pip, so the algorithm does indeed terminate with nonempty \\(\\mathcal{M}\\) . \n\nA cartoon of this picture is shown below.\n\n\n\n \n\nWe commit to assigning each of person in \\(\\mathcal{M}\\) their matched arc (in particular if there are no bad sets at all, the problem is already solved). Now we finish the problem by induction on \\(n\\) (for the remaining people) by simply deleting the arcs used up by \\(\\mathcal{M}\\) . \n\nTo see why this deletion- induction works, consider any particular person Quinn not in \\(\\mathcal{M}\\) . By definition, Quinn is not happy with any of the arcs in \\(\\mathcal{M}\\) . So when an arc \\(\\mathcal{A}\\) of \\(\\mathcal{M}\\) is deleted, it had value less than 1 for Quinn so in particular it couldn't contain entirely any of Quinn's arcs. Hence at most one endpoint among Quinn's arcs was in the deleted arc \\(\\mathcal{A}\\) . When this happens, this causes two arcs of Quinn to merge, and the merged value is \n\n\\[(\\geq 1) + (\\geq 1) - (\\leq 1)\\qquad \\geq \\qquad 1\\] \n\nmeaning the induction is OK. See below for a cartoon of the deletion, where Pip's arcs are drawn in blue while Quinn's arcs and scores are drawn in red (in this example \\(n = 3\\) ). \n\n\n \n\nRemark. This deletion argument can be thought of in some special cases even before the realization of Hall, in the case where \\(\\mathcal{M}\\) has only one person (Pip). This amounts to saying that if one of Pip's arcs isn't liked by anybody, then that arc can be deleted and the induction carries through.\n\n\n\nRemark. Conversely, it should be reasonable to expect Hall's theorem to be helpful even before finding the deletion argument. While working on this problem, one of the first things I said was: \n\n\"We should let Hall do the heavy lifting for us: find a way to make \\(n\\) groups that satisfy Hall's condition, rather than an assignment of \\(n\\) groups to \\(n\\) people.\" \n\nAs a general heuristic, for any type of \"compatible matching\" problem, Hall's condition is usually the go- to tool. (It is much easier to verify Hall's condition than actually find the matching yourself.) Actually in most competition problems, if one realizes one is in a Hall setting, one is usually close to finishing the problem. This is a relatively rare example in which one needs an additional idea to go alongside Hall's theorem.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-2025-notes.jsonl", "problem_match": "6. ", "solution_match": "## \\(\\S 2.3\\) USAMO 2025/6, proposed by Cheng-Yin Chang and Hung-Hsun Yu \n"}}
diff --git a/USA_TST/md/en-sols-TST-IMO-2024.md b/USA_TST/md/en-sols-TST-IMO-2024.md
index fcc16fc08fa58b363e1ff15a5b5cfa992d74171c..4d24ab5895c7e60aaa137014a87f6c47ad282426 100644
--- a/USA_TST/md/en-sols-TST-IMO-2024.md
+++ b/USA_TST/md/en-sols-TST-IMO-2024.md
@@ -144,7 +144,7 @@ We show several approaches.
First solution, by author.
-
+
Claim — We have \(B P = B Q\) .
@@ -199,7 +199,7 @@ First note that \(\angle DFE = 45^{\circ} - \angle A / 4\) , so it suffices to s
Now \(\angle BUV = \angle AEU = 90^{\circ} - \angle A / 2 = \angle BVP\) , so \(\overline{PUV}\) are collinear.
-
+
From the tangency condition, we have that \(\angle AQB = 180^{\circ} - \angle ABI\) , which implies that
@@ -214,7 +214,7 @@ This forces \(\overline{PQ} \parallel \overline{DF}\) , as desired.
\(\P\) Third solution by Pitchayut Saengrungkongka and Maxim Li. We provide yet another proof that \(BP = BQ\) .
-
+
Let the incircle be \(\omega\) and touch \(BC\) and \(AB\) at point \(U\) and \(W\) . Let the tangent to \(\omega\) at \(D\) meet \(UW\) at \(T\) . Notice that \(T\) is the pole of \(BD\) with respect to \(\omega\) , so \(IT \perp BD\) . Now, we make the following critical claim, which in particular implies \(BP = BQ\) .
@@ -342,7 +342,7 @@ To isolate the geometry component of the problem, we rewrite the claim in the fo
Suppose \(A\) and \(B\) are two points such that \(\max (P_{1}A, P_{1}B, P_{2}A, P_{2}B) \leq 1\) , and moreover \(A\) and \(B\) lie on the same side of line \(P_{1}P_{2}\) . Further assume no three of \(\{P_{1}, P_{2}, A, B\}\) are collinear. Then \(AB < 1\) .
-
+
Proof of lemma. Suppose for the sake of contradiction that \(A\) and \(B\) lie on the same side of \(P_{1}P_{2}\) . The convex hull of these four points is either a quadrilateral or a triangle.
diff --git a/USA_TST/md/en-sols-TST-IMO-2025.md b/USA_TST/md/en-sols-TST-IMO-2025.md
index dd0bc63109c72a6032dc027e2c341c763ec0b847..1c6eeff5b7384f8fcbd8329e0a706068e6fcfd74 100644
--- a/USA_TST/md/en-sols-TST-IMO-2025.md
+++ b/USA_TST/md/en-sols-TST-IMO-2025.md
@@ -184,7 +184,7 @@ Proof. This is a standard circumcenter fact in disguise but we will prove it her
The second equality is analogous.
-
+
## Lemma
@@ -199,7 +199,7 @@ Therefore \(P P_{i}^{n} P_{i + 1}^{n} \stackrel{\perp}{\sim} P P_{i}^{0} P_{i +
To solve part (a), we just need to show that \(\mathcal{P}_{0}\) and \(\mathcal{Q}_{0}\) are cyclic. This is where we need the isogonal conjugate property, and we can generalize from analogous facts for the \(n = 3\) case.
-
+
@@ -256,7 +256,7 @@ Let \(A B C\) be a triangle, and let \(X\) , \(Y\) , and \(Z\) be collinear poin
## \(\P\) Solution 1 (Pitchayut Saengrungkonga)
-
+
Let \(S\) denote the circumcenter of \(\triangle X^{\prime}Y^{\prime}Z^{\prime}\) . Observe that \(A Y = A Z = A Y^{\prime} = A Z^{\prime}\) , so \(Y Z Y^{\prime}Z^{\prime}\) is cyclic and \(A S\perp Y^{\prime}Z^{\prime}\) . Similarly, \(B S\perp Z^{\prime}X^{\prime}\) and \(C S\perp X^{\prime}Y^{\prime}\) .
@@ -283,7 +283,7 @@ Let \(A_{0}X^{\prime}\) meet \((A B C)\) again at \(P\) . Then
Therefore \(P\) lies on \(B_{0}Y^{\prime}\) , and likewise \(C_{0}Z^{\prime}\) by symmetry.
-
+
Let \(X_{0}\) , \(Y_{0}\) , and \(Z_{0}\) be the reflections of \(P\) across \(B C\) , \(C A\) , and \(A B\) respectively. By reflection, \(X_{0}\) lies on \(A^{\prime}X\) and \(P X^{\prime} = X_{0}X\) . We also know that if \(H\) is the orthocenter of \(A B C\) , then \(X_{0}Y_{0}Z_{0}H\) is the Steiner line of \(P\) . Let \(D\) be the reflection of \(H\) across \(B C\) , which is also the antipode of \(A_{0}\) on \((A B C)\) . Then \(\angle A^{\prime}X_{0}H = \angle A_{0}P D = 90^{\circ}\) , so \(X_{0}Y_{0}Z_{0}H\) is perpendicular to \(A^{\prime}X_{0}X\) and parallel to \(X Y Z\) .
diff --git a/USA_TST/segmented/en-sols-TST-IMO-2024.jsonl b/USA_TST/segmented/en-sols-TST-IMO-2024.jsonl
index d98e13ca340f775fea5eec334a07761bd5faf5ce..da159e4e7e0ad1850a899e805326df1fa1fd1507 100644
--- a/USA_TST/segmented/en-sols-TST-IMO-2024.jsonl
+++ b/USA_TST/segmented/en-sols-TST-IMO-2024.jsonl
@@ -1,6 +1,6 @@
{"year": "2024", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "USA_TST", "problem": "Find the smallest constant \\(C > 1\\) such that the following statement holds: for every integer \\(n\\geq 2\\) and sequence of non-integer positive real numbers \\(a_{1}\\) , \\(a_{2}\\) , ..., \\(a_{n}\\) satisfying \n\n\\[\\frac{1}{a_{1}} +\\frac{1}{a_{2}} +\\dots +\\frac{1}{a_{n}} = 1,\\] \n\nit's possible to choose positive integers \\(b_{i}\\) such that \n\n(i) for each \\(i = 1,2,\\ldots ,n\\) , either \\(b_{i} = \\lfloor a_{i}\\rfloor\\) or \\(b_{i} = \\lfloor a_{i}\\rfloor +1\\) ; and \n\n(ii) we have \n\n\\[1< \\frac{1}{b_{1}} +\\frac{1}{b_{2}} +\\dots +\\frac{1}{b_{n}}\\leq C.\\]", "solution": "Answer. The answer is \\(C = \\frac{3}{2}\\) . \n\nLower bound. Note that if \\(a_{1} = \\frac{4n - 3}{2n - 1}\\) and \\(a_{i} = \\frac{4n - 3}{2n - 1}\\) for \\(i > 1\\) , then we must have \\(b_{1} \\in \\{1,2\\}\\) and \\(b_{i} \\in \\{2n - 2,2n - 1\\}\\) for \\(i > 1\\) . If we take \\(b_{1} = 2\\) then we obtain \n\n\\[\\frac{1}{b_{1}} +\\frac{1}{b_{2}} +\\dots +\\frac{1}{b_{n}}\\leq \\frac{1}{2} +(n - 1)\\cdot \\frac{1}{2n - 2} = 1,\\] \n\nwhereas if we take \\(b_{1} = 1\\) then we obtain \n\n\\[\\frac{1}{b_{1}} +\\frac{1}{b_{2}} +\\dots +\\frac{1}{b_{n}}\\geq 1 + (n - 1)\\cdot \\frac{1}{2n - 1} = \\frac{3n - 2}{2n - 1}.\\] \n\nThis shows that \\(C \\geq \\frac{3n - 2}{2n - 1}\\) , and as \\(n \\to \\infty\\) this shows that \\(C \\geq \\frac{3}{2}\\) . \n\nUpper bound. For \\(0 \\leq k \\leq n\\) , define \n\n\\[c_{i} = \\sum_{i = 1}^{k}\\frac{1}{\\lfloor a_{i}\\rfloor} +\\sum_{i = k + 1}^{n}\\frac{1}{\\lfloor a_{i}\\rfloor + 1}.\\] \n\nNote that \\(c_{0} < c_{1} < \\dots < c_{n}\\) and \n\n\\[c_{0} < \\frac{1}{a_{1}} +\\frac{1}{a_{2}} +\\dots +\\frac{1}{a_{n}} = 1 < c_{n}.\\] \n\nThis means there exists a unique value of \\(k\\) for which \\(c_{k - 1} < 1 < c_{k}\\) . For this \\(k\\) we have \n\n\\[1< c_{k} = c_{k - 1} + \\frac{1}{(\\lfloor a_{k}\\rfloor)(\\lfloor a_{k}\\rfloor + 1)} < 1 + \\frac{1}{1\\cdot 2} = \\frac{3}{2}.\\] \n\nTherefore we may choose \\(b_{i} = \\lfloor a_{i}\\rfloor\\) for \\(i \\leq k\\) and \\(b_{i} = \\lfloor a_{i}\\rfloor +1\\) for \\(i > k\\) .\n\n\n\nRemark. The solution can be phrased in the following “motion- based” way. Imagine starting with all floors (corresponding to \\(c_{0}\\) ), then changing each floor to a ceiling one by one until after \\(n\\) steps every floor is a ceiling (arriving at \\(c_{n}\\) ). As we saw, \\(c_{0} < 1 < c_{n}\\) , but \\(c_{0} < \\dots < c_{n}\\) . Moreover, each discrete step increases the sum by at most \n\n\\[\\frac{1}{\\lfloor a_{i}\\rfloor} -\\frac{1}{\\lfloor a_{i}\\rfloor + 1}\\leq \\frac{1}{2}\\] \n\nand so the changing sum must be in the interval \\([1,3 / 2]\\) at some point. \n\n\\(\\P\\) Upper bound (alternate). First suppose \\(a_{i}< 2\\) for some \\(i\\) . Assume without loss of generality \\(i = 1\\) here. Let \\(b_{1} = 1\\) and \\(b_{i} = \\lfloor a_{i}\\rfloor +1\\) for all other \\(i\\) . Then \n\n\\[1< \\frac{1}{b_{1}} +\\dots +\\frac{1}{b_{n}} = 1 + \\frac{1}{\\lfloor a_{2}\\rfloor +1} +\\dots +\\frac{1}{\\lfloor a_{n}\\rfloor +1}\\] \\[< \\left(\\frac{1}{2} +\\frac{1}{a_{1}}\\right) + \\frac{1}{a_{2}} +\\dots +\\frac{1}{a_{n}} = \\frac{3}{2}.\\] \n\nNow suppose \\(a_{i} > 2\\) always. Then \\(\\frac{a_{i}}{\\lfloor a_{i}\\rfloor} < \\frac{3}{2}\\) , so \n\n\\[1 = \\frac{1}{a_{1}} +\\dots +\\frac{1}{a_{n}} < \\frac{1}{\\lfloor a_{1}\\rfloor} +\\dots +\\frac{1}{\\lfloor a_{n}\\rfloor} < \\frac{3}{2}\\left(\\frac{1}{a_{1}} +\\dots +\\frac{1}{a_{n}}\\right) = \\frac{3}{2}.\\] \n\nTherefore we may let \\(b_{i} = \\lfloor a_{i}\\rfloor\\) for all \\(i\\) . \n\nRemark. The original proposal asked to find the optimal \\(C\\) for a fixed \\(n\\) . The answer is \\(\\frac{3n - 2}{2n - 1}\\) , i.e. the lower bound construction in the solution is optimal.", "metadata": {"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2024.jsonl", "problem_match": "1. ", "solution_match": "## \\(\\S 1.1\\) USA TST 2024/1, proposed by Merlijn Staps \n"}}
-{"year": "2024", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "USA_TST", "problem": "Let \\(A B C\\) be a triangle with incenter \\(I\\) . Let segment \\(A I\\) intersect the incircle of triangle \\(A B C\\) at point \\(D\\) . Suppose that line \\(B D\\) is perpendicular to line \\(A C\\) . Let \\(P\\) be a point such that \\(\\angle B P A = \\angle P A I = 90^{\\circ}\\) . Point \\(Q\\) lies on segment \\(B D\\) such that the circumcircle of triangle \\(A B Q\\) is tangent to line \\(B I\\) . Point \\(X\\) lies on line \\(P Q\\) such that \\(\\angle I A X = \\angle X A C\\) . Prove that \\(\\angle A X P = 45^{\\circ}\\) .", "solution": "We show several approaches. \n\nFirst solution, by author. \n\n\n \n\nClaim — We have \\(B P = B Q\\) . \n\nProof. For readability, we split the proof into three unconditional parts. \n\n- We translate the condition \\(\\overline{B D} \\perp \\overline{A C}\\) . It gives \\(\\angle D B A = 90^{\\circ} - A\\) , so that \n\n\\[\\angle D B I = \\left|\\frac{B}{2} -(90^{\\circ} - A)\\right| = \\frac{|A - C|}{2}\\] \\[\\angle B D I = \\angle D B A + \\angle B A D = (90^{\\circ} - A) + \\frac{A}{2} = 90^{\\circ} - \\frac{A}{2}.\\]\n\n\n\nHence, letting \\(r\\) denote the inradius, we can translate \\(\\overline{BD} \\perp \\overline{AC}\\) into the following trig condition: \n\n\\[\\sin \\frac{B}{2} = \\frac{r}{BI} = \\frac{DI}{BI} = \\frac{\\sin \\angle DBI}{\\sin \\angle BDI} = \\frac{\\sin \\frac{|A - C|}{2}}{\\sin \\left(90^{\\circ} - \\frac{A}{2}\\right)}.\\] \n\n- The length of \\(BP\\) is given from right triangle \\(APB\\) as \n\n\\[BP = BA \\cdot \\sin \\angle PAB = BA \\cdot \\sin \\left(90^{\\circ} - \\frac{A}{2}\\right).\\] \n\n- The length of \\(BQ\\) is given from the law of sines on triangle \\(ABQ\\) . The tangency gives \\(\\angle BAQ = \\angle DBI\\) and \\(\\angle BQA = 180^{\\circ} - \\angle ABI = 180^{\\circ} - \\angle IBE\\) and thus \n\n\\[BQ = BA \\cdot \\frac{\\sin \\angle BAQ}{\\sin \\angle AQB} = BA \\cdot \\frac{\\sin \\angle DBI}{\\sin \\angle ABI} = BA \\cdot \\frac{\\sin \\frac{|A - C|}{2}}{\\sin \\frac{B}{2}}.\\] \n\nThe first bullet implies the expressions in the second and third bullet for \\(BP\\) and \\(BQ\\) are equal, as needed. \\(\\square\\) \n\nRemark. In the above proof, one dos not actually need to compute \\(\\angle DBI = \\frac{|A - C|}{2}\\) . The proof works equally leaving that expression intact as \\(\\sin \\angle DBI\\) in place of \\(\\sin \\frac{|A - C|}{2}\\) . \n\nNow we can finish by angle chasing. We have \n\n\\[\\angle PBQ = \\angle PBA + \\angle ABD = \\frac{A}{2} +90^{\\circ} - A = 90^{\\circ} - \\frac{A}{2}.\\] \n\nThen \n\n\\[\\angle BPQ = \\frac{180^{\\circ} - \\angle PBQ}{2} = 45^{\\circ} + \\frac{A}{4},\\] \n\nso \\(\\angle APQ = 90^{\\circ} - \\angle BPQ = 45^{\\circ} - \\frac{A}{4}\\) . Also, if we let \\(J\\) be the incenter of \\(IAC\\) , then \\(\\angle PAJ = 90^{\\circ} + \\frac{A}{4}\\) , and clearly \\(X\\) lies on line \\(AJ\\) . Then \\(\\angle APQ + \\angle PAJ = 135^{\\circ} < 180^{\\circ}\\) , so \\(X\\) lies on the same side of \\(AP\\) as \\(Q\\) and \\(J\\) (by the parallel postulate). Therefore \\(\\angle AXP = 180^{\\circ} - 135^{\\circ} = 45^{\\circ}\\) , as desired. \n\nRemark. The problem was basically written backwards by starting from the \\(BD \\perp AC\\) condition, turning that into trig, and then contriving \\(P\\) and \\(Q\\) so that the \\(BD \\perp AC\\) condition implied \\(BP = BQ\\) . \n\nSecond solution, by Jeffrey Kwan. We prove the following restatement: \n\nConsider isosceles triangle \\(AEF\\) with \\(AE = AF\\) and incenter \\(D\\) . Let \\(B\\) be the point on ray \\(AE\\) such that \\(BD \\perp AF\\) , and let \\(P\\) be the projection of \\(B\\) onto the line through \\(A\\) parallel to \\(EF\\) . Let \\(I\\) be the point diametrically opposite \\(A\\) in the circumcircle of \\(AEF\\) , and let \\(Q\\) be the point on line \\(BD\\) such that \\(BI\\) is tangent to the circumcircle of \\(AQB\\) . Then \\(\\angle APQ = 45^{\\circ} - \\angle A / 4\\) . \n\nFirst note that \\(\\angle DFE = 45^{\\circ} - \\angle A / 4\\) , so it suffices to show that \\(\\overline{PQ} \\parallel \\overline{DF}\\) . Let \\(U = \\overline{BD} \\cap \\overline{EF}\\) , and let \\(V = BI \\cap (AEF)\\) . Observe that:\n\n\n\n- \\(P\\) and \\(V\\) both lie on the circle with diameter \\(AB\\) , so \\(\\angle BVP = \\angle PAB = 90^{\\circ} - \\angle A / 2\\) . \n\n- We have \\(\\angle EVB = \\angle EVI = \\angle A / 2 = \\angle DUF = \\angle BUE\\) . Hence \\(BEUV\\) is cyclic. \n\nNow \\(\\angle BUV = \\angle AEU = 90^{\\circ} - \\angle A / 2 = \\angle BVP\\) , so \\(\\overline{PUV}\\) are collinear. \n\n\n \n\nFrom the tangency condition, we have that \\(\\angle AQB = 180^{\\circ} - \\angle ABI\\) , which implies that \n\n\\[\\angle AQU + \\angle APU = \\angle AQB + \\angle APV = (180^{\\circ} - \\angle ABI) + \\angle ABI = 180^{\\circ},\\] \n\nand so \\(APUQ\\) is cyclic. Finally, note that \\(D\\) is the orthocenter of \\(\\triangle AUVF\\) , which implies that \n\n\\[\\angle APQ = \\angle AUQ = \\angle AUD = \\angle AFD = \\angle DFE.\\] \n\nThis forces \\(\\overline{PQ} \\parallel \\overline{DF}\\) , as desired. \n\n\\(\\P\\) Third solution by Pitchayut Saengrungkongka and Maxim Li. We provide yet another proof that \\(BP = BQ\\) . \n\n\n \n\nLet the incircle be \\(\\omega\\) and touch \\(BC\\) and \\(AB\\) at point \\(U\\) and \\(W\\) . Let the tangent to \\(\\omega\\) at \\(D\\) meet \\(UW\\) at \\(T\\) . Notice that \\(T\\) is the pole of \\(BD\\) with respect to \\(\\omega\\) , so \\(IT \\perp BD\\) . Now, we make the following critical claim, which in particular implies \\(BP = BQ\\) . \n\nClaim — Quadrilaterals \\(DIWT\\) and \\(PBQA\\) are inversely similar.\n\n\n\nProof. This follows from four angle relations. \n\n\\(\\angle IDT = \\angle BPA = 90^{\\circ}\\) . \\(\\angle TIW = \\angle ABQ\\) . \\(\\angle DIT = \\angle IAC = \\angle BAI = \\angle ABP\\) . \\(\\angle ITW = \\angle QBI = \\angle QAB\\) . \n\nWith \\(BP = BQ\\) obtained, one finishes with the same angle chasing as in the first solution.", "metadata": {"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2024.jsonl", "problem_match": "2. ", "solution_match": "## \\(\\S 1.2\\) USA TST 2024/2, proposed by Luke Robitaille \n"}}
+{"year": "2024", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "USA_TST", "problem": "Let \\(A B C\\) be a triangle with incenter \\(I\\) . Let segment \\(A I\\) intersect the incircle of triangle \\(A B C\\) at point \\(D\\) . Suppose that line \\(B D\\) is perpendicular to line \\(A C\\) . Let \\(P\\) be a point such that \\(\\angle B P A = \\angle P A I = 90^{\\circ}\\) . Point \\(Q\\) lies on segment \\(B D\\) such that the circumcircle of triangle \\(A B Q\\) is tangent to line \\(B I\\) . Point \\(X\\) lies on line \\(P Q\\) such that \\(\\angle I A X = \\angle X A C\\) . Prove that \\(\\angle A X P = 45^{\\circ}\\) .", "solution": "We show several approaches. \n\nFirst solution, by author. \n\n\n \n\nClaim — We have \\(B P = B Q\\) . \n\nProof. For readability, we split the proof into three unconditional parts. \n\n- We translate the condition \\(\\overline{B D} \\perp \\overline{A C}\\) . It gives \\(\\angle D B A = 90^{\\circ} - A\\) , so that \n\n\\[\\angle D B I = \\left|\\frac{B}{2} -(90^{\\circ} - A)\\right| = \\frac{|A - C|}{2}\\] \\[\\angle B D I = \\angle D B A + \\angle B A D = (90^{\\circ} - A) + \\frac{A}{2} = 90^{\\circ} - \\frac{A}{2}.\\]\n\n\n\nHence, letting \\(r\\) denote the inradius, we can translate \\(\\overline{BD} \\perp \\overline{AC}\\) into the following trig condition: \n\n\\[\\sin \\frac{B}{2} = \\frac{r}{BI} = \\frac{DI}{BI} = \\frac{\\sin \\angle DBI}{\\sin \\angle BDI} = \\frac{\\sin \\frac{|A - C|}{2}}{\\sin \\left(90^{\\circ} - \\frac{A}{2}\\right)}.\\] \n\n- The length of \\(BP\\) is given from right triangle \\(APB\\) as \n\n\\[BP = BA \\cdot \\sin \\angle PAB = BA \\cdot \\sin \\left(90^{\\circ} - \\frac{A}{2}\\right).\\] \n\n- The length of \\(BQ\\) is given from the law of sines on triangle \\(ABQ\\) . The tangency gives \\(\\angle BAQ = \\angle DBI\\) and \\(\\angle BQA = 180^{\\circ} - \\angle ABI = 180^{\\circ} - \\angle IBE\\) and thus \n\n\\[BQ = BA \\cdot \\frac{\\sin \\angle BAQ}{\\sin \\angle AQB} = BA \\cdot \\frac{\\sin \\angle DBI}{\\sin \\angle ABI} = BA \\cdot \\frac{\\sin \\frac{|A - C|}{2}}{\\sin \\frac{B}{2}}.\\] \n\nThe first bullet implies the expressions in the second and third bullet for \\(BP\\) and \\(BQ\\) are equal, as needed. \\(\\square\\) \n\nRemark. In the above proof, one dos not actually need to compute \\(\\angle DBI = \\frac{|A - C|}{2}\\) . The proof works equally leaving that expression intact as \\(\\sin \\angle DBI\\) in place of \\(\\sin \\frac{|A - C|}{2}\\) . \n\nNow we can finish by angle chasing. We have \n\n\\[\\angle PBQ = \\angle PBA + \\angle ABD = \\frac{A}{2} +90^{\\circ} - A = 90^{\\circ} - \\frac{A}{2}.\\] \n\nThen \n\n\\[\\angle BPQ = \\frac{180^{\\circ} - \\angle PBQ}{2} = 45^{\\circ} + \\frac{A}{4},\\] \n\nso \\(\\angle APQ = 90^{\\circ} - \\angle BPQ = 45^{\\circ} - \\frac{A}{4}\\) . Also, if we let \\(J\\) be the incenter of \\(IAC\\) , then \\(\\angle PAJ = 90^{\\circ} + \\frac{A}{4}\\) , and clearly \\(X\\) lies on line \\(AJ\\) . Then \\(\\angle APQ + \\angle PAJ = 135^{\\circ} < 180^{\\circ}\\) , so \\(X\\) lies on the same side of \\(AP\\) as \\(Q\\) and \\(J\\) (by the parallel postulate). Therefore \\(\\angle AXP = 180^{\\circ} - 135^{\\circ} = 45^{\\circ}\\) , as desired. \n\nRemark. The problem was basically written backwards by starting from the \\(BD \\perp AC\\) condition, turning that into trig, and then contriving \\(P\\) and \\(Q\\) so that the \\(BD \\perp AC\\) condition implied \\(BP = BQ\\) . \n\nSecond solution, by Jeffrey Kwan. We prove the following restatement: \n\nConsider isosceles triangle \\(AEF\\) with \\(AE = AF\\) and incenter \\(D\\) . Let \\(B\\) be the point on ray \\(AE\\) such that \\(BD \\perp AF\\) , and let \\(P\\) be the projection of \\(B\\) onto the line through \\(A\\) parallel to \\(EF\\) . Let \\(I\\) be the point diametrically opposite \\(A\\) in the circumcircle of \\(AEF\\) , and let \\(Q\\) be the point on line \\(BD\\) such that \\(BI\\) is tangent to the circumcircle of \\(AQB\\) . Then \\(\\angle APQ = 45^{\\circ} - \\angle A / 4\\) . \n\nFirst note that \\(\\angle DFE = 45^{\\circ} - \\angle A / 4\\) , so it suffices to show that \\(\\overline{PQ} \\parallel \\overline{DF}\\) . Let \\(U = \\overline{BD} \\cap \\overline{EF}\\) , and let \\(V = BI \\cap (AEF)\\) . Observe that:\n\n\n\n- \\(P\\) and \\(V\\) both lie on the circle with diameter \\(AB\\) , so \\(\\angle BVP = \\angle PAB = 90^{\\circ} - \\angle A / 2\\) . \n\n- We have \\(\\angle EVB = \\angle EVI = \\angle A / 2 = \\angle DUF = \\angle BUE\\) . Hence \\(BEUV\\) is cyclic. \n\nNow \\(\\angle BUV = \\angle AEU = 90^{\\circ} - \\angle A / 2 = \\angle BVP\\) , so \\(\\overline{PUV}\\) are collinear. \n\n\n \n\nFrom the tangency condition, we have that \\(\\angle AQB = 180^{\\circ} - \\angle ABI\\) , which implies that \n\n\\[\\angle AQU + \\angle APU = \\angle AQB + \\angle APV = (180^{\\circ} - \\angle ABI) + \\angle ABI = 180^{\\circ},\\] \n\nand so \\(APUQ\\) is cyclic. Finally, note that \\(D\\) is the orthocenter of \\(\\triangle AUVF\\) , which implies that \n\n\\[\\angle APQ = \\angle AUQ = \\angle AUD = \\angle AFD = \\angle DFE.\\] \n\nThis forces \\(\\overline{PQ} \\parallel \\overline{DF}\\) , as desired. \n\n\\(\\P\\) Third solution by Pitchayut Saengrungkongka and Maxim Li. We provide yet another proof that \\(BP = BQ\\) . \n\n\n \n\nLet the incircle be \\(\\omega\\) and touch \\(BC\\) and \\(AB\\) at point \\(U\\) and \\(W\\) . Let the tangent to \\(\\omega\\) at \\(D\\) meet \\(UW\\) at \\(T\\) . Notice that \\(T\\) is the pole of \\(BD\\) with respect to \\(\\omega\\) , so \\(IT \\perp BD\\) . Now, we make the following critical claim, which in particular implies \\(BP = BQ\\) . \n\nClaim — Quadrilaterals \\(DIWT\\) and \\(PBQA\\) are inversely similar.\n\n\n\nProof. This follows from four angle relations. \n\n\\(\\angle IDT = \\angle BPA = 90^{\\circ}\\) . \\(\\angle TIW = \\angle ABQ\\) . \\(\\angle DIT = \\angle IAC = \\angle BAI = \\angle ABP\\) . \\(\\angle ITW = \\angle QBI = \\angle QAB\\) . \n\nWith \\(BP = BQ\\) obtained, one finishes with the same angle chasing as in the first solution.", "metadata": {"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2024.jsonl", "problem_match": "2. ", "solution_match": "## \\(\\S 1.2\\) USA TST 2024/2, proposed by Luke Robitaille \n"}}
