extrac base64 images to parquet
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- APMO/md/en-apmo2025_sol.md +1 -1
- APMO/segmented/en-apmo2025_sol.jsonl +1 -1
- Balkan_MO/md/en-2025-BMO-type1.md +0 -0
- Balkan_MO/segmented/en-2025-BMO-type1.jsonl +0 -0
- Benelux_MO/md/Benelux_en-olympiad_en-bxmo-problems-2025-zz.md +0 -0
- Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2025-zz.jsonl +0 -0
- CANADA_MO/md/en-CMO2025-solutions.md +3 -3
- CANADA_MO/segmented/en-CMO2025-solutions.jsonl +2 -2
- Dutch_TST/md/nl-2025-B2025_uitwerkingen.md +3 -3
- Dutch_TST/md/nl-2025-C2025_uitwerkingen.md +3 -3
- Dutch_TST/md/nl-2025-D2025_uitwerkingen.md +1 -1
- Dutch_TST/md/nl-2025-E2025_uitwerkingen.md +1 -1
- Dutch_TST/segmented/nl-2025-B2025_uitwerkingen.jsonl +0 -0
- Dutch_TST/segmented/nl-2025-C2025_uitwerkingen.jsonl +0 -0
- Dutch_TST/segmented/nl-2025-D2025_uitwerkingen.jsonl +0 -0
- Dutch_TST/segmented/nl-2025-E2025_uitwerkingen.jsonl +1 -1
- EGMO/md/en-2025-solutions.md +0 -0
- EGMO/segmented/en-2025-solutions.jsonl +0 -0
- HarvardMIT/md/en-282-2025-feb-comb-solutions.md +5 -5
- HarvardMIT/md/en-282-2025-feb-geo-solutions.md +0 -0
- HarvardMIT/md/en-282-2025-feb-guts-solutions.md +0 -0
- HarvardMIT/md/en-282-2025-feb-team-solutions.md +0 -0
- HarvardMIT/md/en-284-tournaments-2025-hmic-solutions.md +2 -2
- HarvardMIT/segmented/en-282-2025-feb-comb-solutions.jsonl +0 -0
- HarvardMIT/segmented/en-282-2025-feb-geo-solutions.jsonl +0 -0
- HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl +0 -0
- HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl +0 -0
- HarvardMIT/segmented/en-284-tournaments-2025-hmic-solutions.jsonl +2 -2
- IMO/md/en-IMO-2005-notes.md +3 -3
- IMO/md/en-IMO-2006-notes.md +1 -1
- IMO/md/en-IMO-2007-notes.md +3 -3
- IMO/md/en-IMO-2008-notes.md +2 -2
- IMO/md/en-IMO-2009-notes.md +2 -2
- IMO/md/en-IMO-2010-notes.md +2 -2
- IMO/md/en-IMO-2011-notes.md +1 -1
- IMO/md/en-IMO-2012-notes.md +1 -1
- IMO/md/en-IMO-2013-notes.md +0 -0
- IMO/md/en-IMO-2014-notes.md +0 -0
- IMO/md/en-IMO-2015-notes.md +3 -3
- IMO/md/en-IMO-2016-notes.md +3 -3
- IMO/md/en-IMO-2017-notes.md +3 -3
- IMO/md/en-IMO-2018-notes.md +0 -0
- IMO/md/en-IMO-2019-notes.md +0 -0
- IMO/md/en-IMO-2020-notes.md +0 -0
- IMO/md/en-IMO-2021-notes.md +0 -0
- IMO/md/en-IMO-2022-notes.md +1 -1
- IMO/md/en-IMO-2023-notes.md +0 -0
- IMO/md/en-IMO-2024-notes.md +0 -0
- IMO/md/en-IMO-2025-notes.md +0 -0
- IMO/segmented/en-IMO-2005-notes.jsonl +3 -3
APMO/md/en-apmo2025_sol.md
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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.
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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
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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.
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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
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APMO/segmented/en-apmo2025_sol.jsonl
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{"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"}}
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{"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"}}
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{"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"}}
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{"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"}}
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{"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"}}
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{"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"}}
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{"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"}}
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{"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"}}
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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.
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<center>Figure 1: A possible diagram of all the arcs. </center>
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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.
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<center>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. </center>
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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.
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<center>Figure 3: Possible ant moves going up or right. </center>
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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.
