diff --git "a/HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl" "b/HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl"
--- "a/HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl"
+++ "b/HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl"
@@ -2,22 +2,22 @@
 {"year": "2025", "tier": "T4", "problem_label": "1", "problem_type": null, "exam": "HMMT", "problem": "Let \\(a\\) , \\(b\\) , and \\(c\\) be pairwise distinct positive integers such that \\(\\frac{1}{a}\\) , \\(\\frac{1}{b}\\) , \\(\\frac{1}{c}\\) is an increasing arithmetic sequence in that order. Prove that \\(\\gcd (a,b) > 1\\) .", "solution": "Observe that \\(\\frac{2}{b} -\\frac{1}{a} = \\frac{1}{c}\\) , so \\((2a - b)c = ab\\) and thus \\(2a - b \\mid ab\\) . If we assume that \\(\\gcd (a,b) = 1\\) , then \\(\\gcd (2a - b,a) = 1\\) , so \\(2a - b \\mid b\\) . Then \\(2a - b \\mid (2a - b) + b = 2a\\) , so \\(2a - b \\mid \\gcd (2a,b) \\leq 2\\) . Thus \\(2a - b \\leq 2\\) . But \\(a > b\\) , contradiction. Thus, \\(\\gcd (a,b) > 1\\) , as desired.", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n1. [20]", "solution_match": "\nSolution 2: "}}
 {"year": "2025", "tier": "T4", "problem_label": "2", "problem_type": null, "exam": "HMMT", "problem": "A polyomino is a connected figure constructed by joining one or more unit squares edge-to-edge. Determine, with proof, the number of non-congruent polyominoes with no holes, perimeter 180, and area 2024.", "solution": "Answer: 2", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n2. [25]", "solution_match": "\nProposed by: Albert Wang  \n\n"}}
 {"year": "2025", "tier": "T4", "problem_label": "2", "problem_type": null, "exam": "HMMT", "problem": "A polyomino is a connected figure constructed by joining one or more unit squares edge-to-edge. Determine, with proof, the number of non-congruent polyominoes with no holes, perimeter 180, and area 2024.", "solution": "Define the bounding box of a polyomino to be the smallest axis-aligned rectangle that contains the entire polyomino. Suppose a polyomino satisfying the given conditions has a bounding box with dimensions \\(w \\times h\\) .  \n\nClaim 1. \\(w + h \\leq 90\\) .  \n\nProof. The polyomino has at least \\(2w\\) horizontal edges and at least \\(2h\\) vertical edges. Moreover, it has a perimeter of 180. Therefore, \\(2w + 2h \\leq 180\\) , so \\(w + h \\leq 90\\) . \\(\\square\\)  \n\nClaim 2. The dimensions of the bounding box are either \\(44 \\times 46\\) , \\(45 \\times 45\\) , or \\(46 \\times 44\\) .  \n\nProof. Note that \\(hw \\geq 2024\\) since it contains the polyomino with area 2024. Suppose for sake of contradiction that \\(h + w \\leq 89\\) . Then,  \n\n\\[(h - w)^{2} = (h + w)^{2} - 4hw \\leq 89^{2} - 4 \\cdot 2024 = -175,\\]  \n\ncontradiction. Therefore, \\(h + w = 90\\) , so we can let \\((h,w) = (45 + x,45 - x)\\) . Then, \\(2025 - x^{2} = hw \\geq 2024\\) implies that \\(x \\in \\{- 1,0,1\\}\\) , as desired. \\(\\square\\)  \n\nIn the first and third cases, the bounding box has area 2024, so it must be the entire polyomino, giving us the \\(44 \\times 46\\) rectangle (and its rotation) as a possible answer. In the second case, the bounding box has area 2025, so one cell must be removed to form the polyomino. Removing the corner cell yields a polyomino with perimeter 180, and removing any other cells yields a polyomino with perimeter greater than 180. Therefore, the only other possibility is a \\(45 \\times 45\\) square missing a corner. Thus the answer is \\(\\boxed{2}\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n2. [25]", "solution_match": "\nSolution: "}}
-{"year": "2025", "tier": "T4", "problem_label": "3", "problem_type": null, "exam": "HMMT", "problem": "Let \\(\\omega_{1}\\) and \\(\\omega_{2}\\) be two circles intersecting at distinct points \\(A\\) and \\(B\\) . Point \\(X\\) varies along \\(\\omega_{1}\\) , and point \\(Y\\) on \\(\\omega_{2}\\) is chosen such that \\(AB\\) bisects the angle \\(\\angle XAY\\) . Prove that as \\(X\\) varies along \\(\\omega_{1}\\) , the circumcenter of \\(\\triangle AXY\\) (if it exists) varies along a fixed line.", "solution": "Solution 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 \n\nLet \\(O_{1}\\) , \\(O_{2}\\) , and \\(O\\) be the centers of \\(\\omega_{1}\\) , \\(\\omega_{2}\\) , and the circumcircle of \\(\\triangle AXY\\) , respectively.  \n\nWe claim that triangle \\(O O_{1}O_{2}\\) is isosceles with \\(O O_{1} = O O_{2}\\) , and thus in particular \\(O\\) always lies on the perpendicular bisector of \\(O_{1}O_{2}\\) .  \n\nTo this end, observe that \\(O O_{1} \\perp A X\\) and \\(O_{1}O_{2} \\perp A B\\) , so \\(\\angle O O_{1}O_{2} = \\angle X A B\\) . Analogously, \\(\\angle O O_{2}O_{1} = \\angle Y A B\\) . So indeed \\(O O_{1}O_{2}\\) is isosceles, and we are done.  \n\nRemark. One may also consider the antipodes of \\(A\\) on \\(\\omega_{1}\\) and \\(\\omega_{2}\\) for an equivalent but more natural angle- chase.", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n3. [30]", "solution_match": "\nProposed by: Pitchayut Saengrungkonkka  \n\n"}}