{"year": "2024", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "USA_TST", "problem": "Let \\(n > k\\geq 1\\) be integers and let \\(p\\) be a prime dividing \\(\\textstyle{\\binom{n}{k}}\\) . Prove that the \\(k\\) -element subsets of \\(\\{1,\\ldots ,n\\}\\) can be split into \\(p\\) classes of equal size, such that any two subsets with the same sum of elements belong to the same class.", "solution": "## Problem statem \n\nLet \\(\\sigma (S)\\) denote the sum of the elements of \\(S\\) , so that \n\n\\[P(x):= \\sum_{S\\subseteq \\{1,\\ldots ,n\\}}x^{\\sigma (S)}\\] \n\nis the generating function for the sums of \\(k\\) - element subsets of \\(\\{1, \\ldots , n\\}\\) . \n\nBy Legendre's formula, \n\n\\[\\nu_{p}\\left(\\binom{n}{k}\\right) = \\sum_{r = 1}^{\\infty}\\left(\\left\\lfloor \\frac{n}{p^{r}}\\right\\rfloor -\\left\\lfloor \\frac{k}{p^{r}}\\right\\rfloor -\\left\\lfloor \\frac{n - k}{p^{r}}\\right\\rfloor\\right)\\] \n\nso there exists a positive integer \\(r\\) with \n\n\\[\\left\\lfloor \\frac{n}{p^{r}}\\right\\rfloor -\\left\\lfloor \\frac{k}{p^{r}}\\right\\rfloor -\\left\\lfloor \\frac{n - k}{p^{r}}\\right\\rfloor >0.\\] \n\nThe main claim is the following: \n\nClaim — \\(P(x)\\) is divisible by \n\n\\[\\Phi_{p^{r}}(x) = x^{(p - 1)p^{r - 1}} + \\dots +x^{p^{r - 1}} + 1.\\] \n\nBefore proving this claim, we will show how it solves the problem. It implies that there exists a polynomial \\(Q\\) with integer coefficients satisfying \n\n\\[P(x) = \\Phi_{p^{r}}(x)Q(x)\\] \\[\\qquad = (x^{(p - 1)p^{r - 1}} + \\dots +x^{p^{r - 1}} + 1)Q(x).\\] \n\nLet \\(c_{0}\\) , \\(c_{1}\\) , ... denote the coefficients of \\(P\\) , and define \n\n\\[s_{i} = \\sum_{j\\equiv i\\pmod{p^{r}}}c_{j}.\\] \n\nThen it's easy to see that \n\n\\[s_{0} = s_{p^{r - 1}} = \\dots = s_{(p - 1)p^{r - 1}}\\] \\[s_{1} = s_{p^{r - 1} + 1} = \\dots = s_{(p - 1)p^{r - 1} + 1}\\] \\[\\qquad \\vdots\\] \\[s_{p^{r - 1} - 1} = s_{2p^{r - 1} - 1} = \\dots = s_{p^{r} - 1}.\\] \n\nThis means we can construct the \\(p\\) classes by placing a set with sum \\(z\\) in class \\(\\left\\lfloor \\frac{z \\mod p^{r}}{p^{r - 1}}\\right\\rfloor\\) . Now we present two ways to prove the claim.\n\n\n\nFirst proof of claim. There's a natural bijection between \\(k\\) - element subsets of \\(\\{1,\\ldots ,n\\}\\) and binary strings consisting of \\(k\\) zeroes and \\(\\ell\\) ones: the set \\(\\{a_{1},\\ldots ,a_{k}\\}\\) corresponds to the string which has zeroes at positions \\(a_{1}\\) , ..., \\(a_{k}\\) . Moreover, the inversion count of this string is simply \\((a_{1} + \\dots +a_{k}) - \\frac{1}{2} k(k + 1)\\) , so we only deal with these inversion counts (equivalently, we are factoring \\(x^{\\frac{k(k + 1)}{2}}\\) out of \\(P\\) ). \n\nRecall that the generating function for these inversion counts is given by the \\(q\\) - binomial coefficient \n\n\\[P(x) = \\frac{(x - 1)\\cdot\\cdot\\cdot(x^{k + \\ell} - 1)}{[(x - 1)\\cdot\\cdot\\cdot(x^{k} - 1)]\\times[(x - 1)\\cdot\\cdot\\cdot(x^{\\ell} - 1)]}.\\] \n\nBy choice of \\(r\\) , the numerator of \\(P(x)\\) has more factors of \\(\\Phi_{p^{r}}(x)\\) than the denominator, so \\(\\Phi_{p^{r}}(x)\\) divides \\(P(x)\\) . \n\nRemark. Here is a proof that \\(P(x)\\) is divisible by \\(\\Phi_{p^{r}}(x)\\) for some \\(r\\) using the \\(q\\) - binomial formula, without explicitly identifying \\(r\\) . We know that \\(P(x)\\) is the product of several cyclotomic polynomials, and that \\(P(1)\\) is a multiple of \\(p\\) . Thus there is a factor \\(\\Phi_{q}(x)\\) for which \\(\\Phi_{q}(1)\\) is a multiple of \\(p\\) , which is equivalent to \\(q\\) being a power of \\(p\\) . \n\nSecond proof of claim. Note that \\(P(x)\\) is the coefficient of \\(y^{k}\\) in the polynomial \n\n\\[Q(x,y):= (1 + xy)(1 + x^{2}y)\\cdot \\cdot \\cdot (1 + x^{n}y).\\] \n\nLet \\(a\\) be the remainder when \\(n\\) is divided by \\(p^{r}\\) , and let \\(b\\) be the remainder when \\(k\\) is divided by \\(p^{r}\\) ; then we have \\(a< b\\) by the choice of \\(r\\) . Let \\(q = \\lfloor n / p^{r}\\rfloor\\) so \\(n = p^{r}q + a\\) . Consider taking \\(x\\) to be a primitive \\(p^{r}\\) th root of unity, say \\(\\omega\\) . Then \n\n\\[Q(\\omega ,y) = \\left[(1 + \\omega y)(1 + \\omega^{2}y)\\cdot \\cdot \\cdot (1 + \\omega^{p^{r}}y)\\right]^{q}(1 + \\omega y)(1 + \\omega^{2}y)\\cdot \\cdot \\cdot (1 + \\omega^{a}y).\\] \n\nNow \\(\\omega\\) , \\(\\omega^{2}\\) , ..., \\(\\omega^{p^{r}}\\) are all the \\(p^{r}\\) th roots of unity, each exactly once; then we can see that \n\n\\[(1 + \\omega y)(1 + \\omega^{2}y)\\cdot \\cdot \\cdot (1 + \\omega^{p^{r}}y)\\] \\[= (1 - \\omega (-y))(1 - \\omega^{2}(-y))\\cdot \\cdot \\cdot (1 - \\omega^{p^{r}}(-y))\\] \\[= 1 - (-y)^{p^{r}},\\] \n\nso \n\n\\[Q(\\omega ,y) = (1 - (-y)^{p^{r}})^{q}(1 + \\omega y)(1 + \\omega^{2}y)\\cdot \\cdot \\cdot (1 + \\omega^{a}y).\\] \n\nIn particular, for any \\(m\\) , if the coefficient of \\(y^{m}\\) in \\(Q(w,x)\\) is nonzero, then \\(m\\) must be congruent to one of \\(0,1,\\ldots ,a\\) (mod \\(p^{r}\\) ). Therefore the coefficient of \\(y^{k}\\) in \\(Q(\\omega ,y)\\) is zero. This means that \\(P(\\omega) = 0\\) whenever \\(\\omega\\) is a primitive \\(p^{r}\\) th root of unity, which proves the claim.\n\n\n\n## \\(\\S 2\\) Solutions to Day 2", "metadata": {"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2024.jsonl", "problem_match": "3. ", "solution_match": "## \\(\\S 1.3\\) USA TST 2024/3, proposed by Ankan Bhattacharya \n"}}
-{"year": "2024", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "USA_TST", "problem": "Find all integers \\(n\\geq 2\\) for which there exists a sequence of \\(2n\\) pairwise distinct points \\((P_{1},\\ldots ,P_{n},Q_{1},\\ldots ,Q_{n})\\) in the plane satisfying the following four conditions: \n\n(i) no three of the \\(2n\\) points are collinear; \n\n(ii) \\(P_{i}P_{i + 1}\\geq 1\\) for all \\(i = 1,2,\\ldots ,n\\) , where \\(P_{n + 1} = P_{1}\\) \n\n(iii) \\(Q_{i}Q_{i + 1}\\geq 1\\) for all \\(i = 1,2,\\ldots ,n\\) , where \\(Q_{n + 1} = Q_{1}\\) ; and \n\n(iv) \\(P_{i}Q_{j}\\leq 1\\) for all \\(i = 1,2,\\ldots ,n\\) and \\(j = 1,2,\\ldots ,n\\)", "solution": "Find alln\\) . \n\nAnswer. Even integers only. \n\nProof that even \\(n\\) work. If we ignore the conditions that the points are pairwise distinct and form no collinear triples, we may take \n\n\\[P_{2i + 1} = (0.51,0), \\quad P_{2i} = (-0.51,0), \\quad Q_{2i + 1} = (0,0.51), \\quad Q_{2i} = (0, -0.51).\\] \n\nThe distances \\(P_{i}P_{i + 1}\\) and \\(Q_{i}Q_{i + 1}\\) are \\(1.02 > 1\\) , while the distances \\(P_{i}Q_{j}\\) are \\(0.51\\sqrt{2} < 1\\) . We may then perturb each point by a small amount to ensure that the distance inequalities still hold and have the points in general position. \n\nProof that odd \\(n\\) do not work. The main claim is the following. \n\nClaim — For \\(1 \\leq i \\leq n\\) , points \\(Q_{i}\\) and \\(Q_{i + 1}\\) must lie on opposite sides of line \\(P_{1}P_{2}\\) . \n\nTo isolate the geometry component of the problem, we rewrite the claim in the following contrapositive form, without referencing the points \\(Q_{i}\\) and \\(Q_{i + 1}\\) : \n\n## Lemma \n\nSuppose \\(A\\) and \\(B\\) are two points such that \\(\\max (P_{1}A, P_{1}B, P_{2}A, P_{2}B) \\leq 1\\) , and moreover \\(A\\) and \\(B\\) lie on the same side of line \\(P_{1}P_{2}\\) . Further assume no three of \\(\\{P_{1}, P_{2}, A, B\\}\\) are collinear. Then \\(AB < 1\\) .\n\n\n\n \n\nProof of lemma. Suppose for the sake of contradiction that \\(A\\) and \\(B\\) lie on the same side of \\(P_{1}P_{2}\\) . The convex hull of these four points is either a quadrilateral or a triangle. \n\n- If the convex hull is a quadrilateral, assume WLOG that the vertices are \\(P_{1}P_{2}AB\\) in order. Let \\(X\\) denote the intersection of segments \\(P_{1}A\\) and \\(P_{2}B\\) . Then \n\n\\[1 + AB = P_{1}P_{2} + AB< P_{1}X + XP_{2} + AX + XB = P_{1}A + P_{2}B\\leq 2.\\] \n\n- Otherwise, assume WLOG that \\(B\\) is in the interior of triangle \\(P_{1}P_{2}A\\) . Since \\(\\angle P_{1}BA + \\angle P_{2}BA = 360^{\\circ} - \\angle P_{1}BP_{2} > 180^{\\circ}\\) , at least one of \\(\\angle P_{1}BA\\) and \\(\\angle P_{2}BA\\) is obtuse. Assume WLOG the former angle is obtuse; then \\(AB< P_{1}A\\leq 1\\) . \\(\\square\\) \n\nRemark. Another proof of the lemma can be found by replacing segment \\(AB\\) with the intersection of this line on the boundary of the blue region above, which does not decrease the distance. In other words, one can assume WLOG that \\(A\\) and \\(B\\) lie on either segment \\(AB\\) or one of the two circular arcs. One then proves that \\(AB\\leq 1\\) , and that for equality to occur, one of \\(A\\) and \\(B\\) must lie on segment \\(P_{1}P_{2}\\) However, this approach seems to involve a fair bit more calculation. \n\nYet another clever approach uses the trivia- fact that a Reuleaux triangle happens to have constant width. \n\nIn any case, it's important to realize that this claim is not trivial; while it looks like it is easy to prove, it is not, owing to the two near- equality cases. \n\nIt follows from the claim that \\(Q_{i}\\) is on the same side of line \\(P_{1}P_{2}\\) as \\(Q_{1}\\) if \\(i\\) is odd, and on the opposite side if \\(i\\) is even. Since \\(Q_{1} = Q_{n + 1}\\) , this means the construction is not possible when \\(n\\) is odd. \n\nRemark. The fact that \\(n\\) cannot be odd follows from Theorem 3 of EPTAS for Max Clique on Disks and Unit Balls. In the language of that paper, if \\(G\\) is a unit ball graph, then the induced odd cycle parking number of \\(\\bar{G}\\) is at most 1. \n\nIn earlier versions of the proposed problem, the points were not necessarily distinct to make the even \\(n\\) case nicer, but this resulted in annoying boundary conditions for the odd \\(n\\) case.", "metadata": {"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2024.jsonl", "problem_match": "4. ", "solution_match": "## \\(\\S 2.1\\) USA TST 2024/4, proposed by Ray Li \n"}}
+{"year": "2024", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "USA_TST", "problem": "Find all integers \\(n\\geq 2\\) for which there exists a sequence of \\(2n\\) pairwise distinct points \\((P_{1},\\ldots ,P_{n},Q_{1},\\ldots ,Q_{n})\\) in the plane satisfying the following four conditions: \n\n(i) no three of the \\(2n\\) points are collinear; \n\n(ii) \\(P_{i}P_{i + 1}\\geq 1\\) for all \\(i = 1,2,\\ldots ,n\\) , where \\(P_{n + 1} = P_{1}\\) \n\n(iii) \\(Q_{i}Q_{i + 1}\\geq 1\\) for all \\(i = 1,2,\\ldots ,n\\) , where \\(Q_{n + 1} = Q_{1}\\) ; and \n\n(iv) \\(P_{i}Q_{j}\\leq 1\\) for all \\(i = 1,2,\\ldots ,n\\) and \\(j = 1,2,\\ldots ,n\\)", "solution": "Find alln\\) . \n\nAnswer. Even integers only. \n\nProof that even \\(n\\) work. If we ignore the conditions that the points are pairwise distinct and form no collinear triples, we may take \n\n\\[P_{2i + 1} = (0.51,0), \\quad P_{2i} = (-0.51,0), \\quad Q_{2i + 1} = (0,0.51), \\quad Q_{2i} = (0, -0.51).\\] \n\nThe distances \\(P_{i}P_{i + 1}\\) and \\(Q_{i}Q_{i + 1}\\) are \\(1.02 > 1\\) , while the distances \\(P_{i}Q_{j}\\) are \\(0.51\\sqrt{2} < 1\\) . We may then perturb each point by a small amount to ensure that the distance inequalities still hold and have the points in general position. \n\nProof that odd \\(n\\) do not work. The main claim is the following. \n\nClaim — For \\(1 \\leq i \\leq n\\) , points \\(Q_{i}\\) and \\(Q_{i + 1}\\) must lie on opposite sides of line \\(P_{1}P_{2}\\) . \n\nTo isolate the geometry component of the problem, we rewrite the claim in the following contrapositive form, without referencing the points \\(Q_{i}\\) and \\(Q_{i + 1}\\) : \n\n## Lemma \n\nSuppose \\(A\\) and \\(B\\) are two points such that \\(\\max (P_{1}A, P_{1}B, P_{2}A, P_{2}B) \\leq 1\\) , and moreover \\(A\\) and \\(B\\) lie on the same side of line \\(P_{1}P_{2}\\) . Further assume no three of \\(\\{P_{1}, P_{2}, A, B\\}\\) are collinear. Then \\(AB < 1\\) .\n\n\n\n \n\nProof of lemma. Suppose for the sake of contradiction that \\(A\\) and \\(B\\) lie on the same side of \\(P_{1}P_{2}\\) . The convex hull of these four points is either a quadrilateral or a triangle. \n\n- If the convex hull is a quadrilateral, assume WLOG that the vertices are \\(P_{1}P_{2}AB\\) in order. Let \\(X\\) denote the intersection of segments \\(P_{1}A\\) and \\(P_{2}B\\) . Then \n\n\\[1 + AB = P_{1}P_{2} + AB< P_{1}X + XP_{2} + AX + XB = P_{1}A + P_{2}B\\leq 2.\\] \n\n- Otherwise, assume WLOG that \\(B\\) is in the interior of triangle \\(P_{1}P_{2}A\\) . Since \\(\\angle P_{1}BA + \\angle P_{2}BA = 360^{\\circ} - \\angle P_{1}BP_{2} > 180^{\\circ}\\) , at least one of \\(\\angle P_{1}BA\\) and \\(\\angle P_{2}BA\\) is obtuse. Assume WLOG the former angle is obtuse; then \\(AB< P_{1}A\\leq 1\\) . \\(\\square\\) \n\nRemark. Another proof of the lemma can be found by replacing segment \\(AB\\) with the intersection of this line on the boundary of the blue region above, which does not decrease the distance. In other words, one can assume WLOG that \\(A\\) and \\(B\\) lie on either segment \\(AB\\) or one of the two circular arcs. One then proves that \\(AB\\leq 1\\) , and that for equality to occur, one of \\(A\\) and \\(B\\) must lie on segment \\(P_{1}P_{2}\\) However, this approach seems to involve a fair bit more calculation. \n\nYet another clever approach uses the trivia- fact that a Reuleaux triangle happens to have constant width. \n\nIn any case, it's important to realize that this claim is not trivial; while it looks like it is easy to prove, it is not, owing to the two near- equality cases. \n\nIt follows from the claim that \\(Q_{i}\\) is on the same side of line \\(P_{1}P_{2}\\) as \\(Q_{1}\\) if \\(i\\) is odd, and on the opposite side if \\(i\\) is even. Since \\(Q_{1} = Q_{n + 1}\\) , this means the construction is not possible when \\(n\\) is odd. \n\nRemark. The fact that \\(n\\) cannot be odd follows from Theorem 3 of EPTAS for Max Clique on Disks and Unit Balls. In the language of that paper, if \\(G\\) is a unit ball graph, then the induced odd cycle parking number of \\(\\bar{G}\\) is at most 1. \n\nIn earlier versions of the proposed problem, the points were not necessarily distinct to make the even \\(n\\) case nicer, but this resulted in annoying boundary conditions for the odd \\(n\\) case.", "metadata": {"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2024.jsonl", "problem_match": "4. ", "solution_match": "## \\(\\S 2.1\\) USA TST 2024/4, proposed by Ray Li \n"}}
{"year": "2024", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "USA_TST", "problem": "Suppose \\(a_{1}< a_{2}< \\dots < a_{2024}\\) is an arithmetic sequence of positive integers, and \\(b_{1}< b_{2}< \\dots < b_{2024}\\) is a geometric sequence of positive integers. Find the maximum possible number of integers that could appear in both sequences, over all possible choices of the two sequences.", "solution": "Answer. 11 terms. \n\nConstruction. Let \\(a_{i} = i\\) and \\(b_{i} = 2^{i - 1}\\) \n\nBound. We show a \\(\\nu_{p}\\) - based approach communicated by Derek Liu, which seems to be the shortest one. At first, we completely ignore the geometric sequence \\(b_{i}\\) and focus only on the arithmetic sequence. \n\nClaim — Let \\(p\\) be any prime, and consider the sequence \n\n\\[\\nu_{p}(a_{1}), \\nu_{p}(a_{2}), \\ldots , \\nu_{p}(a_{2024}).\\] \n\nSet \\(C := \\lfloor \\log_{p}(2023) \\rfloor\\) . Then there are at most \\(C + 2\\) different values in this sequence. \n\nProof. By scaling, assume the \\(a_{i}\\) do not have a common factor, so that \\(a_{i} = a + d i\\) where \\(\\gcd (a,d) = 1\\) . \n\n- If \\(p \\mid d\\) , then \\(p \\nmid a\\) and \\(\\nu_{p}(a_{i})\\) is constant. \n\n- Otherwise, assume \\(p \\nmid d\\) . We will in fact prove that every term in the sequence is contained in \\(\\{0,1,\\ldots ,C\\}\\) with at most one exception. \n\nDefine \\(M := \\max_{i} \\nu_{p}(a_{i})\\) . If \\(M \\leq C\\) , there is nothing to prove. Otherwise, fix some index \\(m\\) such that \\(\\nu_{p}(a_{m}) = M\\) . We know \\(\\nu_{p}(i - m) \\leq C\\) since \\(|i - m| \\leq 2023\\) . But now for any other index \\(i \\neq m\\) ; \n\n\\[\\nu_{p}(d(i - m)) = \\nu_{p}(i - m)\\leq C< M = \\nu_{p}(a_{m})\\] \\[\\implies \\nu_{p}(a_{i}) = \\nu_{p}(a_{m} + d(i - m)) = \\nu_{p}(i - m)\\leq C.\\] \n\nso \\(\\nu_{p}(a_{m})\\) is the unique exceptional term of the sequence exceeding \\(C\\) . \n\nRemark. The bound in the claim is best possible by taking \\(a_{i} = p^{M} + (i - 1)\\) for any \\(M > C\\) . Then indeed, the sequence \\(\\nu_{p}(a_{i})\\) takes on values in \\(\\{0,1,\\ldots ,C\\}\\) for \\(i > 1\\) while \\(\\nu_{p}(a_{1}) = M\\) . \n\nBack to the original problem with \\((b_{i})\\) . Consider the common ratio \\(r \\in \\mathbb{Q}\\) of the geometric sequence \\((b_{i})\\) . If \\(p\\) is any prime with \\(\\nu_{p}(r) \\neq 0\\) , then every term of \\((b_{i})\\) has a different \\(\\nu_{p}\\) . So there are two cases to think about.\n\n\n\nSuppose any \\(p \\geq 3\\) has \\(\\nu_{p}(r) \\neq 0\\) . Then there are at most \n\n\\[2 + \\log_{p}(2023) = 2 + \\log_{3}(2023) \\approx 8.929 < 11\\] \n\noverlapping terms, as needed. \n\nOtherwise, suppose \\(r\\) is a power of 2 (in particular, \\(r \\geq 2\\) is an integer). We already have an upper bound of 12; we need to improve it to 11. \n\nAs in the proof before, we may assume WLOG by scaling down by a power of 2 that the common difference of \\(a_{i}\\) is odd. (This may cause some \\(b_{i}\\) to become rational non- integers, but that's OK. However, unless \\(\\nu_{2}(a_{i})\\) is constant, the \\(a_{i}\\) will still be integers.) \n\nThen in order for the bound of 12 to be achieved, the sequence \\(\\nu_{2}(a_{i})\\) must be contained in \\(\\{0,1,\\ldots ,10,M\\}\\) for some \\(M \\geq 11\\) . In particular, we only need to work with \\(r = 2\\) . \n\nDenote by \\(b\\) the unique odd- integer term in the geometric sequence, which must appear among \\((a_{i})\\) . Then \\(2b\\) appears too, so the common difference of \\(a_{i}\\) is at most \\(b\\) . \n\nBut if \\((a_{i})\\) is an arithmetic progression of integers that contains \\(b\\) and has common difference at most \\(b\\) , then no term of the sequence can ever exceed \\(b + 2023 \\cdot b = 2024b\\) . Hence \\(2^{M}b\\) cannot appear for any \\(M \\geq 11\\) . This completes the proof. \n\nRemark. There are several other approaches to the problem, but most take some time to execute. The primary issue is that the common difference of the \\(a_{i}\\) 's could share prime factors with the common ratio in the \\(b_{i}\\) 's, which means that merely trying to write out a lot of modular arithmetic equations leads to a lot of potential technical traps that are not pleasant to defuse. \n\nOne unusual thing is that many solutions end up proving a bound of 12 (in the case the common ratio is 2) and then having to adjust it to 11 later.", "metadata": {"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2024.jsonl", "problem_match": "5. ", "solution_match": "## \\(\\S 2.2\\) USA TST 2024/5, proposed by Ray Li \n"}}
{"year": "2024", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "USA_TST", "problem": "Solve over \\(\\mathbb{R}\\) the functional equation \n\n\\[f(x f(y)) + f(y) = f(x + y) + f(x y).\\]", "solution": "In addition to all constant functions, \\(f(x)\\equiv x + 1\\) clearly works too. We prove these are the only solutions. The solution that follows is by the original proposer. \n\nLet \\(P(x,y)\\) denote the given assertion. \n\nClaim 2.1 — If \\(f\\) is periodic, then \\(f\\) is constant. \n\nProof. Let \\(f\\) have period \\(d\\neq 0\\) . From \\(P(x,y + d)\\) , we have \n\n\\[f(x(y + d)) = f(x + y + d) - f(y + d) - f(x f(y + d)) = f(x + y) - f(y) - f(x f(y)).\\] \n\nApplying \\(P(x,y)\\) gives \n\n\\[f(x(y + d)) = f(x y).\\] \n\nIn particular, taking \\(y = 0\\) yields that \\(f(d x) = f(0)\\) . Thus \\(f\\) is constant, as \\(d\\neq 0\\) . \n\nClaim 2.2 — For all real numbers \\(x\\) and \\(y\\) , we have \\(f(f(x) + y) = f(f(y) + x)\\) . \n\nProof. Applying \\(P(f(x),y)\\) and then \\(P(y,x)\\) gives us \n\n\\[f(f(x)f(y)) = f(f(x) + y) + f(f(x)y) - f(y)\\] \\[\\qquad = f(f(x) + y) + f(x + y) + f(xy) - f(x) - f(y).\\] \n\nNow swapping \\(x\\) and \\(y\\) gives us \\(f(f(x) + y) = f(f(y) + x)\\) . \n\nClaim 2.3 — If \\(f\\) is nonconstant, then \\(f(f(x) + y) = f(x) + f(y)\\) for all reals \\(x,y\\) . \n\nProof. Let \\(x,y\\in \\mathbb{R}\\) and let \\(d:= f(f(x) + y) - f(x) - f(y)\\) . Let \\(z\\) be an arbitrary real number. By repeatedly applying Claim 2.2, we have \n\n\\[f(z + f(f(x) + y)) = f(f(z) + f(x) + y)\\] \\[\\qquad = f(f(x) + f(z) + y)\\] \\[\\qquad = f(x + f(f(z) + y))\\] \\[\\qquad = f(x + f(f(y) + z))\\] \\[\\qquad = f(f(x) + f(y) + z)\\] \\[\\qquad = f(z + f(x) + f(y)).\\] \n\nIf \\(d\\neq 0\\) , then \\(f\\) is periodic with period \\(d\\) , contradicting Claim 2.1.\n\n\n\nClaim 2.4 — If \\(f\\) is nonconstant, then \\(f(0) = 1\\) and \\(f(x + 1) = f(x) + 1\\) . \n\nProof. From \\(P(z,0)\\) we have \\(f(zf(0)) = f(z)\\) for all real \\(z\\) . Then \\(P(xf(0),y)\\) and \\(P(x,y)\\) give us \n\n\\[f(xf(0) + y) = f(y) + f(xf(0)f(y)) - f(xf(0)y)\\] \\[\\qquad = f(y) + f(xf(y)) - f(xy) = f(x + y).\\] \n\nIf \\(f(0)\\neq 1\\) , then \\(x f(0) - x\\) is a period of \\(f\\) for all \\(x\\) , violating Claim 2.1. So we must have \\(f(0) = 1\\) . \n\nNow putting \\(x = 0\\) in Claim 2.3 gives \\(f(x + 1) = f(x) + 1\\) . \n\nClaim 2.5 — If \\(f\\) is nonconstant, then \\(f(x) + f(y) = f(x + y) + 1\\) . \n\nProof. From \\(P(x + 1,y)\\) and Claim 2.4, we have \n\n\\[f((x + 1)f(y)) = f(x + y + 1) + f(xy + y) - f(y) = f(x + y) + f(xy + y) - f(y) + 1.\\] \n\nAlso, from Claim 2.3 and \\(P(x,y)\\) , we have \n\n\\[f((x + 1)f(y)) = f(xf(y)) + f(y) = f(x + y) + f(xy).\\] \n\nThus \\(f(xy) = f(xy + y) - f(y) + 1\\) . Replacing \\(x\\) with \\(\\frac{x}{y}\\) gives the claim for all \\(y\\neq 0\\) (whereas \\(y = 0\\) follows from Claim 2.4). \\(\\square\\) \n\nClaim 2.6 — If \\(f\\) is nonconstant, then \\(f(x)\\equiv x + 1\\) . \n\nProof. We apply Claim 2.3, Claim 2.5, and Claim 2.4: \n\n\\[f(f(x) + y) = f(x) + f(y) = f(x + y) + 1 = f(x + y + 1).\\] \n\nIf \\(f(x)\\neq x + 1\\) for some \\(x\\) , then Claim 2.1 again gives a contradiction. \\(\\square\\)", "metadata": {"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2024.jsonl", "problem_match": "6. ", "solution_match": "## \\(\\S 2.3\\) USA TST 2024/6, proposed by Milan Haiman \n"}}
diff --git a/USA_TST/segmented/en-sols-TST-IMO-2025.jsonl b/USA_TST/segmented/en-sols-TST-IMO-2025.jsonl
index 00f65e80cb7063c602e4450f52f4e28e6051abe4..4531a8527500ca9bc5d0bcd5fe91a60fd9755083 100644
--- a/USA_TST/segmented/en-sols-TST-IMO-2025.jsonl
+++ b/USA_TST/segmented/en-sols-TST-IMO-2025.jsonl
@@ -1,6 +1,6 @@
{"year": "2025", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "Let \\(n\\) be a positive integer. Ana and Banana play a game. Banana thinks of a function \\(f\\colon \\mathbb{Z}\\to \\mathbb{Z}\\) and a prime number \\(p\\) . He tells Ana that \\(f\\) is nonconstant, \\(p< 100\\) and \\(f(x + p) = f(x)\\) for all integers \\(x\\) . Ana's goal is to determine the value of \\(p\\) . She writes down \\(n\\) integers \\(x_{1},\\ldots ,x_{n}\\) . After seeing this list, Banana writes down \\(f(x_{1}),\\ldots ,f(x_{n})\\) in order. Ana wins if she can determine the value of \\(p\\) from this information. Find the smallest value of \\(n\\) for which Ana has a winning strategy.", "solution": "The answer is \\(n = 83 + 89 - 1 = 171\\) \n\nIn general, if Ana has to decide between periods from the set \\(\\mathcal{P}:= \\{p_{1} > p_{2} > \\dots >p_{r}\\}\\) of pairwise distinct relatively prime positive integers for \\(r\\geq 3\\) , the answer is \\(p_{2} + p_{3} - 1\\) \n\n## Bound \n\nSuppose for the sake of contradiction that Ana has a winning sequence of integers \\(x_{1},\\ldots ,x_{n}\\) with \\(n\\leq p_{2} + p_{3} - 2\\) . We will generate contradictions by providing two primes \\(p,q\\in \\mathcal{P}\\) and associated nonconstant functions \\(f_{p},f_{q}:\\mathbb{Z}\\to \\mathbb{Z}\\) with periods \\(p\\) and \\(q\\) respectively such that \\(f_{p}(x_{i}) = f_{q}(x_{i})\\) for all \\(i\\) \n\nClaim — There exists a prime \\(r\\in \\mathcal{P}\\) such that for all primes \\(p\\in \\mathcal{P}\\setminus \\{r\\}\\) , the set of integers \\(\\{x_{1},\\ldots ,x_{n}\\}\\) forms a complete residue class modulo \\(p\\) (i.e. for all \\(t\\) , there exists \\(i\\) such that \\(x_{i}\\equiv t\\) (mod \\(p\\) )). \n\nClaim — Suppose for the sake of contradiction that such \\(r\\) didn't exist, and there were in fact two primes \\(p,q\\in \\mathcal{P}\\) such that \\(\\{x_{1}\\ldots ,x_{n}\\}\\) did not form a complete residue class modulo either \\(p\\) or \\(q\\) . Concretely, consider \\(t,s\\) such that there is no \\(i\\) with \\(x_{i}\\equiv t\\) (mod \\(p\\) ) and not \\(j\\) with \\(x_{j}\\equiv s\\) (mod \\(q\\) ). \n\nConstruct the functions \\(f_{p},f_{q}:\\mathbb{Z}\\to \\mathbb{Z}\\) as \n\n \n\nand \n\n \n\nWe have \\(f_{p}(x_{i}) = f_{q}(x_{i}) = 0\\) for all \\(i\\) , which is the desired contradiction. \n\nLet \\(r\\) be the prime from the above claim. Let \\(p,q\\) be the largest two primes in \\(\\mathcal{P}\\setminus \\{r\\}\\) , so \\(n\\leq p + q - 2\\) . Construct the graph \\(G_{pq}\\) with vertex set \\(\\{x_{1},\\ldots ,x_{n}\\}\\) and edge \\(x_{i}\\sim x_{j}\\) if \\(p\\mid x_{i} - x_{j}\\) or \\(q\\mid x_{i} - x_{j}\\) . The following claim allows us to construct a pair of bad functions \\(f_{p},f_{q}\\) . \n\nClaim — The graph \\(G_{pq}\\) is disconnected. \n\nProof. Let \\(G_{p}\\) be the graph on vertex set \\(\\{x_{1},\\ldots ,x_{n}\\}\\) with edge \\(x_{i}\\sim x_{j}\\) if \\(p\\mid x_{i} - x_{j}\\) . Note that \\(G_{p}\\) is a collection of \\(p\\) disjoint cliques, one for each residue class modulo \\(p\\) .\n\n\n\nPrune the graph into \\(G_{p}^{\\prime}\\) , where each clique \\(K_{r}\\) is replaced by a path of edge- length \\(r - 1\\) . Define \\(G_{q}^{\\prime}\\) similarly, and let \\(G_{p q}^{\\prime}\\) be the union of \\(G_{p}^{\\prime}\\) and \\(G_{q}^{\\prime}\\) . \n\nNote that \\(G_{p q}\\) and \\(G_{p q}^{\\prime}\\) have the exact same connectivity properties. We have \n\n\\[|E(G_{p q}^{\\prime})|\\leq |E(G_{p}^{\\prime})| + |E(G_{q}^{\\prime})| = (n - p) + (n - q)\\leq n - 2,\\] \n\nso \\(G_{p q}^{\\prime}\\) is disconnected, as desired. \n\nSuppose \\(A\\sqcup B = \\{x_{1},\\ldots ,x_{n}\\}\\) are sets of disjoint vertices in \\(G_{p q}\\) . Construct the functions \\(f_{p},f_{q}:\\mathbb{Z}\\to \\mathbb{Z}\\) as \n\n \n\nand \n\n \n\nThese are well defined due to the fact that \\(p,q\\nmid a - b\\) for \\(a\\in A\\) and \\(b\\in B\\) , and the fact that \\(A\\sqcup B\\) forms a complete residue class modulo \\(p\\) . Again, we have \\(f_{p}(x_{i}) = f_{q}(x_{i})\\) for all \\(i\\) , which is the desired contradiction. \n\n## Construction \n\nLet \\(n = p_{2} + p_{3} - 1\\) . We claim that Ana has a winning strategy with the selection \\(x_{i} = p_{1}(i - 1)\\) . Indeed, suppose that Banana writes down the values \\(y_{1},\\ldots ,y_{n}\\) in order. We will show that Ana can always reconstruct \\(p\\) . \n\nClaim — If \\(y_{1} = \\dots = y_{n}\\) , then Ana can correctly guess \\(p = p_{1}\\) . \n\nProof. Suppose for the sake of contradiction that \\(p< p_{1}\\) . Then, since \\(x_{1},\\ldots ,x_{n}\\) forms a complete residue class modulo \\(p\\) , \\(f\\) must be a constant function, which is the desired contradiction. \n\nWe can now assume that \\(y_{1},\\ldots ,y_{n}\\) are not all equal, which means \\(p\\neq p_{1}\\) . Suppose for the sake of contradiction that there are two primes \\(q,r< p_{1}\\) with associated nonconstant functions \\(f_{p},f_{q}:\\mathbb{Z}\\to \\mathbb{Z}\\) with periods \\(q\\) and \\(r\\) respectively, such that \\(f_{q}(x_{i}) = f_{r}(x_{i}) = y_{i}\\) for all \\(i\\) . \n\nThe following claim shows that \\(y_{1},\\ldots ,y_{n}\\) must all be equal, which is the desired contradiction. \n\nClaim — Let \\(G\\) be the graph on vertex set \\(\\{0,\\ldots ,q + r - 2\\}\\) with edge \\(i\\sim j\\) if \\(|i - j|\\in \\{q,r\\}\\) . The graph \\(G\\) is connected. \n\nProof. Note that \\(G\\) has \\(q + r - 2\\) edges, so it suffices to show that it has no cycles. Suppose for the sake of contradiction it had a cycle \\(c_{1},\\ldots ,c_{k}\\) with \\(k\\geq 3\\) and indices taken mod \\(k\\) . \n\nSuppose first that \\(c_{i + 1} - c_{i} = q\\) . Then, \\(c_{i + 2} - c_{i + 1}\\) cannot be \\(- q\\) (else \\(c_{i} = c_{i + 2}\\) ), it cannot be \\(r\\) (else \\(c_{i + 2} > q + r - 2\\) ), so \\(c_{i + 2} - c_{i + 1}\\in \\{q, - r\\}\\) . \n\nThe same logic shows the following collated results: \n\n\\[c_{i + 1} - c_{i} = q\\Longrightarrow c_{i + 2} - c_{i + 1}\\in \\{q, - r\\}\\]\n\n\n\n\\[c_{i + 1} - c_{i} = -q \\Rightarrow c_{i + 2} - c_{i + 1} \\in \\{-q, r\\}\\] \\[c_{i + 1} - c_{i} = r \\Rightarrow c_{i + 2} - c_{i + 1} \\in \\{-q, r\\}\\] \\[c_{i + 1} - c_{i} = -r \\Rightarrow c_{i + 2} - c_{i + 1} \\in \\{q, -r\\}.\\] \n\nThus, either all consecutive differences of vertices in the cycle are in \\(\\{q, - r\\}\\) , or all in \\(\\{- q, r\\}\\) . Assume the first case, proof is similar for second case. \n\nLet \\(a\\) be the number of consecutive differences that are \\(q\\) , and \\(b\\) be the number that are \\(- r\\) . We see that \\(a + b = k\\) and \\(qa - rb = 0\\) . The second condition implies that \\(a \\geq r\\) and \\(b \\geq q\\) , so we have \\(k \\geq q + r\\) , which is the desired contradiction since \\(G\\) has only \\(q + r - 1\\) vertices. \\(\\square\\)", "metadata": {"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2025.jsonl", "problem_match": "Problem 1. ", "solution_match": "## \\(\\S 1.1\\) Solution to TST 1, by Anthony Wang \n"}}