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<center>Figure 1: A possible diagram of all the arcs. </center>
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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.
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<center>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. </center>
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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.
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<center>Figure 3: Possible ant moves going up or right. </center>
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{"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"}}
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{"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"}}
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{"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"}}
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{"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\n<center>Figure 1: A possible diagram of all the arcs. </center> \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"}}
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{"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\n<center>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. </center> \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\n<center>Figure 3: Possible ant moves going up or right. </center> \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"}}
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{"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"}}
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{"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"}}
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{"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"}}
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{"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"}}
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{"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\n<center>Figure 1: A possible diagram of all the arcs. </center> \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"}}
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{"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\n<center>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. </center> \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\n<center>Figure 3: Possible ant moves going up or right. </center> \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"}}
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{"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"}}
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Dutch_TST/md/nl-2025-B2025_uitwerkingen.md
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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 \(b_i\).
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- Er is één computer superconnected met zowel \(a_i\) als \(a_{i+2}\), namelijk de computer \(a_{i+1}\).
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Bewijs dat \(ND\), \(MF\) en \(PE\) concurrent zijn.
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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 \(b_i\).
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- Er is één computer superconnected met zowel \(a_i\) als \(a_{i+2}\), namelijk de computer \(a_{i+1}\).
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Dutch_TST/md/nl-2025-C2025_uitwerkingen.md
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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.
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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.
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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.)
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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 \(4|XY|\).
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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.
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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.
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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.)
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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 \(4|XY|\).
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Dutch_TST/md/nl-2025-D2025_uitwerkingen.md
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Bewijs dat de lijn \(KT\) door \(O\) gaat.
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Dutch_TST/md/nl-2025-E2025_uitwerkingen.md
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Bewijs dat \(\angle ACB = \angle DCF\).
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{"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."}}
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{"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."}}
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{"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."}}
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{"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."}}
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{"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."}}
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{"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."}}
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{"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."}}
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{"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."}}
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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 \(i \in [0, 5]\) .
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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.
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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.
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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.
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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 \(\overline{{C D}}\) .
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 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 \(\frac{1}{2}\left(1 - 4\cdot \frac{1}{12}\right) = \frac{1}{3}\) .
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 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 \(3\cdot \frac{4}{27} = \left[\frac{4}{9}\right]\) .
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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 \(i \in [0, 5]\) .
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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.
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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.
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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.
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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 \(\overline{{C D}}\) .
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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 \(\frac{1}{2}\left(1 - 4\cdot \frac{1}{12}\right) = \frac{1}{3}\) .
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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 \(3\cdot \frac{4}{27} = \left[\frac{4}{9}\right]\) .
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Proposed by: Albert Wang
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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.
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Proposed by: Derek Liu
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Solution: In what follows, all angles are directed.
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Proposed by: Albert Wang
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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.
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Proposed by: Derek Liu
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Solution: In what follows, all angles are directed.
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{"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"}}
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{"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: "}}
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{"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: "}}
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{"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"}}
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{"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: "}}
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{"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: "}}
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{"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"}}
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{"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: "}}
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{"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"}}
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{"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: "}}
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{"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"}}
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{"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: "}}
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{"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: "}}
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{"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"}}
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{"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: "}}
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{"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: "}}
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{"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"}}
|
| 8 |
{"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: "}}
|
| 9 |
{"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"}}
|
| 10 |
{"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: "}}
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IMO/md/en-IMO-2005-notes.md
CHANGED
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@@ -73,7 +73,7 @@ Thus
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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.
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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.
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@@ -108,7 +108,7 @@ From this we deduce \(a_{n + 1} \in \{k, k + n + 1\}\) as desired.
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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.
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@@ -193,7 +193,7 @@ Let \(A B C D\) be a fixed convex quadrilateral with \(B C = D A\) and \(\overli
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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.
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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).
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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.
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+
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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.
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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.
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+
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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.
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+
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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).
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IMO/md/en-IMO-2006-notes.md
CHANGED
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@@ -251,7 +251,7 @@ The main work is the proof of the lemma.
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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 \(N = 8\) .
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Claim — Every point \(X\) in the polygon is contained in the wingspan of some butterfly.
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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 \(N = 8\) .
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+
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Claim — Every point \(X\) in the polygon is contained in the wingspan of some butterfly.