-{"year": "2025", "tier": "T4", "problem_label": "3", "problem_type": null, "exam": "HMMT", "problem": "Let \\(\\omega_{1}\\) and \\(\\omega_{2}\\) be two circles intersecting at distinct points \\(A\\) and \\(B\\) . Point \\(X\\) varies along \\(\\omega_{1}\\) , and point \\(Y\\) on \\(\\omega_{2}\\) is chosen such that \\(AB\\) bisects the angle \\(\\angle XAY\\) . Prove that as \\(X\\) varies along \\(\\omega_{1}\\) , the circumcenter of \\(\\triangle AXY\\) (if it exists) varies along a fixed line.", "solution": 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 \n\nLet \\(A^{\\prime}\\) be the \\(A\\) - antipode in circle \\((A X Y)\\) . It suffices to show that \\(A^{\\prime}\\) lies on a fixed line. We will show that this line is one that is parallel to \\(A B\\) .  \n\nLet \\(M\\) be the second intersection of line \\(A B\\) with circle \\((A X Y)\\) , and let \\(N\\) be the antipode of \\(M\\) on this circle. Since \\(A M A^{\\prime}N\\) is a rectangle with \\(N\\) lying on the line through \\(A\\) perpendicular to \\(A B\\) , it suffices to show that \\(N\\) is fixed (independent of \\(X\\) and \\(Y\\) ).  \n\nTo this end, take an inversion at \\(A\\) with arbitrary radius, denoting images with \\(\\bullet \\mapsto \\bullet^{*}\\) .  \n\nObserve that \\(X^{*}\\) and \\(Y^{*}\\) lie on the fixed lines \\(\\ell_{1} = \\omega_{1}^{*}\\) and \\(\\ell_{2} = \\omega_{2}^{*}\\) . Let \\(\\ell\\) be the line through \\(A\\) perpendicular to \\(A B^{*}\\) , and suppose that \\(\\ell_{1}\\) and \\(\\ell_{2}\\) intersect \\(\\ell\\) at \\(P\\) and \\(Q\\) , respectively.  \n\nSince \\(\\angle X A B = \\angle B A Y\\) , we have \\(\\angle X^{*}A B^{*} = \\angle B^{*}A Y^{*}\\) . Circles \\(\\omega_{1}\\) , \\(\\omega_{2}\\) , and \\((A X Y)\\) are mapped to lines \\(B^{*}X^{*}\\) , \\(B^{*}Y^{*}\\) , and \\(X^{*}Y^{*}\\) . As \\(A N \\perp A B\\) , it follows that \\(N^{*}\\) is the intersection of \\(X^{*}Y^{*}\\) and \\(\\ell\\) .\n\n\n\nFinally, observe that \\((N^{*},A;P,Q)\\stackrel {B^{*}}{=}(N^{*},M^{*};X,Y)\\) is a harmonic bundle, as \\(A M^{*}\\) bisects \\(\\angle X^{*}A Y^{*}\\) and \\(\\angle M^{*}A N^{*} = 90^{\\circ}\\) . Since \\(A\\) , \\(P\\) , and \\(Q\\) are fixed, so is \\(N^{*}\\) . Thus \\(N\\) is fixed, and \\(O\\) lies on the perpendicular bisector of \\(A N\\) , which is also fixed.  \n\nRemark. An alternative approach to the last paragraph is to recall Blanchet's theorem, which states that \\(P Y^{*}\\) , \\(Q X^{*}\\) , and \\(A B^{*}\\) are concurrent. By Ceva and Menelaus, we get that \\((N^{*},A;P,Q) = - 1\\) .  \n\nRemark. If one projects the kite/harmonic quadrilateral \\(N X M Y\\) from \\(A^{\\prime}\\) onto the line through the antipodes defined in the previous remark, we obtain that \\(A^{\\prime}N\\) (a line parallel to \\(A B\\) ) passes through the midpoint of the two antipodes, directly finishing.", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n3. [30]", "solution_match": "\nSolution 2:  \n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "3", "problem_type": null, "exam": "HMMT", "problem": "Let \\(\\omega_{1}\\) and \\(\\omega_{2}\\) be two circles intersecting at distinct points \\(A\\) and \\(B\\) . Point \\(X\\) varies along \\(\\omega_{1}\\) , and point \\(Y\\) on \\(\\omega_{2}\\) is chosen such that \\(AB\\) bisects the angle \\(\\angle XAY\\) . Prove that as \\(X\\) varies along \\(\\omega_{1}\\) , the circumcenter of \\(\\triangle AXY\\) (if it exists) varies along a fixed line.", "solution": "Solution 1:\n\n\n![md5:d98306254a4727f82e4d13898cfcfd32](d98306254a4727f82e4d13898cfcfd32.jpeg)\n  \n\nLet \\(O_{1}\\) , \\(O_{2}\\) , and \\(O\\) be the centers of \\(\\omega_{1}\\) , \\(\\omega_{2}\\) , and the circumcircle of \\(\\triangle AXY\\) , respectively.  \n\nWe claim that triangle \\(O O_{1}O_{2}\\) is isosceles with \\(O O_{1} = O O_{2}\\) , and thus in particular \\(O\\) always lies on the perpendicular bisector of \\(O_{1}O_{2}\\) .  \n\nTo this end, observe that \\(O O_{1} \\perp A X\\) and \\(O_{1}O_{2} \\perp A B\\) , so \\(\\angle O O_{1}O_{2} = \\angle X A B\\) . Analogously, \\(\\angle O O_{2}O_{1} = \\angle Y A B\\) . So indeed \\(O O_{1}O_{2}\\) is isosceles, and we are done.  \n\nRemark. One may also consider the antipodes of \\(A\\) on \\(\\omega_{1}\\) and \\(\\omega_{2}\\) for an equivalent but more natural angle- chase.", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n3. [30]", "solution_match": "\nProposed by: Pitchayut Saengrungkonkka  \n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "3", "problem_type": null, "exam": "HMMT", "problem": "Let \\(\\omega_{1}\\) and \\(\\omega_{2}\\) be two circles intersecting at distinct points \\(A\\) and \\(B\\) . Point \\(X\\) varies along \\(\\omega_{1}\\) , and point \\(Y\\) on \\(\\omega_{2}\\) is chosen such that \\(AB\\) bisects the angle \\(\\angle XAY\\) . Prove that as \\(X\\) varies along \\(\\omega_{1}\\) , the circumcenter of \\(\\triangle AXY\\) (if it exists) varies along a fixed line.", "solution": "![md5:ae48eed2043117eba1dbe2e28f2c5e1a](ae48eed2043117eba1dbe2e28f2c5e1a.jpeg)\n  \n\nLet \\(A^{\\prime}\\) be the \\(A\\) - antipode in circle \\((A X Y)\\) . It suffices to show that \\(A^{\\prime}\\) lies on a fixed line. We will show that this line is one that is parallel to \\(A B\\) .  \n\nLet \\(M\\) be the second intersection of line \\(A B\\) with circle \\((A X Y)\\) , and let \\(N\\) be the antipode of \\(M\\) on this circle. Since \\(A M A^{\\prime}N\\) is a rectangle with \\(N\\) lying on the line through \\(A\\) perpendicular to \\(A B\\) , it suffices to show that \\(N\\) is fixed (independent of \\(X\\) and \\(Y\\) ).  \n\nTo this end, take an inversion at \\(A\\) with arbitrary radius, denoting images with \\(\\bullet \\mapsto \\bullet^{*}\\) .  \n\nObserve that \\(X^{*}\\) and \\(Y^{*}\\) lie on the fixed lines \\(\\ell_{1} = \\omega_{1}^{*}\\) and \\(\\ell_{2} = \\omega_{2}^{*}\\) . Let \\(\\ell\\) be the line through \\(A\\) perpendicular to \\(A B^{*}\\) , and suppose that \\(\\ell_{1}\\) and \\(\\ell_{2}\\) intersect \\(\\ell\\) at \\(P\\) and \\(Q\\) , respectively.  \n\nSince \\(\\angle X A B = \\angle B A Y\\) , we have \\(\\angle X^{*}A B^{*} = \\angle B^{*}A Y^{*}\\) . Circles \\(\\omega_{1}\\) , \\(\\omega_{2}\\) , and \\((A X Y)\\) are mapped to lines \\(B^{*}X^{*}\\) , \\(B^{*}Y^{*}\\) , and \\(X^{*}Y^{*}\\) . As \\(A N \\perp A B\\) , it follows that \\(N^{*}\\) is the intersection of \\(X^{*}Y^{*}\\) and \\(\\ell\\) .\n\n\n\nFinally, observe that \\((N^{*},A;P,Q)\\stackrel {B^{*}}{=}(N^{*},M^{*};X,Y)\\) is a harmonic bundle, as \\(A M^{*}\\) bisects \\(\\angle X^{*}A Y^{*}\\) and \\(\\angle M^{*}A N^{*} = 90^{\\circ}\\) . Since \\(A\\) , \\(P\\) , and \\(Q\\) are fixed, so is \\(N^{*}\\) . Thus \\(N\\) is fixed, and \\(O\\) lies on the perpendicular bisector of \\(A N\\) , which is also fixed.  \n\nRemark. An alternative approach to the last paragraph is to recall Blanchet's theorem, which states that \\(P Y^{*}\\) , \\(Q X^{*}\\) , and \\(A B^{*}\\) are concurrent. By Ceva and Menelaus, we get that \\((N^{*},A;P,Q) = - 1\\) .  \n\nRemark. If one projects the kite/harmonic quadrilateral \\(N X M Y\\) from \\(A^{\\prime}\\) onto the line through the antipodes defined in the previous remark, we obtain that \\(A^{\\prime}N\\) (a line parallel to \\(A B\\) ) passes through the midpoint of the two antipodes, directly finishing.", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n3. [30]", "solution_match": "\nSolution 2:  \n\n"}}
 {"year": "2025", "tier": "T4", "problem_label": "4", "problem_type": null, "exam": "HMMT", "problem": "Jerry places at most one rook in each cell of a \\(2025 \\times 2025\\) grid of cells. A rook attacks another rook if the two rooks are in the same row or column and there are no other rooks between them.  \n\nDetermine, with proof, the maximum number of rooks Jerry can place on the grid such that no rook attacks 4 other rooks.", "solution": "Answer: 8096", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n4. [35]", "solution_match": "\nProposed by: Arul Kolla  \n\n"}}
 {"year": "2025", "tier": "T4", "problem_label": "4", "problem_type": null, "exam": "HMMT", "problem": "Jerry places at most one rook in each cell of a \\(2025 \\times 2025\\) grid of cells. A rook attacks another rook if the two rooks are in the same row or column and there are no other rooks between them.  \n\nDetermine, with proof, the maximum number of rooks Jerry can place on the grid such that no rook attacks 4 other rooks.", "solution": "The answer is \\(2024 \\times 4 = 8096\\) . More generally, for an \\(n \\times n\\) grid, the answer is \\(4n - 4\\) . Call a rook that attacks at most 3 other rooks good.  \n\nWe use the following observation in both parts of the solution: a rook on the border of the grid must be good.  \n\nLower Bound: Place rooks on all \\(4n - 4\\) border cells of the grid. By the above observation, every rook is good.  \n\nUpper Bound: Consider any valid placement of rooks, and assume there exists a rook that is not on the border. We can move this rook to the border via a usual rook move, since this rook is good and thus the path to one of the four border cells in its row or column must be empty.  \n\nAfter this move, we claim the placement of rooks is still valid. Indeed:  \n\n- the moved rook is now on the border, so by the observation above, it must be good;  \n\n- any rook that used to attack this rook cannot attack more rooks after the move, so such rooks must still be good;  \n\n- any rook attacked in the final position must be either:  \n\n- opposite the direction moved, in which case it attacked the moved rook both before and after the move (so is still good), or \n- perpendicular to the direction moved, in which case it is a border rook and must always be good.  \n\nBy repeating the above process, we can always move from any good position to one where all rooks are on the border. This implies that the number of rooks in any good position is at most \\(4n - 4\\) . When \\(n = 2024\\) , the answer is \\(4n - 4 = \\left\\lfloor 8096 \\right\\rfloor\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n4. [35]", "solution_match": "\nSolution 1: "}}