{"year": "2025", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "Let \\(a_{1},a_{2},\\ldots\\) and \\(b_{1},b_{2},\\ldots\\) be sequences of real numbers for which \\(a_{1} > b_{1}\\) and \n\n\\[a_{n + 1} = a_{n}^{2} - 2b_{n\\] \\[b_{n + 1} = b_{n}^{2} - 2a_{n\\] \n\nfor all positive integers \\(n\\) . Prove that \\(a_{1},a_{2},\\ldots\\) is eventually increasing (that is, there exists a positive integer \\(N\\) for which \\(a_{k}< a_{k + 1}\\) for all \\(k > N\\) ).", "solution": "Let \\(r\\) , \\(s\\) , and \\(t\\) be the complex roots of the polynomial \\(p(\\lambda) = \\lambda^{3} - a_{1}\\lambda^{2} + b_{1}\\lambda - 1\\) . By Vieta's formulas, \n\n\\[a_{1} = r + s + t\\] \\[b_{1} = 1 / r + 1 / s + 1 / t\\] \\[1 = rst.\\] \n\nClaim — For every positive integer \\(n\\) , \n\n\\[a_{n} = r^{2^{n - 1}} + s^{2^{n - 1}} + t^{2^{n - 1}}\\] \n\nand \n\n\\[b_{n} = (1 / r)^{2^{n - 1}} + (1 / s)^{2^{n - 1}} + (1 / t)^{2^{n - 1}}.\\] \n\nProof. The base case follows from Vieta's formulas above. For the inductive step, observe that \\(rst = 1\\) , so \n\n\\[a_{n + 1} = a_{n}^{2} - 2b_{n\\] \\[\\qquad = \\left(r^{2^{n - 1}} + s^{2^{n - 1}} + t^{2^{n - 1}}\\right)^{2} - 2\\left((1 / r)^{2^{n - 1}} + (1 / s)^{2^{n - 1}} + (1 / t)^{2^{n - 1}}\\right)\\] \\[\\qquad = \\left(r^{2^{n - 1}} + s^{2^{n - 1}} + t^{2^{n - 1}}\\right)^{2} - 2\\left((st)^{2^{n - 1}} + (tr)^{2^{n - 1}} + (rs)^{2^{n - 1}}\\right)\\] \\[\\qquad = r^{2^{n}} + s^{2^{n}} + t^{2^{n}}\\] \n\nand similarly for \\(b_{n + 1}\\) . \n\nSince \\(p(1) = b_{1} - a_{1}< 0\\) , \\(p\\) has a real root greater than 1; let \\(r\\) be the largest such root. \n\n- If \\(s\\) and \\(t\\) are real, let \\(m = \\max (|r|,|s|,|t|) > 1\\) be the largest magnitude of the roots and \\(k\\in \\{1,2,3\\}\\) be the number of roots with that magnitude. Then asymptotically \n\n\\[a_{n} = r^{2^{n - 1}} + s^{2^{n - 1}} + t^{2^{n - 1}}\\approx k m^{2^{n - 1}}\\] \n\nwhich implies that \\(\\{a_{n}\\}\\) is eventually increasing. \n\n- If \\(s\\) and \\(t\\) are not real, they must be complex conjugates of each other, each with magnitude \\(\\frac{1}{\\sqrt{r}} < 1\\) . Therefore \n\n\\[r^{2^{n - 1}} - 2< a_{n}< r^{2^{n - 1}} + 2,\\] \n\nso \\(\\{a_{n}\\}\\) is eventually increasing.", "metadata": {"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2025.jsonl", "problem_match": "Problem 2. ", "solution_match": "## \\(\\S 1.2\\) Solution to TST 2, by Holden Mui \n"}}
-{"year": "2025", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "Let \\(A_{1}A_{2}\\dots A_{2025}\\) be a convex 2025- gon, and let \\(A_{i} = A_{i + 2025}\\) for all integers \\(i\\) . Distinct points \\(P\\) and \\(Q\\) lie in its interior such that \\(\\angle A_{i - 1}A_{i}P = \\angle QA_{i}A_{i + 1}\\) for all \\(i\\) . Define points \\(P_{i}^{j}\\) and \\(Q_{i}^{j}\\) for integers \\(i\\) and positive integers \\(j\\) as follows: \n\nFor all \\(i\\) \\(P_{i}^{1} = Q_{i}^{1} = A_{i}\\) For all \\(i\\) and \\(j\\) \\(P_{i}^{j + 1}\\) and \\(Q_{i}^{j + 1}\\) are the circumcenters of \\(P P_{i}^{j}P_{i + 1}^{j}\\) and \\(Q Q_{i}^{j}Q_{i + 1}^{j}\\) respectively. \n\nLet \\(\\mathcal{P}\\) and \\(\\mathcal{Q}\\) be the polygons \\(P_{1}^{2025}P_{2}^{2025}\\ldots P_{2025}^{2025}\\) and \\(Q_{1}^{2025}Q_{2}^{2025}\\ldots Q_{2025}^{2025}\\) , respectively. \n\n(a) Prove that \\(\\mathcal{P}\\) and \\(\\mathcal{Q}\\) are cyclic. \n\n(b) Let \\(O_{P}\\) and \\(O_{Q}\\) be the circumcenters of \\(\\mathcal{P}\\) and \\(\\mathcal{Q}\\) , respectively. Assuming that \\(O_{P}\\neq O_{Q}\\) , show that \\(O_{P}O_{Q}\\) is parallel to \\(P Q\\) .", "solution": "Let \\(n = 2025\\) . Let \\(\\mathcal{P}_{i}\\) and \\(\\mathcal{Q}_{i}\\) denote the polygons \\(P_{1}^{i}\\cdot \\cdot \\cdot P_{n}^{i}\\) and \\(Q_{1}^{i}\\cdot \\cdot \\cdot Q_{n}^{i}\\) . In this notation, \\(\\mathcal{P} = \\mathcal{P}_{n}\\) , \\(\\mathcal{Q} = \\mathcal{Q}_{n}\\) , and \\(\\mathcal{P}_{1} = \\mathcal{Q}_{1} = A_{1}\\cdot \\cdot \\cdot A_{n}\\) . \n\nThe angle condition for \\(P\\) and \\(Q\\) just says that they are isogonal conjugates in \\(\\mathcal{P}_{1}\\) . We will first find some properties that do not depend on \\(P\\) having an isogonal conjugate. \n\nNote that \\(P_{i - 1}^{j + 1}P_{i}^{j + 1}\\) is the perpendicular bisector of \\(P P_{i + 1}^{j}\\) , so we can go backwards from \\(\\mathcal{P}_{j}\\) to \\(\\mathcal{P}_{j - 1}\\) by reflecting over the sides. Use this to extend the points backwards to \\(\\mathcal{P}_{0}\\) , i.e. define \\(P_{i}^{0}\\) to be the reflection of \\(P\\) over \\(P_{i - 1}^{1}P_{i}^{1}\\) . \n\n## Lemma 1.1 \n\nFor integers \\(j \\geq 0\\) and \\(i\\) , \n\n\\[\\angle P P_{i}^{j}P_{i + 1}^{j} = \\angle P P_{i}^{j + 1}P_{i + 1}^{j + 1}\\quad \\mathrm{and}\\] \\[\\angle P P_{i + 1}^{j}P_{i}^{j} = \\angle P P_{i}^{j + 1}P_{i - 1}^{j + 1}.\\] \n\nProof. This is a standard circumcenter fact in disguise but we will prove it here for completeness. If \\(X\\) is the antipode of \\(P\\) on \\((P P_{i}^{j}P_{i + 1}^{j})\\) (equivalently, the reflection of \\(P\\) over \\(P P_{i}^{j + 1}\\) ) and \\(M\\) is the midpoint of \\(P P_{i + 1}^{j}\\) then \n\n\\[\\angle P P_{i}^{j}P_{i + 1}^{j} = \\angle P X P_{i + 1}^{j} = \\angle P P_{i}^{j + 1}M = \\angle P P_{i}^{j + 1}P_{i + 1}^{j + 1}.\\] \n\nThe second equality is analogous.\n\n\n\n \n\n## Lemma \n\nThe following similarity holds: \\(\\mathcal{P}_{n} \\cup P \\stackrel{\\perp}{\\sim} \\mathcal{P}_{0} \\cup P\\) . \n\nProof. By Lemma 1.1, \n\n\\[\\begin{aligned} \\angle P P_{i}^{0} P_{i + 1}^{0} & = \\angle P P_{i}^{1} P_{i + 1}^{1} = \\dots = \\angle P_{i}^{n} P_{i + 1}^{n} \\quad \\text{and} \\\\ \\angle P P_{i + 1}^{0} P_{i}^{0} & = \\angle P P_{i}^{1} P_{i - 1}^{1} = \\dots = \\angle P_{i - n + 1}^{n} P_{i - n}^{n} = \\angle P_{i + 1}^{n} P_{i}^{n}. \\end{aligned}\\] \n\nTherefore \\(P P_{i}^{n} P_{i + 1}^{n} \\stackrel{\\perp}{\\sim} P P_{i}^{0} P_{i + 1}^{0}\\) . Combining these similarities for all \\(i\\) shows that \\(\\mathcal{P}_{n} \\cup P \\stackrel{\\perp}{\\sim} \\mathcal{P}_{0} \\cup P\\) . \\(\\square\\) \n\nTo solve part (a), we just need to show that \\(\\mathcal{P}_{0}\\) and \\(\\mathcal{Q}_{0}\\) are cyclic. This is where we need the isogonal conjugate property, and we can generalize from analogous facts for the \\(n = 3\\) case. \n\n\n\n\n\n\nWe have \n\n\\[\\angle P P_{i + 1}^{0}P_{i}^{0} = \\angle P A_{i}A_{i - 1} = \\angle A_{i + 1}A_{i}Q\\] \n\nso \n\n\\[90^{\\circ} = \\angle (P P_{i + 1}^{0},A_{i}A_{i + 1}) = \\angle (P_{i + 1}^{0}P_{i}^{0},A Q).\\] \n\nSince \\(A_{i}P_{i}^{0} = A_{i}P_{i + 1}^{0}\\) , it follows that \\(A_{i}Q\\) is the perpendicular bisector of \\(P_{i}^{0}P_{i + 1}^{0}\\) and \\(Q P_{i}^{0} = Q P_{i + 1}^{0}\\) . By applying this equality for all \\(i\\) , it follows that \\(\\mathcal{P}_{0}\\) is cyclic with circumcenter \\(Q\\) . This completes the solution for part (a). \n\nFor part (b), we will analyze the angle of rotation and scale factor for the similarity \\(\\mathcal{P}_{n}\\cup P\\stackrel {\\star}{\\sim}\\mathcal{P}_{0}\\cup P\\) . Assume without loss of generality that \\(A_{1}A_{2}\\dots A_{n}\\) is labeled in clockwise order as in the earlier diagrams. \n\nClaim — The spiral similarity at \\(P\\) sending \\(\\mathcal{P}_{0}\\cup Q\\) to \\(\\mathcal{P}_{n}\\cup O_{P}\\) is the composition of a clockwise rotation by \\(\\theta_{P}\\) and a dilation with factor \\(r_{P}\\) , where \n\n\\[\\theta_{P} = \\frac{n\\pi}{2} -\\sum_{i = 1}^{n}\\angle P A_{i}A_{i - 1}\\quad \\mathrm{and}\\quad r_{P} = \\prod_{i = 1}^{n}\\frac{1}{2\\sin\\angle P A_{i}A_{i - 1}}.\\] \n\nAnalogously, the spiral similarity at \\(Q\\) sending \\(\\mathcal{Q}_{0}\\cup P\\) to \\(\\mathcal{Q}_{n}\\cup O_{Q}\\) is the composition of a clockwise rotation by \\(\\theta_{Q}\\) and a dilation with factor \\(r_{Q}\\) , where \n\n\\[\\theta_{Q} = \\frac{n\\pi}{2} -\\sum_{i = 1}^{n}\\angle Q A_{i}A_{i - 1}\\quad \\mathrm{and}\\quad r_{Q} = \\prod_{i = 1}^{n}\\frac{1}{2\\sin\\angle Q A_{i}A_{i + 1}}.\\] \n\nProof. Note that \n\n\\[\\angle P P_{0}^{i}P P_{0}^{i + 1} = \\frac{\\pi}{2} -\\angle P P_{1}^{i}P_{0}^{i} = \\frac{\\pi}{2} -\\angle P P_{1}^{i}P_{i - 1}^{1} = \\frac{\\pi}{2} -\\angle P A_{i}A_{i - 1}\\] \n\nby circumcenter properties and Lemma 1.1. Summing over \\(0\\leq i< n\\) yields the claimed formula for \\(\\theta_{P}\\) , since the left hand side adds up to the rotation angle from \\(P P_{0}^{0}\\) to \\(P P_{0}^{n}\\) . \n\nBy the law of sines, \n\n\\[\\prod_{i = 1}^{n}\\frac{P P_{i}^{j + 1}}{P_{i}^{j}} = \\prod_{i = 1}^{n}\\frac{1}{2\\sin\\angle P P_{i + 1}^{j}P_{i}^{j}} = \\prod_{i = 1}^{n}\\frac{1}{2\\sin\\angle P A_{i}A_{i - 1}}\\] \n\nwhere in the last equality we use Lemma 1.1 again. Multiply over \\(0\\leq j< n\\) and raise to the power \\(\\frac{1}{n}\\) to obtain \n\n\\[\\left(\\prod_{i = 1}^{n}\\frac{P P_{i}^{n}}{P_{i}^{0}}\\right)^{\\frac{1}{n}} = \\prod_{i = 1}^{n}\\frac{1}{2\\sin\\angle P A_{i}A_{i - 1}}.\\] \n\nThis proves the formula for \\(r_{P}\\) because the left hand side gives the scale factor. The argument for \\(Q\\) is similar. Note that we reversed the angle in the formula for \\(r_{Q}\\) but not \\(\\theta_{Q}\\) because \\(\\theta_{Q}\\) depends on orientation. \n\nBy the given angle conditions on \\(P\\) and \\(Q\\) we have \\(r_{P} = r_{Q}\\) . Meanwhile, \n\n\\[\\theta_{P} + \\theta_{Q} = n\\pi -\\sum_{i = 1}^{n}(\\angle P A_{i}A_{i - 1} + \\angle Q A_{i}A_{i - 1}) = n\\pi -\\sum_{i = 1}^{n}\\angle A_{i - 1}A_{i}A_{i + 1} = 2\\pi .\\] \n\nThis means that clockwise rotation by \\(\\theta_{Q}\\) is just counterclockwise rotation by \\(\\theta_{P}\\) . The combination of these two implies that \\(P Q O_{Q}O_{P}\\) is an isosceles trapezoid with \\(O_{P}O_{Q}\\) parallel to \\(P Q\\) , which proves part (b).", "metadata": {"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2025.jsonl", "problem_match": "Problem 3. ", "solution_match": "## \\(\\S 1.3\\) Solution to TST 3, by Ruben Carpenter \n"}}
-{"year": "2025", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "Let \\(ABC\\) be a triangle, and let \\(X\\) , \\(Y\\) , and \\(Z\\) be collinear points such that \\(AY = AZ\\) , \\(BZ = BX\\) , and \\(CX = CY\\) . Points \\(X'\\) , \\(Y'\\) , and \\(Z'\\) are the reflections of \\(X\\) , \\(Y\\) , and \\(Z\\) over \\(BC\\) , \\(CA\\) , and \\(AB\\) , respectively. Prove that if \\(XY'Z'\\) is a nondegenerate triangle, then its circumcenter lies on the circumcircle of \\(ABC\\) .", "solution": "## \\(\\P\\) Solution 1 (Pitchayut Saengrungkonga) \n\n\n \n\nLet \\(S\\) denote the circumcenter of \\(\\triangle X^{\\prime}Y^{\\prime}Z^{\\prime}\\) . Observe that \\(A Y = A Z = A Y^{\\prime} = A Z^{\\prime}\\) , so \\(Y Z Y^{\\prime}Z^{\\prime}\\) is cyclic and \\(A S\\perp Y^{\\prime}Z^{\\prime}\\) . Similarly, \\(B S\\perp Z^{\\prime}X^{\\prime}\\) and \\(C S\\perp X^{\\prime}Y^{\\prime}\\) . \n\nThe rest is angle chasing. Let \\(\\angle \\ell\\) denote the angle between line \\(\\ell\\) and a fixed line. Then, we have \n\n\\[\\angle A S = 90^{\\circ} + \\angle Y^{\\prime}Z^{\\prime} = 90^{\\circ} + \\angle Y Y^{\\prime} + \\angle Z Z^{\\prime} - \\angle Y Z\\] \\[\\qquad = 90^{\\circ} + \\angle C A + \\angle A B - \\angle Y Z.\\] \n\nAnalogously, we get \n\n\\[\\angle B S = 90^{\\circ} + \\angle A B + \\angle B C - \\angle X Z,\\] \n\nso subtracting these gives \\(\\angle A S B = \\angle A C B\\) , as desired. \n\nRemark. There are some other angle chasing solutions that use the fact that \\(X X^{\\prime}\\) , \\(Y Y^{\\prime}\\) , and \\(Z Z^{\\prime}\\) meet at a point on \\((X^{\\prime}Y^{\\prime}Z^{\\prime})\\) . This one is featured as it does not require any additional points.\n\n\n\n\\(\\P\\) Solution 2 (author) Let \\(A^{\\prime}B^{\\prime}C^{\\prime}\\) be the anticomplimentary triangle of \\(A B C\\) . Note that \\(X\\) , \\(Y\\) , and \\(Z\\) are the projections of \\(A^{\\prime}\\) , \\(B^{\\prime}\\) , and \\(C^{\\prime}\\) onto line \\(X Y Z\\) . Let \\(A_{0}\\) , \\(B_{0}\\) , and \\(C_{0}\\) be the reflections of \\(A^{\\prime}\\) , \\(B^{\\prime}\\) , and \\(C^{\\prime}\\) over \\(B C\\) , \\(C A\\) , and \\(A B\\) , respectively. If we take \\(A^{\\prime}B^{\\prime}C^{\\prime}\\) as our reference triangle, we see that \\(A_{0}B_{0}C_{0}\\) is the orthic triangle so \\((A B C A_{0}B_{0}C_{0})\\) is the nine- point circle of \\(A^{\\prime}B^{\\prime}C^{\\prime}\\) . \n\nLet \\(A_{0}X^{\\prime}\\) meet \\((A B C)\\) again at \\(P\\) . Then \n\n\\[{\\angle B B_{0}P=\\angle A A_{0}P+\\angle B C A}\\] \\[{\\quad=\\angle A A_{0}X+\\angle B C A}\\] \\[{\\quad=\\angle(A^{\\prime}X,B C)+\\angle B C A}\\] \\[{\\quad=\\angle(B^{\\prime}Y,C A)}\\] \\[{\\quad=\\angle B B_{0}Y^{\\prime}.}\\] \n\nTherefore \\(P\\) lies on \\(B_{0}Y^{\\prime}\\) , and likewise \\(C_{0}Z^{\\prime}\\) by symmetry. \n\n\n \n\nLet \\(X_{0}\\) , \\(Y_{0}\\) , and \\(Z_{0}\\) be the reflections of \\(P\\) across \\(B C\\) , \\(C A\\) , and \\(A B\\) respectively. By reflection, \\(X_{0}\\) lies on \\(A^{\\prime}X\\) and \\(P X^{\\prime} = X_{0}X\\) . We also know that if \\(H\\) is the orthocenter of \\(A B C\\) , then \\(X_{0}Y_{0}Z_{0}H\\) is the Steiner line of \\(P\\) . Let \\(D\\) be the reflection of \\(H\\) across \\(B C\\) , which is also the antipode of \\(A_{0}\\) on \\((A B C)\\) . Then \\(\\angle A^{\\prime}X_{0}H = \\angle A_{0}P D = 90^{\\circ}\\) , so \\(X_{0}Y_{0}Z_{0}H\\) is perpendicular to \\(A^{\\prime}X_{0}X\\) and parallel to \\(X Y Z\\) . \n\nThis means that \\(X X_{0} = Y Y_{0} = Z Z_{0}\\) . Combined with \\(P X^{\\prime} = X X_{0}\\) and analogous statements, we have \\(P X^{\\prime} = P Y^{\\prime} = P Z^{\\prime}\\) so \\(P\\) is the circumcenter of \\(X^{\\prime}Y^{\\prime}Z^{\\prime}\\) . This solves the problem. \n\nRemark. If the condition that \\(X\\) , \\(Y\\) , and \\(Z\\) are collinear is removed, then in general the circumcenters of \\(X Y Z\\) and \\(X^{\\prime}Y^{\\prime}Z^{\\prime}\\) are isogonal conjugates with respect to \\(A B C\\) .", "metadata": {"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2025.jsonl", "problem_match": "Problem 4. ", "solution_match": "## \\(\\S 2.1\\) Solution to TST 4, by Michael Ren \n"}}
+{"year": "2025", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "Let \\(A_{1}A_{2}\\dots A_{2025}\\) be a convex 2025- gon, and let \\(A_{i} = A_{i + 2025}\\) for all integers \\(i\\) . Distinct points \\(P\\) and \\(Q\\) lie in its interior such that \\(\\angle A_{i - 1}A_{i}P = \\angle QA_{i}A_{i + 1}\\) for all \\(i\\) . Define points \\(P_{i}^{j}\\) and \\(Q_{i}^{j}\\) for integers \\(i\\) and positive integers \\(j\\) as follows: \n\nFor all \\(i\\) \\(P_{i}^{1} = Q_{i}^{1} = A_{i}\\) For all \\(i\\) and \\(j\\) \\(P_{i}^{j + 1}\\) and \\(Q_{i}^{j + 1}\\) are the circumcenters of \\(P P_{i}^{j}P_{i + 1}^{j}\\) and \\(Q Q_{i}^{j}Q_{i + 1}^{j}\\) respectively. \n\nLet \\(\\mathcal{P}\\) and \\(\\mathcal{Q}\\) be the polygons \\(P_{1}^{2025}P_{2}^{2025}\\ldots P_{2025}^{2025}\\) and \\(Q_{1}^{2025}Q_{2}^{2025}\\ldots Q_{2025}^{2025}\\) , respectively. \n\n(a) Prove that \\(\\mathcal{P}\\) and \\(\\mathcal{Q}\\) are cyclic. \n\n(b) Let \\(O_{P}\\) and \\(O_{Q}\\) be the circumcenters of \\(\\mathcal{P}\\) and \\(\\mathcal{Q}\\) , respectively. Assuming that \\(O_{P}\\neq O_{Q}\\) , show that \\(O_{P}O_{Q}\\) is parallel to \\(P Q\\) .", "solution": "Let \\(n = 2025\\) . Let \\(\\mathcal{P}_{i}\\) and \\(\\mathcal{Q}_{i}\\) denote the polygons \\(P_{1}^{i}\\cdot \\cdot \\cdot P_{n}^{i}\\) and \\(Q_{1}^{i}\\cdot \\cdot \\cdot Q_{n}^{i}\\) . In this notation, \\(\\mathcal{P} = \\mathcal{P}_{n}\\) , \\(\\mathcal{Q} = \\mathcal{Q}_{n}\\) , and \\(\\mathcal{P}_{1} = \\mathcal{Q}_{1} = A_{1}\\cdot \\cdot \\cdot A_{n}\\) . \n\nThe angle condition for \\(P\\) and \\(Q\\) just says that they are isogonal conjugates in \\(\\mathcal{P}_{1}\\) . We will first find some properties that do not depend on \\(P\\) having an isogonal conjugate. \n\nNote that \\(P_{i - 1}^{j + 1}P_{i}^{j + 1}\\) is the perpendicular bisector of \\(P P_{i + 1}^{j}\\) , so we can go backwards from \\(\\mathcal{P}_{j}\\) to \\(\\mathcal{P}_{j - 1}\\) by reflecting over the sides. Use this to extend the points backwards to \\(\\mathcal{P}_{0}\\) , i.e. define \\(P_{i}^{0}\\) to be the reflection of \\(P\\) over \\(P_{i - 1}^{1}P_{i}^{1}\\) . \n\n## Lemma 1.1 \n\nFor integers \\(j \\geq 0\\) and \\(i\\) , \n\n\\[\\angle P P_{i}^{j}P_{i + 1}^{j} = \\angle P P_{i}^{j + 1}P_{i + 1}^{j + 1}\\quad \\mathrm{and}\\] \\[\\angle P P_{i + 1}^{j}P_{i}^{j} = \\angle P P_{i}^{j + 1}P_{i - 1}^{j + 1}.\\] \n\nProof. This is a standard circumcenter fact in disguise but we will prove it here for completeness. If \\(X\\) is the antipode of \\(P\\) on \\((P P_{i}^{j}P_{i + 1}^{j})\\) (equivalently, the reflection of \\(P\\) over \\(P P_{i}^{j + 1}\\) ) and \\(M\\) is the midpoint of \\(P P_{i + 1}^{j}\\) then \n\n\\[\\angle P P_{i}^{j}P_{i + 1}^{j} = \\angle P X P_{i + 1}^{j} = \\angle P P_{i}^{j + 1}M = \\angle P P_{i}^{j + 1}P_{i + 1}^{j + 1}.\\] \n\nThe second equality is analogous.\n\n\n\n \n\n## Lemma \n\nThe following similarity holds: \\(\\mathcal{P}_{n} \\cup P \\stackrel{\\perp}{\\sim} \\mathcal{P}_{0} \\cup P\\) . \n\nProof. By Lemma 1.1, \n\n\\[\\begin{aligned} \\angle P P_{i}^{0} P_{i + 1}^{0} & = \\angle P P_{i}^{1} P_{i + 1}^{1} = \\dots = \\angle P_{i}^{n} P_{i + 1}^{n} \\quad \\text{and} \\\\ \\angle P P_{i + 1}^{0} P_{i}^{0} & = \\angle P P_{i}^{1} P_{i - 1}^{1} = \\dots = \\angle P_{i - n + 1}^{n} P_{i - n}^{n} = \\angle P_{i + 1}^{n} P_{i}^{n}. \\end{aligned}\\] \n\nTherefore \\(P P_{i}^{n} P_{i + 1}^{n} \\stackrel{\\perp}{\\sim} P P_{i}^{0} P_{i + 1}^{0}\\) . Combining these similarities for all \\(i\\) shows that \\(\\mathcal{P}_{n} \\cup P \\stackrel{\\perp}{\\sim} \\mathcal{P}_{0} \\cup P\\) . \\(\\square\\) \n\nTo solve part (a), we just need to show that \\(\\mathcal{P}_{0}\\) and \\(\\mathcal{Q}_{0}\\) are cyclic. This is where we need the isogonal conjugate property, and we can generalize from analogous facts for the \\(n = 3\\) case. \n\n\n\n\n\n\nWe have \n\n\\[\\angle P P_{i + 1}^{0}P_{i}^{0} = \\angle P A_{i}A_{i - 1} = \\angle A_{i + 1}A_{i}Q\\] \n\nso \n\n\\[90^{\\circ} = \\angle (P P_{i + 1}^{0},A_{i}A_{i + 1}) = \\angle (P_{i + 1}^{0}P_{i}^{0},A Q).\\] \n\nSince \\(A_{i}P_{i}^{0} = A_{i}P_{i + 1}^{0}\\) , it follows that \\(A_{i}Q\\) is the perpendicular bisector of \\(P_{i}^{0}P_{i + 1}^{0}\\) and \\(Q P_{i}^{0} = Q P_{i + 1}^{0}\\) . By applying this equality for all \\(i\\) , it follows that \\(\\mathcal{P}_{0}\\) is cyclic with circumcenter \\(Q\\) . This completes the solution for part (a). \n\nFor part (b), we will analyze the angle of rotation and scale factor for the similarity \\(\\mathcal{P}_{n}\\cup P\\stackrel {\\star}{\\sim}\\mathcal{P}_{0}\\cup P\\) . Assume without loss of generality that \\(A_{1}A_{2}\\dots A_{n}\\) is labeled in clockwise order as in the earlier diagrams. \n\nClaim — The spiral similarity at \\(P\\) sending \\(\\mathcal{P}_{0}\\cup Q\\) to \\(\\mathcal{P}_{n}\\cup O_{P}\\) is the composition of a clockwise rotation by \\(\\theta_{P}\\) and a dilation with factor \\(r_{P}\\) , where \n\n\\[\\theta_{P} = \\frac{n\\pi}{2} -\\sum_{i = 1}^{n}\\angle P A_{i}A_{i - 1}\\quad \\mathrm{and}\\quad r_{P} = \\prod_{i = 1}^{n}\\frac{1}{2\\sin\\angle P A_{i}A_{i - 1}}.\\] \n\nAnalogously, the spiral similarity at \\(Q\\) sending \\(\\mathcal{Q}_{0}\\cup P\\) to \\(\\mathcal{Q}_{n}\\cup O_{Q}\\) is the composition of a clockwise rotation by \\(\\theta_{Q}\\) and a dilation with factor \\(r_{Q}\\) , where \n\n\\[\\theta_{Q} = \\frac{n\\pi}{2} -\\sum_{i = 1}^{n}\\angle Q A_{i}A_{i - 1}\\quad \\mathrm{and}\\quad r_{Q} = \\prod_{i = 1}^{n}\\frac{1}{2\\sin\\angle Q A_{i}A_{i + 1}}.\\] \n\nProof. Note that \n\n\\[\\angle P P_{0}^{i}P P_{0}^{i + 1} = \\frac{\\pi}{2} -\\angle P P_{1}^{i}P_{0}^{i} = \\frac{\\pi}{2} -\\angle P P_{1}^{i}P_{i - 1}^{1} = \\frac{\\pi}{2} -\\angle P A_{i}A_{i - 1}\\] \n\nby circumcenter properties and Lemma 1.1. Summing over \\(0\\leq i< n\\) yields the claimed formula for \\(\\theta_{P}\\) , since the left hand side adds up to the rotation angle from \\(P P_{0}^{0}\\) to \\(P P_{0}^{n}\\) . \n\nBy the law of sines, \n\n\\[\\prod_{i = 1}^{n}\\frac{P P_{i}^{j + 1}}{P_{i}^{j}} = \\prod_{i = 1}^{n}\\frac{1}{2\\sin\\angle P P_{i + 1}^{j}P_{i}^{j}} = \\prod_{i = 1}^{n}\\frac{1}{2\\sin\\angle P A_{i}A_{i - 1}}\\] \n\nwhere in the last equality we use Lemma 1.1 again. Multiply over \\(0\\leq j< n\\) and raise to the power \\(\\frac{1}{n}\\) to obtain \n\n\\[\\left(\\prod_{i = 1}^{n}\\frac{P P_{i}^{n}}{P_{i}^{0}}\\right)^{\\frac{1}{n}} = \\prod_{i = 1}^{n}\\frac{1}{2\\sin\\angle P A_{i}A_{i - 1}}.\\] \n\nThis proves the formula for \\(r_{P}\\) because the left hand side gives the scale factor. The argument for \\(Q\\) is similar. Note that we reversed the angle in the formula for \\(r_{Q}\\) but not \\(\\theta_{Q}\\) because \\(\\theta_{Q}\\) depends on orientation. \n\nBy the given angle conditions on \\(P\\) and \\(Q\\) we have \\(r_{P} = r_{Q}\\) . Meanwhile, \n\n\\[\\theta_{P} + \\theta_{Q} = n\\pi -\\sum_{i = 1}^{n}(\\angle P A_{i}A_{i - 1} + \\angle Q A_{i}A_{i - 1}) = n\\pi -\\sum_{i = 1}^{n}\\angle A_{i - 1}A_{i}A_{i + 1} = 2\\pi .\\] \n\nThis means that clockwise rotation by \\(\\theta_{Q}\\) is just counterclockwise rotation by \\(\\theta_{P}\\) . The combination of these two implies that \\(P Q O_{Q}O_{P}\\) is an isosceles trapezoid with \\(O_{P}O_{Q}\\) parallel to \\(P Q\\) , which proves part (b).", "metadata": {"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2025.jsonl", "problem_match": "Problem 3. ", "solution_match": "## \\(\\S 1.3\\) Solution to TST 3, by Ruben Carpenter \n"}}
+{"year": "2025", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "Let \\(ABC\\) be a triangle, and let \\(X\\) , \\(Y\\) , and \\(Z\\) be collinear points such that \\(AY = AZ\\) , \\(BZ = BX\\) , and \\(CX = CY\\) . Points \\(X'\\) , \\(Y'\\) , and \\(Z'\\) are the reflections of \\(X\\) , \\(Y\\) , and \\(Z\\) over \\(BC\\) , \\(CA\\) , and \\(AB\\) , respectively. Prove that if \\(XY'Z'\\) is a nondegenerate triangle, then its circumcenter lies on the circumcircle of \\(ABC\\) .", "solution": "## \\(\\P\\) Solution 1 (Pitchayut Saengrungkonga) \n\n\n \n\nLet \\(S\\) denote the circumcenter of \\(\\triangle X^{\\prime}Y^{\\prime}Z^{\\prime}\\) . Observe that \\(A Y = A Z = A Y^{\\prime} = A Z^{\\prime}\\) , so \\(Y Z Y^{\\prime}Z^{\\prime}\\) is cyclic and \\(A S\\perp Y^{\\prime}Z^{\\prime}\\) . Similarly, \\(B S\\perp Z^{\\prime}X^{\\prime}\\) and \\(C S\\perp X^{\\prime}Y^{\\prime}\\) . \n\nThe rest is angle chasing. Let \\(\\angle \\ell\\) denote the angle between line \\(\\ell\\) and a fixed line. Then, we have \n\n\\[\\angle A S = 90^{\\circ} + \\angle Y^{\\prime}Z^{\\prime} = 90^{\\circ} + \\angle Y Y^{\\prime} + \\angle Z Z^{\\prime} - \\angle Y Z\\] \\[\\qquad = 90^{\\circ} + \\angle C A + \\angle A B - \\angle Y Z.\\] \n\nAnalogously, we get \n\n\\[\\angle B S = 90^{\\circ} + \\angle A B + \\angle B C - \\angle X Z,\\] \n\nso subtracting these gives \\(\\angle A S B = \\angle A C B\\) , as desired. \n\nRemark. There are some other angle chasing solutions that use the fact that \\(X X^{\\prime}\\) , \\(Y Y^{\\prime}\\) , and \\(Z Z^{\\prime}\\) meet at a point on \\((X^{\\prime}Y^{\\prime}Z^{\\prime})\\) . This one is featured as it does not require any additional points.\n\n\n\n\\(\\P\\) Solution 2 (author) Let \\(A^{\\prime}B^{\\prime}C^{\\prime}\\) be the anticomplimentary triangle of \\(A B C\\) . Note that \\(X\\) , \\(Y\\) , and \\(Z\\) are the projections of \\(A^{\\prime}\\) , \\(B^{\\prime}\\) , and \\(C^{\\prime}\\) onto line \\(X Y Z\\) . Let \\(A_{0}\\) , \\(B_{0}\\) , and \\(C_{0}\\) be the reflections of \\(A^{\\prime}\\) , \\(B^{\\prime}\\) , and \\(C^{\\prime}\\) over \\(B C\\) , \\(C A\\) , and \\(A B\\) , respectively. If we take \\(A^{\\prime}B^{\\prime}C^{\\prime}\\) as our reference triangle, we see that \\(A_{0}B_{0}C_{0}\\) is the orthic triangle so \\((A B C A_{0}B_{0}C_{0})\\) is the nine- point circle of \\(A^{\\prime}B^{\\prime}C^{\\prime}\\) . \n\nLet \\(A_{0}X^{\\prime}\\) meet \\((A B C)\\) again at \\(P\\) . Then \n\n\\[{\\angle B B_{0}P=\\angle A A_{0}P+\\angle B C A}\\] \\[{\\quad=\\angle A A_{0}X+\\angle B C A}\\] \\[{\\quad=\\angle(A^{\\prime}X,B C)+\\angle B C A}\\] \\[{\\quad=\\angle(B^{\\prime}Y,C A)}\\] \\[{\\quad=\\angle B B_{0}Y^{\\prime}.}\\] \n\nTherefore \\(P\\) lies on \\(B_{0}Y^{\\prime}\\) , and likewise \\(C_{0}Z^{\\prime}\\) by symmetry. \n\n\n \n\nLet \\(X_{0}\\) , \\(Y_{0}\\) , and \\(Z_{0}\\) be the reflections of \\(P\\) across \\(B C\\) , \\(C A\\) , and \\(A B\\) respectively. By reflection, \\(X_{0}\\) lies on \\(A^{\\prime}X\\) and \\(P X^{\\prime} = X_{0}X\\) . We also know that if \\(H\\) is the orthocenter of \\(A B C\\) , then \\(X_{0}Y_{0}Z_{0}H\\) is the Steiner line of \\(P\\) . Let \\(D\\) be the reflection of \\(H\\) across \\(B C\\) , which is also the antipode of \\(A_{0}\\) on \\((A B C)\\) . Then \\(\\angle A^{\\prime}X_{0}H = \\angle A_{0}P D = 90^{\\circ}\\) , so \\(X_{0}Y_{0}Z_{0}H\\) is perpendicular to \\(A^{\\prime}X_{0}X\\) and parallel to \\(X Y Z\\) . \n\nThis means that \\(X X_{0} = Y Y_{0} = Z Z_{0}\\) . Combined with \\(P X^{\\prime} = X X_{0}\\) and analogous statements, we have \\(P X^{\\prime} = P Y^{\\prime} = P Z^{\\prime}\\) so \\(P\\) is the circumcenter of \\(X^{\\prime}Y^{\\prime}Z^{\\prime}\\) . This solves the problem. \n\nRemark. If the condition that \\(X\\) , \\(Y\\) , and \\(Z\\) are collinear is removed, then in general the circumcenters of \\(X Y Z\\) and \\(X^{\\prime}Y^{\\prime}Z^{\\prime}\\) are isogonal conjugates with respect to \\(A B C\\) .", "metadata": {"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2025.jsonl", "problem_match": "Problem 4. ", "solution_match": "## \\(\\S 2.1\\) Solution to TST 4, by Michael Ren \n"}}