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IMO/md/en-IMO-2007-notes.md
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@@ -113,7 +113,7 @@ Let \(M\) , \(N\) , \(P\) denote the midpoints of \(\overline{CF}\) , \(\overlin
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By a homothety at \(C\) with ratio \(\frac{1}{2}\) , we find \(\overline{MNP}\) is the image of line \(\ell \equiv \overline{AGF}\) .
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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 \(\overline{BCD}\) .
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@@ -150,7 +150,7 @@ So let's consider the situation
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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 \(2r = 6\) .
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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.
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@@ -203,7 +203,7 @@ Thus \(OP = OQ\) . Since \(OC = OR\) as well, we get the conclusion.
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Denote by \(X\) and \(Y\) the feet from \(R\) to \(\overline{CA}\) and \(\overline{CB}\) , so \(\triangle CXR \cong \triangle CYR\) . Then, let \(t = \frac{CQ}{CR} = 1 - \frac{CP}{CR}\) .
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Then it follows that
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By a homothety at \(C\) with ratio \(\frac{1}{2}\) , we find \(\overline{MNP}\) is the image of line \(\ell \equiv \overline{AGF}\) .
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| 116 |
+
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| 117 |
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| 118 |
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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 \(\overline{BCD}\) .
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| 150 |
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| 151 |
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 \(2r = 6\) .
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| 152 |
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+
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| 155 |
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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.
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Denote by \(X\) and \(Y\) the feet from \(R\) to \(\overline{CA}\) and \(\overline{CB}\) , so \(\triangle CXR \cong \triangle CYR\) . Then, let \(t = \frac{CQ}{CR} = 1 - \frac{CP}{CR}\) .
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+
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| 208 |
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Then it follows that
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IMO/md/en-IMO-2008-notes.md
CHANGED
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@@ -73,7 +73,7 @@ We show two solutions.
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| 73 |
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| 74 |
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 \(\Gamma_{C}\) .
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-
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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.
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@@ -240,7 +240,7 @@ Let \(\overline{{P Q}}\) and \(\overline{{S T}}\) be diameters of \(\omega_{1}\)
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Now orient \(A C\) horizontally and let \(K\) be the "uppermost" point of \(\omega\) , as shown.
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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 \(\omega_{2}\) .
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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 \(\Gamma_{C}\) .
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+
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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.
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Now orient \(A C\) horizontally and let \(K\) be the "uppermost" point of \(\omega\) , as shown.
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+
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| 245 |
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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 \(\omega_{2}\) .
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IMO/md/en-IMO-2009-notes.md
CHANGED
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@@ -101,7 +101,7 @@ Let \(ABC\) be a triangle with circumcenter \(O\) . The points \(P\) and \(Q\) a
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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 \(AQ \cdot QB = AP \cdot PC\) .
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-
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As \(\overline{ML} \parallel \overline{AC}\) and \(\overline{MK} \parallel \overline{AB}\) we have that
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@@ -192,7 +192,7 @@ Let \(I\) be the incenter of \(A B C\) , and set \(\angle D A C = 2x\) (so that
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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 \(I E C\) .
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By the law of sines, we obtain
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| 101 |
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| 102 |
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 \(AQ \cdot QB = AP \cdot PC\) .
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| 103 |
|
| 104 |
+
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|
| 106 |
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| 107 |
As \(\overline{ML} \parallel \overline{AC}\) and \(\overline{MK} \parallel \overline{AB}\) we have that
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| 192 |
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| 193 |
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 \(I E C\) .
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| 195 |
+
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| 197 |
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By the law of sines, we obtain
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IMO/md/en-IMO-2010-notes.md
CHANGED
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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 \(\overline{{I F}}\) .
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@@ -187,7 +187,7 @@ We present two solutions using harmonic bundles.
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\(\P\) 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 \(\triangle MPN\) ).
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| 189 |
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| 191 |
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| 192 |
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| 193 |
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 \(\Gamma\) .
|
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|
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| 130 |
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| 131 |
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 \(\overline{{I F}}\) .
|
| 132 |
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| 133 |
+
|
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| 135 |
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| 136 |
|
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|
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| 187 |
|
| 188 |
\(\P\) 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 \(\triangle MPN\) ).
|
| 189 |
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| 190 |
+
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| 192 |
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| 193 |
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 \(\Gamma\) .
|
IMO/md/en-IMO-2011-notes.md
CHANGED
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@@ -253,7 +253,7 @@ Let \(A B C\) be an acute triangle with circumcircle \(\Gamma\) . Let \(\ell\) b
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|
| 253 |
|
| 254 |
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 \(C_{2}\) .