 {"year": "2025", "tier": "T4", "problem_label": "4", "problem_type": null, "exam": "HMMT", "problem": "Jerry places at most one rook in each cell of a \\(2025 \\times 2025\\) grid of cells. A rook attacks another rook if the two rooks are in the same row or column and there are no other rooks between them.  \n\nDetermine, with proof, the maximum number of rooks Jerry can place on the grid such that no rook attacks 4 other rooks.", "solution": "Consider the set of all rooks which are either the leftmost or rightmost in their row, or the topmost or bottommost in their column. Note that this set must include every rook, as any rook not in this set attacks a rook in all 4 directions.  \n\nEach column contributes at most 2 rooks to this set, and each row contributes at most 2 rooks. We can safely ignore the top and bottom rows in this count, as any rook in the top or bottom row is already the topmost or bottommost rook in its column. Thus the number of rooks in the set is at most \\(2 \\cdot (2025 + 2025 - 2) = \\left\\lfloor 8096 \\right\\rfloor\\) , which can be constructed as seen before.", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n4. [35]", "solution_match": "\nSolution 2: "}}
-{"year": "2025", "tier": "T4", "problem_label": "5", "problem_type": null, "exam": "HMMT", "problem": "Let \\(\\triangle ABC\\) be an acute triangle with orthocenter \\(H\\) . Points \\(E\\) and \\(F\\) are on segments \\(\\overline{AC}\\) and \\(\\overline{AB}\\) , respectively, such that \\(\\angle EHF = 90^{\\circ}\\) . Let \\(X\\) be the foot of the altitude from \\(H\\) to \\(\\overline{EF}\\) . Prove that \\(\\angle BXC = 90^{\\circ}\\) .", "solution": "Solution 1:  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 \n\nWe use \\(\\angle\\) to denote directed angles. Let \\(Y\\) and \\(Z\\) be the feet of the altitudes from \\(B\\) and \\(C\\) to \\(AC\\) and \\(AB\\) , respectively. Then \\(\\angle HZF = \\angle HXF = 90^{\\circ}\\) , so \\(HZFX\\) is cyclic. Similarly, \\(HYEX\\) is cyclic. Therefore,  \n\n\\[\\angle BYX = \\angle HYX = \\angle HEX = \\angle FHX = \\angle FZX = \\angle BZX.\\]  \n\nHence, \\(BZXY\\) is cyclic. A symmetric argument shows \\(C\\) lies on this circle as well. It follows that \\(\\angle BXC = \\angle BYC = 90^{\\circ}\\) , as desired.", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n5. [35]", "solution_match": "\nProposed by: Pitchayut Saengrungkonka  \n\n"}}
-{"year": "2025", "tier": "T4", "problem_label": "5", "problem_type": null, "exam": "HMMT", "problem": "Let \\(\\triangle ABC\\) be an acute triangle with orthocenter \\(H\\) . Points \\(E\\) and \\(F\\) are on segments \\(\\overline{AC}\\) and \\(\\overline{AB}\\) , respectively, such that \\(\\angle EHF = 90^{\\circ}\\) . Let \\(X\\) be the foot of the altitude from \\(H\\) to \\(\\overline{EF}\\) . Prove that \\(\\angle BXC = 90^{\\circ}\\) .", "solution": "Let \\(T\\) be the foot of altitude from \\(A\\) to \\(BC\\) . For any point \\(X\\) , let \\(X'\\) denote the image of \\(X\\) under the negative inversion at \\(H\\) with radius \\(\\sqrt{HA \\cdot HT}\\) . Then \\(B'\\) and \\(C'\\) are the feet of the altitudes from \\(B\\) and \\(C\\) to sides \\(AC\\) and \\(AB\\) , respectively.  \n\nClaim 1. \\(\\angle BX'C = 90^{\\circ}\\) .  \n\nProof. Because \\(HX \\perp EF\\) and \\(HE \\perp HF\\) , the quadrilateral \\(HE'X'F'\\) is a rectangle. Note that \\(\\angle BE'H = \\angle EB'H = 90^{\\circ}\\) and \\(\\angle X'E'H = 90^{\\circ}\\) . Consequently, \\(X', B\\) , and \\(E'\\) are collinear. Similarly, \\(X', C\\) , and \\(F'\\) are collinear. Then, \\(\\angle BX'C = \\angle E'X'F' = 90^{\\circ}\\) , as desired. \\(\\square\\)  \n\nFrom the claim, \\(X'\\) lies on the circle with diameter \\(BC\\) (which \\(B'\\) and \\(C'\\) also lie on). Since this circle is invariant under the inversion, \\(X\\) lies on the circle with diameter \\(BC\\) as well, and \\(\\angle BXC = 90^{\\circ}\\) .\n\n\n![](data:image/jpeg;base64,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"metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n5. [35]", "solution_match": "\nSolution 2: "}}
-{"year": "2025", "tier": "T4", "problem_label": "5", "problem_type": null, "exam": "HMMT", "problem": "Let \\(\\triangle ABC\\) be an acute triangle with orthocenter \\(H\\) . Points \\(E\\) and \\(F\\) are on segments \\(\\overline{AC}\\) and \\(\\overline{AB}\\) , respectively, such that \\(\\angle EHF = 90^{\\circ}\\) . Let \\(X\\) be the foot of the altitude from \\(H\\) to \\(\\overline{EF}\\) . Prove that \\(\\angle BXC = 90^{\\circ}\\) .", "solution": "We begin by proving the following lemma.  \n\nLemma 2. Let \\(A B C D\\) be a quadrilateral and \\(P\\) be a point such that \\(\\angle A P B + \\angle C P D = 180^{\\circ}\\) . Then, the feet of the altitudes from \\(P\\) to each side of \\(A B C D\\) are concyclic.  \n\nProof. Let \\(P_{A},P_{B},P_{C},P_{D}\\) the feet of the altitudes from \\(P\\) to \\(A B\\) , \\(B C\\) , \\(C D\\) , and \\(D A\\) respectively. Note that quadrilateral \\(P_{A}P P_{B}B\\) is cyclic. By angle chasing,  \n\n\\[\\angle P_{D}P_{A}P_{B} + \\angle P_{B}P_{C}P_{D} = \\angle P_{D}P_{A}P + \\angle P P_{A}P_{B} + \\angle P_{B}P_{C}P + \\angle P P_{C}P_{D}\\] \\[\\qquad = \\angle P_{D}A P + \\angle P B P_{B} + \\angle P_{B}C P + \\angle P D P_{D}\\] \\[\\qquad = (180^{\\circ} - \\angle A P D) + (180^{\\circ} - \\angle B P C)\\] \\[\\qquad = \\angle B P A + \\angle D P C\\] \\[\\qquad = 180^{\\circ}.\\]  \n\nTherefore, \\(P_{A}P_{B}P_{C}P_{D}\\) is cyclic as desired.  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)\n\n\n\n\nLet \\(P_{\\infty}\\) be the point at infinity on line \\(A C\\) . Let \\(Y\\) and \\(Z\\) be the feet of the altitudes from \\(H\\) to \\(A B\\) and \\(A C\\) , respectively. Note that \\(\\angle E H F + \\angle B H P_{\\infty} = 90^{\\circ} + 90^{\\circ} = 180^{\\circ}\\) . Thus, the feet of the altitudes from \\(H\\) to \\(E F\\) , \\(E B\\) , \\(B P_{\\infty}\\) , and \\(C P_{\\infty}\\) are concyclic. In other words, \\(X Y Z B\\) is cyclic. Since \\(B C Y Z\\) is a cyclic quadrilateral, we conclude \\(X\\) lies on this circle, giving us that \\(\\angle B X C = 90^{\\circ}\\) as desired.  \n\nRemark. Here's another way to prove the lemma.  \n\nIt is well known that, with the provided condition, there is a point \\(P^{\\prime}\\) that is the isogonal conjugate of \\(P\\) with respect to quadrilateral \\(A B C D\\) . Let \\(P_{A}\\) , \\(P_{B}\\) , \\(P_{C}\\) , and \\(P_{D}\\) be the feet of the altitudes from \\(P\\) to \\(A B\\) , \\(B C\\) , \\(C D\\) , and \\(D A\\) , respectively, and let \\(Q\\) be the foot of the altitude from \\(P^{\\prime}\\) to \\(A B\\) . Because \\(P\\) and \\(P^{\\prime}\\) are isogonal conjugates with respect to the triangle formed by lines \\(A B\\) , \\(B C\\) , and \\(C D\\) , we have \\(P_{A}P_{B}P_{C}Q\\) is cyclic. Similarly, because \\(P\\) and \\(P^{\\prime}\\) are also isogonal conjugate with respect to the triangle formed by lines \\(D A\\) , \\(A B\\) , and \\(B C\\) , we have \\(P_{D}P_{A}P_{B}Q\\) is cyclic. Consequently, \\(P_{A}P_{B}P_{C}P_{C}\\) is cyclic as desired.", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n5. [35]", "solution_match": "\nSolution 3: "}}
+{"year": "2025", "tier": "T4", "problem_label": "5", "problem_type": null, "exam": "HMMT", "problem": "Let \\(\\triangle ABC\\) be an acute triangle with orthocenter \\(H\\) . Points \\(E\\) and \\(F\\) are on segments \\(\\overline{AC}\\) and \\(\\overline{AB}\\) , respectively, such that \\(\\angle EHF = 90^{\\circ}\\) . Let \\(X\\) be the foot of the altitude from \\(H\\) to \\(\\overline{EF}\\) . Prove that \\(\\angle BXC = 90^{\\circ}\\) .", "solution": "Solution 1:  \n\n![md5:034f5986e2970ee1d6cd8c5f3ebb9d68](034f5986e2970ee1d6cd8c5f3ebb9d68.jpeg)\n  \n\nWe use \\(\\angle\\) to denote directed angles. Let \\(Y\\) and \\(Z\\) be the feet of the altitudes from \\(B\\) and \\(C\\) to \\(AC\\) and \\(AB\\) , respectively. Then \\(\\angle HZF = \\angle HXF = 90^{\\circ}\\) , so \\(HZFX\\) is cyclic. Similarly, \\(HYEX\\) is cyclic. Therefore,  \n\n\\[\\angle BYX = \\angle HYX = \\angle HEX = \\angle FHX = \\angle FZX = \\angle BZX.\\]  \n\nHence, \\(BZXY\\) is cyclic. A symmetric argument shows \\(C\\) lies on this circle as well. It follows that \\(\\angle BXC = \\angle BYC = 90^{\\circ}\\) , as desired.", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n5. [35]", "solution_match": "\nProposed by: Pitchayut Saengrungkonka  \n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "5", "problem_type": null, "exam": "HMMT", "problem": "Let \\(\\triangle ABC\\) be an acute triangle with orthocenter \\(H\\) . Points \\(E\\) and \\(F\\) are on segments \\(\\overline{AC}\\) and \\(\\overline{AB}\\) , respectively, such that \\(\\angle EHF = 90^{\\circ}\\) . Let \\(X\\) be the foot of the altitude from \\(H\\) to \\(\\overline{EF}\\) . Prove that \\(\\angle BXC = 90^{\\circ}\\) .", "solution": "Let \\(T\\) be the foot of altitude from \\(A\\) to \\(BC\\) . For any point \\(X\\) , let \\(X'\\) denote the image of \\(X\\) under the negative inversion at \\(H\\) with radius \\(\\sqrt{HA \\cdot HT}\\) . Then \\(B'\\) and \\(C'\\) are the feet of the altitudes from \\(B\\) and \\(C\\) to sides \\(AC\\) and \\(AB\\) , respectively.  \n\nClaim 1. \\(\\angle BX'C = 90^{\\circ}\\) .  \n\nProof. Because \\(HX \\perp EF\\) and \\(HE \\perp HF\\) , the quadrilateral \\(HE'X'F'\\) is a rectangle. Note that \\(\\angle BE'H = \\angle EB'H = 90^{\\circ}\\) and \\(\\angle X'E'H = 90^{\\circ}\\) . Consequently, \\(X', B\\) , and \\(E'\\) are collinear. Similarly, \\(X', C\\) , and \\(F'\\) are collinear. Then, \\(\\angle BX'C = \\angle E'X'F' = 90^{\\circ}\\) , as desired. \\(\\square\\)  \n\nFrom the claim, \\(X'\\) lies on the circle with diameter \\(BC\\) (which \\(B'\\) and \\(C'\\) also lie on). Since this circle is invariant under the inversion, \\(X\\) lies on the circle with diameter \\(BC\\) as well, and \\(\\angle BXC = 90^{\\circ}\\) .\n\n\n![md5:fce69e9afe523959d3e7279c83798c3e](fce69e9afe523959d3e7279c83798c3e.jpeg)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n5. [35]", "solution_match": "\nSolution 2: "}}