{"year": "2025", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "A pond has 2025 lily pads arranged in a circle. Two frogs, Alice and Bob, begin on different lily pads. A frog jump is a jump which travels 2, 3, or 5 positions clockwise. Alice and Bob each make a series of frog jumps, and each frog ends on the same lily pad that it started from. Given that each lily pad is the destination of exactly one jump, prove that each frog completes exactly two laps around the pond (i.e. travels 4050 positions in total).", "solution": "Let \\(\\pi :\\mathbb{Z} / 2025\\mathbb{Z}\\to \\mathbb{Z} / 2025\\mathbb{Z}\\) be the permutation where the jump with source \\(i\\) has destination \\(\\pi (i)\\) . We know that \\(\\pi\\) has exactly two cycles, corresponding to the paths of each frog. \n\nSuppose that the frogs complete a total of \\(\\ell\\) laps around the circle, so the sum of the lengths of all the jumps made is \\(2025\\ell\\) . Thus the average jump moves \\(\\ell\\) positions, so \\(\\ell \\in \\{2,3,4,5\\}\\) . We now split into cases. \n\n\\(\\P\\) Case 1: \\(\\ell \\in \\{2,5\\}\\) If \\(\\ell = 2\\) , then every jump travels 2 positions, so \\(\\pi (i) = i + 2\\) for all \\(i\\) . This permutation has exactly one cycle, contradiction. If \\(\\ell = 5\\) , then every jump travels 5 positions, so \\(\\pi (i) = i + 5\\) for all \\(i\\) . This permutation has exactly five cycles, contradiction. \n\n\\(\\P\\) Case 2: \\(\\ell = 3\\) The key idea is to consider the cycle decomposition of the auxiliary permutation \\(\\pi^{\\prime}:\\mathbb{Z} / 2025\\mathbb{Z}\\to \\mathbb{Z} / 2025\\mathbb{Z}\\) by \\(\\pi^{\\prime}(i) = \\pi (i) - 3\\) . The average jump in \\(\\pi^{\\prime}\\) (i.e. the distance from \\(i\\) to \\(\\pi^{\\prime}(i)\\) traveling clockwise) travels 0 positions. Thus, either each cycle of \\(\\pi^{\\prime}\\) either travels net zero positions, or there is some cycle that travels net negative positions. Since \\(\\pi^{\\prime}(i)\\in \\{i - 1,i,i + 2\\}\\) , the latter case can only happen if \\(\\pi^{\\prime}(i) = i - 1\\) for all \\(i\\) , which has average jump \\(- 1\\) , not 0, contradiction. Thus, each cycle travels net 0 positions. It is easy to see that the only such cycles are \\(i\\mapsto i\\) and \\(i\\mapsto i + 2\\mapsto i + 1\\mapsto i\\) \n\nSay that \\(i\\) is a unicycle start if \\(i\\mapsto i\\) is a cycle in \\(\\pi^{\\prime}\\) , a tricycle start if \\(i\\mapsto i + 2\\mapsto\\) \\(i + 1\\mapsto i\\) is a cycle in \\(\\pi^{\\prime}\\) , and a cycle start if it is either a unicycle start or a tricycle start. It is easy to see that if \\(i\\) is a unicycle start, then \\(i + 1\\) must be a cycle start, and if \\(i\\) is a tricycle start, then \\(i + 3\\) must be a cycle start. \n\nGiven this structural classification of \\(\\pi^{\\prime}\\) , we will now generate a contradiction by showing that \\(\\pi\\) can't have exactly two cycles. Place auxillary frogs \\(C\\) , \\(D\\) , and \\(E\\) on lily pads \\(i\\) , \\(i + 1\\) , and \\(i + 2\\) respectively, where \\(i\\) is a cycle start. If \\(i\\) is a unicycle start, then have \\(C\\) jump to \\(\\pi (i) = i + 3\\) . If \\(i\\) is a tricycle start, then have \\(C\\) , \\(D\\) , and \\(E\\) jump to \\(\\pi (i) = i + 5\\) , \\(\\pi (i + 1) = i + 3\\) , and \\(\\pi (i + 2) = i + 4\\) , respectively. Note that \\(C\\) , \\(D\\) , and \\(E\\) are still on consecutive lily pads in some order, with the first lily pad number being a cycle start. We can repeat this process, having either the first frog or all three frogs jump according to permutation \\(\\pi\\) , depending on whether the first lily pad number is a unicycle start or a tricycle start. When all of the cycles of \\(\\pi^{\\prime}\\) have been exhausted in this manner, \\(C\\) , \\(D\\) , and \\(E\\) are back to positions \\(i\\) , \\(i + 1\\) , and \\(i + 2\\) , in some order. Note that each step of the above process permutes \\(C\\) , \\(D\\) , and \\(E\\) with an even permutation, so the final permutation of \\(C\\) , \\(D\\) , and \\(E\\) must be \\(CDE\\) , \\(DEC\\) , or \\(ECD\\) . The first case implies that \\(\\pi\\) has three cycles, while the other two cases imply that \\(\\pi\\) has only one cycle, a contradiction.\n\n\n\nCase 3: \\(\\ell = 4\\) Now consider the auxillary permutation \\(\\pi^{\\prime \\prime}:\\mathbb{Z} / 2025\\mathbb{Z}\\to \\mathbb{Z} / 2025\\mathbb{Z}\\) given by \\(\\pi^{\\prime \\prime}(i) = \\pi (i) - 4\\) . Note that \\(\\pi^{\\prime \\prime}(i)\\in \\{i - 2,i - 1,i + 1\\}\\) , so a similar analysis as the \\(\\ell = 3\\) case shows that the only two possible types of cycles in \\(\\pi^{\\prime \\prime}\\) are \\(i\\to i + 1\\to i\\) and \\(i\\to i + 1\\to i + 2\\to i\\) . Again, define \\(i\\) to be a bicycle start if \\(i\\mapsto i + 1\\mapsto i\\) is a cycle in \\(\\pi^{\\prime \\prime}\\) a tricycle start if \\(i\\mapsto i + 1\\mapsto i + 2\\mapsto i\\) is a cycle in \\(\\pi^{\\prime \\prime}\\) , and a cycle start if it is either a bicycle or tricycle start. It is easy to see that if \\(i\\) is a bicycle start, then \\(i + 2\\) must be a cycle start, and if \\(i\\) is a tricycle start, then \\(i + 3\\) must be a cycle start. \n\nNow, place auxillary frogs \\(C\\) , \\(D\\) , \\(E\\) , and \\(F\\) on lily pads \\(i\\) , \\(i + 1\\) , \\(i + 2\\) , and \\(i + 3\\) respectively, where \\(i\\) is a cycle start. If \\(i\\) is a bicycle start, have \\(C\\) jump to \\(\\pi (i) = i + 5\\) and \\(D\\) jump to \\(\\pi (i + 1) = i + 4\\) , and if \\(i\\) is a tricycle start, have \\(C\\) jump to \\(\\pi (i) = i + 5\\) , \\(D\\) jump to \\(\\pi (i + 1) = i + 6\\) , and \\(E\\) jump to \\(\\pi (i + 2) = i + 4\\) . Note that \\(C\\) , \\(D\\) , \\(E\\) , and \\(F\\) are still on consecutive lily pads in some order, with the first lily pad number being a cycle start. If \\(i\\) is a bicycle start, then it causes the frogs to permute according to \\(CDEF\\mapsto EFDC\\) , and if \\(i\\) is a tricycle start, then it causes the frogs to permute according to \\(CDEF\\mapsto FECD\\) . \n\nThus, when all of the cycles of \\(\\pi^{\\prime \\prime}\\) have been exhausted in this manner, \\(C\\) , \\(D\\) , \\(E\\) , and \\(F\\) are back to positions \\(i\\) , \\(i + 1\\) , \\(i + 2\\) , and \\(i + 3\\) , in one of the permutations \\(CDEF\\) , \\(EFDC\\) , \\(DCFE\\) , or \\(FECD\\) . The first case implies that \\(\\pi\\) has four cycles, while the second and fourth cases imply that \\(\\pi\\) has only one cycle, a contradiction, so we must be in the third case, i.e. the frogs end up on lily pads \\(i\\) , \\(i + 1\\) , \\(i + 2\\) , and \\(i + 3\\) in the order \\(DCFE\\) . Therefore, one of Alice or Bob travels the combined path of \\(C\\) and \\(D\\) , while the other travels the combined path of \\(E\\) and \\(F\\) , each of which completes two laps around the circle, as desired. \n\nRemark. 2025 can be replaced with any odd number (sufficiently large as to avoid confusion about what a \"lap\" is), and the problem statement will still be true. If 2025 is replaced with an even number, then the problem statement is false only due to the \\(\\ell = 2\\) case. \n\nRemark. Here are some variations of the problem (not checked thoroughly): \n\nIf 2025 is replaced by any odd number and the allowed distances are changed from 2, 3, 5 to 1, 2, 4, then you can conclude that one frog completes one lap and the other completes two laps. \n\nIf 2025 is replaced by any number and the allowed distances are changed to 1, 3, 4, then you can conclude that each frog completes only one lap.", "metadata": {"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2025.jsonl", "problem_match": "Problem 5. ", "solution_match": "## \\(\\S 2.2\\) Solution to TST 5, by Linus Tang \n"}}
{"year": "2025", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "Prove that there exists a real number \\(\\epsilon > 0\\) such that there are infinitely many sequences of integers \\(0 < a_{1} < a_{2} < \\ldots < a_{2025}\\) satisfying \n\n\\[\\gcd (a_{1}^{2} + 1,a_{2}^{2} + 1,\\ldots ,a_{2025}^{2} + 1) > a_{2025}^{1 + \\epsilon}.\\]", "solution": "\\(\\P\\) Solution 1 (Alexander Wang) By the Chinese Remainder Theorem for \\(\\mathbb{Q}[x]\\) , for any choice of \\(2^{12}\\) signs there exists a unique polynomial \\(f_{i}\\) of degree at most 23 such that \n\n\\[f_{i}\\equiv \\pm cx\\pmod {c^{2}x^{2} + 1}\\] \n\nfor \\(c = 1\\) , 2, ..., 12. Furthermore, these polynomials are multiples of \\(x\\) and come in pairs which sum to zero, so we can pick 2048 of these polynomials which have positive leading coefficients and label them \\(f_{1}\\) , \\(f_{2}\\) , ..., \\(f_{2048}\\) . \n\nLet \\(N\\) be a positive integer such that any coefficient of the polynomials is \\(\\frac{1}{N}\\) times an integer. For a suffiently large positive integer \\(x\\) , take \\(a_{i} = f_{i}(Nx)\\) for \\(i = 1\\) , 2, ..., 2025, which will be positive integers. Then, \n\n\\[\\gcd (a_{1}^{2} + 1,a_{2}^{2} + 1,\\ldots ,a_{2025}^{2} + 1)\\geq \\prod_{c = 1}^{12}(c^{2}x^{2} + 1)\\] \n\nbecause \\(a_{i}^{2} + 1\\equiv 0\\) (mod \\(c^{2}x^{2} + 1\\) ) for any \\(i\\) by construction. The right hand side is asymptotically \\(x^{24}\\) while \\(a_{2025}\\) is \\(x^{23}\\) up to constant factors, so any \\(\\epsilon < \\frac{1}{23}\\) works. \n\nRemark. In terms of \\(n = 2025\\) , this solution achieves \\(\\epsilon = \\Omega \\left(\\frac{1}{\\log n}\\right)\\) which is the best that we know of. Solution 2 achieves \\(\\epsilon = \\Omega \\left(\\frac{1}{n^{\\log_{2}(3)}}\\right)\\) while solutions 3 and 4 achieve \\(\\epsilon = \\Omega \\left(\\frac{1}{n}\\right)\\) . \n\n\\(\\P\\) Solution 2 (Luke Robitaille) Define the sequence of polynomials \\(P_{i}\\) by \n\n\\[P_{1}(x) = x^{2} + 1,\\] \\[P_{n}(x) = \\left(\\prod_{i = 1}^{n - 1}P_{i}(x)\\right)^{2} + 1\\mathrm{~for~}n > 1.\\] \n\nDue to the recurrence, we have \\(\\gcd (P_{i}(x),P_{j}(x)) = 1\\) for \\(i\\neq j\\) . Let \n\n\\[P(x) = P_{1}(x)P_{2}(x)\\ldots P_{m}(x),\\] \n\nwhich is a polynomial of degree \\(2\\cdot 3^{m}\\) . For \\(i\\geq 2\\) we have \n\n\\[\\left(\\frac{P(x)}{P_{i}(x)}\\right)^{2} = \\left(\\prod_{i = 1}^{n - 1}P_{i}(x)\\right)^{2}\\times (P_{i + 1}(x))^{2}\\times \\cdot \\cdot \\times (P_{m}(x))^{2}\\] \\[\\equiv -1\\times 1\\times \\cdot \\cdot \\times 1\\equiv -1\\pmod {P_{i}(x)}.\\]\n\n\n\nand additionally \n\n\\[\\left(\\frac{x P(x)}{P_{1}(x)}\\right)^{2} = x^{2}\\times \\left(\\prod_{i = 2}^{m}P_{i}(x)\\right)^{2}\\] \\[\\equiv -1\\pmod {P_{1}(x)}.\\] \n\nNow consider all \\(2^{m - 1}\\) polynomials of the form \n\n\\[Q(x) = \\frac{x P(x)}{P_{1}(x)}\\pm \\frac{P(x)}{P_{2}(x)}\\pm \\frac{P(x)}{P_{3}(x)}\\pm \\dots \\pm \\frac{P(x)}{P_{m}(x)}.\\] \n\nEach such polynomial has leading coefficient 1 and degree \\(2\\cdot 3^{m} - 1\\) , and they are all distinct (the terms are in decreasing order of degree from left to right). Furthermore, each \\(Q(x)\\) satisfies \\(Q(x)^{2}\\equiv - 1\\) (mod \\(P(x)\\) ) by the Chinese Remainder Theorem. \n\nNow we can construct the solution. Let \\(m = 12\\) (so \\(2^{m - 1} > 2025\\) ), \\(\\epsilon < \\frac{1}{2\\cdot 3^{m - 1} - 1}\\) , \\(x\\) be a (large) positive integer and take \\(a_{1},\\ldots ,a_{2025}\\) to be distinct values of \\(Q(x)\\) as described above. Then \\(\\gcd (a_{1}^{2} + 1,a_{2}^{2} + 1,\\ldots ,a_{2025}^{2} + 1)\\geq P(x)\\) and \n\n\\[a_{2025}^{1 + \\epsilon}< a_{2025}^{2\\cdot 3^{m} - 1}< x^{2\\cdot 3^{m}}< P(x)\\] \n\nfor sufficiently large \\(x\\) . Since there are infinitely many such \\(x\\) , we are done. \n\nRemark. This solution takes some inspiration from the other solutions that follow, but stands out for not having to deal with any polynomials outside of \\(\\mathbb{Z}[x]\\) . \n\n\\(\\P\\) Solution 3 (author) We first explain the general idea of the construction. The idea is to work in polynomial ring \\(\\mathbb{Q}[x]\\) . Suppose that the target \\(\\mathrm{gcd}\\) is \\(g(x)\\) , which will have a lot of factors. Then, the equation \\(f(x)^{2}\\equiv - 1\\) (mod \\(g(x)\\) ) will have multiple solutions \\(f_{1},f_{2},\\ldots ,f_{n}\\) , and we will have \\(\\deg f_{i}< \\deg g\\) . Thus, we may take \\(\\begin{array}{r}{\\epsilon = \\frac{1}{\\deg g}} \\end{array}\\) . However, one also needs to find \\(x\\) for which \\(f_{1}(x),\\ldots ,f_{n}(x)\\) are integers. The way we get around this is to find some explicit choice of \\(f_{i}\\) and \\(g\\) as described above, and then pick \\(x\\) manually to ensure integrality. \n\nLet \\(p_{1},p_{2},\\ldots ,p_{n}\\equiv 1\\) (mod 4) be distinct primes that we will constrain later. Let \\(N = p_{1}p_{2}\\ldots p_{n}\\) . Our \\(g\\) will be \n\n\\[g(x) = x^{2N} + 1.\\] \n\nThen, we claim that \n\nClaim — the polynomial \n\n\\[f_{i}(x) = \\frac{2p_{i}}{N}\\cdot \\frac{x^{p_{i}}(x^{2N} + 1)}{x^{2p_{i}} + 1} -x^{N}\\] \n\nsatisfies \\(f_{i}^{2}\\equiv - 1\\) (mod \\(g\\) ). \n\nProof. A direct computation shows that \n\n\\[f_{i}\\equiv x^{p_{i}}\\pmod{x^{2p_{i}} + 1\\] \\[f_{i}\\equiv -x^{N}\\pmod{\\frac{x^{2N} + 1}{x^{2p_{i}} + 1}}\\]\n\n\n\nIn the next claim, we will show that one may choose \\(p_{1},\\ldots ,p_{n}\\) so that \\(f_{1}(2)\\) , ..., \\(f_{n}(2)\\) are all integers. This will imply that if \\(a\\equiv 2\\) (mod \\(N\\) ), then \\(f_{1}(a)\\) , ..., \\(f_{n}(a)\\) are integers all divisible by \\(\\frac{g(a)}{N}\\) . By taking \\(a\\) large, this gives a construction for \\(\\epsilon < \\frac{1}{2N}\\) . \n\nTo finish the problem, we prove the claim. \n\nClaim — There exists \\(n\\) distinct primes \\(p_{1},p_{2},\\ldots ,p_{n}\\equiv 1\\) (mod 4) such that for any indices \\(i\\neq j\\) , \n\n\\[p_{j}\\mid \\frac{4^{p_{1}p_{2}\\ldots p_{n}} + 1}{4^{p_{i}} + 1}.\\] \n\nProof. First, we will establish a prime sequence which \\(p_{1}p_{2}\\ldots p_{n}\\mid 4^{p_{1}p_{2}\\ldots p_{n}} + 1\\) . This is essentially IMO 2000 P5: begin with \\(p_{1} = 5\\) and for each \\(i\\) , assign (by Zsigmondy) \\(p_{i}\\) a primitive prime divisor of \\(4^{p_{1}p_{2}\\ldots p_{i - 1}} + 1\\) ; it's easy to see that this works as advertised. \n\nNow, for each \\(i,j\\) , the divisibility relation is verified unless \\(p_{j}\\mid 4^{p_{i}} + 1\\) , in which case we can apply Lifting the Exponent lemma to complete the proof. Finally, note that \\(p_{i}\\equiv 1\\) (mod 4) because each divide \\(4^{p_{1}p_{2}\\ldots p_{n}} + 1\\) , a perfect square plus one. \\(\\square\\) \n\n\\(\\P\\) Solution 4 (Carl Schildkraut, Mihir Singhal, Brandon Wang) We show the result for each \\(n = 2^{m - 1}\\) . For positive integers \\(i\\) , define polynomials \\(h_{i}(x) = x^{2^{i - 1}}\\) and \\(f_{i}(x) = h_{i}(x)^{2} + 1 = x^{2^{i}} + 1\\) . The main claim is the following: \n\nClaim — For every positive integer \\(m\\) , there exist polynomials \\(g_{1},\\ldots ,g_{m}\\in \\mathbb{Z}[x]\\) satisfying \n\n\\[\\sum_{i = 1}^{m}\\frac{g_{i}(x)}{f_{i}(x)} = \\frac{2^{m - 1}}{f_{1}(x)\\cdot\\cdot\\cdot f_{m}(x)}.\\] \n\nProof. We prove the result by induction on \\(m\\) . In the base case of \\(m = 1\\) , we may simply take \\(g_{1}(x) = 1\\) . \n\nFor the inductive step, suppose \\(m\\geq 2\\) , and the result holds for \\(m - 1\\) . Let \\(p_{1},\\ldots ,p_{m - 1}\\) be polynomials for which \n\n\\[\\sum_{i = 1}^{m - 1}\\frac{p_{i}(x)}{f_{i}(x)} = \\frac{2^{m - 2}}{f_{1}(x)\\cdot\\cdot\\cdot f_{m - 1}(x)}.\\] \n\nWrite \\(p_{0}(x) = p_{m}(x)\\) for notational simplicity. Since \\(f_{i + 1}(x) = f_{i}(x^{2})\\) , we also have \n\n\\[\\sum_{i = 1}^{m - 1}\\frac{p_{i}(x^{2})}{f_{i + 1}(x)} = \\frac{2^{m - 2}}{f_{2}(x)\\cdot\\cdot\\cdot f_{m}(x)}.\\] \n\nDefining \\(g_{i}(x) = p_{i}(x) - p_{i - 1}(x^{2})\\cdot \\frac{x^{2^{m}} - 1}{x^{2} + 1}\\) , we have \n\n\\[\\sum_{i = 1}^{m}\\frac{g_{i}(x)}{f_{i}(x)} = \\sum_{i = 1}^{m}\\frac{p_{i}(x)}{f_{i}(x)} -\\frac{x^{2^{m}} - 1}{x^{2} + 1}\\sum_{i = 1}^{m}\\frac{p_{i - 1}(x^{2})}{f_{i}(x)}\\] \\[\\qquad = \\frac{2^{m - 2}}{f_{1}\\cdot\\cdot\\cdot f_{m - 1}} -\\frac{f_{m} - 2}{f_{1}}\\cdot \\frac{2^{m - 2}}{f_{2}\\cdot\\cdot\\cdot f_{m}} = \\frac{2^{m - 2}(f_{m} - (f_{m} - 2))}{f_{1}\\cdot\\cdot\\cdot f_{m}} = \\frac{2^{m - 1}}{f_{1}\\cdot\\cdot\\cdot f_{m}},\\] \n\nas desired.\n\n\n\nFix \\(m\\) , and let \\(g_{1},\\ldots ,g_{m}\\) be as guaranteed by the lemma. Now, for each \\(\\theta \\in \\{- 1,1\\}^{m}\\) , define a polynomial \\(q_{\\theta}(x)\\) by \n\n\\[q_{\\theta}(x) = \\sum_{i = 1}^{m}\\theta_{i}g_{i}(x)h_{i}(x)\\prod_{\\stackrel{j = 1}{j\\neq i}}^{m}f_{i}(x).\\] \n\nFor each \\(i\\) , we have \n\n\\[q_{\\theta}(x)\\equiv \\theta_{i}h_{i}(x)\\left(g_{i}(x)\\prod_{\\stackrel{j = 1}{j\\neq i}}^{m}f_{i}(x)\\right)\\equiv 2^{m - 1}\\theta_{i}h_{i}(x)\\pmod {f_{i}(x)}.\\] \n\nAs a result, since \\(h_{i}^{2}\\equiv - 1\\) (mod \\(f_{i}\\) ), we have \n\n\\[q_{\\theta}(x)^{2} + 2^{2m - 2}\\equiv 2^{2m - 2}\\left(\\theta_{i}^{2}h_{i}(x)^{2} + 1\\right) = 2^{2m - 2}f_{i}(x)\\equiv 0\\pmod {f_{i}}\\] \n\nfor each \\(i\\) . The polynomials \\(f_{i}\\) are relatively prime, so we conclude in fact that \n\n\\[f_{1}\\cdot \\cdot \\cdot f_{m}\\mid q_{\\theta}^{2} + 2^{2m - 2}. \\quad (*)\\] \n\nLet \\(p_{\\theta}\\) be the remainder when \\(q_{\\theta}\\) is divided by the product \\(f_{1}\\cdot \\cdot \\cdot f_{m}\\) , so that \\(\\deg p_{\\theta}<\\) \\(\\deg (f_{1}\\cdot \\cdot \\cdot f_{m})\\) ; since \\(f_{1}\\cdot \\cdot \\cdot f_{m}\\) is monic, \\(p_{\\theta}\\) has integer coefficients. Moreover, there exists a polynomial \\(s_{\\theta}\\) so that \n\n\\[p_{\\theta} = s_{\\theta}\\cdot f_{1}\\cdot \\cdot \\cdot f_{m} + \\sum_{i = 1}^{m}\\theta_{i}g_{i}h_{i}\\prod_{\\stackrel{j = 1}{j\\neq i}}^{m}f_{i}.\\] \n\nIn particular, since \\(f_{i}(X)\\) is even for any \\(i\\) and any odd integer \\(X\\) , \\(p_{\\theta}(X)\\) is a multiple of \\(2^{m - 1}\\) . Let \\(X\\) be an odd integer, and define \\(a_{\\theta ,X} = p_{\\theta}(X) / 2^{m - 1}\\in \\mathbb{Z}\\) . Since \\((\\star)\\) holds for \\(p_{\\theta}\\) as well as \\(q_{\\theta}\\) , there is a monic polynomial \\(t_{\\theta}\\) for which \n\n\\[p_{\\theta}^{2} + 2^{2m - 2} = t_{\\theta}f_{1}\\cdot \\cdot \\cdot f_{m},\\] \n\nwhence \n\n\\[a_{\\theta ,X}^{2} + 1 = \\frac{1}{2^{2m - 2}}\\left(p_{\\theta}(X)^{2} + 1\\right) = t_{\\theta}(x)\\prod_{i = 1}^{m}\\frac{f_{i}(X)}{2}.\\] \n\nTherefore \n\n\\[\\prod_{i = 1}^{m}\\frac{f_{i}(X)}{2}\\mathrm{~divides~}\\gcd \\left(\\{a_{\\theta ,X}:\\theta \\in \\{-1,1\\}^{m}\\}\\right).\\] \n\nAs \\(X\\) grows, the gcd above is at least \\(2^{- m}X^{\\deg f_{1}\\cdot \\cdot \\cdot f_{m}}\\) . But each \\(p_{\\theta}\\) has degree at most \\((\\deg f_{1}\\cdot \\cdot \\cdot f_{m}) - 1\\) , so each \\(a_{\\theta ,X}\\) is at most some constant \\(C\\) times \\(X^{\\deg (f_{1}\\cdot \\cdot \\cdot f_{m}) - 1}\\) . Therefore, for any \\(0< \\epsilon < \\frac{1}{\\deg(f_{1} + \\cdot \\cdot \\cdot f_{m}) - 1}\\) , we have for large enough \\(X\\) that \n\n\\[\\gcd \\left(\\{a_{\\theta ,X}:\\theta \\in \\{-1,1\\}^{m}\\} \\right)\\geq \\max \\left(\\left\\{|a_{\\theta ,X}|:\\theta \\in \\{-1,1\\}^{m}\\right\\} \\right)^{1 + \\epsilon}.\\] \n\nAs \\(a_{\\theta ,X} = - a_{- \\theta ,X}\\) , at least half of the \\(a_{\\theta ,X}\\) are nonnegative. We conclude the result for \\(n = 2^{m - 1}\\) and any \\(\\epsilon < \\frac{1}{2^{m + 1} - 3}\\) .", "metadata": {"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2025.jsonl", "problem_match": "Problem 6. ", "solution_match": "## \\(\\S 2.3\\) Solution to TST 6, by Pitchayut Saengrungkonga \n"}}
diff --git a/USA_TSTST/md/en-sols-TSTST-2024.md b/USA_TSTST/md/en-sols-TSTST-2024.md
index eebc2bbe13fcd43bd86bb5cea544f4d6bef9ecb6..c476b4bbe34ae270776baca7a1e2796e063608f6 100644
--- a/USA_TSTST/md/en-sols-TSTST-2024.md
+++ b/USA_TSTST/md/en-sols-TSTST-2024.md
@@ -351,7 +351,7 @@ Let \(ABCD\) be a quadrilateral inscribed in a circle with center \(O\) and \(E\
\(\P\) Solution 1. Let \(R = \overline{AD} \cap \overline{BC}\) (possibly at infinity, but we'll see it's an Euclidean point later).
-
+
Claim — \(ACRP\) is an isosceles trapezoid with \(\overline{AC} \parallel \overline{PR}\) . Consequently, \(R\) is an Euclidean point, and \(OP = OR\) .
@@ -372,7 +372,7 @@ Similarly \(OQ = OR\) , so we're done.
Solution 2.
-
+
Let \(PA\) intersect (ABCD) again at \(A^{\prime}\) and let \(QB\) intersect (ABCD) again at \(B^{\prime}\) The problem follows from the following claim.
@@ -749,7 +749,7 @@ Let \(ABC\) be a scalene triangle, and let \(D\) be a point on side \(BC\) satis
Claim — Points \(X\) and \(Y\) are isogonal conjugates with respect to \(\triangle ABC\) .
-
+
Here are two proofs of the claim.
diff --git a/USA_TSTST/md/en-sols-TSTST-2025.md b/USA_TSTST/md/en-sols-TSTST-2025.md
index 018ca7622940ffd5ef185ecd86bd7f77544127c9..93df6e863d8f7d8a714a5687825ff836ecfa5832 100644
--- a/USA_TSTST/md/en-sols-TSTST-2025.md
+++ b/USA_TSTST/md/en-sols-TSTST-2025.md
@@ -143,12 +143,12 @@ A tetrahedron \(A B C D\) is said to be angelic if it has nonzero volume and sat
Across all angelic tetrahedrons, what is the maximum number of distinct lengths that could appear in the set \(\{A B,A C,A D,B C,B D,C D\}\) ?
-
+
We claim the maximum cardinality is \(\boxed{4}\) . This is attained by taking a non- square rectangle (or parallelogram) \(A C B D\) and folding it along diagonal \(A B\) , creating edges \(A B\) and \(C D\) in the process. Here, \(A C = B D\) and \(A D = B C\) , while every other length is distinct in the general case. This tetrahedron is congruent to itself under the permutation of vertices \((A,B,C,D)\mapsto (B,A,D,C)\) , and we can verify that the required angle conditions follow from this symmetry.
-
+
In the other direction, let \(f(X)\) denote the sum of the angles at \(X\) , so the conditions of the problem statement can be written as \(f(A) = f(B)\) and \(f(C) = f(D)\) . Unfold the three faces that meet at \(D\) to create a net of the tetrahedron. Along with the face \(\triangle A B C\) , we also create the faces \(\triangle A B D_{1}\) , \(\triangle B C D_{2}\) , and \(\triangle A C D_{3}\) . Note that
@@ -325,7 +325,7 @@ An analogous proof works for the other side.
Since \(N B_{2} = N C_{2}\) , it follows that \(N\) lies on the radical axis of \((B B_{1}B_{2})\) and \((C C_{1}C_{2})\) . Thus we want to show \(X\) , the foot of \(N\) onto \(O_{B}O_{C}\) , lies on the nine- point circle.
-
+
Let \(D\) , \(E\) , \(F\) be the feet of the altitudes in \(\triangle A B C\) , and let \(M_{B}\) , \(M_{C}\) be the midpoints of \(B H\) , \(C H\) . Let \(B_{3}\) , \(C_{3}\) be the second intersections of \((B_{2}C_{2}H)\) with \((B B_{1}B_{2})\) , \((C C_{1}C_{2})\) .
@@ -365,7 +365,7 @@ Proof. Move \(\ell\) linearly. Then \(\triangle BB_{1}B_{2}\) is dilating at \(B
which by symmetry means \(\overline{B O_{B}} \parallel \overline{C O_{C}}\) . Thus we need to find a point \(Y\) on the nine- point circle such that \(O_{B} - O_{C} - Y\) in 2 cases.
-
+
Suppose \(\ell\) passes through \(H\) . In that case, let \(Q = (BB_{1}H)\cap (CC_{1}H)\) . We get that \(Q\) lies on \((ABC)\) by chasing
@@ -381,7 +381,7 @@ so \(\overline{AY_{1}}\perp \ell\) . If \(Y\) is the midpoint of \(HY_{1}\) , th
Let \(P_{\infty}\) be the point at infinity of \(\ell\) , and let \(P = r\cap \ell\) . Since \(PB_{1}\cdot PB_{2} = PC_{1}\cdot PC_{2}\) (with signed lengths), we have that \((P,P_{\infty}),(B_{1},B_{2}),(C_{1},C_{2})\) are 3 pairs under the same involution on \(\ell\) . Then by DIT on the 4 points \(A,H,E,F\) and the line \(\ell\) , if \(\mathcal{C}\) is the conic through \(A,H,E,F,P_{\infty}\) , then \(P\) also lies on \(\mathcal{C}\) . Now we do moving points.
-
+
- Let \(P_{\infty}\) be a sufficiently general fixed point at infinity (proving the problem for sufficiently general \(P_{\infty}\) suffices by continuity.