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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.
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| 253 |
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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 \(C_{2}\) .
|
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+
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|
| 258 |
|
| 259 |
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.
|
IMO/md/en-IMO-2012-notes.md
CHANGED
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@@ -242,7 +242,7 @@ Let \(A B C\) be a triangle with \(\angle B C A = 90^{\circ}\) , and let \(D\) b
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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 \(\omega\) .
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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 \(M K = M L\) .
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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 \(\omega\) .
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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 \(M K = M L\) .
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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.
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, 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
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In the inverted statement, we want line \(ML\) to be tangent to \((AQL)\) .
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Claim — \(\overline{LM} \parallel \overline{AQ}\) .
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Let line \(FG\) meet \((BDF)\) and \((CGE)\) again at \(F_{2}\) and \(G_{2}\) .
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Claim — Quadrilaterals \(FBDF_{2}\) and \(G_{2}ECG\) are similar, actually homothetic through \(\overline{FG} \cap \overline{BC}\) .
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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.
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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
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In the inverted statement, we want line \(ML\) to be tangent to \((AQL)\) .
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Claim — \(\overline{LM} \parallel \overline{AQ}\) .
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Let line \(FG\) meet \((BDF)\) and \((CGE)\) again at \(F_{2}\) and \(G_{2}\) .
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Claim — Quadrilaterals \(FBDF_{2}\) and \(G_{2}ECG\) are similar, actually homothetic through \(\overline{FG} \cap \overline{BC}\) .
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These three facts, together with \(F\) lying inside \(\triangle ABD\) , are enough to imply the result. \(\square\)
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Using these observations as the anchor for everything that follows, we now prove several claims about \(X\) and \(E\) in succession.
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Claim — Point \(E\) is the midpoint of arc \(\overline{AD}\) in \((ABMD)\) , and hence lies on ray \(BF\) .
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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 \(P_{i}P_{i + n}\) .
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Then:
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These three facts, together with \(F\) lying inside \(\triangle ABD\) , are enough to imply the result. \(\square\)
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Using these observations as the anchor for everything that follows, we now prove several claims about \(X\) and \(E\) in succession.
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Claim — Point \(E\) is the midpoint of arc \(\overline{AD}\) in \((ABMD)\) , and hence lies on ray \(BF\) .
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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 \(P_{i}P_{i + n}\) .
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Then:
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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.
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 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 \(HX = HY\) .
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The two similarities are equivalent because \(RS = ST\) the SAS gives \(KR \cdot TA = RS \cdot RT = TS \cdot TR\) .
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Remark. The problem is actually symmetric with respect to two circles; \(\overline{RA}\) is tangent at \(R\) if and only if \(\overline{TK}\) at \(T\) .
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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.
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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 \(G_{k}\)
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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.
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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 \(HX = HY\) .
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The two similarities are equivalent because \(RS = ST\) the SAS gives \(KR \cdot TA = RS \cdot RT = TS \cdot TR\) .
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Remark. The problem is actually symmetric with respect to two circles; \(\overline{RA}\) is tangent at \(R\) if and only if \(\overline{TK}\) at \(T\) .
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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.
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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 \(G_{k}\)
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Define \(K = \overline{CT} \cap \overline{AE}\) , \(L = \overline{DT} \cap \overline{AB}\) , \(X = \overline{BT} \cap \overline{AE}\) , \(Y = \overline{ET} \cap \overline{BY}\) .
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Claim (Main claim) — We have
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Define \(K = \overline{CT} \cap \overline{AE}\) , \(L = \overline{DT} \cap \overline{AB}\) , \(X = \overline{BT} \cap \overline{AE}\) , \(Y = \overline{ET} \cap \overline{BY}\) .
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Claim (Main claim) — We have
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IMO/segmented/en-IMO-2005-notes.jsonl
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{"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"}}
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{"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"}}
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{"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"}}
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{"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"}}
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{"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"}}
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{"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"}}
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{"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"}}
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{"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"}}
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{"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"}}
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{"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"}}
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{"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"}}
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{"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"}}
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