+{"year": "2025", "tier": "T4", "problem_label": "5", "problem_type": null, "exam": "HMMT", "problem": "Let \\(\\triangle ABC\\) be an acute triangle with orthocenter \\(H\\) . Points \\(E\\) and \\(F\\) are on segments \\(\\overline{AC}\\) and \\(\\overline{AB}\\) , respectively, such that \\(\\angle EHF = 90^{\\circ}\\) . Let \\(X\\) be the foot of the altitude from \\(H\\) to \\(\\overline{EF}\\) . Prove that \\(\\angle BXC = 90^{\\circ}\\) .", "solution": "We begin by proving the following lemma.  \n\nLemma 2. Let \\(A B C D\\) be a quadrilateral and \\(P\\) be a point such that \\(\\angle A P B + \\angle C P D = 180^{\\circ}\\) . Then, the feet of the altitudes from \\(P\\) to each side of \\(A B C D\\) are concyclic.  \n\nProof. Let \\(P_{A},P_{B},P_{C},P_{D}\\) the feet of the altitudes from \\(P\\) to \\(A B\\) , \\(B C\\) , \\(C D\\) , and \\(D A\\) respectively. Note that quadrilateral \\(P_{A}P P_{B}B\\) is cyclic. By angle chasing,  \n\n\\[\\angle P_{D}P_{A}P_{B} + \\angle P_{B}P_{C}P_{D} = \\angle P_{D}P_{A}P + \\angle P P_{A}P_{B} + \\angle P_{B}P_{C}P + \\angle P P_{C}P_{D}\\] \\[\\qquad = \\angle P_{D}A P + \\angle P B P_{B} + \\angle P_{B}C P + \\angle P D P_{D}\\] \\[\\qquad = (180^{\\circ} - \\angle A P D) + (180^{\\circ} - \\angle B P C)\\] \\[\\qquad = \\angle B P A + \\angle D P C\\] \\[\\qquad = 180^{\\circ}.\\]  \n\nTherefore, \\(P_{A}P_{B}P_{C}P_{D}\\) is cyclic as desired.  \n\n![md5:566f3a0c1030d06209b8a6b669b29550](566f3a0c1030d06209b8a6b669b29550.jpeg)\n\n\n\n\nLet \\(P_{\\infty}\\) be the point at infinity on line \\(A C\\) . Let \\(Y\\) and \\(Z\\) be the feet of the altitudes from \\(H\\) to \\(A B\\) and \\(A C\\) , respectively. Note that \\(\\angle E H F + \\angle B H P_{\\infty} = 90^{\\circ} + 90^{\\circ} = 180^{\\circ}\\) . Thus, the feet of the altitudes from \\(H\\) to \\(E F\\) , \\(E B\\) , \\(B P_{\\infty}\\) , and \\(C P_{\\infty}\\) are concyclic. In other words, \\(X Y Z B\\) is cyclic. Since \\(B C Y Z\\) is a cyclic quadrilateral, we conclude \\(X\\) lies on this circle, giving us that \\(\\angle B X C = 90^{\\circ}\\) as desired.  \n\nRemark. Here's another way to prove the lemma.  \n\nIt is well known that, with the provided condition, there is a point \\(P^{\\prime}\\) that is the isogonal conjugate of \\(P\\) with respect to quadrilateral \\(A B C D\\) . Let \\(P_{A}\\) , \\(P_{B}\\) , \\(P_{C}\\) , and \\(P_{D}\\) be the feet of the altitudes from \\(P\\) to \\(A B\\) , \\(B C\\) , \\(C D\\) , and \\(D A\\) , respectively, and let \\(Q\\) be the foot of the altitude from \\(P^{\\prime}\\) to \\(A B\\) . Because \\(P\\) and \\(P^{\\prime}\\) are isogonal conjugates with respect to the triangle formed by lines \\(A B\\) , \\(B C\\) , and \\(C D\\) , we have \\(P_{A}P_{B}P_{C}Q\\) is cyclic. Similarly, because \\(P\\) and \\(P^{\\prime}\\) are also isogonal conjugate with respect to the triangle formed by lines \\(D A\\) , \\(A B\\) , and \\(B C\\) , we have \\(P_{D}P_{A}P_{B}Q\\) is cyclic. Consequently, \\(P_{A}P_{B}P_{C}P_{C}\\) is cyclic as desired.", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n5. [35]", "solution_match": "\nSolution 3: "}}
 {"year": "2025", "tier": "T4", "problem_label": "6", "problem_type": null, "exam": "HMMT", "problem": "Complex numbers \\(\\omega_{1}\\) , ..., \\(\\omega_{n}\\) each have magnitude 1. Let \\(z\\) be a complex number distinct from \\(\\omega_{1}\\) , ..., \\(\\omega_{n}\\) such that  \n\n\\[\\frac{z + \\omega_{1}}{z - \\omega_{1}} +\\dots +\\frac{z + \\omega_{n}}{z - \\omega_{n}} = 0.\\]  \n\nProve that \\(|z| = 1\\)", "solution": "Solution 1: We show that no solutions \\(z\\) not on the unit circle can exist. First, we eliminate \\(|z| > 1\\)  \n\nClaim 1. For all \\(j\\) and \\(|z| > 1\\) , the real part of \\(\\frac{z + \\omega_{j}}{z - \\omega_{j}}\\) is positive.  \n\nProof. We use geometry. Note that \\(\\omega_{j}\\) and \\(- \\omega_{j}\\) are antipodes on the unit circle. Since \\(z\\) lies outside the unit circle, it follows that \\(\\angle \\omega_{j}z(- \\omega_{j})\\) is acute. But this means the complex number \\(\\frac{z + \\omega_{j}}{z - \\omega_{j}}\\) lies strictly in the first or fourth quadrant of the complex plane and thus has positive real part, as desired. \\(\\square\\)  \n\nIt is then clear that whenever \\(|z| > 1\\) , the sum \\(\\sum_{j = 1}^{n} \\frac{z + \\omega_{j}}{z - \\omega_{j}}\\) has positive real part and thus cannot be 0. The case where \\(|z| < 1\\) is analogous, except \\(\\angle \\omega_{j}z(- \\omega_{j})\\) is obtuse instead, so \\(\\frac{z + \\omega_{j}}{z - \\omega_{j}}\\) has negative real part for all \\(j\\) . Therefore, all solutions \\(z\\) to the original equation must satisfy \\(|z| = 1\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n6. [40]", "solution_match": "\nProposed by: Karthik Venkata Vedula  \n\n"}}