diff --git a/USA_TSTST/segmented/en-sols-TSTST-2024.jsonl b/USA_TSTST/segmented/en-sols-TSTST-2024.jsonl
index b187ed3eca749b4c853161ed10ecc66b1b5ed3d6..d6ec46b004fab181afbfd98f44546b94d3c99efc 100644
--- a/USA_TSTST/segmented/en-sols-TSTST-2024.jsonl
+++ b/USA_TSTST/segmented/en-sols-TSTST-2024.jsonl
@@ -1,9 +1,9 @@
{"year": "2024", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "USA_TSTST", "problem": "For every ordered pair of integers \\((i,j)\\) , not necessarily positive, we wish to select a point \\(P_{i,j}\\) in the Cartesian plane whose coordinates lie inside the unit square defined by \n\n\\[i< x< i + 1,\\qquad j< y< j + 1.\\] \n\nFind all real numbers \\(c > 0\\) for which it's possible to choose these points such that for all integers \\(i\\) and \\(j\\) , the (possibly concave or degenerate) quadrilateral \\(P_{i,j}P_{i + 1,j}P_{i + 1,j + 1}P_{i,j + 1}\\) has perimeter strictly less than \\(c\\) .", "solution": "Answer. \\(c\\geq 4\\) \n\nProof \\(c< 4\\) is not possible. Let \\(n\\) be an arbitrary positive integer. We take an \\(n\\times n\\) subgrid of unit squares (i.e. \\(P_{i,j}\\) for \\(1\\leq i,j\\leq n\\) ), and compute a lower bound on the average of all possible quadrilaterals from this subgrid. \n\nConsider the average length of the \"top side\" of all possible quadrilaterals in this grid. Note that this is equal to: \n\n\\[\\frac{1}{(n - 1)^2}\\sum_{i = 1}^{n - 1}\\sum_{j = 1}^{n - 1}P_{i,j}P_{i + 1,j}\\geq \\frac{1}{(n - 1)^2}\\sum_{j = 1}^{n - 1}P_{1,j}P_{n,j} > \\frac{n - 2}{n - 1}.\\] \n\nWe can apply this bound to all four sides of the quadrilateral (the left, right, bottom, and top sides) to find that the average perimeter of all possible quadrilaterals is greater than \n\n\\[\\frac{4(n - 2)}{n - 1} = 4 - \\frac{4}{n - 1}.\\] \n\nThis means we can always find a quadrilateral whose perimeter is at least \\(4 - \\frac{4}{n - 1}\\) . By taking sufficiently large \\(n\\) , this lower bound will exceed \\(c\\) . \n\nProof \\(c = 4\\) is possible. We'll place point \\(P_{i,j}\\) at the coordinates \\((f(i),f(j))\\) for some function \\(f:\\mathbb{Z}\\to \\mathbb{R}\\) . The perimeter of \\(P_{i,j}P_{i + 1,j}P_{i + 1,j + 1}P_{i,j + 1}\\) is then \n\n\\[2\\big(|f(i + 1) - f(i)| + |f(j + 1) - f(j)|\\big).\\] \n\nTherefore we have a valid construction for \\(c = 4\\) if \\(f\\) satisfies \\(n< f(n)< n + 1\\) and \\(|f(n + 1) - f(n)|< 1\\) for all \\(n\\) . This is achieved by \n\n\\[f(n) = n + 0.5 + \\left\\{ \\begin{array}{ll} - \\sum_{i = 1}^{n}\\frac{1}{10^{i}} & \\mathrm{if~}n\\geq 0,\\\\ \\sum_{i = 1}^{-n}\\frac{1}{10^{i}} & \\mathrm{if~}n< 0. \\end{array} \\right.\\] \n\nLet's check the conditions. The sum is bounded by \\(\\sum_{i = 1}^{\\infty}\\frac{1}{10^{i}} = \\frac{1}{9}\\) in magnitude, so \\(n< f(n)< n + 1\\) . Furthermore, we can verify that \n\n\\[f(n + 1) - f(n) = 1 - \\left\\{ \\begin{array}{ll}\\frac{1}{10^{n + 1}} & \\mathrm{if~}n\\geq 0,\\\\ \\frac{1}{10^{-n}} & \\mathrm{if~}n< 0 \\end{array} \\right.< 1.\\]", "metadata": {"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2024.jsonl", "problem_match": "1. ", "solution_match": "## \\(\\S 1.1\\) TSTST 2024/1, proposed by Karthik Vedula \n"}}
{"year": "2024", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "USA_TSTST", "problem": "Let \\(p\\) be an odd prime number. Suppose \\(P\\) and \\(Q\\) are polynomials with integer coefficients such that \\(P(0) = Q(0) = 1\\) , there is no nonconstant polynomial dividing both \\(P\\) and \\(Q\\) , and \n\n\\[1 + \\frac{x}{1 + \\frac{2x}{1 + \\frac{\\cdots}{1 + (p - 1)x}}} = \\frac{P(x)}{Q(x)}.\\] \n\nShow that all coefficients of \\(P\\) except for the constant coefficient are divisible by \\(p\\) , and all coefficients of \\(Q\\) are not divisible by \\(p\\) .", "solution": "\\(\\P\\) Solution 1. We first make some general observations about rational functions represented through continued fractions. \n\nClaim — Let \\(a_{1}, a_{2}, \\ldots\\) , be a sequence of nonzero integers. Define the sequence of polynomials \\(P_{1}(x) = 1\\) , \\(P_{2}(x) = 1 + a_{1}x\\) , and \n\n\\[P_{k + 1}(x) = P_{k}(x) + a_{k}x P_{k - 1}(x)\\] \n\nfor \\(k \\geq 1\\) . Then the following properties hold for all \\(k \\geq 0\\) : \n\n- \\(P_{k}(0) = 1\\) , \n\n- \\(\\gcd (P_{k + 1}, P_{k}) = 1\\) , and \n\n- \\(\\deg P_{k} = \\lfloor k / 2 \\rfloor\\) . \n\n\\[1 + \\frac{a_{k}x}{1 + \\frac{a_{k - 1}x}{\\ddots}} = \\frac{P_{k + 1}(x)}{P_{k}(x)}.\\] \\[1 + \\frac{\\ddots}{1 + a_{1}x}\\] \n\nProof. These all follow by induction. \n\nWith the setup of the claim, the polynomials \\(P\\) and \\(Q\\) in the problem are exactly \\(P_{p}\\) and \\(P_{p}\\) for the sequence \\(a_{1} = p - 1\\) , \\(a_{2} = p - 2\\) , ..., \\(a_{p - 1} = 1\\) . From here, we will define \\(P\\) and \\(Q\\) in terms of this recurrence and make two transformations on the \\(a_{i}\\) . \n\nFirst, subtract \\(p\\) from each \\(a_{i}\\) so we now have \\(a_{1} = - 1\\) , \\(a_{2} = - 2\\) , ..., \\(a_{p - 1} = -(p - 1)\\) . The coefficients of \\(P\\) and \\(Q\\) only have changed by a multiple of \\(p\\) . The claim also shows that \\(P(0) = Q(0) = 1\\) , \\(\\gcd (P, Q) = 1\\) , and the degrees of \\(P\\) and \\(Q\\) have not changed, i.e. no leading terms have been added or removed. Therefore it is equivalent to work with this sequence instead.\n\n\n\nSimilarly, we can multiply \\(a_{i}\\) by \\(- 1\\) to get \\(a_{1} = 1\\) , \\(a_{2} = 2\\) , ..., \\(a_{p - 1} = p - 1\\) . This is equivalent to replacing \\(x\\) with \\(- x\\) , so some coefficients of \\(P\\) and \\(Q\\) are negated by this. The key conditions still remain unchanged. \n\nRemark. For rigor, it's important to prove the facts about continued fractions first rather than going straight to the coefficient replacement as the requirements imposed by the problem statement on \\(P\\) and \\(Q\\) are not always preserved modulo \\(p\\) . \n\nAt this point, we are ready to directly calculate the polynomials \\(P_{n}\\) . \n\nClaim — For all \\(n\\) , we have \n\n\\[P_{n}(x) = \\sum_{k\\geq 0}\\frac{n!}{2^{k}k!(n - 2k)!} x^{k}.\\] \n\nHere, we define \\(\\frac{n!}{(n - 2k)!} = n(n - 1)\\ldots (n - 2k + 1)\\) so the sum actually stops at \\(k = \\lfloor n / 2\\rfloor\\) . \n\nProof. We use induction. The base cases are clear, and \n\n\\[P_{n} + n x P_{n - 1} = \\sum_{k\\geq 0}\\frac{1}{2^{k}k!}\\frac{n!}{(n - 2k)!} x^{k} + n x\\sum_{k\\geq 0}\\frac{1}{2^{k}k!}\\frac{(n - 1)!}{(n - 2k - 1)!} x^{k}\\] \\[\\qquad = \\sum_{k\\geq 0}\\frac{1}{2^{k}k!}\\frac{n!}{(n - 2k)!} x^{k} + \\sum_{k\\geq 0}\\frac{1}{2^{k}k!}\\frac{n!}{(n - 2k - 1)!} x^{k + 1}\\] \\[\\qquad = \\sum_{k\\geq 0}\\frac{1}{2^{k}k!}\\frac{n!}{(n - 2k)!} x^{k} + \\sum_{k\\geq 1}\\frac{1}{2^{k - 1}(k - 1)!}\\frac{n!}{(n - 2k + 1)!} x^{k}\\] \\[\\qquad = 1 + \\sum_{k\\geq 1}\\left[\\frac{1}{2^{k}k!}\\frac{n!}{(n - 2k)!} +\\frac{1}{2^{k - 1}(k - 1)!}\\frac{n!}{(n - 2k + 1)!}\\right]x^{k}\\] \\[\\qquad = 1 + \\sum_{k\\geq 1}\\frac{1}{2^{k}k!}\\frac{n!}{(n - 2k + 1)!}\\left[(n - 2k + 1) + 2k\\right]x^{k}\\] \\[\\qquad = 1 + \\sum_{k\\geq 1}\\frac{1}{2^{k}k!}\\frac{(n + 1)!}{(n - 2k + 1)!} x^{k}\\] \\[\\qquad = \\sum_{k\\geq 0}\\frac{1}{2^{k}k!}\\frac{(n + 1)!}{(n - 2k + 1)!} x^{k}\\] \\[\\qquad = P_{n + 1}.\\] \n\nAt this point we can directly check the coefficients of \\(P\\) and \\(Q\\) . We have \n\n\\[P(x) = P_{p}(x) = \\sum_{k\\geq 0}\\frac{p!}{2^{k}k!(p - 2k)!} x^{k}.\\] \n\nFor \\(k = 0\\) , we get a coefficient of 1. For \\(k \\geq 1\\) , the denominator is not a multiple of \\(p\\) , so the remaining coefficients are multiples of \\(p\\) . Meanwhile, the coefficients for \\(Q\\) are \\(\\frac{(p - 1)!}{2^{k}k!(p - 1 - 2k)!}\\) . None of these are divisible by \\(p\\) because \\((p - 1)!\\) is not divisible by \\(p\\) .\n\n\n\nRemark. This problem was created from the identity \n\n\\[\\frac{1}{1 - \\frac{x}{1 - \\frac{2x}{1 - \\frac{3x}{1 - \\cdots}}}}\\] \n\nYou can use this to solve the problem. However, the proof of the identity as well as the details to convert this into the given problem are very lengthy so this is only left as a remark. \n\n## Solution 2. \n\n## Lemma \n\nDefine a series of polynomials \\(A_{n}\\) , \\(B_{n}\\) , \\(C_{n}\\) , \\(D_{n}\\) by letting \\(A_{1} = 1\\) , \\(B_{1} = x\\) , \\(C_{1} = 1\\) , and \\(D_{1} = 0\\) , and for \\(n \\geq 2\\) , \n\n\\[A_{n} = A_{n - 1} + B_{n - 1} B_{n} = nxA_{n - 1\\] \\[C_{n} = C_{n - 1} + D_{n - 1} D_{n} = nxC_{n - 1}.\\] \n\nThen, if we define \\(f_{k}(t) = 1 + \\frac{kx}{t}\\) , then \n\n\\[(f_{1}\\circ f_{2}\\circ \\cdot \\cdot \\cdot \\circ f_{n})(t) = \\frac{A_{n}t + B_{n}}{C_{n}t + D_{n}}.\\] \n\nProof. Straightforward by induction: the \\(n = 1\\) case is true and for \\(n \\geq 2\\) , \n\n\\[\\frac{A_{n - 1}f_{n}(t) + B_{n - 1}}{C_{n - 1}f_{n}(t) + D_{n - 1}} = \\frac{A_{n - 1}(t + nx) + B_{n - 1}t}{C_{n - 1}(t + nx) + D_{n - 1}t} = \\frac{(A_{n - 1} + B_{n - 1})t + nxA_{n - 1}}{(C_{n - 1} + D_{n - 1})t + nxD_{n - 1}}.\\] \n\nIt is possible to solve these recurrences algebraically, but there is in fact a nice combinatorial interpretation of these polynomials: \n\n## Lemma \n\nLet \\(\\mathcal{T}_{n}\\) be the set of involutions on \\(\\{1,2,\\ldots ,n\\}\\) and for \\(\\pi \\in \\mathcal{T}_{n}\\) , let \\(a(\\pi)\\) denote the number of transpositions in \\(\\pi\\) . Let \\(\\mathcal{J}_{n} \\subseteq \\mathcal{T}_{n}\\) be the set of involutions \\(\\pi\\) where the minimal fixed point is less than the minimal \\(x\\) with \\(\\pi (x) < x\\) . Then, for all \\(k \\geq 1\\) , we have \n\n\\[A_{n} = \\sum_{\\pi \\in \\mathcal{T}_{n}}x^{a(\\pi)} B_{n} = \\sum_{\\pi \\in \\mathcal{T}_{n + 1}\\atop \\pi (n + 1)\\neq n + 1}x^{a(\\pi)}\\] \\[C_{n} = \\sum_{\\pi \\in \\mathcal{J}_{n}}x^{a(\\pi)} D_{n} = \\sum_{\\pi \\in \\mathcal{J}_{n + 1}\\atop \\pi (n + 1)\\neq n + 1}x^{a(\\pi)}.\\] \n\nProof. We use induction on \\(n\\) , with the case \\(n = 1\\) being easily verified. For \\(n \\geq 2\\) , note that \n\n\\[A_{n} - B_{n - 1} = \\sum_{\\substack{\\pi \\in \\mathcal{I}_{k + 1} \\\\ \\pi (k + 1) = k + 1}}x^{a(\\pi)} = A_{n - 1}.\\]\n\n\n\nMoreover, every \\(\\pi \\in \\mathcal{I}_{n + 1}\\) swapping \\(n + 1\\) with something else can be characterized by one of \\(n\\) choices for \\(\\pi (n + 1)\\) and an involution on the remaining \\(n - 1\\) elements. Therefore \\(B_{n} = n x A_{n - 1}\\) . A similar proof holds for the \\(C_{n}\\) and \\(D_{n}\\) . \\(\\square\\) \n\nWe now note that \n\n\\[\\frac{P(x)}{Q(x)} = (f_{1}\\circ f_{2}\\circ \\cdot \\cdot \\cdot \\circ f_{p - 1})(1) = \\frac{A_{p - 1} + B_{p - 1}}{C_{p - 1} + D_{p - 1}} = \\frac{A_{p}}{C_{p}}.\\] \n\nIt suffices to show that all nonconstant coefficients of \\(A_{p}\\) are divisible by \\(p\\) and that all coefficients of \\(C_{p}\\) are not divisible by \\(p\\) . If we show this, then any nonconstant factor dividing \\(A_{p}\\) must have a leading coefficient divisible by \\(p\\) (proof: reduce the factorization mod \\(p\\) ), then any nonconstant factor dividing \\(C_{p}\\) cannot have a leading coefficient divisible by \\(p\\) . Moreover, since \\(A_{p}\\) and \\(C_{p}\\) have constant terms of 1, they are relatively prime and thus \\(P = A_{p}\\) and \\(Q = C_{p}\\) (up to sign). \n\nWe now evaluate the coefficients of \\(A_{p}\\) and \\(C_{p}\\) . We observe that \n\n\\[A_{p} = \\sum_{k = 0}^{\\frac{p - 1}{2}}\\binom{p}{2k}(2k - 1)!!x^{k},\\] \n\nand it is easy to see that \\(A_{p} \\equiv 1\\) (mod \\(p\\) ). Computing the coefficients of \\(C_{n}\\) is a bit harder. Consider elements of \\(\\mathcal{J}_{p}\\) with \\(0 \\leq k \\leq \\frac{p - 1}{2}\\) swaps and a minimal fixed point of \\(a + 1\\) (for \\(0 \\leq a \\leq k\\) ). Everything in \\(\\{1, \\ldots , a\\}\\) must be paired with something greater than \\(a + 1\\) , and there are \\(k - a\\) swaps among the remaining elements. It follows that \n\n\\[C_{p} = \\sum_{k = 0}^{\\frac{p - 1}{2}}\\sum_{a = 0}^{k}\\frac{(p - a - 1)!}{(p - 2a - 1)!}\\binom{p - 2a - 1}{2k - 2a}(2k - 2a - 1)!!\\cdot x^{k}\\] \\[\\quad = \\sum_{k = 0}^{\\frac{p - 1}{2}}\\sum_{a = 0}^{k}\\frac{(p - a - 1)!}{(p - 2k - 1)!2^{k - a}(k - a)!}x^{k}\\] \\[\\quad = \\sum_{k = 0}^{\\frac{p - 1}{2}}\\frac{(p - 1)!}{(p - 2k - 1)!}\\left(\\sum_{a = 0}^{k}\\frac{1}{2^{k - a}(k - a)!}\\prod_{j = 1}^{a}(p - j)\\right)x^{k}\\] \\[\\quad \\equiv \\sum_{k = 0}^{\\frac{p - 1}{2}}\\frac{(p - 1)!}{(p - 2k - 1)!}\\left(\\sum_{a = 0}^{k}\\frac{(-1)^{a}}{2^{k - a}(k - a)!a!}\\right)x^{k}\\] \\[\\quad = \\sum_{k = 0}^{\\frac{p - 1}{2}}\\frac{(p - 1)!}{(p - 2k - 1)!k!}\\left(\\sum_{a = 0}^{k}\\binom{k}{a}\\frac{(-1)^{a}}{2^{k - a}}\\right)x^{k}\\] \\[\\quad = \\sum_{k = 0}^{\\frac{p - 1}{2}}\\frac{(p - 1)!}{(p - 2k - 1)!k!}\\left(-1 + \\frac{1}{2}\\right)^{k}x^{k}.\\] \n\nThe result follows. \n\nRemark. It is not hard to see that \\(C_{p}\\) simplifies further mod \\(p\\) to \n\n\\[\\sum_{k = 0}^{\\frac{p - 1}{2}}(-1)^{k}(2k - 1)!!x^{k}.\\]", "metadata": {"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2024.jsonl", "problem_match": "2. ", "solution_match": "## \\(\\S 1.2\\) TSTST 2024/2, proposed by Andrew Gu \n"}}
{"year": "2024", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "USA_TSTST", "problem": "Let \\(A = \\{a_{1},\\ldots ,a_{2024}\\}\\) be a set of 2024 pairwise distinct real numbers. Assume that there exist positive integers \\(b_{1},b_{2},\\ldots ,b_{2024}\\) such that \n\n\\[a_{1}b_{1} + a_{2}b_{2} + \\cdot \\cdot \\cdot +a_{2024}b_{2024} = 0.\\] \n\nProve that one can choose \\(a_{2025},a_{2026},a_{2027},\\ldots\\) such that \\(a_{k}\\in A\\) for all \\(k\\geq 2025\\) and, for every positive integer \\(d\\) , there exist infinitely many positive integers \\(n\\) satisfying \n\n\\[\\sum_{k = 1}^{n}a_{k}k^{d} = 0.\\]", "solution": "It will be convenient to use 0- based indexing here, i.e. \\(A = \\{a_{0},\\ldots ,a_{2023}\\}\\) and so on. Let \\(m = \\sum_{i = 0}^{2023}b_{i}\\) . By appending \\(b_{i} - 1\\) copies of \\(a_{i}\\) for each \\(i\\) , we may extend the sequence to \\(a_{0},\\ldots ,a_{m - 1}\\) such that \\(a_{0} + \\cdot \\cdot \\cdot +a_{m - 1} = 0\\) . \n\n\\(\\P\\) Solution by direct construction. Let \\(s_{m}(n)\\) be the sum of the base- \\(m\\) digits of \\(n\\) , reduced modulo \\(m\\) . We now claim that \\(a_{k} = a_{s_{m}(k)}\\) works. It is clear that this extends the original sequence in a valid way. \n\nClaim — Fix \\(d\\) . Then the sum \n\n\\[\\sum_{k = 0}^{n - 1}a_{k}(k + 1)^{d}\\] \n\nis zero when \\(n = cm^{d + 1}\\) for any positive integer \\(c\\) . \n\nIf we prove this, then the problem is solved as there are infinitely many possible \\(c\\) . We have two proofs of this. \n\nProof of claim by expansion (Andrew Gu). We only need to show that each block of \\(m^{d + 1}\\) terms sums to zero. Each block takes the form \n\n\\[S = \\sum_{k = cm^{d + 1}}^{(c + 1)m^{d + 1} - 1}a_{s_{m}(k)}(k + 1)^{d}.\\] \n\nWe group terms based on the value of \\(s_{m}(k)\\) . The values of \\(k\\) in this sum are given by \n\n\\[\\left\\{cm^{d + 1} + \\sum_{i = 0}^{d}e_{i}m^{i}\\mid (e_{0},\\ldots ,e_{d})\\in \\{0,\\ldots ,m - 1\\}^{d + 1}\\right\\} .\\]\n\n\n\nIn this form, the sum of the base- \\(m\\) digits is \\(s_{m}(c) + e_{0} + \\dots +e_{d}\\) . Therefore we can write \n\n\\[S = \\sum_{i = 0}^{m - 1}a_{i}\\left(\\underbrace{\\sum_{(e_{0},\\dots,e_{d})\\in\\{0,\\dots,m - 1\\}^{d + 1}}}_{e_{0} + \\dots +e_{d}\\equiv i - s_{m}(c)}\\underbrace{\\left(cm^{d + 1} + 1 + \\sum_{j = 0}^{d}e_{j}m^{j}\\right)^{d}}_{(2)}\\right).\\] \n\nSince \\(\\sum a_{i} = 0\\) , it suffices to show that the value of the coefficient given by (1) is independent of \\(i\\) . \n\nExpand the term (2) with the multinomial theorem. Each monomial in the expansion is a product of terms \\(c\\) , \\(m\\) , \\(e_{j}\\) . Since the exponent is \\(d\\) , no monomial is divisible by the whole product \\(e_{0}e_{1}\\dots e_{d}\\) . Since the tuples \\((e_{0},\\ldots ,e_{d})\\) are uniformly distributed over all possible tuples when one or more coordinates are removed, that means that the sum of each monomial term is independent of \\(i\\) . We conclude that the sum (1) is independent of \\(i\\) as well. \\(\\square\\) \n\nProof of claim by Fourier transform. Let \\(\\omega \\neq 1\\) be an \\(m^{\\mathrm{th}}\\) root of unity. Then, define the operator \n\n\\[\\Delta_{a}^{(\\omega)}f(x) = \\sum_{i = 0}^{m - 1}\\omega^{i}f(x + ia).\\] \n\nBy looking at the leading coefficients, one can see that \\(\\Delta_{a}^{(\\omega)}\\) lowers the degree of a polynomial by at least \\(1\\) . As a consequence, by letting \\(f(x) = (x + 1)^{d}\\) we conclude that \n\n\\[\\sum_{k\\in [0,m_{e})}\\omega^{s_{m}(k)}(k + 1)^{d} = (\\Delta_{m^{e - 1}}^{(\\omega)}\\cdot \\cdot \\cdot \\Delta_{m}^{(\\omega)}\\Delta_{1}^{(\\omega)}f)(0) = 0.\\] \n\nSince \\(a_{1} + \\dots +a_{m} = 0\\) , by a discrete Fourier transform we may write \\(a_{i} = \\sum_{j = 1}^{m - 1}c_{j}e^{2\\pi ij / m}\\) for some \\(c_{1},\\ldots ,c_{m - 1}\\in \\mathbb{C}\\) . Therefore \n\n\\[\\sum_{k = 0}^{m^{e} - 1}a_{k}(k + 1)^{d} = \\sum_{j = 1}^{m - 1}c_{j}\\sum_{k = 0}^{m^{e} - 1}(e^{2\\pi ij / m})^{s_{m}(k)}(k + 1)^{d} = 0.\\] \n\nInductive solution. Define \\(m\\) and extend the sequence to \\(a_{m - 1}\\) as before. For \\(d\\geq 0\\) , let \\(\\Sigma_{d}(S)\\) denote \\(\\textstyle \\sum_{k\\in S}k^{d}\\) . Also, for \\(d\\geq 0\\) call a tuple of sets \\((S_{0},\\ldots ,S_{m - 1})\\) to be \\(d\\) - uniform if for all \\(0\\leq d^{\\prime}\\leq d\\) we have \\(\\Sigma_{d^{\\prime}}(S_{0}) = \\Sigma_{d^{\\prime}}(S_{1}) = \\dots = \\Sigma_{d^{\\prime}}(S_{m - 1})\\) \n\n## Lemma 1.1 \n\nSuppose \\((S_{0},\\ldots ,S_{m - 1})\\) is \\(d\\) - uniform. Then for all \\(0\\leq d^{\\prime}\\leq d + 1\\) , \\(x\\in \\mathbb{R}\\) , and \\(0\\leq i,j< m\\) , the quantity \\(\\Sigma_{d^{\\prime}}(S_{i} + x) - \\Sigma_{d^{\\prime}}(S_{j} + x)\\) is independent of \\(x\\) . \n\nProof. We may expand \n\n\\[\\Sigma_{d^{\\prime}}(S_{i} + x) = \\sum_{k = 0}^{d^{\\prime}}\\binom{d^{\\prime}}{k} x^{d^{\\prime} - k}\\Sigma_{k}(S_{i}).\\] \n\nBy \\(d\\) - uniformity, the terms with \\(k\\leq d\\) are independent of \\(i\\) and go away in the subtraction. The only term left is when \\(k = d^{\\prime} = d + 1\\) . However, this term is independent of \\(x\\) , as desired. \\(\\square\\)\n\n\n\n## Lemma 1.2 \n\nSuppose \\((S_{0},\\ldots ,S_{m - 1})\\) is a \\(d\\) - uniform partition of \\([0,n)\\) . Then there exists a \\((d + 1)\\) uniform partition \\((S_{0}^{\\prime},\\ldots ,S_{m - 1}^{\\prime})\\) of \\([0,mn)\\) such that \\(S_{i}\\subseteq S_{i}^{\\prime}\\) for all \\(i\\) \n\nProof. We let \\(S_{i}^{\\prime} = \\bigcup_{k = 0}^{m - 1}(S_{(i + k)\\bmod m} + kn)\\) . Then, for every \\(0\\leq d^{\\prime}\\leq d + 1\\) and \\(0\\leq i,j< m\\) , we have \n\n\\[\\Sigma_{d^{\\prime}}(S_{i}^{\\prime}) - \\Sigma_{d^{\\prime}}(S_{j}^{\\prime}) = \\sum_{k = 0}^{m - 1}\\Sigma_{d^{\\prime}}(S_{(i + k)\\bmod m} + kn) - \\Sigma_{d^{\\prime}}(S_{(j + k)\\bmod m} + kn)\\] \\[\\qquad = \\sum_{k = 0}^{m - 1}\\Sigma_{d^{\\prime}}(S_{(i + k)\\bmod m}) - \\Sigma_{d^{\\prime}}(S_{(j + k)\\bmod m})\\] \\[\\qquad = 0.\\] \n\nConsider the 0- uniform partition of \\([0,m)\\) given by \\((\\{0\\} ,\\{1\\} ,\\ldots ,\\{m - 1\\})\\) . By repeatedly applying Lemma 1.2, we get \\(d\\) - uniform partitions \\((S_{0}^{(d)},\\ldots ,S_{m - 1}^{(d)})\\) of \\([0,m^{d + 1})\\) such that \\(S_{i}^{(d)}\\subseteq S_{i}^{(d + 1)}\\) for all \\(d\\) and \\(i\\) . For every \\(k\\) , let \\(i\\) be the unique index such that \\(k\\in S_{i}^{(d)}\\) for some \\(d\\) , and let \\(a_{k} = a_{i}\\) . (This doesn't redefine \\(a_{0}\\) , ..., \\(a_{m - 1}\\) as for \\(k< m\\) , we have \\(k\\in S_{k}^{(0)}\\) .) Then, for each \\(e\\geq d\\) , we find that \n\n\\[\\sum_{k = 0}^{m^{e + 1} - 1}a_{k}(k + 1)^{d} = \\sum_{i = 0}^{m - 1}a_{i}\\sum_{k\\in S_{i}^{(e)}}(k + 1)^{d} = \\sum_{i = 0}^{m - 1}a_{i}\\sum_{j = 1}^{d}\\binom{d}{j}\\Sigma_{j}(S_{i}^{(e)}) = 0.\\] \n\nRemark. The construction here is essentially the same as the sum- of- digits construction (in fact it would be exactly the same if we replace \\((i + k)\\) mod \\(m\\) in the proof of Lemma 1.2 with \\((i - k)\\) mod \\(m\\) ). However, it hints at some flexibility in the construction which is less apparent with other approaches. For example, if \\(G\\) is any transitive subgroup of permutations on \\([0,m)\\) and \\(\\pi_{0} = \\mathrm{id}\\) , \\(\\pi_{1},\\ldots ,\\pi_{|G| - 1}\\) is an enumeration of the elements of \\(G\\) , one can turn a \\(d\\) - uniform partition \\((S_{0},\\ldots ,S_{m - 1})\\) of \\([0,n)\\) into a \\((d + 1)\\) - uniform partition \\((S_{0}^{\\prime},\\ldots ,S_{m - 1}^{\\prime})\\) of \\([0,|G|n)\\) by letting \\(S_{i}^{\\prime} = \\bigcup_{k = 0}^{|G| - 1}(S_{\\pi_{k}(i)} + kn)\\) . Here we used the special case \\(\\pi_{k}(i) = (i + k)\\) mod \\(m\\) ; it may also be natural to let \\(G\\) be the set of all \\(m!\\) permutations. \n\nRemark. The construction in the solution has appeared a number of times in the literature; see \n\n- D. H. Lehmer, The Tarry-Escott problem, Scripta Math. 13, 37-41, 1947; \n\n- E. Prouhet, Mémoire sur quelques relations entre les puissances des nombres, C. R. Acad. Sci. Paris 33, 225, 1851; \n\n- E. M. Wright, Equal sums of like powers, Proc. Edinburgh Math. Soc. 2nd series 8, 138-142, 1949; \n\n- E. M. Wright, Prouhet's 1851 solution of the Tarry-Escott problem of 1910, Amer. Math. Monthly 66, 199-201, 1959.", "metadata": {"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2024.jsonl", "problem_match": "3. ", "solution_match": "## \\(\\S 1.3\\) TSTST 2024/3, proposed by Daniel Zhu \n"}}
-{"year": "2024", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "USA_TSTST", "problem": "Let \\(ABCD\\) be a quadrilateral inscribed in a circle with center \\(O\\) and \\(E\\) be the intersection of segments \\(AC\\) and \\(BD\\) . Let \\(\\omega_{1}\\) be the circumcircle of \\(ADE\\) and \\(\\omega_{2}\\) be the circumcircle of \\(BCE\\) . The tangent to \\(\\omega_{1}\\) at \\(A\\) and the tangent to \\(\\omega_{2}\\) at \\(C\\) meet at \\(P\\) . The tangent to \\(\\omega_{1}\\) at \\(D\\) and the tangent to \\(\\omega_{2}\\) at \\(B\\) meet at \\(Q\\) . Show that \\(OP = OQ\\) .", "solution": "\\(\\P\\) Solution 1. Let \\(R = \\overline{AD} \\cap \\overline{BC}\\) (possibly at infinity, but we'll see it's an Euclidean point later). \n\n\n \n\nClaim — \\(ACRP\\) is an isosceles trapezoid with \\(\\overline{AC} \\parallel \\overline{PR}\\) . Consequently, \\(R\\) is an Euclidean point, and \\(OP = OR\\) . \n\nProof. It is equivalent to show that \\(\\triangle PAC \\cong \\triangle RCA\\) , or without using \\(R\\) , that \n\n\\[\\angle PAC = \\angle ACR = \\angle ACB, \\quad \\text{and} \\quad \\angle PCA = \\angle CAR = \\angle CAD.\\] \n\nIndeed, we have \n\n\\[\\angle PAC = \\angle ADE = \\angle ADB = \\angle ACB\\] \n\nand likewise \\(\\angle PCA = \\angle CAD\\) , as requested. \n\nSimilarly \\(OQ = OR\\) , so we're done.\n\n\n\nSolution 2. \n\n\n \n\nLet \\(PA\\) intersect (ABCD) again at \\(A^{\\prime}\\) and let \\(QB\\) intersect (ABCD) again at \\(B^{\\prime}\\) The problem follows from the following claim. \n\nClaim — \\(OA^{\\prime}PC \\cong ODQB^{\\prime}\\) . \n\nProof. We use directed angles mod \\(\\pi\\) . Note that \n\n\\[\\angle ODQ = \\angle ODA + \\angle ADQ = \\angle DAO + \\angle PAD = \\angle PAO = \\angle OA^{\\prime}P,\\] \n\nand similarly \\(\\angle OCP = \\angle OB^{\\prime}Q\\) . Furthermore, \n\n\\[\\angle A^{\\prime}AD = \\angle AED = \\angle CEB = \\angle CBB^{\\prime},\\] \n\nso \\(A^{\\prime}D = CB^{\\prime}\\) , i.e. \\(A^{\\prime}CDB^{\\prime}\\) is an isosceles trapezoid. This means \\(\\angle A^{\\prime}OC = \\angle DOB^{\\prime}\\) . Combined with \\(OA^{\\prime} = OB^{\\prime} = OD = DC\\) , these three angle equalities imply the congruency. \\(\\square\\) \n\nSolution 3 using complex numbers (Noah Walsh). Use complex numbers with (ABCD) as the unit circle. Because \\(PA\\) is tangent to (ADE), we have \\(\\angle PAC = \\angle ADB = \\frac{1}{2}\\angle AOB\\) , and therefore \n\n\\[\\frac{\\frac{p - a}{c - a}}{\\frac{p - a - 1}{c - a - 1}} = \\frac{a}{b\\] \\[\\Rightarrow b\\frac{p - a}{c - a} = a\\frac{ac\\overline{p} - c}{a - c\\] \\[\\Rightarrow bp + a^2 c\\overline{p} = ab + ac.\\] \n\nThe condition that \\(PC\\) is tangent to (BCE) can be expressed by swapping \\(a\\) with \\(c\\) and \\(b\\) with \\(d\\) , so we get \n\n\\[b p + a^{2}c\\overline{{p}} = a b + a c\\] \\[d p + a c^{2}\\overline{{p}} = c d + a c\\] \\[\\Rightarrow p(a d - b c) = a c d + a^{2}c - a b c - a c^{2}\\] \\[\\Rightarrow p = \\frac{a c(a + d - b - c)}{a d - b c}.\\]\n\n\n\nSimilarly, \n\n\\[q = \\frac{bd(a + d - b - c)}{ad - bc}.\\] \n\nIt is now obvious that \\(p\\) and \\(q\\) have the same magnitude. \n\nRemark. One possible trick one might use to compute \\(p\\) is to compute the coordinates of the points \\(A^{\\prime},C^{\\prime}\\in (ABCD)\\) such that \\(A A^{\\prime}\\) and \\(C C^{\\prime}\\) are tangent to \\((A D E)\\) and \\((B C E)\\) respectively, and then use the chords intersection formula. \n\nIf one does this, one gets \\(a^{\\prime} = a c / b\\) and \\(c^{\\prime} = a c / d\\) . At this point, one can simply recognize that \\(A^{\\prime}\\) and \\(C^{\\prime}\\) are the reflection of \\(B\\) and \\(D\\) over the perpendicular bisector of \\(A C\\) , reflect everything over said perpendicular bisector, and suddenly realize that one has solved the problem synthetically. \n\n\\(\\boxed{ \\begin{array}{r l} \\end{array} }\\) Solution 4 by linearity (Noah Walsh). If \\(A E = D E\\) the problem is true by symmetry, so we assume \\(A E\\neq D E\\) \n\nFix \\(A\\) \\(D\\) , and \\(E\\) , and move \\(B\\) linearly along line \\(D E\\) . By e.g. Reim's theorem, \\(C\\) moves linearly as well. \n\nHow do we compute \\(O\\) ? It lies on the perpendicular bisector of \\(A D\\) , which is fixed. It also lies on the perpendicular bisector of \\(B C\\) , which is parallel to a fixed line, and passes through the midpoint of \\(B C\\) , which moves linearly. Therefore, \\(O\\) moves linearly as well. \n\nPoint \\(P\\) is the intersection of the tangent to \\((A D E)\\) at \\(A\\) , which is fixed, and the tangent to \\((B C E)\\) at \\(C\\) , which moves linearly, and therefore is linear. Similarly, \\(Q\\) moves linearly. Therefore, to verify that \\(\\overrightarrow{O P}\\times \\overrightarrow{O P} = \\overrightarrow{O Q}\\times \\overrightarrow{O Q}\\) , it suffices to check three cases. \n\nThe two cases where \\(E A = E B\\) are trivial by symmetry. We can also check that \\(\\lim_{B\\to \\infty_{D E}}\\frac{\\overrightarrow{O P}\\times\\overrightarrow{O P}}{\\overrightarrow{O Q}\\times\\overrightarrow{O Q}} = 1\\) \n\nIndeed, obviously \\(\\lim_{B\\to \\infty_{D E}}\\frac{|O P|}{|A P|} = \\lim_{B\\to \\infty_{D E}}\\frac{|O Q|}{|D Q|} = 1\\) . In addition, \n\n\\[\\frac{d A P}{d D Q} = \\frac{d A P}{d A C}\\times \\frac{d A C}{d B D}\\times \\frac{d B D}{d B Q}\\] \\[\\qquad = \\frac{\\sin\\angle A C P}{\\sin\\angle A P C}\\times \\frac{d E C}{d E B}\\times \\frac{\\sin\\angle B Q D}{\\sin\\angle B D Q}\\] \\[\\qquad = \\frac{\\sin\\angle A C P}{\\sin(\\angle A C P + \\angle C A P)}\\times \\frac{\\sin\\angle E B C}{\\sin\\angle E C B}\\times \\frac{\\sin(\\angle B D Q + \\angle D B Q)}{\\sin\\angle B D Q}\\] \\[\\qquad = \\frac{\\sin\\angle B C E}{\\sin(\\angle B C E + \\angle A D E)}\\times \\frac{\\sin\\angle D A E}{\\sin\\angle A D E}\\times \\frac{\\sin(\\angle D A E + \\angle E C B)}{\\sin\\angle E A D}\\] \\[\\qquad = \\frac{\\sin(\\angle D A E + \\angle E C B)}{\\sin(\\angle B C E + \\angle A D E)}\\] \\[\\qquad = 1.\\] \n\nIt follows that \\(\\lim_{B\\to \\infty_{D E}}\\frac{|A P|}{|D Q|} = 1\\) . Therefore, \n\n\\[\\lim_{B\\to \\infty_{D E}}\\frac{|O P|}{|O Q|} = \\lim_{B\\to \\infty_{D E}}\\frac{|O P|}{|A P|}\\times \\lim_{B\\to \\infty_{D E}}\\frac{|A P|}{|D Q|}\\times \\lim_{B\\to \\infty_{D E}}\\frac{|D Q|}{|O Q|}\\] \\[\\qquad = 1,\\] \n\nwhich completes the \\(B\\to \\infty_{D E}\\) case, and we are done.", "metadata": {"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2024.jsonl", "problem_match": "4. ", "solution_match": "## \\(\\S 2.1\\) TSTST 2024/4, proposed by Merlijn Staps \n"}}
+{"year": "2024", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "USA_TSTST", "problem": "Let \\(ABCD\\) be a quadrilateral inscribed in a circle with center \\(O\\) and \\(E\\) be the intersection of segments \\(AC\\) and \\(BD\\) . Let \\(\\omega_{1}\\) be the circumcircle of \\(ADE\\) and \\(\\omega_{2}\\) be the circumcircle of \\(BCE\\) . The tangent to \\(\\omega_{1}\\) at \\(A\\) and the tangent to \\(\\omega_{2}\\) at \\(C\\) meet at \\(P\\) . The tangent to \\(\\omega_{1}\\) at \\(D\\) and the tangent to \\(\\omega_{2}\\) at \\(B\\) meet at \\(Q\\) . Show that \\(OP = OQ\\) .", "solution": "\\(\\P\\) Solution 1. Let \\(R = \\overline{AD} \\cap \\overline{BC}\\) (possibly at infinity, but we'll see it's an Euclidean point later). \n\n\n \n\nClaim — \\(ACRP\\) is an isosceles trapezoid with \\(\\overline{AC} \\parallel \\overline{PR}\\) . Consequently, \\(R\\) is an Euclidean point, and \\(OP = OR\\) . \n\nProof. It is equivalent to show that \\(\\triangle PAC \\cong \\triangle RCA\\) , or without using \\(R\\) , that \n\n\\[\\angle PAC = \\angle ACR = \\angle ACB, \\quad \\text{and} \\quad \\angle PCA = \\angle CAR = \\angle CAD.\\] \n\nIndeed, we have \n\n\\[\\angle PAC = \\angle ADE = \\angle ADB = \\angle ACB\\] \n\nand likewise \\(\\angle PCA = \\angle CAD\\) , as requested. \n\nSimilarly \\(OQ = OR\\) , so we're done.\n\n\n\nSolution 2. \n\n\n \n\nLet \\(PA\\) intersect (ABCD) again at \\(A^{\\prime}\\) and let \\(QB\\) intersect (ABCD) again at \\(B^{\\prime}\\) The problem follows from the following claim. \n\nClaim — \\(OA^{\\prime}PC \\cong ODQB^{\\prime}\\) . \n\nProof. We use directed angles mod \\(\\pi\\) . Note that \n\n\\[\\angle ODQ = \\angle ODA + \\angle ADQ = \\angle DAO + \\angle PAD = \\angle PAO = \\angle OA^{\\prime}P,\\] \n\nand similarly \\(\\angle OCP = \\angle OB^{\\prime}Q\\) . Furthermore, \n\n\\[\\angle A^{\\prime}AD = \\angle AED = \\angle CEB = \\angle CBB^{\\prime},\\] \n\nso \\(A^{\\prime}D = CB^{\\prime}\\) , i.e. \\(A^{\\prime}CDB^{\\prime}\\) is an isosceles trapezoid. This means \\(\\angle A^{\\prime}OC = \\angle DOB^{\\prime}\\) . Combined with \\(OA^{\\prime} = OB^{\\prime} = OD = DC\\) , these three angle equalities imply the congruency. \\(\\square\\) \n\nSolution 3 using complex numbers (Noah Walsh). Use complex numbers with (ABCD) as the unit circle. Because \\(PA\\) is tangent to (ADE), we have \\(\\angle PAC = \\angle ADB = \\frac{1}{2}\\angle AOB\\) , and therefore \n\n\\[\\frac{\\frac{p - a}{c - a}}{\\frac{p - a - 1}{c - a - 1}} = \\frac{a}{b\\] \\[\\Rightarrow b\\frac{p - a}{c - a} = a\\frac{ac\\overline{p} - c}{a - c\\] \\[\\Rightarrow bp + a^2 c\\overline{p} = ab + ac.\\] \n\nThe condition that \\(PC\\) is tangent to (BCE) can be expressed by swapping \\(a\\) with \\(c\\) and \\(b\\) with \\(d\\) , so we get \n\n\\[b p + a^{2}c\\overline{{p}} = a b + a c\\] \\[d p + a c^{2}\\overline{{p}} = c d + a c\\] \\[\\Rightarrow p(a d - b c) = a c d + a^{2}c - a b c - a c^{2}\\] \\[\\Rightarrow p = \\frac{a c(a + d - b - c)}{a d - b c}.\\]\n\n\n\nSimilarly, \n\n\\[q = \\frac{bd(a + d - b - c)}{ad - bc}.\\] \n\nIt is now obvious that \\(p\\) and \\(q\\) have the same magnitude. \n\nRemark. One possible trick one might use to compute \\(p\\) is to compute the coordinates of the points \\(A^{\\prime},C^{\\prime}\\in (ABCD)\\) such that \\(A A^{\\prime}\\) and \\(C C^{\\prime}\\) are tangent to \\((A D E)\\) and \\((B C E)\\) respectively, and then use the chords intersection formula. \n\nIf one does this, one gets \\(a^{\\prime} = a c / b\\) and \\(c^{\\prime} = a c / d\\) . At this point, one can simply recognize that \\(A^{\\prime}\\) and \\(C^{\\prime}\\) are the reflection of \\(B\\) and \\(D\\) over the perpendicular bisector of \\(A C\\) , reflect everything over said perpendicular bisector, and suddenly realize that one has solved the problem synthetically. \n\n\\(\\boxed{ \\begin{array}{r l} \\end{array} }\\) Solution 4 by linearity (Noah Walsh). If \\(A E = D E\\) the problem is true by symmetry, so we assume \\(A E\\neq D E\\) \n\nFix \\(A\\) \\(D\\) , and \\(E\\) , and move \\(B\\) linearly along line \\(D E\\) . By e.g. Reim's theorem, \\(C\\) moves linearly as well. \n\nHow do we compute \\(O\\) ? It lies on the perpendicular bisector of \\(A D\\) , which is fixed. It also lies on the perpendicular bisector of \\(B C\\) , which is parallel to a fixed line, and passes through the midpoint of \\(B C\\) , which moves linearly. Therefore, \\(O\\) moves linearly as well. \n\nPoint \\(P\\) is the intersection of the tangent to \\((A D E)\\) at \\(A\\) , which is fixed, and the tangent to \\((B C E)\\) at \\(C\\) , which moves linearly, and therefore is linear. Similarly, \\(Q\\) moves linearly. Therefore, to verify that \\(\\overrightarrow{O P}\\times \\overrightarrow{O P} = \\overrightarrow{O Q}\\times \\overrightarrow{O Q}\\) , it suffices to check three cases. \n\nThe two cases where \\(E A = E B\\) are trivial by symmetry. We can also check that \\(\\lim_{B\\to \\infty_{D E}}\\frac{\\overrightarrow{O P}\\times\\overrightarrow{O P}}{\\overrightarrow{O Q}\\times\\overrightarrow{O Q}} = 1\\) \n\nIndeed, obviously \\(\\lim_{B\\to \\infty_{D E}}\\frac{|O P|}{|A P|} = \\lim_{B\\to \\infty_{D E}}\\frac{|O Q|}{|D Q|} = 1\\) . In addition, \n\n\\[\\frac{d A P}{d D Q} = \\frac{d A P}{d A C}\\times \\frac{d A C}{d B D}\\times \\frac{d B D}{d B Q}\\] \\[\\qquad = \\frac{\\sin\\angle A C P}{\\sin\\angle A P C}\\times \\frac{d E C}{d E B}\\times \\frac{\\sin\\angle B Q D}{\\sin\\angle B D Q}\\] \\[\\qquad = \\frac{\\sin\\angle A C P}{\\sin(\\angle A C P + \\angle C A P)}\\times \\frac{\\sin\\angle E B C}{\\sin\\angle E C B}\\times \\frac{\\sin(\\angle B D Q + \\angle D B Q)}{\\sin\\angle B D Q}\\] \\[\\qquad = \\frac{\\sin\\angle B C E}{\\sin(\\angle B C E + \\angle A D E)}\\times \\frac{\\sin\\angle D A E}{\\sin\\angle A D E}\\times \\frac{\\sin(\\angle D A E + \\angle E C B)}{\\sin\\angle E A D}\\] \\[\\qquad = \\frac{\\sin(\\angle D A E + \\angle E C B)}{\\sin(\\angle B C E + \\angle A D E)}\\] \\[\\qquad = 1.\\] \n\nIt follows that \\(\\lim_{B\\to \\infty_{D E}}\\frac{|A P|}{|D Q|} = 1\\) . Therefore, \n\n\\[\\lim_{B\\to \\infty_{D E}}\\frac{|O P|}{|O Q|} = \\lim_{B\\to \\infty_{D E}}\\frac{|O P|}{|A P|}\\times \\lim_{B\\to \\infty_{D E}}\\frac{|A P|}{|D Q|}\\times \\lim_{B\\to \\infty_{D E}}\\frac{|D Q|}{|O Q|}\\] \\[\\qquad = 1,\\] \n\nwhich completes the \\(B\\to \\infty_{D E}\\) case, and we are done.", "metadata": {"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2024.jsonl", "problem_match": "4. ", "solution_match": "## \\(\\S 2.1\\) TSTST 2024/4, proposed by Merlijn Staps \n"}}
{"year": "2024", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "USA_TSTST", "problem": "For a positive integer \\(k\\) , let \\(s(k)\\) denote the number of 1s in the binary representation of \\(k\\) . Prove that for any positive integer \\(n\\) , \n\n\\[\\sum_{i = 1}^{n}(-1)^{s(3i)} > 0.\\]", "solution": "\\(\\P\\) Solution 1. Given a set of positive integers \\(S\\) , define \n\n\\[f(S) = \\sum_{k\\in S}(-1)^{s(k)}.\\] \n\nWe also define \n\n\\[S_{\\mathrm{even}} = \\{k\\in S\\mid k\\mathrm{~is~even}\\}\\] \\[S_{\\mathrm{odd}} = \\{k\\in S\\mid k\\mathrm{~is~odd}\\}\\] \n\nand apply functions on sets pointwise, e.g. \n\n\\[\\frac{S - 1}{2} = \\left\\{\\frac{k - 1}{2}\\mid k\\in S\\right\\} .\\] \n\nThe problem follows from the first bullet of the following claim. \n\nClaim — For every positive integer \\(n\\) , \n\n\\[f(\\{3,6,\\ldots ,3n\\}) > 0\\] \\[f(\\{1,4,\\ldots ,3n - 2\\})< 0\\] \\[f(\\{2,5,\\ldots ,3n - 1\\})\\leq 0\\] \n\nProof. Induct on \\(n\\) . The base case \\(n = 1\\) is easy to check. \n\nFor \\(S = \\{3,6,\\ldots ,3n\\}\\) \n\n\\[f(S) = f(S_{\\mathrm{even}}) + f(S_{\\mathrm{odd}})\\] \\[\\qquad = f(S_{\\mathrm{even}} / 2) - f((S_{\\mathrm{odd}} - 1) / 2) > 0\\] \n\nsince the elements of \\(S_{\\mathrm{even}} / 2\\) are 0 mod 3 and the elements of \\((S_{\\mathrm{odd}} - 1) / 2\\) are 1 mod 3. \n\nFor \\(S = \\{1,4,\\ldots ,3n - 2\\}\\) \n\n\\[f(S) = f(S_{\\mathrm{even}}) + f(S_{\\mathrm{odd}})\\] \\[\\qquad = f(S_{\\mathrm{even}} / 2) - f((S_{\\mathrm{odd}} - 1) / 2)< 0\\] \n\nsince the elements of \\(S_{\\mathrm{even}} / 2\\) are 2 mod 3 and the elements of \\((S_{\\mathrm{odd}} - 1) / 2\\) are 0 mod 3. (Note that \\((S_{\\mathrm{odd}} - 1) / 2\\) has multiples of 3 starting at 0 rather than 3, but this is fine because \\(f(\\{3,\\ldots ,3n\\}) > 0\\) clearly implies \\(f(\\{0,\\ldots ,3n\\}) > 0\\) as well.)\n\n\n\n- For \\(S = \\{2, 5, \\ldots , 3n - 1\\}\\) , there are two cases. \n\n- Case 1: \\(2^{2k - 1} \\leq 3n - 1 < 2^{2k}\\) . Define \n\n\\[S_{\\mathrm{small}} = \\{s \\in S \\mid s < 2^{2k - 1}\\}\\] \\[S_{\\mathrm{big}} = \\{s \\in S \\mid s \\geq 2^{2k - 1}\\} .\\] \n\nThen \n\n\\[f(S) = f(S_{\\mathrm{small}}) + f(S_{\\mathrm{big}})\\] \\[\\qquad = f(S_{\\mathrm{small}}) - f(S_{\\mathrm{big}} - 2^{2k - 1})< 0\\] \n\nsince \\(S_{\\mathrm{big}} - 2^{2k - 1} = \\{0, \\ldots , 3n - 1 - 2^{2k - 1}\\}\\) consists of a prefix of the 0 (mod 3) positive integers in addition to 0. \n\n- Case 2: \\(2^{2k} < 3n - 1 < 2^{2k + 1}\\) . Note that if \\(a + b = 2^{2k + 1} - 1\\) , then \\(f(\\{a, b\\}) = 0\\) . If we use this to cancel out all pairs of numbers in \\(S\\) which add to \\(2^{2k + 1} - 1\\) , then we find \n\n\\[f(S) = f(\\{2, 5, \\ldots , 2^{2k + 1} - 3n - 2\\}) \\leq 0.\\] \n\nThis completes the induction. \n\n\\(\\P\\) Solution 2. Let \n\n\\[f(n) = \\sum_{i = 0}^{n}(-1)^{s(3i)},\\qquad g(n) = \\sum_{i = 0}^{n}(-1)^{s(i)}.\\] \n\nNote that the problem is equivalent to showing \\(f(n) \\geq 2\\) for all \\(n \\geq 1\\) . \n\n## Lemma \n\n\\(|g(n)| \\leq 1\\) for all \\(n\\) . \n\nProof. Note that \\(s(2i + 1) = - s(2i)\\) , so \\(g(2n + 1) = 0\\) and \\(g(2n) = (- 1)^{s(2n)}\\) by pairing terms. \\(\\square\\) \n\nWe will now prove that \\(f(n) \\geq 2\\) for all \\(n \\geq 1\\) by strong induction. We can manually verify \\(n \\leq 5\\) as a base case. For the inductive step, let \\(m = \\left\\lfloor \\frac{3n - 3}{4} \\right\\rfloor\\) so that \\(4m + 3 \\leq n\\) . Let \\(S = \\{0, 3, \\ldots , 3n\\}\\) . Given \\(0 \\leq k \\leq m\\) , we can append two digits to the binary representation of \\(k\\) (equivalently, \\(4k + r\\) for \\(0 \\leq r < 4\\) ) to get a number in \\(S\\) . If \\(k\\) is a multiple of 3, we can append 00 or 11. Otherwise, we can append one of 01 or 10 depending on \\(k\\) (mod 3). Almost all elements of \\(S\\) can be covered uniquely in this way, except for at most one missing element. Note that appending 00 or 11 will preserve the parity of the number of 1s while appending 01 or 10 will change the parity. Therefore we can write \n\n\\[f(n) \\geq 2 \\sum_{\\substack{0 \\leq i \\leq m \\\\ i \\equiv 0 \\pmod{3}}} (-1)^{s(i)} - \\sum_{\\substack{0 \\leq i \\leq m \\\\ i \\not\\equiv 0 \\pmod{3}}} (-1)^{s(i)} - 1\\] \n\n(the \\(- 1\\) is to account for the possible missing element). Let \n\n\\[x = \\sum_{\\substack{0 \\leq i \\leq m \\\\ i \\equiv 0 \\pmod{3}}} (-1)^{s(i)}, \\qquad y = \\sum_{\\substack{0 \\leq i \\leq m \\\\ i \\not\\equiv 0 \\pmod{3}}} (-1)^{s(i)}.\\]\n\n\n\nAccording to the lemma, \\(|x + y| \\leq 1\\) . By the inductive hypothesis, \\(x \\geq 2\\) . Therefore \n\n\\[f(n) \\geq 2x - y - 1 = 3x - (x + y) - 1 \\geq 6 - 1 - 1 > 2\\] \n\nas desired. \n\n\\(\\P\\) Solution 3. For \\(r \\in \\{0,1,2\\}\\) , define \n\n\\[S_{r}(a,b) = \\sum_{\\substack{i\\in [a,b] \\\\ i\\equiv r\\bmod 3}}(-1)^{s(i)},\\] \n\nwhere \\(s(i)\\) denotes the number of 1s in \\(i\\) 's binary representation. The problem is equivalent to proving \\(S_{0}(0,3n + 1) > 1\\) for all \\(n \\geq 1\\) . \n\n## Lemma \n\nFor any integer \\(d\\) , \\(S_{r}(0,2^{d})\\) is given by: \n\n\\[d \\mid S_{0}(0,2^{d}) S_{1}(0,2^{d}) S_{2}(0,2^{d})\\] \\[\\mathrm{d o d d} 3^{\\frac{d - 1}{2}} -3^{\\frac{d - 1}{2}} 0\\] \\[\\mathrm{d e v e n} 2\\cdot 3^{\\frac{d - 2}{2}} -3^{\\frac{d - 2}{2}} -3^{\\frac{d - 2}{3}}\\] \n\nProof. Induction with \\(d \\leq 2\\) being clear. For the inductive step, observe that \n\n\\[S_{0}(0,2^{d}) = S_{0}(0,2^{d - 1}) + S_{0}(2^{d - 1},2^{d}) = S_{0}(0,2^{d - 1}) - S_{2^{d - 1}}(0,2^{d - 1})\\] \\[S_{1}(0,2^{d}) = S_{1}(0,2^{d - 1}) + S_{1}(2^{d - 1},2^{d}) = S_{1}(0,2^{d - 1}) - S_{1 + 2^{d - 1}}(0,2^{d - 1})\\] \\[S_{2}(0,2^{d}) = S_{2}(0,2^{d - 1}) + S_{2}(2^{d - 1},2^{d}) = S_{2}(0,2^{d - 1}) - S_{2 + 2^{d - 1}}(0,2^{d - 1}).\\] \n\nUsing \n\n\\[2^{d - 1} \\equiv \\begin{cases} 1 \\bmod 3 & d \\text{odd} \\\\ 2 \\bmod 3 & d \\text{even} \\end{cases}\\] \n\nand applying the inductive hypothesis gives the desired result. \n\nThis proves the problem for powers of 2. To prove the problem for general \\(n\\) , split \\([0,3n + 1)\\) into blocks \n\n\\[[0,3n + 1) = [0,2^{d_{1}})\\sqcup [2^{d_{1}},2^{d_{1}} + 2^{d_{2}})\\sqcup [2^{d_{1}} + 2^{d_{2}},2^{d_{1}} + 2^{d_{2}} + 2^{d_{3}}),\\dots\\] \n\nwhose lengths are decreasing powers of 2. By the lemma, \n\n\\[S_{0}(0,2^{d_{1}}) \\geq \\begin{cases} 3^{\\frac{d_{1} - 1}{2}} & d_{1} \\text{odd} \\\\ 2 \\cdot 3^{\\frac{d_{1} - 2}{2}} & d_{1} \\text{even} \\end{cases}\\] \n\n\\[S_{0}(2^{d_{1}},2^{d_{1}} + 2^{d_{2}}) = -S_{2^{d_{1}}}(0,2^{d_{2}}) \\geq 0\\] \n\nand for all \\(i\\) , \n\n\\[S_{0}(2^{d_{1}} + \\dots +2^{d_{i - 1}},2^{d_{1}} + \\dots +2^{d_{i}}) = (-1)^{i - 1}S_{2^{d_{1}} + \\dots +2^{d_{i - 1}}}(0,2^{d_{i}})\\] \\[\\qquad \\geq \\begin{cases} -2 \\cdot 3^{\\frac{d_{i} - 2}{2}} & d_{i} \\text{even} \\\\ -1 \\cdot 3^{\\frac{d_{i} - 1}{2}} & d_{i} \\text{odd}. \\end{cases}\\]\n\n\n\nLet \\(D = d_{1}\\) ; assume \\(D \\geq 4\\) since checking \\(D \\in \\{1, 2, 3\\}\\) is easy. Summing over all intervals gives \n\n\\[S_{0}(0,3n + 1) \\geq 2 \\cdot 3^{\\frac{D - 2}{2}} + 0 - \\left(2 \\cdot 3^{\\frac{D - 4}{2}} + 3^{\\frac{D - 4}{2}} + 2 \\cdot 3^{\\frac{D - 6}{2}} + 3^{\\frac{D - 6}{2}} + \\dots\\right) = 3^{\\frac{D - 2}{2}} > 1\\] \n\nwhen \\(D\\) is even, and \n\n\\[S_{0}(0,3n + 1) \\geq 3^{\\frac{D - 1}{2}} + 0 - \\left(3^{\\frac{D - 3}{2}} + 2 \\cdot 3^{\\frac{D - 5}{2}} + 3^{\\frac{D - 5}{2}} + 2 \\cdot 3^{\\frac{D - 7}{2}} + \\dots\\right) = \\frac{1}{2} \\cdot 3^{\\frac{D - 3}{2}} > 1\\] \n\nwhen \\(D\\) is odd. \n\nRemark. The following Python code calculates \\(\\sum_{i = 0}^{n}(- 1)^{s(3i)}\\) in \\(O(\\log n)\\) time using digit DP (and in \\(O(K \\log n)\\) time if 3 is replaced by \\(K\\) ): \n\nimport functools \n\nK = 3 \n\ndef sgn(n: int) - > int: \n\nreturn - 1 if bin(n).count('1') % 2 else 1 \n\n@functools.lru_cache(maxsize=None) \n\ndef f(n: int) - > list[int]: \n\n# Returns an array of size K where arr[i] = sum of sgn(j) over 0 \n\n<= j <= n with j % K = i \n\nans = [0] * K \n\nif n == 0: \n\nans[0] += 1 \n\nreturn ans \n\nx = f(n // 2) \n\nfor i in range(K): \n\nans[2 * i % K] += x[i] \n\nans[(2 * i + 1) % K] -= x[i] \n\nif n % 2 == 0: \n\nans[(n + 1) % K] -= sgn(n + 1) \n\nreturn ans \n\nRemark. Define \n\n\\[f(n) = \\sum_{i = 0}^{\\lfloor n / 3\\rfloor}(-1)^{s(3i)}.\\] \n\nThen there exist constants \\(0 < c_{1} < c_{2}\\) such that \n\n\\[\\lim_{n\\to \\infty}\\inf_{n\\log_{4}(3)}\\frac{f(n)}{n^{\\log_{4}(3)}} = c_{1},\\qquad \\lim_{n\\to \\infty}\\sup_{n^{\\log_{4}(3)}}\\frac{f(n)}{n^{\\log_{4}(3)}} = c_{2}.\\] \n\nThe upper constant is \\(c_{2} = \\frac{2}{3}\\) , achieved when \\(n\\) is a power of 4.", "metadata": {"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2024.jsonl", "problem_match": "5. ", "solution_match": "## \\(\\S 2.2\\) TSTST 2024/5, proposed by Holden Mui \n"}}
{"year": "2024", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "USA_TSTST", "problem": "Determine whether there exists a function \\(f\\colon \\mathbb{Z}_{>0}\\to \\mathbb{Z}_{>0}\\) such that for all positive integers \\(m\\) and \\(n\\) , \n\n\\[f(m + n f(m)) = f(n)^{m} + 2024!\\cdot m.\\]", "solution": "The answer is no. Let \\(P(m,n)\\) denote the given FE. \n\n\\(\\P\\) Solution 1 (Gopal Goel). Suppose there was a function \\(f\\) , and let \\(r = f(1)\\) . Note that \\(P(1,n)\\) gives \n\n\\[f(1 + r n) = f(n) + 2024!.\\] \n\nIterating this result gives \n\n\\[f(1 + r + \\dots +r^{k}) = r + k\\cdot 2024!\\] \n\nfor all \\(k\\in \\mathbb{Z}_{\\geq 0}\\) . If \\(r = 1\\) , this implies \\(f(k + 1) = 1 + k\\cdot 2024!\\) , which isn't a valid solution, so \\(r\\geq 2\\) \n\nLet \\(m = 1 + r + \\dots +r^{r^{2}(r - 1)}\\) . Note that \n\n\\[1 + r + \\dots +r^{k} = m + n f(m)\\iff \\frac{r^{k + 1} - r^{r^{2}(r - 1) + 1}}{r - 1} = n[r + r^{2}(r - 1)2024!].\\] \n\nThis has a positive integer solution for \\(n\\) as long as \\(k\\) is sufficiently large in terms of \\(r\\) and \n\n\\[k\\equiv r^{2}(r - 1)\\pmod {\\phi (1 + r(r - 1)\\cdot 2024!)}\\] \n\nFor such \\(k\\) , \\(P(m,n)\\) implies \n\n\\[r + k\\cdot 2024! = f(n)^{m} + 2024!\\cdot m.\\] \n\nIn particular, \\(r + (k - m)\\cdot 2024!\\) is a perfect \\(m\\) th power. This has to be true for all \\(k\\) sufficiently large in an arithmetic progression, which is clearly impossible, as desired. \n\n\\(\\P\\) Solution 2 (Carl Schildkraut). As in the previous solution, we have \n\n\\[f(1 + r + \\dots +r^{k}) = r + k\\cdot 2024!\\] \n\nfor all \\(k\\in \\mathbb{Z}_{\\geq 0}\\) , where \\(r = f(1)\\geq 2\\) \n\nClaim — We have \\(f(a)\\geq \\frac{2^{a} - 1}{2024! \\cdot a}\\) for \\(a\\in \\mathbb{Z}_{>0}\\) . \n\nProof. Let \\(c = a + af(a)\\) , \\(b = f(c) + a\\) , and \\(d = f(a)\\) , so that \n\n\\[a + bf(a) = a + af(a) + f(a)f(c) = c + df(c).\\] \n\nComparing \\(P(a,b)\\) and \\(P(c,d)\\) tells us \n\n\\[f(b)^{a} + 2024!\\cdot a = f(d)^{c} + 2024!\\cdot c,\\]\n\n\n\nso \n\n\\[f(b)^{a} - \\left(f(d)^{1 + f(a)}\\right)^{a} = 2024! \\cdot af(a).\\] \n\nSince the left side of the above equation is positive, it must be at least \\(2^{a} - 1\\) , which implies the claim. \\(\\square\\) \n\nPlugging in \\(a = 1 + r + \\dots +r^{k}\\) into the above claim immediately gives the desired contradiction, for sufficiently large \\(k\\) . \n\n\\(\\P\\) Solution 3 (students). If \\(f\\) existed, then for any \\(k\\in \\mathbb{Z}_{>0}\\) we would have \n\n\\[f(f(3))^{3 + k f(3)} + 2024!(3 + k f(3)) = f(3 + k f(3) + f(3)f(3 + k f(3)))\\] \\[\\qquad = f(3 + (k + f(3 + k f(3))f(3)))\\] \\[\\qquad = f(k + f(3 + f(3)k))^{3} + 2024!\\cdot 3.\\] \n\nChoosing \\(k = 27(2024!)^{2}f(3)^{2}\\) gives \n\n\\[f(k + f(3 + f(3)k))^{3} = \\left(f(f(3))^{1 + 9(2024!)^{2}f(3)^{3}}\\right)^{3} + (3\\cdot 2024!f(3))^{3}\\] \n\nwhich contradicts Fermat's last theorem (no two perfect cubes may sum to a perfect cube). Hence \\(f\\) cannot exist.", "metadata": {"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2024.jsonl", "problem_match": "6. ", "solution_match": "## \\(\\S 2.3\\) TSTST 2024/6, proposed by Jaedon Whyte \n"}}
{"year": "2024", "tier": "T0", "problem_label": "7", "problem_type": null, "exam": "USA_TSTST", "problem": "An infinite sequence \\(a_{1},a_{2},a_{3},\\ldots\\) of real numbers satisfies \n\n\\[a_{2n - 1} + a_{2n} > a_{2n + 1} + a_{2n + 2}\\qquad \\mathrm{and}\\qquad a_{2n} + a_{2n + 1}< a_{2n + 2} + a_{2n + 3}\\] \n\nfor every positive integer \\(n\\) . Prove that there exists a real number \\(C\\) such that \\(a_{n}a_{n + 1}< C\\) for every positive integer \\(n\\) .", "solution": "Aer \\(n\\) . \n\nIt suffices to solve the problem for sufficiently large \\(n\\) . Let \\(d_{n} = (- 1)^{n - 1}(a_{n + 2} - a_{n})\\) . The assertion simply says that \\(d_{1}, d_{2}, \\ldots\\) is strictly increasing. \n\nWe consider the following cases. \n\n- Suppose that \\(d_{k} > 0\\) for some \\(k\\) . Then, \n\n\\[a_{2n + 1} = a_{1} + (d_{1} + d_{3} + \\dots +d_{2n - 1})\\] \n\nclearly diverges to \\(+\\infty\\) , and \n\n\\[a_{2n} = a_{2} - (d_{2} + d_{4} + \\dots +d_{2n - 2})\\] \n\nclearly diverges to \\(-\\infty\\) , so taking \\(C = 0\\) works. \n\n- Now assume that \\(d_{k} \\leq 0\\) for all \\(k\\) . This implies that \\((a_{2n + 1})_{n \\geq 0}\\) is weakly decreasing and \\((a_{2n})_{n \\geq 1}\\) is weakly increasing. Adding the two expressions above together, we see that \n\n\\[d_{1} + a_{1} + a_{2}< a_{2n} + a_{2n + 1}< a_{1} + a_{2}.\\] \n\nSince \\((a_{2n + 1})_{n \\geq 0}\\) is weakly decreasing, it either diverges to \\(-\\infty\\) or has a finite limit. Similarly, \\((a_{2n})_{n \\geq 1}\\) either diverges to \\(+\\infty\\) or has a finite limit. \n\n- If \\((a_{2n + 1})_{n \\geq 0}\\) diverges to \\(-\\infty\\) , then the first inequality above implies that \\((a_{2n})_{n \\geq 1}\\) must diverge to \\(+\\infty\\) , in which case \\(C = 0\\) works. \n\n- If \\((a_{2n})_{n \\geq 1}\\) diverges to \\(+\\infty\\) , then the second inequality implies that \\((a_{2n + 1})_{n \\geq 0}\\) must diverge to \\(-\\infty\\) , so again \\(C = 0\\) works. \n\n- Finally assume that both \\(\\lim_{n \\to \\infty} a_{2n + 1}\\) and \\(\\lim_{n \\to \\infty} a_{2n}\\) exist, so the limit \\(L = \\lim_{n \\to \\infty} a_{2n} a_{2n + 1}\\) also exists. The result follows taking \\(C > L\\) .", "metadata": {"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2024.jsonl", "problem_match": "7. ", "solution_match": "## \\(\\S 3.1\\) TSTST 2024/7, proposed by Merlijn Staps \n"}}
-{"year": "2024", "tier": "T0", "problem_label": "8", "problem_type": null, "exam": "USA_TSTST", "problem": "Let \\(ABC\\) be a scalene triangle, and let \\(D\\) be a point on side \\(BC\\) satisfying \\(\\angle BAD = \\angle DAC\\) . Suppose that \\(X\\) and \\(Y\\) are points inside \\(ABC\\) such that triangles \\(ABX\\) and \\(ACY\\) are similar and quadrilaterals \\(ACDX\\) and \\(ABDY\\) are cyclic. Let lines \\(BX\\) and \\(CY\\) meet at \\(S\\) and lines \\(BY\\) and \\(CX\\) meet at \\(T\\) . Prove that lines \\(DS\\) and \\(AT\\) are parallel.", "solution": "\\(\\P\\) Solution by characterizing \\(X\\) and \\(Y\\) . We first state an important property of \\(X\\) and \\(Y\\) . \n\nClaim — Points \\(X\\) and \\(Y\\) are isogonal conjugates with respect to \\(\\triangle ABC\\) . \n\n\n \n\nHere are two proofs of the claim. \n\nFirst proof of claim by Maxim Li. We prove an equivalent statement that \\(S\\) and \\(T\\) are isogonal conjugates with respect to \\(\\triangle ABC\\) . \n\nFirst, we note that \\(\\angle YBC = \\angle YAD = \\angle XAD = \\angle XCB\\) , so \\(BT = TC\\) . Now, let \\(T'\\) be the isogonal conjugate of \\(S\\) w.r.t. \\(\\triangle ABC\\) . Since \\(\\angle ABS = \\angle ACS\\) , it follows that \\(T'B = T'C\\) . Thus, both \\(T\\) and \\(T'\\) lies on perpendicular bisector of \\(BC\\) . \n\nMoreover, since \\(AX\\) and \\(AY\\) are isogonal with respect to \\(\\angle BAC\\) , by DDIT on point \\(A\\) and \\(BXCY\\) , we get that \\(AS\\) and \\(AT\\) are isogonal with respect to \\(\\angle BAC\\) . Hence, \\(A, T, T'\\) are collinear. Since \\(\\triangle ABC\\) is scalene, combining this with the previous paragraph gives \\(T = T'\\) , or \\(S\\) and \\(T\\) are isogonal conjugates. \\(\\square\\)\n\n\n\nA second different proof of the claim. Let \\(\\gamma_{B}\\) be the circle through \\(B\\) and \\(C\\) tangent to \\(AB\\) , and define \\(\\gamma_{C}\\) similarly. We claim that \\(X\\in \\gamma_{B}\\) and \\(Y\\in \\gamma_{C}\\) . \n\nLet \\(E\\) be the second intersection of \\(\\gamma_{B}\\) with \\(A C\\) , and let \\(F\\) be the intersection of \\(B E\\) and \\(A D\\) . Consider the transformation \\(\\tau\\) that is the composition of a reflection across \\(A D\\) and a homothety at \\(A\\) with ratio \\(\\frac{A B}{A C}\\) . Note that \\(\\tau\\) maps \\(Y\\mapsto X\\) , \\(B\\mapsto E\\) , and \\(D\\mapsto F\\) . Thus, \\(X\\) lies on \\(\\tau (\\odot (A B D)) = \\odot (A E F)\\) , so \\(X\\) is the second intersection of \\(\\odot (A E F)\\) and \\(\\odot (A D C)\\) , i.e. it is the Miquel point of quadrilateral \\(C D F E\\) . Thus, \\(X\\) lies on \\(\\odot (B E C) = \\gamma_{B}\\) , as claimed. \n\nFinally, the main claim follows from \\(\\angle B C X = \\angle A B X = \\angle A C Y\\) . \n\nHere are two ways to finish after the claim. \n\nFinish with isosceles trapezoid. We compute \n\n\\[\\angle A X B = 360^{\\circ} - \\angle B X C - \\angle C X A = 360^{\\circ} - \\angle B E C - \\angle C D A = 180^{\\circ} - \\frac{\\angle A}{2}.\\] \n\nThus, if \\(B X\\) meets \\(\\odot (A C D)\\) again at \\(C^{\\prime}\\) , then \n\n\\[\\angle A D C^{\\prime} = \\angle A X C^{\\prime} = \\frac{\\angle A}{2} = \\angle C A D,\\] \n\nso \\(A D C C^{\\prime}\\) is an isosceles trapezoid. Similarly, if \\(C Y\\) meets \\(\\odot (A B D)\\) again at \\(B^{\\prime}\\) , then \\(A D B B^{\\prime}\\) is an isosceles trapezoid. Hence, \\(B C C^{\\prime}B^{\\prime}\\) is an isosceles trapezoid whose diagonals meet at \\(S\\) , so we have that \\(S A = S D\\) . \n\nHowever, by the claim, \\(S\\) and \\(T\\) are isogonal conjugates in \\(A B C\\) . Thus, \n\n\\[\\angle A D S = \\angle S A D = \\angle D A T,\\] \n\nwhich completes the proof. \n\nFinish with angle chasing (Pitchayut Saengrungkongka). By the claim, let \\(\\angle D A X =\\) \\(\\angle D A Y = \\angle A B X = \\angle A C Y = \\theta\\) . Then, \n\n\\[\\angle X S Y = \\angle A + 2\\theta\\] \\[\\angle X D Y = \\angle B D Y + \\angle C D X - 180^{\\circ} = 180^{\\circ} - (\\angle B A Y + \\angle C A X)\\] \\[\\qquad = 180^{\\circ} - (\\angle A + 2\\theta),\\] \n\nso \\(X S Y D\\) is cyclic. Moreover, \n\n\\[\\angle X T Y = 180^{\\circ} - \\angle Y B C - \\angle X C B = 180^{\\circ} - 2\\theta = 180^{\\circ} - \\angle X A Y,\\] \n\nso \\(A X T Y\\) is also cyclic. Finally, \n\n\\[\\angle (A T,A X) = \\angle T Y X = \\angle B Y D + \\angle D Y X\\] \\[\\qquad = \\angle B A D + \\angle D S X = \\angle B X A + \\angle D S X\\] \\[\\qquad = \\angle (D S,A X).\\] \n\n\\(\\P\\) Solution 2 (Ruben Carpenter). Let \\(M\\) be the midpoint of \\(B C\\) , and \\(S^{\\prime}\\) , \\(T^{\\prime}\\) the reflections of \\(S\\) , \\(T\\) over \\(M\\) . In what follows, DDIT is short for the dual of Desargues' involution theorem.\n\n\n\nClaim — \\(S^{\\prime}\\) lies on \\(A T\\) . \n\nProof. From \\(\\triangle A B X \\sim \\triangle A Y C\\) , reflection along the angle bisector of \\(\\angle A B C\\) is an involution on the pencil of lines through \\(A\\) with pairs \\((A B,A C)\\) , \\((A X,A Y)\\) and \\((A\\infty_{B S},A\\infty_{C S})\\) . By DDIT from \\(A\\) onto complete quadrilateral \\(X Y S T B C\\) , \\(A S\\) is sent to \\(A T\\) . By DDIT from \\(A\\) onto \\(B C \\infty_{B X} \\infty_{C Y} S S^{\\prime}\\) , \\((A S,A S^{\\prime})\\) is also an involutive pair, so \\(A S^{\\prime} \\equiv A T\\) . \\(\\square\\) \n\nClaim — \\(T^{\\prime}\\) lies on \\(D S\\) . \n\nProof. Using the circles \\((A C D X)\\) , \\((A B D Y)\\) we obtain \n\n\\[\\angle B D T = \\angle D A Y = \\angle D A C - \\angle Y A C = \\angle B A D - \\angle B A X = \\angle X A D = \\angle T C B,\\] \n\nso \\(T\\) lies on the perpendicular bisector of \\(B C\\) . Furthermore \n\n\\[\\angle C D Y = \\angle B A Y = \\angle X A C = \\angle X D B,\\] \n\nso by DDIT from \\(D\\) onto \\(X S Y T B C\\) shows \\(\\angle C D T = \\angle S D B\\) . The conclusion follows. \\(\\square\\) \n\nFinally, since \\(S T S^{\\prime}T^{\\prime}\\) is a parallelogram we immediately have \\(D S \\parallel A T\\) . \n\nRemark. This problem was discovered by taking a degenerate case of IMO 2018/6 in which three of the vertices of the quadrilateral are collinear. Nevertheless, this origin is very obscured in the problem statement and does not seem to help with the solution at all. The original statement asked to show that the midpoints of \\(A D\\) , \\(B C\\) , and \\(X Y\\) are collinear. However, it was found that this is a straightforward consequence of a high- powered result about isogonal conjugates, so the statement was changed.", "metadata": {"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2024.jsonl", "problem_match": "8. ", "solution_match": "## \\(\\S 3.2\\) TSTST 2024/8, proposed by Michael Ren \n"}}