 {"year": "2025", "tier": "T4", "problem_label": "6", "problem_type": null, "exam": "HMMT", "problem": "Complex numbers \\(\\omega_{1}\\) , ..., \\(\\omega_{n}\\) each have magnitude 1. Let \\(z\\) be a complex number distinct from \\(\\omega_{1}\\) , ..., \\(\\omega_{n}\\) such that  \n\n\\[\\frac{z + \\omega_{1}}{z - \\omega_{1}} +\\dots +\\frac{z + \\omega_{n}}{z - \\omega_{n}} = 0.\\]  \n\nProve that \\(|z| = 1\\)", "solution": "We show more generally that for any positive integers \\(k\\) , \\(a_{1}\\) , ..., \\(a_{k}\\) , and distinct \\(\\omega_{j}\\) on the unit circle, the equation  \n\n\\[\\sum_{j = 1}^{k} a_{j} \\left(\\frac{z + \\omega_{i}}{z - \\omega_{j}}\\right) = 0\\]  \n\nhas \\(k\\) distinct solutions on the unit circle. The original problem then follows upon consolidating duplicate \\(\\omega_{j}\\) 's. Without loss of generality, assume that \\(\\omega_{1}\\) , ..., \\(\\omega_{k}\\) are in this order going clockwise around the unit circle.  \n\nClaim 2. There is a solution on the (clockwise) arc from \\(\\omega_{j}\\) to \\(\\omega_{j + 1}\\) for all \\(j\\) (where \\(\\omega_{k + 1} = \\omega_{1}\\) ).  \n\nProof. First, \\(\\omega_{j}\\) and \\(- \\omega_{j}\\) are antipodes on the unit circle, so if \\(z\\) is on the unit circle, \\(\\angle \\omega_{j}z(- \\omega_{j}) = 90^{\\circ}\\) This means \\(\\frac{z + \\omega_{j}}{z - \\omega_{j}}\\) is purely imaginary. Now consider the imaginary part of the left hand side of the equation, which is a real and continuous function on the arc strictly between \\(\\omega_{j}\\) and \\(\\omega_{j + 1}\\) for each \\(j\\) . In particular, as \\(z\\) approaches \\(\\omega_{j + 1}\\) from the clockwise direction, this function approaches \\(\\infty\\) . On the other hand, as \\(z\\) approaches \\(\\omega_{j}\\) from the counterclockwise direction, this function approaches \\(-\\infty\\) . By the Intermediate Value Theorem, there must be a solution on this arc, as desired. \\(\\square\\)\n\n\n\nIt follows that there are at least \\(k\\) solutions on the unit circle. But the equation is equivalent to a polynomial of degree \\(k\\) . Hence, there are exactly \\(k\\) solutions, all of which lie on the unit circle.  \n\nRemark. The coefficients \\(a_{j}\\) 's are introduced to handle the case where some of \\(\\omega_{1},\\ldots ,\\omega_{n}\\) are equal. Another way to get around this case is to utilize the fact that roots of polynomials are continuous, so we can take the limit where several \\(\\omega_{j}\\) 's approach each other.  \n\nRemark. The Möbius transformation \\(z\\mapsto i\\frac{z - 1}{z + 1}\\) sends the unit circle to a real line. One can rephrase both solutions as working on the real line instead of the unit circle.", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n6. [40]", "solution_match": "\nSolution 2: "}}
 {"year": "2025", "tier": "T4", "problem_label": "7", "problem_type": null, "exam": "HMMT", "problem": "Determine, with proof, whether a square can be dissected into finitely many (not necessarily congruent) triangles, each of which has interior angles \\(30^{\\circ}\\) , \\(75^{\\circ}\\) , and \\(75^{\\circ}\\) .", "solution": "Answer: No", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n7. [45]", "solution_match": "\nProposed by: Derek Liu  \n\n"}}
 {"year": "2025", "tier": "T4", "problem_label": "7", "problem_type": null, "exam": "HMMT", "problem": "Determine, with proof, whether a square can be dissected into finitely many (not necessarily congruent) triangles, each of which has interior angles \\(30^{\\circ}\\) , \\(75^{\\circ}\\) , and \\(75^{\\circ}\\) .", "solution": "Assume for sake of contradiction that such a dissection exists. It has exactly half as many \\(30^{\\circ}\\) angles as \\(75^{\\circ}\\) angles.  \n\nAround any intersection point except the square's vertices, the only angles that can appear are \\(30^{\\circ}\\) , \\(75^{\\circ}\\) , and \\(180^{\\circ}\\) . The only combinations of these that sum to \\(180^{\\circ}\\) or \\(360^{\\circ}\\) are  \n\n\\[6\\cdot 30^{\\circ} = 180^{\\circ},\\] \\[30^{\\circ} + 2\\cdot 75^{\\circ} = 180^{\\circ},\\] \\[180^{\\circ} = 180^{\\circ},\\] \\[12\\cdot 30^{\\circ} = 360^{\\circ},\\] \\[7\\cdot 30^{\\circ} + 2\\cdot 75^{\\circ} = 360^{\\circ},\\] \\[2\\cdot 30^{\\circ} + 4\\cdot 75^{\\circ} = 360^{\\circ},\\] \\[6\\cdot 30^{\\circ} + 180^{\\circ} = 360^{\\circ},\\] \\[30^{\\circ} + 2\\cdot 75^{\\circ} + 180^{\\circ} = 360^{\\circ},\\] \\[180^{\\circ} + 180^{\\circ} = 360^{\\circ}.\\]  \n\nIn particular, around any such point, there are at least half as many \\(30^{\\circ}\\) angles as \\(75^{\\circ}\\) angles.  \n\nHowever, the square's vertices must each be surrounded by three \\(30^{\\circ}\\) angles and zero \\(75^{\\circ}\\) angles, as there is no other way to get a sum of \\(90^{\\circ}\\) . Thus the total number of \\(30^{\\circ}\\) angles in the dissection must be at least 12 more than half the number of \\(75^{\\circ}\\) angles, contradiction.  \n\nThus no such dissection exists.", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n7. [45]", "solution_match": "\nSolution 1: "}}