+{"year": "2024", "tier": "T0", "problem_label": "8", "problem_type": null, "exam": "USA_TSTST", "problem": "Let \\(ABC\\) be a scalene triangle, and let \\(D\\) be a point on side \\(BC\\) satisfying \\(\\angle BAD = \\angle DAC\\) . Suppose that \\(X\\) and \\(Y\\) are points inside \\(ABC\\) such that triangles \\(ABX\\) and \\(ACY\\) are similar and quadrilaterals \\(ACDX\\) and \\(ABDY\\) are cyclic. Let lines \\(BX\\) and \\(CY\\) meet at \\(S\\) and lines \\(BY\\) and \\(CX\\) meet at \\(T\\) . Prove that lines \\(DS\\) and \\(AT\\) are parallel.", "solution": "\\(\\P\\) Solution by characterizing \\(X\\) and \\(Y\\) . We first state an important property of \\(X\\) and \\(Y\\) . \n\nClaim — Points \\(X\\) and \\(Y\\) are isogonal conjugates with respect to \\(\\triangle ABC\\) . \n\n\n \n\nHere are two proofs of the claim. \n\nFirst proof of claim by Maxim Li. We prove an equivalent statement that \\(S\\) and \\(T\\) are isogonal conjugates with respect to \\(\\triangle ABC\\) . \n\nFirst, we note that \\(\\angle YBC = \\angle YAD = \\angle XAD = \\angle XCB\\) , so \\(BT = TC\\) . Now, let \\(T'\\) be the isogonal conjugate of \\(S\\) w.r.t. \\(\\triangle ABC\\) . Since \\(\\angle ABS = \\angle ACS\\) , it follows that \\(T'B = T'C\\) . Thus, both \\(T\\) and \\(T'\\) lies on perpendicular bisector of \\(BC\\) . \n\nMoreover, since \\(AX\\) and \\(AY\\) are isogonal with respect to \\(\\angle BAC\\) , by DDIT on point \\(A\\) and \\(BXCY\\) , we get that \\(AS\\) and \\(AT\\) are isogonal with respect to \\(\\angle BAC\\) . Hence, \\(A, T, T'\\) are collinear. Since \\(\\triangle ABC\\) is scalene, combining this with the previous paragraph gives \\(T = T'\\) , or \\(S\\) and \\(T\\) are isogonal conjugates. \\(\\square\\)\n\n\n\nA second different proof of the claim. Let \\(\\gamma_{B}\\) be the circle through \\(B\\) and \\(C\\) tangent to \\(AB\\) , and define \\(\\gamma_{C}\\) similarly. We claim that \\(X\\in \\gamma_{B}\\) and \\(Y\\in \\gamma_{C}\\) . \n\nLet \\(E\\) be the second intersection of \\(\\gamma_{B}\\) with \\(A C\\) , and let \\(F\\) be the intersection of \\(B E\\) and \\(A D\\) . Consider the transformation \\(\\tau\\) that is the composition of a reflection across \\(A D\\) and a homothety at \\(A\\) with ratio \\(\\frac{A B}{A C}\\) . Note that \\(\\tau\\) maps \\(Y\\mapsto X\\) , \\(B\\mapsto E\\) , and \\(D\\mapsto F\\) . Thus, \\(X\\) lies on \\(\\tau (\\odot (A B D)) = \\odot (A E F)\\) , so \\(X\\) is the second intersection of \\(\\odot (A E F)\\) and \\(\\odot (A D C)\\) , i.e. it is the Miquel point of quadrilateral \\(C D F E\\) . Thus, \\(X\\) lies on \\(\\odot (B E C) = \\gamma_{B}\\) , as claimed. \n\nFinally, the main claim follows from \\(\\angle B C X = \\angle A B X = \\angle A C Y\\) . \n\nHere are two ways to finish after the claim. \n\nFinish with isosceles trapezoid. We compute \n\n\\[\\angle A X B = 360^{\\circ} - \\angle B X C - \\angle C X A = 360^{\\circ} - \\angle B E C - \\angle C D A = 180^{\\circ} - \\frac{\\angle A}{2}.\\] \n\nThus, if \\(B X\\) meets \\(\\odot (A C D)\\) again at \\(C^{\\prime}\\) , then \n\n\\[\\angle A D C^{\\prime} = \\angle A X C^{\\prime} = \\frac{\\angle A}{2} = \\angle C A D,\\] \n\nso \\(A D C C^{\\prime}\\) is an isosceles trapezoid. Similarly, if \\(C Y\\) meets \\(\\odot (A B D)\\) again at \\(B^{\\prime}\\) , then \\(A D B B^{\\prime}\\) is an isosceles trapezoid. Hence, \\(B C C^{\\prime}B^{\\prime}\\) is an isosceles trapezoid whose diagonals meet at \\(S\\) , so we have that \\(S A = S D\\) . \n\nHowever, by the claim, \\(S\\) and \\(T\\) are isogonal conjugates in \\(A B C\\) . Thus, \n\n\\[\\angle A D S = \\angle S A D = \\angle D A T,\\] \n\nwhich completes the proof. \n\nFinish with angle chasing (Pitchayut Saengrungkongka). By the claim, let \\(\\angle D A X =\\) \\(\\angle D A Y = \\angle A B X = \\angle A C Y = \\theta\\) . Then, \n\n\\[\\angle X S Y = \\angle A + 2\\theta\\] \\[\\angle X D Y = \\angle B D Y + \\angle C D X - 180^{\\circ} = 180^{\\circ} - (\\angle B A Y + \\angle C A X)\\] \\[\\qquad = 180^{\\circ} - (\\angle A + 2\\theta),\\] \n\nso \\(X S Y D\\) is cyclic. Moreover, \n\n\\[\\angle X T Y = 180^{\\circ} - \\angle Y B C - \\angle X C B = 180^{\\circ} - 2\\theta = 180^{\\circ} - \\angle X A Y,\\] \n\nso \\(A X T Y\\) is also cyclic. Finally, \n\n\\[\\angle (A T,A X) = \\angle T Y X = \\angle B Y D + \\angle D Y X\\] \\[\\qquad = \\angle B A D + \\angle D S X = \\angle B X A + \\angle D S X\\] \\[\\qquad = \\angle (D S,A X).\\] \n\n\\(\\P\\) Solution 2 (Ruben Carpenter). Let \\(M\\) be the midpoint of \\(B C\\) , and \\(S^{\\prime}\\) , \\(T^{\\prime}\\) the reflections of \\(S\\) , \\(T\\) over \\(M\\) . In what follows, DDIT is short for the dual of Desargues' involution theorem.\n\n\n\nClaim — \\(S^{\\prime}\\) lies on \\(A T\\) . \n\nProof. From \\(\\triangle A B X \\sim \\triangle A Y C\\) , reflection along the angle bisector of \\(\\angle A B C\\) is an involution on the pencil of lines through \\(A\\) with pairs \\((A B,A C)\\) , \\((A X,A Y)\\) and \\((A\\infty_{B S},A\\infty_{C S})\\) . By DDIT from \\(A\\) onto complete quadrilateral \\(X Y S T B C\\) , \\(A S\\) is sent to \\(A T\\) . By DDIT from \\(A\\) onto \\(B C \\infty_{B X} \\infty_{C Y} S S^{\\prime}\\) , \\((A S,A S^{\\prime})\\) is also an involutive pair, so \\(A S^{\\prime} \\equiv A T\\) . \\(\\square\\) \n\nClaim — \\(T^{\\prime}\\) lies on \\(D S\\) . \n\nProof. Using the circles \\((A C D X)\\) , \\((A B D Y)\\) we obtain \n\n\\[\\angle B D T = \\angle D A Y = \\angle D A C - \\angle Y A C = \\angle B A D - \\angle B A X = \\angle X A D = \\angle T C B,\\] \n\nso \\(T\\) lies on the perpendicular bisector of \\(B C\\) . Furthermore \n\n\\[\\angle C D Y = \\angle B A Y = \\angle X A C = \\angle X D B,\\] \n\nso by DDIT from \\(D\\) onto \\(X S Y T B C\\) shows \\(\\angle C D T = \\angle S D B\\) . The conclusion follows. \\(\\square\\) \n\nFinally, since \\(S T S^{\\prime}T^{\\prime}\\) is a parallelogram we immediately have \\(D S \\parallel A T\\) . \n\nRemark. This problem was discovered by taking a degenerate case of IMO 2018/6 in which three of the vertices of the quadrilateral are collinear. Nevertheless, this origin is very obscured in the problem statement and does not seem to help with the solution at all. The original statement asked to show that the midpoints of \\(A D\\) , \\(B C\\) , and \\(X Y\\) are collinear. However, it was found that this is a straightforward consequence of a high- powered result about isogonal conjugates, so the statement was changed.", "metadata": {"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2024.jsonl", "problem_match": "8. ", "solution_match": "## \\(\\S 3.2\\) TSTST 2024/8, proposed by Michael Ren \n"}}
{"year": "2024", "tier": "T0", "problem_label": "9", "problem_type": null, "exam": "USA_TSTST", "problem": "Let \\(n \\geq 2\\) be a fixed integer. The cells of an \\(n \\times n\\) table are filled with the integers from 1 to \\(n^2\\) with each number appearing exactly once. Let \\(N\\) be the number of unordered quadruples of cells on this board which form an axis-aligned rectangle, with the two smaller integers being on opposite vertices of this rectangle. Find the largest possible value of \\(N\\) .", "solution": "L \n\nThe largest possible value of \\(N\\) is \\(\\frac{1}{12} n^2 (n^2 - 1)\\) . Call these rectangles wobbly. We defer the construction until the proof is complete, since the proof suggests the construction. \n\nProof of bound. Call a triple of integers \\((a, b, c)\\) an elbow if \\(a\\) and \\(b\\) are in the same row, \\(b\\) and \\(c\\) are in the same column, and \\(a < b > c\\) . Observe that the wobbly rectangles are exactly the ones with 2 elbows. \n\nClaim — Every axis- aligned rectangle has at least 1 elbow. \n\nProof. The smallest integer in any rectangle is the center of an elbow. \n\nRemark. In fact, it is true that any rectangle must have exactly 1 or 2 elbows. However, this fact is not needed in the proof. \n\nLet \\(E\\) be the number of elbows and \\(M\\) be the number of non- wobbly rectangles. Then, the number of elbows is at least \\(M + 2N = \\binom{n}{2}^2 + N\\) , so \n\n\\[E\\geq \\binom{n}{2}^{2} + N\\iff N\\leq E - \\binom{n}{2}^{2}.\\] \n\nTo this end, we will provide an upper bound on \\(E\\) . For each cell \\(c\\) , define: \n\n- \\(f(c)\\) to be the number of cells in \\(c\\) 's row which are smaller than \\(c\\) , and \n\n- \\(g(c)\\) to be the number of cells in \\(c\\) 's column which are smaller than \\(c\\) . \n\nThen, the number of elbows centered at \\(c\\) equals \\(f(c)g(c)\\) . Thus, the number of elbows satisfies \n\n\\[E = \\sum_{c}f(c)g(c)\\] \\[\\leq \\sum_{c}\\frac{1}{2}\\big[f(c)^{2} + g(c)^{2}\\big]\\] \\[= n\\big(0^{2} + \\cdot \\cdot \\cdot +(n - 1)^{2}\\big)\\] \\[= \\frac{n^{2}(n - 1)(2n - 1)}{6}.\\] \n\nIt follows that the number of wobbly rectangles satisfies \n\n\\[N\\leq E - \\binom{n}{2}^{2}\\]\n\n\n\n\\[= \\frac{n^{2}(n - 1)(2n - 1)}{6} -\\frac{n^{2}(n - 1)^{2}}{4}\\] \\[= \\frac{n^{2}(n^{2} - 1)}{12}\\]\n\nas desired.\n\nConstruction. From the proof above, equality holds if and only if \\(f(c)=g(c)\\) everywhere.\n\nSelect any \\(n\\times n\\) Latin square on symbols 0,..., \\(n-1\\) , and replace the \\(n\\) copies of symbol \\(k\\) with the integers \\(kn+1,\\ldots ,kn+n\\) in some order. This construction is valid,because for each cell \\(c\\) , \\(f(c)\\) and \\(g(c)\\) both equal the symbol originally placed in \\(c\\) .\n\nHere is an example for \\(n=6\\) . The left grid is the Latin square, and the right grid is one possible table that can be derived from it.\n\n| 0 | 1 | 2 | 3 | 4 | 5 |
| 5 | 0 | 1 | 2 | 3 | 4 |
| 4 | 5 | 0 | 1 | 2 | 3 |
| 3 | 4 | 5 | 0 | 1 | 2 |
| 2 | 3 | 4 | 5 | 0 | 1 |
| 1 | 2 | 3 | 4 | 5 | 0 |
\n\n| 1 | 7 | 13 | 19 | 25 | 31 |
| 32 | 2 | 8 | 14 | 20 | 26 |
| 27 | 33 | 3 | 9 | 15 | 21 |
| 22 | 28 | 34 | 4 | 10 | 16 |
| 17 | 23 | 29 | 35 | 5 | 11 |
| 12 | 18 | 24 | 30 | 36 | 6 |
\n\n**Remark** (Structure of optimal tables). In fact, any optimal table must lead to a Latin square in a similar way, by reversing the above idea: replace each cell \\(c\\) with \\(f(c)=g(c)\\) .\n\nThis suggests how to construct all optimal tables: choose any Latin square on 0,..., \\(n-1\\) , and fill in the cells so that the relative order of cells in every row and column is preserved.\n\n**Remark.** If the grid is filled out in a random order, the expected number of wobbly rectangles is \\(\\frac {1}{12}n^{2}(n-1)^{2}\\) . This is asymptotically equal to the maximum value.", "metadata": {"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2024.jsonl", "problem_match": "9. ", "solution_match": "## §3.3 TSTST 2024/9 \n"}}
diff --git a/USA_TSTST/segmented/en-sols-TSTST-2025.jsonl b/USA_TSTST/segmented/en-sols-TSTST-2025.jsonl
index 3b72eb84b2673a6493c3b82e0af17e789c622e06..85759f4c7fd14d468847a6b140a33fa6392293cd 100644
--- a/USA_TSTST/segmented/en-sols-TSTST-2025.jsonl
+++ b/USA_TSTST/segmented/en-sols-TSTST-2025.jsonl
@@ -2,8 +2,8 @@
{"year": "2025", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "Find all sets \\(S \\subseteq \\mathbb{Z}\\) for which there exists a function \\(f: \\mathbb{R} \\to \\mathbb{Z}\\) such that \n\n- \\(f(x - y) - 2f(x) + f(x + y) \\geq -1\\) for all \\(x, y \\in \\mathbb{R}\\) , and \n\n- \\(S = \\{f(z) \\mid z \\in \\mathbb{R}\\}\\) .", "solution": "The answer is \\(\\{a\\}\\) , \\(\\{a,a + 1\\}\\) , \\(\\{a,a + 1,a + 2,\\ldots \\}\\) , and \\(\\mathbb{Z}\\) , for arbitrary \\(a\\in \\mathbb{Z}\\) . For constructions, it is not hard to show that if \\(g\\colon \\mathbb{R}\\to \\mathbb{R}\\) is a convex function, then \\(\\lfloor g\\rfloor\\) satisfies the functional equation. Thus \\(f(x) = a\\) , \\(f(x) = \\lfloor x\\rfloor\\) , and \\(f(x) = \\lfloor x^{2}\\rfloor +a\\) work, covering the first, fourth, and third class of answers respectively. Furthermore, it is not hard to show that \\(f(x) = a + \\mathbf{1}_{x > 0}\\) also works to cover the second class. \n\nLet \\(P(x,y)\\) denote the condition. To prove that nothing else works, the key result is to prove an \"intermediate value theorem\": if \\(a\\) and \\(b\\) are in the range of \\(f\\) , then so is every integer between \\(a\\) and \\(b\\) . Let's first see how this finishes. If we assume the intermediate value theorem, then all we need to show is that if the range of the range of \\(f\\) is at least 2, then the range of \\(f\\) is unbounded above. Indeed, if \\(f(x) - f(y)\\geq 2\\) , then \\(P(x,y - x)\\) gives us that \\(f(2x - y) > f(x)\\) , so iterating this procedure finishes. \n\nWe will now prove the intermediate value theorem. We will repeatedly use the fact that if \\(f(x)\\) is a solution, so is \\(f(ax + b) + c\\) for \\(a,b\\in \\mathbb{R}\\) and \\(c\\in \\mathbb{Z}\\) . \n\n## Lemma 1.1 \n\nIf \\(f(0)\\leq - 1\\) , then \\(f(2^{k})\\geq 2^{k}f(1)\\) for \\(k\\geq 0\\) \n\nProof. \\(P(2^{k},2^{k})\\) yields that \\(f(2^{k + 1})\\geq 2f(2^{k})\\) \n\n## Lemma 1.2 \n\nIf \\(f(- 1)\\leq - 2\\) and \\(f(0) = 0\\) , then \\(f(2^{k})\\geq 2^{k} - 1\\) for all positive integers \\(k\\) . \n\nProof. Applying Lemma 1.1 to \\(f(x - 1) + 1\\) yields that \\(f(2^{k} - 1)\\geq 2^{k} - 1\\) . Then, applying Lemma 1.1 to \\(f(2^{k} - x) - f(2^{k}) - 1\\) yields that \n\n\\[f(0) - f(2^{k}) - 1\\geq 2^{k}(f(2^{k} - 1) - f(2^{k}) - 1)\\implies f(2^{k}) + 1\\geq \\frac{2^{k}f(2^{k} - 1)}{2^{k} - 1}\\geq 2^{k}.\\] \n\nNow to prove the intermediate value theorem, scale and shift such that \\(f(- 1)\\leq - 2\\) and \\(f(0) = 0\\) ; it suffices to show that there exists some number strictly between \\(f(- 1)\\) and \\(f(0)\\) in the range of \\(f\\) (since by iteration we can then get all values). Suppose not and let \\(a_{k} = f(- 1 / 2^{k})\\) . If \\(k\\) is minimal such that \\(a_{k}\\geq 0\\) , then \\(P(- 1 / 2^{k},1 / 2^{k})\\) yields a contradiction. Thus \\(a_{k}\\leq - 2\\) for all \\(k\\) . However, applying Lemma 1.2 to \\(f(x / 2^{k})\\) yields that \\(a_{k}\\leq - 2\\implies f(1)\\geq 2^{k} - 1\\) , which cannot hold for all \\(k\\) since \\(f(1)\\) is constant.", "metadata": {"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2025.jsonl", "problem_match": "Problem 2. ", "solution_match": "## \\(\\S 1.2\\) Solution to TSTST 2, by Daniel Zhu \n"}}
{"year": "2025", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "Let \\(a_{1}, a_{2}, r\\) , and \\(s\\) be positive integers with \\(r\\) and \\(s\\) odd. The sequence \\(a_{1}, a_{2}, a_{3}, \\ldots\\) is defined by \n\n\\[a_{n + 2} = r a_{n + 1} + s a_{n}\\] \n\nfor all \\(n \\geq 1\\) . Determine the maximum possible number of integers \\(1 \\leq \\ell \\leq 2025\\) such that \\(a_{\\ell}\\) divides \\(a_{\\ell +1}\\) , over all possible choices of \\(a_{1}, a_{2}, r\\) , and \\(s\\) .", "solution": "Answer 1350. \n\n\\(\\P\\) Solution We first provide the upper bound. We start by dividing out any common factors of \\(a_{1}\\) and \\(a_{2}\\) from the whole sequence. Note that since \\(r\\) and \\(s\\) are odd, and \\(a_{1}\\) and \\(a_{2}\\) cannot both be divisible by 2, the sequence \\(a_{1}, a_{2}, a_{3}, \\ldots\\) (mod 2) must be some cyclic shift of the sequence \\(1, 1, 0, 1, 1, 0, 1, 1, 0, \\ldots\\) . This means that exactly \\(\\frac{2025}{3} = 675\\) of the values of \\(\\ell\\) satisfy \\(a_{\\ell} \\equiv 0\\) (mod 2) and \\(a_{\\ell +1} \\equiv 1\\) (mod 2). An even number can never divide an odd number, so we have an upper bound of \\(2025 \\times \\frac{2}{3} = 1350\\) . \n\nNow we provide a construction so that \\(a_{\\ell}\\) divides \\(a_{\\ell +1}\\) for 1350 values of \\(\\ell\\) . Let \\(F_{1}, F_{2}, F_{3}, \\ldots\\) denote the Fibonacci sequence with \\(F_{1} = F_{2} = 1\\) . Let \n\n\\[C = \\frac{F_{1} \\cdot F_{2} \\cdot F_{3} \\cdot F_{4} \\cdots F_{2025}}{F_{3} \\cdot F_{6} \\cdot F_{9} \\cdot F_{12} \\cdots F_{2025}} = F_{1} \\cdot F_{2} \\cdot F_{4} \\cdot F_{5} \\cdots F_{2024}.\\] \n\nNote that \\(C\\) is odd, since it is the product of all odd Fibonacci numbers up to \\(F_{2025}\\) . We let \\(a_{n} = C^{n - 1} F_{n}\\) , which satisfies the recurrence with \\(r = C^{2}\\) and \\(s = C\\) . This gives \n\n\\[\\frac{a_{\\ell +1}}{a_{\\ell}} = C \\cdot \\frac{F_{\\ell +1}}{F_{\\ell}},\\] \n\nwhich is an integer whenever \\(3 \\nmid \\ell\\) for \\(1 \\leq \\ell \\leq 2025\\) .", "metadata": {"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2025.jsonl", "problem_match": "Problem 3. ", "solution_match": "## §1.3 Solution to TSTST 3, by Carlos Rodriguez, Albert Wang, Kevin Wu, Isaac Zhu, Nathan Cho \n"}}
{"year": "2025", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "Let \\(n \\geq 2\\) be a positive integer. Let \\(a_{1}, a_{2}, \\ldots , a_{n}\\) be a sequence of positive integers such that \n\n\\[\\gcd (a_{1},a_{2}),\\gcd (a_{2},a_{3}),\\ldots ,\\gcd (a_{n - 1},a_{n})\\] \n\nis a strictly increasing sequence. Find, in terms of \\(n\\) , the maximum possible value of \n\n\\[\\frac{1}{a_{1}} +\\frac{1}{a_{2}} +\\dots +\\frac{1}{a_{n}}\\] \n\nover all such sequences.", "solution": "We claim the maximum possible value is 2. To see that this is achievable, let the sequence \\((a_{i})_{i = 1}^{n}\\) be 1, 2, 4, ..., \\(2^{n - 2}\\) , \\(2^{n - 2}\\) . Then \\(\\gcd (a_{i},a_{i + 1}) = 2^{i - 1}\\) , which is an increasing sequence, and it is easy to check that \\(\\sum \\frac{1}{a_{i}} = 2\\) . We now show this is the maximum. \n\nLet \\(d_{i} = \\gcd (a_{i},a_{i + 1})\\) . Since \\(d_{i - 1}< d_{i}\\) , we have that \\(\\frac{a_{i}}{d_{i - 1}} >\\frac{a_{i}}{d_{i}}\\) . But these are both integers, so we get that \n\n\\[1\\leq \\frac{a_{i}}{d_{i - 1}} -\\frac{a_{i}}{d_{i}}\\Longrightarrow \\frac{1}{a_{i}}\\leq \\frac{1}{d_{i - 1}} -\\frac{1}{d_{i}}.\\] \n\nAdding everything up, we get that \n\n\\[\\sum_{i = 1}^{n}\\frac{1}{a_{i}}\\leq \\frac{1}{a_{1}} +\\frac{1}{d_{1}} -\\frac{1}{d_{n - 1}} +\\frac{1}{a_{n}}.\\] \n\nBut \\(d_{n - 1}\\mid a_{n}\\) , so \\(d_{n - 1}\\leq a_{n}\\) , and \\(\\frac{1}{d_{n - 1}}\\geq \\frac{1}{a_{n}}\\) . Thus, this sum is at most \\(\\frac{1}{a_{1}} +\\frac{1}{d_{1}}\\leq 2\\) . \\(\\square\\)", "metadata": {"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2025.jsonl", "problem_match": "Problem 4. ", "solution_match": "## \\(\\S 2.1\\) Solution to TSTST 4, by Maxim Li \n"}}
-{"year": "2025", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "A tetrahedron \\(ABCD\\) is said to be angelic if it has nonzero volume and satisfies \n\n\\[\\angle BAC + \\angle CAD + \\angle DAB = \\angle ABC + \\angle CBD + \\angle DBA,\\] \\[\\angle ACB + \\angle BCD + \\angle DCA = \\angle ADB + \\angle BDC + \\angle CDA.\\] \n\nAcross all angelic tetrahedrons, what is the maximum number of distinct lengths that could appear in the set \\(\\{AB, AC, AD, BC, BD, CD\\}\\) ?", "solution": "\n \n\nWe claim the maximum cardinality is \\(\\boxed{4}\\) . This is attained by taking a non- square rectangle (or parallelogram) \\(A C B D\\) and folding it along diagonal \\(A B\\) , creating edges \\(A B\\) and \\(C D\\) in the process. Here, \\(A C = B D\\) and \\(A D = B C\\) , while every other length is distinct in the general case. This tetrahedron is congruent to itself under the permutation of vertices \\((A,B,C,D)\\mapsto (B,A,D,C)\\) , and we can verify that the required angle conditions follow from this symmetry. \n\n\n \n\nIn the other direction, let \\(f(X)\\) denote the sum of the angles at \\(X\\) , so the conditions of the problem statement can be written as \\(f(A) = f(B)\\) and \\(f(C) = f(D)\\) . Unfold the three faces that meet at \\(D\\) to create a net of the tetrahedron. Along with the face \\(\\triangle A B C\\) , we also create the faces \\(\\triangle A B D_{1}\\) , \\(\\triangle B C D_{2}\\) , and \\(\\triangle A C D_{3}\\) . Note that\n\n\n\n\\(f(A) + f(B) + f(C) + f(D)\\) is the sum of all of the angles of all four faces of the tetrahedron, which is \\(720^{\\circ}\\) . Therefore, \\(f(A) = f(B) = 360^{\\circ} - f(C) = 360^{\\circ} - f(D)\\) . Furthermore, we have that \\(\\angle D_{1}AD_{3} = f(A)\\) and \\(\\angle D_{2}CD_{3} = 360^{\\circ} - f(C)\\) . Since \\(AD_{1} = AD_{3}\\) and \\(CD_{2} = CD_{3}\\) by the definition of unfolding, this gives \\(\\triangle D_{3}AD_{1} \\sim \\triangle D_{3}CD_{2}\\) . Thus, due to spiral similarity, we have \\(\\triangle D_{3}AC \\sim \\triangle D_{3}D_{1}D_{2}\\) . \n\nSimilarly, we also have \\(\\triangle D_{2}BD_{1} \\sim \\triangle D_{2}CD_{3} \\Rightarrow \\triangle CBD_{2} \\sim \\triangle D_{3}D_{1}D_{2}\\) . This means that \\(\\triangle CBD_{2} \\sim \\triangle D_{3}AC\\) , and since \\(CD_{3} = CD_{2}\\) , the two triangles are actually congruent. Therefore, \\(AC = BD_{2} = BD\\) and \\(BC = AD_{3} = AD\\) . Since we have two pairs of equal edge lengths, the number of distinct edge lengths is at most 4, as desired.", "metadata": {"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2025.jsonl", "problem_match": "Problem 5. ", "solution_match": "## \\(\\S 2.2\\) Solution to TSTST 5, by Karthik Vedula \n"}}
+{"year": "2025", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "A tetrahedron \\(ABCD\\) is said to be angelic if it has nonzero volume and satisfies \n\n\\[\\angle BAC + \\angle CAD + \\angle DAB = \\angle ABC + \\angle CBD + \\angle DBA,\\] \\[\\angle ACB + \\angle BCD + \\angle DCA = \\angle ADB + \\angle BDC + \\angle CDA.\\] \n\nAcross all angelic tetrahedrons, what is the maximum number of distinct lengths that could appear in the set \\(\\{AB, AC, AD, BC, BD, CD\\}\\) ?", "solution": "\n \n\nWe claim the maximum cardinality is \\(\\boxed{4}\\) . This is attained by taking a non- square rectangle (or parallelogram) \\(A C B D\\) and folding it along diagonal \\(A B\\) , creating edges \\(A B\\) and \\(C D\\) in the process. Here, \\(A C = B D\\) and \\(A D = B C\\) , while every other length is distinct in the general case. This tetrahedron is congruent to itself under the permutation of vertices \\((A,B,C,D)\\mapsto (B,A,D,C)\\) , and we can verify that the required angle conditions follow from this symmetry. \n\n\n \n\nIn the other direction, let \\(f(X)\\) denote the sum of the angles at \\(X\\) , so the conditions of the problem statement can be written as \\(f(A) = f(B)\\) and \\(f(C) = f(D)\\) . Unfold the three faces that meet at \\(D\\) to create a net of the tetrahedron. Along with the face \\(\\triangle A B C\\) , we also create the faces \\(\\triangle A B D_{1}\\) , \\(\\triangle B C D_{2}\\) , and \\(\\triangle A C D_{3}\\) . Note that\n\n\n\n\\(f(A) + f(B) + f(C) + f(D)\\) is the sum of all of the angles of all four faces of the tetrahedron, which is \\(720^{\\circ}\\) . Therefore, \\(f(A) = f(B) = 360^{\\circ} - f(C) = 360^{\\circ} - f(D)\\) . Furthermore, we have that \\(\\angle D_{1}AD_{3} = f(A)\\) and \\(\\angle D_{2}CD_{3} = 360^{\\circ} - f(C)\\) . Since \\(AD_{1} = AD_{3}\\) and \\(CD_{2} = CD_{3}\\) by the definition of unfolding, this gives \\(\\triangle D_{3}AD_{1} \\sim \\triangle D_{3}CD_{2}\\) . Thus, due to spiral similarity, we have \\(\\triangle D_{3}AC \\sim \\triangle D_{3}D_{1}D_{2}\\) . \n\nSimilarly, we also have \\(\\triangle D_{2}BD_{1} \\sim \\triangle D_{2}CD_{3} \\Rightarrow \\triangle CBD_{2} \\sim \\triangle D_{3}D_{1}D_{2}\\) . This means that \\(\\triangle CBD_{2} \\sim \\triangle D_{3}AC\\) , and since \\(CD_{3} = CD_{2}\\) , the two triangles are actually congruent. Therefore, \\(AC = BD_{2} = BD\\) and \\(BC = AD_{3} = AD\\) . Since we have two pairs of equal edge lengths, the number of distinct edge lengths is at most 4, as desired.", "metadata": {"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2025.jsonl", "problem_match": "Problem 5. ", "solution_match": "## \\(\\S 2.2\\) Solution to TSTST 5, by Karthik Vedula \n"}}
{"year": "2025", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "Alice and Bob play a game on \\(n\\) vertices labelled \\(1, 2, \\ldots , n\\) . They take turns adding edges \\(\\{i, j\\}\\) , with Alice going first. Neither player is allowed to make a move that creates a cycle, and the game ends after \\(n - 1\\) total turns. \n\nLet the weight of the edge \\(\\{i, j\\}\\) be \\(|i - j|\\) , and let \\(W\\) be the total weight of all edges at the end of the game. Alice plays to maximize \\(W\\) and Bob plays to minimize \\(W\\) . If both play optimally, what will \\(W\\) be?", "solution": "\\(\\P\\) Solution Let \\(k = \\lceil \\frac{n - 1}{2}\\rceil\\) . The answer is \n\n\\[\\frac{1}{2} (k + 1)(2n - k - 2) = (n - k - 1) + (n - k) + \\dots +(n - 1).\\] \n\nWhen \\(n = 1\\) , this is clear. \n\nConsider now when \\(n\\geq 2\\) . Note that the game consists of \\(n - 1\\) moves, with Alice making \\(k\\) moves and Bob making \\(n - k - 1\\) moves. \n\nWe first show Alice can guarantee a total of at least \\(\\textstyle {\\frac{1}{2}}(k + 1)(2n - k - 2)\\) \n\nClaim — Alice can ensure her \\(i\\) th move is an edge of weight at least \\(n - i\\) \n\nProof. On her first move, Alice chooses the edge \\((1,n)\\) , which has weight \\(n - 1\\) . Now, consider the \\(i\\) th move, for \\(i > 1\\) . \n\nConsider the induced subgraph on vertices \\(\\{1,\\ldots ,i\\} \\sqcup \\{n - i + 1,\\ldots ,n\\}\\) . (Note that \\(i< n - i + 1\\) for all \\(1\\leq i\\leq k\\) .) Before Alice's \\(i\\) th move, exactly \\(2i - 2\\) moves have been made, so since this subgraph has \\(2i\\) vertices, it is disconnected. \n\nTake a vertex \\(j\\) such that \\(j\\) and 1 are in different connected components. Since Alice's first move is \\((1,n)\\) , \\(j\\) is also not connected to \\(n\\) . Now, if \\(j\\in \\{1,\\ldots ,i\\}\\) , Alice adds \\((j,n)\\) , otherwise Alice adds \\((1,j)\\) . \\(\\square\\) \n\nSince each of Bob's edges has weight at least 1, his edges in total have weight at least \\(n - k - 1\\) . Thus, Alice can ensure a total weight of at least \\((n - k - 1) + (n - k) + \\dots +(n - 1)\\) . \n\nWe now give a strategy for Bob. \n\nClaim — Bob can ensure that for each \\(1\\leq i\\leq k\\) , there are at most \\(i\\) edges of weight at least \\(n - i\\) , while only adding edges of weight 1. \n\nProof. For \\(i = 1\\) , this is clear. Consider \\(i > 1\\) . As above, consider the induced subgraph on vertices \\(\\{1,\\ldots ,i\\} \\sqcup \\{n - i + 1,\\ldots ,n\\}\\) before Bob's \\((i - 1)\\) st move. \n\nWe show that there exists some edge of weight 1 in this subgraph that Bob can choose. Suppose not. Then, since none of \\((1,2),\\ldots ,(i - 1,i)\\) are valid moves, the vertices \\(1,\\ldots ,i\\) are connected. Likewise, the vertices \\((n - i + 1,\\ldots ,n)\\) are connected. \n\nHence, there are at least \\(2(i - 1)\\) edges in this subgraph. However, only \\(2i - 3\\) moves have been played up to this point, contradiction. So, Bob can choose an edge of weight 1 in this subgraph. \n\nWhen \\(n\\) is even, Bob makes \\(k - 1\\) moves, so this accounts for all his moves. When \\(n\\) is odd, on the final move, since not all the vertices are connected, there exists some edge of weight 1 that Bob can add.\n\n\n\nThus, Bob can select only edges of weight 1. Now, following the above strategy, for any \\(1 \\leq i \\leq k\\) , Bob chooses at least \\(k - 1\\) edges in the induced subgraph on vertices \\(\\{1, \\ldots , i\\} \\cup \\{n - i + 1, \\ldots , n\\}\\) . Since the induced subgraph contains no cycles, Alice plays at most \\((2i - 1) - (i - 1) = i\\) edges in this subgraph. \n\nNow, note that any edge of weight at least \\(n - i\\) must have both vertices contained in this subgraph. Thus, there are at most \\(i\\) edges of weight at least \\(n - i\\) , as desired. (When \\(n > 2\\) , \\(n - i \\geq n - k > 1\\) , and when \\(n = 2\\) , Bob plays no edges, so the bound still holds.) \\(\\square\\) \n\nUsing the above strategy, the total weight of Bob's edges is \\(n - k - 1\\) . Let \\(a_{i}\\) be the number of edges Alice plays of weight at least \\(i\\) . Then, the total weight of Alice's edges is \n\n\\[\\sum_{i = 1}^{n - 1}a_{i}\\leq k(n - k - 1) + \\sum_{i = n - k}^{n - 1}a_{i}\\] \\[\\qquad \\leq k(n - k - 1) + (k + (k - 1) + \\cdot \\cdot \\cdot +1)\\] \\[\\qquad = (n - 1) + (n - 2) + \\cdot \\cdot \\cdot +(n - k).\\] \n\nThus, Bob can ensure a total weight of at most \\((n - k - 1) + (n - k) + \\dots +(n - 1)\\) .", "metadata": {"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2025.jsonl", "problem_match": "Problem 6. ", "solution_match": "## \\(\\S 2.3\\) Solution to TSTST 6, by Max Lu, Kevin Wu \n"}}
{"year": "2025", "tier": "T0", "problem_label": "7", "problem_type": null, "exam": "IMO", "problem": "For a positive real number \\(c\\) , the sequence \\(a_{1}, a_{2}, \\ldots\\) of real numbers is defined as follows. Let \\(a_{1} = c\\) , and for \\(n \\geq 2\\) , let \n\n\\[a_{n} = \\sum_{i = 1}^{n - 1}(a_{i})^{n - i + 1}.\\] \n\nFind all positive real numbers \\(c\\) such that \\(a_{i} > a_{i + 1}\\) for all positive integers \\(i\\) .", "solution": "\\(\\P\\) Solution (author) The answer is \\(c< \\frac{\\sqrt{5} - 1}{2}\\) \n\nTo show this is necessary, note that \\(a_{2} = c^{2}\\) and \\(a_{3} = c^{3} + c^{4}\\) , so if the sequence is decreasing, we have \\(c^{2} > c^{3} + c^{4}\\) , implying \\(c< \\frac{\\sqrt{5} - 1}{2}\\) \n\nIn the other direction, suppose \\(c\\) is a positive real number with \\(c< \\frac{\\sqrt{5} - 1}{2}\\) . We will apply strong induction. We have \n\n\\[a_1 = c\\] \\[a_2 = c^2\\] \\[a_3 = c^3 +c^4\\] \n\nso \\(a_{1} > a_{2}\\) (since \\(c< 1\\) ) and \\(a_{2} > a_{3}\\) (since \\(c< \\frac{\\sqrt{5} - 1}{2}\\) ). For the inductive step, we'll show \\(a_{n} > a_{n + 1}\\) assuming that \\(n\\geq 3\\) and \\(a_{i} > a_{i + 1}\\) for all \\(i< n\\) . We have \n\n\\[a_{1}^{n} = c^{n} > c^{n}(c + c^{2}) > c^{n}(c + c^{n}) = a_{1}^{n + 1} + a_{2}^{n}.\\] \n\nAlso, since \\(a_{i} > a_{i + 1}\\) for \\(i = 2,\\ldots ,n - 1\\) , we have \n\n\\[\\sum_{i = 2}^{n - 1}a_{i}^{n - i + 1} > \\sum_{i = 2}^{n - 1}a_{i + 1}^{n - i + 1} = \\sum_{j = 3}^{n}a_{j}^{n - j + 2},\\] \n\nwhere the equality comes from shifting indices to take \\(j = i + 1\\) . Thus \n\n\\[a_{n} = a_{1}^{n} + \\sum_{i = 2}^{n - 1}a_{i}^{n - i + 1} > a_{1}^{n + 1} + a_{2}^{n} + \\sum_{j = 3}^{n}a_{j}^{n - j + 2} = a_{n + 1},\\] \n\nas desired. This completes the induction step, finishing the solution. \n\nRemark (author). The behavior of the sequence for other values of \\(c\\) is interesting. In particular, computer experiments seem to indicate that, for \\(c = 0.655736876792\\) , the sequence starting from \\(a_{2}\\) increases to a value barely less than \\(\\frac{1}{2}\\) and then decreases, going to 0 as \\(n\\to \\infty\\) , while for \\(c = 0.655736876793\\) , the sequence starting from \\(a_{2}\\) is increasing and goes to infinity. We conjecture that some form of these patterns hold in general, and that there should exist some value of \\(c\\) with \\(0.655736876792< c< 0.655736876793\\) so that the sequence starting from \\(a_{2}\\) is increasing and \\(\\lim_{n\\to \\infty}a_{n} = \\frac{1}{2}\\) .\n\n\n\n\\(\\parallel\\) Solution (Pitchayut) We get \\(c< \\frac{\\sqrt{5} - 1}{2}\\) in the same way as the first solution. To show that any such \\(c\\) gives a strictly decreasing sequence, we use induction on \\(n\\) . The base cases of \\(n = 1,2\\) are clear. Let \\(t = \\frac{\\sqrt{5} - 1}{2}\\) , then \\(t^{2} + t = 1\\) . For the inductive step, assume \\(n\\geq 3\\) and \\(a_{1} > a_{2} > \\dots >a_{n - 1}\\) . Since \\(t > a_{1}\\) , \\(t\\) is greater than \\(a_{1},a_{2},\\ldots ,a_{n - 1}\\) . Furthermore, \\(t^{2} > a_{1}^{2} = a_{2}\\geq a_{n - 1}\\) . Thus \n\n\\[a_{n} = a_{1}^{n} + a_{2}^{n - 1} + \\dots +a_{n - 1}^{2}\\] \\[\\qquad < t(a_{1}^{n - 1} + a_{2}^{n - 2} + \\dots +a_{n - 2}^{2}) + t^{2}a_{n - 1}\\] \\[\\qquad = t a_{n - 1} + t^{2}a_{n - 1}\\] \\[\\qquad = a_{n - 1},\\] \n\nas desired.", "metadata": {"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2025.jsonl", "problem_match": "Problem 7. ", "solution_match": "## \\(\\S 3.1\\) Solution to TSTST 7, by Luke Robitaille \n"}}
{"year": "2025", "tier": "T0", "problem_label": "8", "problem_type": null, "exam": "IMO", "problem": "Find all polynomials \\(f\\) with integer coefficients such that for all positive integers \\(n\\) , \n\n\\[n \\text{ divides } \\frac{f(f(\\ldots(f(0)) \\ldots) - 1}{n + 1 f^{\\mathrm{s}}}\\]", "solution": "There are three families. \n\n- \\(f(x) = x + 1\\) . \n\n- \\(f(x) = x(x - 1)g(x) + 1\\) for any polynomial \\(g(x)\\) (i.e., any \\(f(x)\\) such that \\(f(0) = f(1) = 1\\) ). \n\n- \\(f(x) = x(x - 1)(x + 1)g(x) + (2x^2 - 1)\\) for any polynomial \\(g(x)\\) (i.e., any \\(f(x)\\) such that \\(f(0) = -1\\) , \\(f(-1) = f(1) = 1\\) ). \n\nThese all clearly work, so we focus on proving that these are all solutions. \n\nClaim — For any prime \\(p\\) , either \n\n(a) \\(f(1) \\equiv 1\\) (mod \\(p\\) ) or \n\n(b) the directed graph of \\(f\\) in \\(\\mathbb{F}_p\\) forms a single cycle of size \\(p\\) . \n\nProof. Work modulo \\(p\\) . Consider the sequence \n\n\\[0,f(0),f(f(0)),\\ldots ,\\] \n\nwhich must be eventually periodic. Clearly 1 must be in the periodic part by taking \\(n\\) to be a large multiple of \\(p\\) . Now, note that since the non- periodic part must have size less than \\(p\\) , we have \n\n\\[\\left. \\begin{array}{l}{f^{p + 1}(0) = 1}\\\\ {f^{2p + 1}(0) = 1} \\end{array} \\right\\} \\Longrightarrow f^{p}(1) = 1,\\] \n\nso the period must divide \\(p\\) and hence must be either 1 or \\(p\\) . If it is 1, then \\(f(1) = 1\\) . Otherwise, (b) holds. \\(\\square\\) \n\nNext, we note that if \\(f(x) - x\\) is non- constant, then by Schur's theorem on \\(f(x) - x\\) , \\(f\\) has a fixed point modulo infinitely many primes \\(p\\) , so (b) fails for infinitely many primes \\(p\\) . This means that (a) holds for infinitely many primes \\(p\\) , so \\(f(1) = 1\\) . Therefore, either \\(f(x) - x\\) is constant or \\(f(1) = 1\\) . \n\nIn the case that \\(f(x) - x\\) a constant \\(c\\) , \\(f^{n + 1}(0) - 1 = (n + 1)c - 1\\) , so \\(n \\mid c - 1\\) for all \\(n\\) and \\(f(x) = x + 1\\) . Henceforth, assume \\(f(1) = 1\\) . \n\nSince \\(f(0) \\mid f^k (0)\\) for all \\(k\\) , plugging in \\(n = |f(0)|\\) gives \n\n\\[f(0) \\mid f^{|f(0)| + 1}(0) - 1 \\Longrightarrow f(0) \\mid 1 \\Longrightarrow f(0) = \\pm 1.\\] \n\nIf \\(f(0) = 1\\) , then we are done. Otherwise, assume \\(f(0) = - 1\\) . Then, \\(f(- 1)\\) is odd because \\(f(1) = 1\\) , so plugging in \\(n = |f(- 1)| = 2k - 1\\) gives \n\n\\[f(-1) \\mid f^{2k}(0) - 1.\\] \n\nWe have that \\(f^2 (0) = 0\\) modulo \\(f(- 1) = f^2 (0)\\) , so \\(f(- 1) \\mid 1\\) . This gives \\(f(- 1) = \\pm 1\\) . If \\(f(- 1) = - 1\\) , then we have \\(f^n (0) = - 1\\) for all \\(n \\geq 3\\) , which makes the divisibility condition fail. Hence, \\(f(- 1) = 1\\) , and we get the third solution set.\n\n\n\nRemark (author). An easier variant is to ask for all polynomials \\(f\\) with integer coefficients such that \\(n \\mid f^n (0)\\) for all positive integers \\(n\\) .", "metadata": {"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2025.jsonl", "problem_match": "Problem 8. ", "solution_match": "## \\(\\S 3.2\\) Solution to TSTST 8, by Pitchayut Saengrungkonga \n"}}
-{"year": "2025", "tier": "T0", "problem_label": "9", "problem_type": null, "exam": "IMO", "problem": "Let acute triangle \\(ABC\\) have orthocenter \\(H\\) . Let \\(B_{1}\\) , \\(C_{1}\\) , \\(B_{2}\\) , and \\(C_{2}\\) be collinear points which lie on lines \\(AB\\) , \\(AC\\) , \\(BH\\) , and \\(CH\\) , respectively. Let \\(\\omega_{B}\\) and \\(\\omega_{C}\\) be the circumcircles of triangles \\(BB_{1}B_{2}\\) and \\(CC_{1}C_{2}\\) , respectively. Prove that the radical axis of \\(\\omega_{B}\\) and \\(\\omega_{C}\\) intersects the line through their centers on the nine- point circle of triangle \\(ABC\\) .", "solution": "\\(\\P\\) Solution (author) The first important step is to introduce \\(N\\) , the circumcenter of \\(H B_{2}C_{2}\\) . \n\nClaim — Lines \\(N B_{2}\\) and \\(N C_{2}\\) are tangent to \\((B B_{1}B_{2})\\) and \\((C C_{1}C_{2})\\) , respectively. \n\nProof. This follows from chasing \n\n\\[\\angle N B_{2}B = \\angle N B_{2}H = 90^{\\circ} - \\angle H C_{2}B_{2} = \\angle B_{2}B_{1}B.\\] \n\nAn analogous proof works for the other side. \n\nSince \\(N B_{2} = N C_{2}\\) , it follows that \\(N\\) lies on the radical axis of \\((B B_{1}B_{2})\\) and \\((C C_{1}C_{2})\\) . Thus we want to show \\(X\\) , the foot of \\(N\\) onto \\(O_{B}O_{C}\\) , lies on the nine- point circle. \n\n\n \n\nLet \\(D\\) , \\(E\\) , \\(F\\) be the feet of the altitudes in \\(\\triangle A B C\\) , and let \\(M_{B}\\) , \\(M_{C}\\) be the midpoints of \\(B H\\) , \\(C H\\) . Let \\(B_{3}\\) , \\(C_{3}\\) be the second intersections of \\((B_{2}C_{2}H)\\) with \\((B B_{1}B_{2})\\) , \\((C C_{1}C_{2})\\) . \n\nClaim — Hexagons \\(O_{B}B_{2}B_{3}M_{B}X N\\) and \\(O_{C}C_{2}C_{3}M_{C}X N\\) are cyclic.\n\n\n\nProof. Clearly \\(B_{2}\\) , \\(B_{3}\\) and \\(X\\) lie on the circle of diameter \\(O_{B}N\\) . To show \\(M_{B}\\) lies in this circle, we first note that \\(BB_{3}HFD\\) is cyclic. Indeed, this follows from chasing \n\n\\[\\angle BB_{3}H = \\angle B_{2}B_{3}H - \\angle B_{2}B_{3}B = \\angle B_{2}C_{2}H - \\angle B_{2}B_{1}B = \\angle B_{1}FC_{2} = \\angle BFF.\\] \n\nThis circle has center \\(M_{B}\\) , so \\(O_{B}M_{B}\\) and \\(M_{B}N\\) are the perpendicular bisectors of \\(BB_{3}\\) , \\(B_{3}H\\) . Since \\(BB_{3} \\perp B_{3}H\\) , \\(O_{B}M_{B} \\perp M_{B}N\\) , as we wanted to show. \\(\\square\\) \n\nNow we are ready to finish. We know \\(\\angle M_{B}DM_{C} = \\angle M_{C}HM_{B} = \\angle BAC\\) . We will now show that \\(\\angle M_{B}XM_{C}\\) gives the same value. On one hand \n\n\\[\\angle M_{B}XO_{B} = \\angle M_{B}B_{2}O_{B} = \\angle BB_{2}O_{B} = 90^{\\circ} - \\angle B_{2}B_{1}B = 90^{\\circ} - \\angle C_{1}B_{1}A.\\] \n\nSimilarly \\(\\angle O_{C}XM_{C} = 90^{\\circ} - \\angle AC_{1}B_{1}\\) , so \n\n\\[\\angle M_{B}XM_{C} = (90^{\\circ} - \\angle C_{1}B_{1}A) + (90^{\\circ} - \\angle AC_{1}B_{1}) = \\angle BAC,\\] \n\nwhich establishes the result. \n\nRemark (author). Since the problem is symmetric under \\(A \\leftrightarrow H\\) , a similar solution can be found by considering the circumcenter of \\(\\triangle AB_{1}C_{1}\\) (instead of \\(N\\) ). Moreover, introducing both circumcenters adds more structure to the diagram, which can make it easier to finish. \n\nThe most important step in this solution is to introduce the point \\(N\\) . This can be motivated by noticing that the circles \\((BB_{1}B_{2})\\) and \\((HB_{2}C_{2})\\) are orthogonal. \n\nMoving Points Solution (Krishna) Let \\(E, F\\) be the feet of the \\(B, C\\) altitudes, as in the first solution. \n\nClaim — As \\(\\ell\\) varies while staying parallel to a fixed line, \\(\\overline{O_{B}O_{C}}\\) passes through a fixed point \\(Y\\) on the nine- point circle. \n\nProof. Move \\(\\ell\\) linearly. Then \\(\\triangle BB_{1}B_{2}\\) is dilating at \\(B\\) linearly, so \\(O_{B}\\) moves linearly. Similarly \\(O_{C}\\) moves linearly. Furthermore, we have that \n\n\\[\\angle \\overline{B O_{B}} = 90 + \\angle B A + \\angle B H - \\angle \\ell = \\angle B A + \\angle B C - \\angle \\ell ,\\] \n\nwhich by symmetry means \\(\\overline{B O_{B}} \\parallel \\overline{C O_{C}}\\) . Thus we need to find a point \\(Y\\) on the nine- point circle such that \\(O_{B} - O_{C} - Y\\) in 2 cases.\n\n\n\n \n\nSuppose \\(\\ell\\) passes through \\(H\\) . In that case, let \\(Q = (BB_{1}H)\\cap (CC_{1}H)\\) . We get that \\(Q\\) lies on \\((ABC)\\) by chasing \n\n\\[\\angle BQC = \\angle BQH + \\angle HQC = \\angle BB_{1}H + \\angle HC_{1}C = \\angle ABB_{1} + \\angle BC_{1}A = \\angle BAC.\\] \n\nThe perpendicular bisector of \\(\\overline{HQ}\\) is \\(\\overline{O_{B}O_{C}}\\) , and the midpoint of \\(HQ\\) lies on the nine- point circle. Let the perpendicular from \\(Q\\) to \\(\\overline{HQ}\\) intersect \\((ABC)\\) at \\(Y_{1}\\) . Then \n\n\\[\\angle BAY_{1} = \\angle BQY_{1} = 90 + \\angle BQH = 90 + \\angle AB_{1}H,\\] \n\nso \\(\\overline{AY_{1}}\\perp \\ell\\) . If \\(Y\\) is the midpoint of \\(HY_{1}\\) , then \\(Y\\) lies on \\(\\overline{O_{B}O_{C}}\\) and the nine- point circle of \\((ABC)\\) . The line between \\(Y\\) and the midpoint of \\(AH\\) is perpendicular to \\(\\ell\\) . Hence when \\(\\ell\\) passes through \\(A\\) , by symmetry we also get \\(O_{B} - O_{C} - Y\\) , as desired (this angle chase is basically done as part of 2019 TSTST 5). \\(\\square\\) \n\nLet \\(P_{\\infty}\\) be the point at infinity of \\(\\ell\\) , and let \\(P = r\\cap \\ell\\) . Since \\(PB_{1}\\cdot PB_{2} = PC_{1}\\cdot PC_{2}\\) (with signed lengths), we have that \\((P,P_{\\infty}),(B_{1},B_{2}),(C_{1},C_{2})\\) are 3 pairs under the same involution on \\(\\ell\\) . Then by DIT on the 4 points \\(A,H,E,F\\) and the line \\(\\ell\\) , if \\(\\mathcal{C}\\) is the conic through \\(A,H,E,F,P_{\\infty}\\) , then \\(P\\) also lies on \\(\\mathcal{C}\\) . Now we do moving points.\n\n\n\n \n\n- Let \\(P_{\\infty}\\) be a sufficiently general fixed point at infinity (proving the problem for sufficiently general \\(P_{\\infty}\\) suffices by continuity. \n\n- Define \\(P\\) as a moving point with degree 2 varying on the conic through \\(A, H, E, F, P_{\\infty}\\) . \n\n- As we let \\(P\\) vary and we let \\(\\ell = \\overline{PP_{\\infty}}\\) , we get that \\(\\overline{PP_{\\infty}}\\) has degree 1 so \\(B_{1}\\) has degree 1. The perpendicular bisector of \\(BB_{1}\\) then has degree 1, and \\(\\overline{BO_{B}}\\) is fixed so \\(O_{B}\\) has degree 1. \n\n- \\(Y\\) has degree 0, and the antipode \\(Y'\\) of \\(Y\\) on the nine-point circle has degree 0. It suffices to show that \\(\\overline{Y'P} \\perp \\overline{OBY}\\) . \n\n- \\(\\overline{Y'P}\\) has degree \\(\\leq 2\\) and \\(\\overline{OBY}\\) has degree 1, so we get a degree 3 statement that needs to be checked at 4 cases. \n\nWe can do \\(P = A, H, E, F\\) . \\(P = H\\) and \\(P = A\\) are true by previous work. By symmetry, to solve \\(P = E\\) and \\(P = F\\) we only need to look at \\(P = E\\) . In this case, we want to show that \\(E - OB = OC\\) . \n\n\\[\\angle EOB = \\angle BA + \\angle \\ell -\\angle BH = 90 + \\angle CH + \\angle \\ell -90 - \\angle AC = 90 - 90 + \\angle EOC = \\angle EOC,\\] \n\nas desired. \n\n\\(\\P\\) Solution (Vivian Loh) Let \\(S_{B} = (BH) \\cap (BB_{1}B_{2})\\) , and \\(S_{C} = (CH) \\cap (CC_{1}C_{2})\\) . Then \\(S_{B}\\) and \\(S_{C}\\) are the Miquel points of complete quadrilaterals \\(\\{B, B_{1}, B_{2}, H, C_{2}, CH \\cap AB\\}\\) and \\(\\{C, C_{1}, C_{2}, H, B_{2}, BH \\cap AC\\}\\) respectively, so \\(S_{B}\\) and \\(S_{C}\\) both lie on \\((HB_{2}C_{2})\\) . Furthermore, by radical center, \\(P = B_{2}S_{B} \\cap C_{2}S_{C}\\) lies on the radical axis of \\(\\omega_{B}\\) , \\(\\omega_{C}\\) . Also note that if \\(H'\\) is the antipode of \\(H\\) with respect to \\((HB_{2}C_{2})\\) , then \\(S_{B}\\) and \\(S_{C}\\) are the projections of \\(H\\) onto \\(BH'\\) and \\(CH'\\) , respectively. Note that the 6 points \\(H, B_{2}, C_{2}, H', S_{B}, S_{C}\\) all lie on \\((HH')\\) .\n\n\n\nClaim 3.1 — If \\(O_{B}\\) and \\(O_{C}\\) are the centers of \\(\\omega_{B}\\) and \\(\\omega_{C}\\) respectively, then \\(O_{B}\\) is the intersection of tangents to \\((HH^{\\prime})\\) at \\(B_{2}\\) and \\(S_{B}\\) , while \\(O_{C}\\) is the intersection of tangents to \\((HH^{\\prime})\\) at \\(C_{2}\\) and \\(S_{C}\\) . \n\nProof. Angle chasing; \\(\\angle B_{2}O_{B}S_{B} = 2\\angle B_{2}BS_{B} = 2(90^{\\circ} - \\angle BHS_{B}) = 180^{\\circ} - 2\\angle BHS_{B}\\) . It works for \\(O_{C}\\) similarly. \\(\\square\\) \n\nClaim 3.2 — \\(O_{B}O_{C}\\) is the polar of \\(P\\) with respect to \\((HH^{\\prime})\\) . \n\nProof. By the definition of poles and polars, \\(B_{2}S_{B}\\) is the polar of \\(O_{B}\\) and \\(C_{2}S_{C}\\) is the polar of \\(O_{C}\\) , and by La Hire's, we know that a point lies on the polar of \\(O_{B}\\) and the polar of \\(O_{C}\\) if and only if \\(O_{B}\\) and \\(O_{C}\\) both lie on its polar, so \\(O_{B}O_{C}\\) is the polar of \\(B_{2}S_{B} \\cap C_{2}S_{C} = P\\) . \\(\\square\\) \n\nSince \\(P\\) lies on the radical axis of \\(\\omega_{B}\\) and \\(\\omega_{C}\\) , the desired point in the problem (the intersection of the radical axis and \\(O_{B}O_{C}\\) ) is the foot from \\(P\\) to \\(O_{B}O_{C}\\) , which is the inverse \\(P^{*}\\) of \\(P\\) about \\((HH^{\\prime})\\) . It suffices to show that this point lies on the nine- point circle of \\(\\triangle ABC\\) . \n\nLet \\(M_{B}\\) and \\(M_{C}\\) be the midpoints of \\(HB\\) and \\(HC\\) . Then the nine- point circle of \\(\\triangle ABC\\) passes through \\(M_{B}\\) , \\(M_{C}\\) , and the foot from \\(H\\) to \\(BC\\) , so \\(P^{*}\\) lies on this nine- point circle if and only if \\(\\angle M_{B}P^{*}M_{C} = \\angle BHC\\) . \n\nClaim 3.3 — \\(M_{B}^{*} \\in B_{2}S_{B}\\) , \\(M_{C}^{*} \\in C_{2}S_{C}\\) . \n\nProof. This is simply angle chasing. We will show that \\(M_{C}^{*} \\in C_{2}S_{C}\\) . Let \\(O\\) be the center of \\((HH^{\\prime})\\) . Then \\(\\angle O M_{C}H = \\angle O H M_{C}^{*}\\) , which equals \\(\\angle O S_{C}M_{C}^{*}\\) since \\(O M_{C} \\perp H S_{C}\\) . However, \\(\\angle O S_{C}C_{2} = \\angle H S_{C}C_{2} - \\angle H S_{C}O = \\angle (H C_{2}, S_{C}H^{\\prime}) = \\angle O M_{C}H\\) , so \\(M_{C}^{*} \\in C_{2}S_{C}\\) , and \\(M_{B}^{*}\\) follows similarly. \\(\\square\\) \n\nThe last part is just angle chasing. We know from before that it is sufficient to show \\(\\angle M_{B}P^{*}M_{C} = \\angle BHC\\) . We have: \\(\\angle M_{B}P^{*}M_{C} = \\angle OP^{*}M_{B} + \\angle OP^{*}M_{C} = \\angle OM_{B}^{*}P + \\angle OM_{C}^{*}P = 360^{\\circ} - \\angle M_{B}^{*}OM_{C}^{*} - \\angle M_{B}^{*}PM_{C}^{*} = 360^{\\circ} - \\angle S_{C}PS_{B} - \\angle S_{C}H^{\\prime}S_{B} = \\angle H^{\\prime}S_{B}P + \\angle H^{\\prime}S_{C}P = \\angle B_{2}HC_{2} = \\angle BHC\\) , as desired.", "metadata": {"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2025.jsonl", "problem_match": "Problem 9. ", "solution_match": "## \\(\\S 3.3\\) Solution to TSTST 9, by Ruben Carpenter \n"}}
+{"year": "2025", "tier": "T0", "problem_label": "9", "problem_type": null, "exam": "IMO", "problem": "Let acute triangle \\(ABC\\) have orthocenter \\(H\\) . Let \\(B_{1}\\) , \\(C_{1}\\) , \\(B_{2}\\) , and \\(C_{2}\\) be collinear points which lie on lines \\(AB\\) , \\(AC\\) , \\(BH\\) , and \\(CH\\) , respectively. Let \\(\\omega_{B}\\) and \\(\\omega_{C}\\) be the circumcircles of triangles \\(BB_{1}B_{2}\\) and \\(CC_{1}C_{2}\\) , respectively. Prove that the radical axis of \\(\\omega_{B}\\) and \\(\\omega_{C}\\) intersects the line through their centers on the nine- point circle of triangle \\(ABC\\) .", "solution": "\\(\\P\\) Solution (author) The first important step is to introduce \\(N\\) , the circumcenter of \\(H B_{2}C_{2}\\) . \n\nClaim — Lines \\(N B_{2}\\) and \\(N C_{2}\\) are tangent to \\((B B_{1}B_{2})\\) and \\((C C_{1}C_{2})\\) , respectively. \n\nProof. This follows from chasing \n\n\\[\\angle N B_{2}B = \\angle N B_{2}H = 90^{\\circ} - \\angle H C_{2}B_{2} = \\angle B_{2}B_{1}B.\\] \n\nAn analogous proof works for the other side. \n\nSince \\(N B_{2} = N C_{2}\\) , it follows that \\(N\\) lies on the radical axis of \\((B B_{1}B_{2})\\) and \\((C C_{1}C_{2})\\) . Thus we want to show \\(X\\) , the foot of \\(N\\) onto \\(O_{B}O_{C}\\) , lies on the nine- point circle. \n\n\n \n\nLet \\(D\\) , \\(E\\) , \\(F\\) be the feet of the altitudes in \\(\\triangle A B C\\) , and let \\(M_{B}\\) , \\(M_{C}\\) be the midpoints of \\(B H\\) , \\(C H\\) . Let \\(B_{3}\\) , \\(C_{3}\\) be the second intersections of \\((B_{2}C_{2}H)\\) with \\((B B_{1}B_{2})\\) , \\((C C_{1}C_{2})\\) . \n\nClaim — Hexagons \\(O_{B}B_{2}B_{3}M_{B}X N\\) and \\(O_{C}C_{2}C_{3}M_{C}X N\\) are cyclic.\n\n\n\nProof. Clearly \\(B_{2}\\) , \\(B_{3}\\) and \\(X\\) lie on the circle of diameter \\(O_{B}N\\) . To show \\(M_{B}\\) lies in this circle, we first note that \\(BB_{3}HFD\\) is cyclic. Indeed, this follows from chasing \n\n\\[\\angle BB_{3}H = \\angle B_{2}B_{3}H - \\angle B_{2}B_{3}B = \\angle B_{2}C_{2}H - \\angle B_{2}B_{1}B = \\angle B_{1}FC_{2} = \\angle BFF.\\] \n\nThis circle has center \\(M_{B}\\) , so \\(O_{B}M_{B}\\) and \\(M_{B}N\\) are the perpendicular bisectors of \\(BB_{3}\\) , \\(B_{3}H\\) . Since \\(BB_{3} \\perp B_{3}H\\) , \\(O_{B}M_{B} \\perp M_{B}N\\) , as we wanted to show. \\(\\square\\) \n\nNow we are ready to finish. We know \\(\\angle M_{B}DM_{C} = \\angle M_{C}HM_{B} = \\angle BAC\\) . We will now show that \\(\\angle M_{B}XM_{C}\\) gives the same value. On one hand \n\n\\[\\angle M_{B}XO_{B} = \\angle M_{B}B_{2}O_{B} = \\angle BB_{2}O_{B} = 90^{\\circ} - \\angle B_{2}B_{1}B = 90^{\\circ} - \\angle C_{1}B_{1}A.\\] \n\nSimilarly \\(\\angle O_{C}XM_{C} = 90^{\\circ} - \\angle AC_{1}B_{1}\\) , so \n\n\\[\\angle M_{B}XM_{C} = (90^{\\circ} - \\angle C_{1}B_{1}A) + (90^{\\circ} - \\angle AC_{1}B_{1}) = \\angle BAC,\\] \n\nwhich establishes the result. \n\nRemark (author). Since the problem is symmetric under \\(A \\leftrightarrow H\\) , a similar solution can be found by considering the circumcenter of \\(\\triangle AB_{1}C_{1}\\) (instead of \\(N\\) ). Moreover, introducing both circumcenters adds more structure to the diagram, which can make it easier to finish. \n\nThe most important step in this solution is to introduce the point \\(N\\) . This can be motivated by noticing that the circles \\((BB_{1}B_{2})\\) and \\((HB_{2}C_{2})\\) are orthogonal. \n\nMoving Points Solution (Krishna) Let \\(E, F\\) be the feet of the \\(B, C\\) altitudes, as in the first solution. \n\nClaim — As \\(\\ell\\) varies while staying parallel to a fixed line, \\(\\overline{O_{B}O_{C}}\\) passes through a fixed point \\(Y\\) on the nine- point circle. \n\nProof. Move \\(\\ell\\) linearly. Then \\(\\triangle BB_{1}B_{2}\\) is dilating at \\(B\\) linearly, so \\(O_{B}\\) moves linearly. Similarly \\(O_{C}\\) moves linearly. Furthermore, we have that \n\n\\[\\angle \\overline{B O_{B}} = 90 + \\angle B A + \\angle B H - \\angle \\ell = \\angle B A + \\angle B C - \\angle \\ell ,\\] \n\nwhich by symmetry means \\(\\overline{B O_{B}} \\parallel \\overline{C O_{C}}\\) . Thus we need to find a point \\(Y\\) on the nine- point circle such that \\(O_{B} - O_{C} - Y\\) in 2 cases.\n\n\n\n \n\nSuppose \\(\\ell\\) passes through \\(H\\) . In that case, let \\(Q = (BB_{1}H)\\cap (CC_{1}H)\\) . We get that \\(Q\\) lies on \\((ABC)\\) by chasing \n\n\\[\\angle BQC = \\angle BQH + \\angle HQC = \\angle BB_{1}H + \\angle HC_{1}C = \\angle ABB_{1} + \\angle BC_{1}A = \\angle BAC.\\] \n\nThe perpendicular bisector of \\(\\overline{HQ}\\) is \\(\\overline{O_{B}O_{C}}\\) , and the midpoint of \\(HQ\\) lies on the nine- point circle. Let the perpendicular from \\(Q\\) to \\(\\overline{HQ}\\) intersect \\((ABC)\\) at \\(Y_{1}\\) . Then \n\n\\[\\angle BAY_{1} = \\angle BQY_{1} = 90 + \\angle BQH = 90 + \\angle AB_{1}H,\\] \n\nso \\(\\overline{AY_{1}}\\perp \\ell\\) . If \\(Y\\) is the midpoint of \\(HY_{1}\\) , then \\(Y\\) lies on \\(\\overline{O_{B}O_{C}}\\) and the nine- point circle of \\((ABC)\\) . The line between \\(Y\\) and the midpoint of \\(AH\\) is perpendicular to \\(\\ell\\) . Hence when \\(\\ell\\) passes through \\(A\\) , by symmetry we also get \\(O_{B} - O_{C} - Y\\) , as desired (this angle chase is basically done as part of 2019 TSTST 5). \\(\\square\\) \n\nLet \\(P_{\\infty}\\) be the point at infinity of \\(\\ell\\) , and let \\(P = r\\cap \\ell\\) . Since \\(PB_{1}\\cdot PB_{2} = PC_{1}\\cdot PC_{2}\\) (with signed lengths), we have that \\((P,P_{\\infty}),(B_{1},B_{2}),(C_{1},C_{2})\\) are 3 pairs under the same involution on \\(\\ell\\) . Then by DIT on the 4 points \\(A,H,E,F\\) and the line \\(\\ell\\) , if \\(\\mathcal{C}\\) is the conic through \\(A,H,E,F,P_{\\infty}\\) , then \\(P\\) also lies on \\(\\mathcal{C}\\) . Now we do moving points.\n\n\n\n \n\n- Let \\(P_{\\infty}\\) be a sufficiently general fixed point at infinity (proving the problem for sufficiently general \\(P_{\\infty}\\) suffices by continuity. \n\n- Define \\(P\\) as a moving point with degree 2 varying on the conic through \\(A, H, E, F, P_{\\infty}\\) . \n\n- As we let \\(P\\) vary and we let \\(\\ell = \\overline{PP_{\\infty}}\\) , we get that \\(\\overline{PP_{\\infty}}\\) has degree 1 so \\(B_{1}\\) has degree 1. The perpendicular bisector of \\(BB_{1}\\) then has degree 1, and \\(\\overline{BO_{B}}\\) is fixed so \\(O_{B}\\) has degree 1. \n\n- \\(Y\\) has degree 0, and the antipode \\(Y'\\) of \\(Y\\) on the nine-point circle has degree 0. It suffices to show that \\(\\overline{Y'P} \\perp \\overline{OBY}\\) . \n\n- \\(\\overline{Y'P}\\) has degree \\(\\leq 2\\) and \\(\\overline{OBY}\\) has degree 1, so we get a degree 3 statement that needs to be checked at 4 cases. \n\nWe can do \\(P = A, H, E, F\\) . \\(P = H\\) and \\(P = A\\) are true by previous work. By symmetry, to solve \\(P = E\\) and \\(P = F\\) we only need to look at \\(P = E\\) . In this case, we want to show that \\(E - OB = OC\\) . \n\n\\[\\angle EOB = \\angle BA + \\angle \\ell -\\angle BH = 90 + \\angle CH + \\angle \\ell -90 - \\angle AC = 90 - 90 + \\angle EOC = \\angle EOC,\\] \n\nas desired. \n\n\\(\\P\\) Solution (Vivian Loh) Let \\(S_{B} = (BH) \\cap (BB_{1}B_{2})\\) , and \\(S_{C} = (CH) \\cap (CC_{1}C_{2})\\) . Then \\(S_{B}\\) and \\(S_{C}\\) are the Miquel points of complete quadrilaterals \\(\\{B, B_{1}, B_{2}, H, C_{2}, CH \\cap AB\\}\\) and \\(\\{C, C_{1}, C_{2}, H, B_{2}, BH \\cap AC\\}\\) respectively, so \\(S_{B}\\) and \\(S_{C}\\) both lie on \\((HB_{2}C_{2})\\) . Furthermore, by radical center, \\(P = B_{2}S_{B} \\cap C_{2}S_{C}\\) lies on the radical axis of \\(\\omega_{B}\\) , \\(\\omega_{C}\\) . Also note that if \\(H'\\) is the antipode of \\(H\\) with respect to \\((HB_{2}C_{2})\\) , then \\(S_{B}\\) and \\(S_{C}\\) are the projections of \\(H\\) onto \\(BH'\\) and \\(CH'\\) , respectively. Note that the 6 points \\(H, B_{2}, C_{2}, H', S_{B}, S_{C}\\) all lie on \\((HH')\\) .\n\n\n\nClaim 3.1 — If \\(O_{B}\\) and \\(O_{C}\\) are the centers of \\(\\omega_{B}\\) and \\(\\omega_{C}\\) respectively, then \\(O_{B}\\) is the intersection of tangents to \\((HH^{\\prime})\\) at \\(B_{2}\\) and \\(S_{B}\\) , while \\(O_{C}\\) is the intersection of tangents to \\((HH^{\\prime})\\) at \\(C_{2}\\) and \\(S_{C}\\) . \n\nProof. Angle chasing; \\(\\angle B_{2}O_{B}S_{B} = 2\\angle B_{2}BS_{B} = 2(90^{\\circ} - \\angle BHS_{B}) = 180^{\\circ} - 2\\angle BHS_{B}\\) . It works for \\(O_{C}\\) similarly. \\(\\square\\) \n\nClaim 3.2 — \\(O_{B}O_{C}\\) is the polar of \\(P\\) with respect to \\((HH^{\\prime})\\) . \n\nProof. By the definition of poles and polars, \\(B_{2}S_{B}\\) is the polar of \\(O_{B}\\) and \\(C_{2}S_{C}\\) is the polar of \\(O_{C}\\) , and by La Hire's, we know that a point lies on the polar of \\(O_{B}\\) and the polar of \\(O_{C}\\) if and only if \\(O_{B}\\) and \\(O_{C}\\) both lie on its polar, so \\(O_{B}O_{C}\\) is the polar of \\(B_{2}S_{B} \\cap C_{2}S_{C} = P\\) . \\(\\square\\) \n\nSince \\(P\\) lies on the radical axis of \\(\\omega_{B}\\) and \\(\\omega_{C}\\) , the desired point in the problem (the intersection of the radical axis and \\(O_{B}O_{C}\\) ) is the foot from \\(P\\) to \\(O_{B}O_{C}\\) , which is the inverse \\(P^{*}\\) of \\(P\\) about \\((HH^{\\prime})\\) . It suffices to show that this point lies on the nine- point circle of \\(\\triangle ABC\\) . \n\nLet \\(M_{B}\\) and \\(M_{C}\\) be the midpoints of \\(HB\\) and \\(HC\\) . Then the nine- point circle of \\(\\triangle ABC\\) passes through \\(M_{B}\\) , \\(M_{C}\\) , and the foot from \\(H\\) to \\(BC\\) , so \\(P^{*}\\) lies on this nine- point circle if and only if \\(\\angle M_{B}P^{*}M_{C} = \\angle BHC\\) . \n\nClaim 3.3 — \\(M_{B}^{*} \\in B_{2}S_{B}\\) , \\(M_{C}^{*} \\in C_{2}S_{C}\\) . \n\nProof. This is simply angle chasing. We will show that \\(M_{C}^{*} \\in C_{2}S_{C}\\) . Let \\(O\\) be the center of \\((HH^{\\prime})\\) . Then \\(\\angle O M_{C}H = \\angle O H M_{C}^{*}\\) , which equals \\(\\angle O S_{C}M_{C}^{*}\\) since \\(O M_{C} \\perp H S_{C}\\) . However, \\(\\angle O S_{C}C_{2} = \\angle H S_{C}C_{2} - \\angle H S_{C}O = \\angle (H C_{2}, S_{C}H^{\\prime}) = \\angle O M_{C}H\\) , so \\(M_{C}^{*} \\in C_{2}S_{C}\\) , and \\(M_{B}^{*}\\) follows similarly. \\(\\square\\) \n\nThe last part is just angle chasing. We know from before that it is sufficient to show \\(\\angle M_{B}P^{*}M_{C} = \\angle BHC\\) . We have: \\(\\angle M_{B}P^{*}M_{C} = \\angle OP^{*}M_{B} + \\angle OP^{*}M_{C} = \\angle OM_{B}^{*}P + \\angle OM_{C}^{*}P = 360^{\\circ} - \\angle M_{B}^{*}OM_{C}^{*} - \\angle M_{B}^{*}PM_{C}^{*} = 360^{\\circ} - \\angle S_{C}PS_{B} - \\angle S_{C}H^{\\prime}S_{B} = \\angle H^{\\prime}S_{B}P + \\angle H^{\\prime}S_{C}P = \\angle B_{2}HC_{2} = \\angle BHC\\) , as desired.", "metadata": {"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2025.jsonl", "problem_match": "Problem 9. ", "solution_match": "## \\(\\S 3.3\\) Solution to TSTST 9, by Ruben Carpenter \n"}}
diff --git a/images.parquet b/images.parquet
new file mode 100644
index 0000000000000000000000000000000000000000..dbddfee0d49f7b6367173eac9034e59a5d05ceef
--- /dev/null
+++ b/images.parquet
@@ -0,0 +1,3 @@
+version https://git-lfs.github.com/spec/v1
+oid sha256:3e7c6c9a7756c9027a0d4236d067e62baf2459cfe58641413430b87ac2df09cb
+size 3932371
diff --git a/images_extract.py b/images_extract.py
new file mode 100644
index 0000000000000000000000000000000000000000..cb2d76281bed3079fef48a83620462c2c451ef06
--- /dev/null
+++ b/images_extract.py
@@ -0,0 +1,65 @@
+import base64
+import re
+import hashlib
+from pathlib import Path
+import sys
+from datasets import Dataset, concatenate_datasets
+
+if __name__ == "__main__":
+ project_root = Path(__file__).parent
+ images_parquet = project_root / "images.parquet"
+
+ pattern = re.compile(
+ r"!\[[^\]]*?\]\(data\:image/(?P\w+)\;base64\,(?P[A-Za-z0-9+/=]+?)\)"
+ )
+ md_files = filter(lambda x: x.name != "README.md", project_root.glob("**/*.md"))
+ json_files = project_root.glob("**/*.jsonl")
+
+ hashs = []
+ images = []
+
+ for md_file in md_files:
+ text = md_file.read_text()
+ matches = pattern.finditer(text)
+
+ for match in matches:
+ b64_data = match.group("data")
+ img_type = match.group("type")
+ img_data = base64.b64decode(b64_data)
+ hash_digest = hashlib.md5(img_data).hexdigest()
+
+ hashs.append(hash_digest)
+ images.append(img_data)
+
+ new_md_link = f""
+ text = text.replace(match.group(0), new_md_link)
+
+ md_file.write_text(text, encoding="utf-8")
+
+ # for json_file in json_files:
+ # text = json_file.read_text()
+ # matches = pattern.finditer(text)
+
+ # for match in matches:
+ # b64_data = match.group("data")
+ # img_type = match.group("type")
+ # img_data = base64.b64decode(b64_data)
+ # hash_digest = hashlib.md5(img_data).hexdigest()
+
+ # new_md_link = f""
+ # text = text.replace(match.group(0), new_md_link)
+
+ # json_file.write_text(text, encoding="utf-8")
+
+ ds = Dataset.from_dict({"hash": hashs, "image": images})
+ if len(images) == 0:
+ print("No images found, exiting.")
+ sys.exit(0)
+
+ if images_parquet.exists():
+ print("Existing images parquet found, loading and concatenating.")
+ old_ds = Dataset.from_parquet(images_parquet)
+ ds = concatenate_datasets([old_ds, ds])
+
+ ds.to_parquet(images_parquet)
+ print(f"Extracted {len(images)} images to {images_parquet}")