 {"year": "2025", "tier": "T4", "problem_label": "7", "problem_type": null, "exam": "HMMT", "problem": "Determine, with proof, whether a square can be dissected into finitely many (not necessarily congruent) triangles, each of which has interior angles \\(30^{\\circ}\\) , \\(75^{\\circ}\\) , and \\(75^{\\circ}\\) .", "solution": "Again assume for sake of contradiction that a dissection exists. Interpret the dissection as a graph \\(G\\) , where the vertices of the graph are the vertices of all the triangles, and edges connect each pair of consecutive vertices along a line segment.  \n\nCall a vertex flat if it is on either the boundary of the square (including its corners) or the interior of an edge of any triangle. Let \\(X\\) be the number of flat vertices and \\(Y\\) be the number of non- flat vertices in \\(G\\) . Let \\(E\\) and \\(F\\) be the number of edges and faces (triangles) in the dissection, respectively. Then \\((X + Y) - E + F = 1\\) .  \n\nObserving the angle combinations in the first solution, we see that any non- flat vertex must have at least 6 incident edges, and any flat vertex must have at least 4. Thus \\(2E \\geq 6Y + 4X\\) , so \\(E \\geq 3Y + 2X\\) .  \n\nThe sum of the angles of all \\(F\\) triangles is \\(\\pi F\\) . Around any non- flat vertex, such angles sum to \\(2\\pi\\) . Around any flat vertex, the angles sum to \\(\\pi\\) , with the exception of the four corners of the square, where they sum to \\(\\pi /2\\) instead. Thus  \n\n\\[F\\pi = (X - 4)\\pi +4(\\pi /2) + Y(2\\pi) = (X + 2Y - 2)\\pi ,\\]\n\n\n\nso \\(F = X + 2Y - 2\\) . This means  \n\n\\[X + Y - E + F\\leq (X + Y) - (3Y + 2X) + (X + 2Y - 2) = -2,\\]  \n\ncontradiction. Thus no dissection exists.", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n7. [45]", "solution_match": "\nSolution 2: "}}
-{"year": "2025", "tier": "T4", "problem_label": "8", "problem_type": null, "exam": "HMMT", "problem": "Let \\(\\triangle ABC\\) be a triangle with incenter \\(I\\) . The incircle of triangle \\(\\triangle ABC\\) touches \\(\\overline{BC}\\) at \\(D\\) . Let \\(M\\) be the midpoint of \\(\\overline{BC}\\) , and let line \\(AI\\) meet the circumcircle of triangle \\(\\triangle ABC\\) again at \\(L \\neq A\\) . Let \\(\\omega\\) be the circle centered at \\(L\\) tangent to \\(AB\\) and \\(AC\\) . If \\(\\omega\\) intersects segment \\(\\overline{AD}\\) at point \\(P\\) , prove that \\(\\angle IPM = 90^{\\circ}\\) .", "solution": "Solution 1: Let \\(X\\) and \\(Y\\) be the bottom and top point on \\(\\omega\\) (i.e., the tangents of \\(X\\) and \\(Y\\) to \\(\\omega\\) are parallel to \\(BC\\) , and \\(Y\\) and \\(A\\) lie on the same side of \\(BC\\) ). Note that \\(A\\) , \\(P\\) , \\(D\\) , and \\(X\\) are collinear by homothety between the incircle and \\(\\omega\\) . The key claim is the following.  \n\nClaim 1. Line \\(IY\\) is tangent to \\(\\omega\\) .  \n\nProof. Let the line through \\(I\\) parallel to \\(BC\\) meet \\(AB\\) and \\(AC\\) at \\(B^{\\prime}\\) and \\(C^{\\prime}\\) , respectively. Notice that \\(B^{\\prime}L\\) is the perpendicular bisector of \\(BI\\) , so \\(B^{\\prime}L\\) externally bisects \\(\\angle AB^{\\prime}C^{\\prime}\\) . Similarly, \\(C^{\\prime}L\\) externally bisects \\(\\angle AC^{\\prime}B^{\\prime}\\) . Hence, \\(L\\) is the excenter of \\(\\triangle AB^{\\prime}C^{\\prime}\\) , which means that \\(B^{\\prime}C^{\\prime}\\) is tangent to \\(\\omega\\) . \\(\\square\\)  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 \n\nNow, we note that \\(LY \\perp BC\\) , so \\(L\\) , \\(Y\\) , and \\(M\\) are collinear (on the perpendicular bisector of \\(BC\\) ). Since \\(\\angle YPX = 90^{\\circ}\\) and \\(\\angle YMD = 90^{\\circ}\\) , \\(PDMY\\) is cyclic. However, \\(IYMD\\) is a rectangle, so \\(IPDMY\\) is a cyclic pentagon. Hence, \\(\\angle IPM = \\angle IDM = 90^{\\circ}\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n8. [50]", "solution_match": "\nProposed by: Pitchayut Saengrungkongka  \n\n"}}
-{"year": "2025", "tier": "T4", "problem_label": "8", "problem_type": null, "exam": "HMMT", "problem": "Let \\(\\triangle ABC\\) be a triangle with incenter \\(I\\) . The incircle of triangle \\(\\triangle ABC\\) touches \\(\\overline{BC}\\) at \\(D\\) . Let \\(M\\) be the midpoint of \\(\\overline{BC}\\) , and let line \\(AI\\) meet the circumcircle of triangle \\(\\triangle ABC\\) again at \\(L \\neq A\\) . Let \\(\\omega\\) be the circle centered at \\(L\\) tangent to \\(AB\\) and \\(AC\\) . If \\(\\omega\\) intersects segment \\(\\overline{AD}\\) at point \\(P\\) , prove that \\(\\angle IPM = 90^{\\circ}\\) .", "solution": "Let the incircle touch \\(AC\\) and \\(AB\\) at \\(E\\) and \\(F\\) , respectively. Let \\(DI\\) intersect \\(EF\\) at \\(X\\) . Let \\(D'\\) be the other intersection of \\(AD\\) and the incircle.\n\n\n\nClaim 2. \\(P M \\parallel D^{\\prime}X\\) .  \n\nProof. Consider the homothety at \\(A\\) that sends \\(\\omega\\) to the incircle. It sends \\(L\\) to \\(I\\) and \\(P\\) to \\(D^{\\prime}\\) . Furthermore, it's well- known that \\(X\\) lies on \\(A M\\) . Because \\(I X \\parallel L M\\) , we also have that the homothety sends \\(M\\) to \\(X\\) . These facts imply that \\(D^{\\prime}X \\parallel P M\\) . \\(\\square\\)  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 \n\nLet \\(T\\) be the antipode of \\(D\\) on the incircle. Let \\(A T\\) intersect the incircle again at \\(T^{\\prime}\\) . Since \\(X\\) lies on the polar of \\(A\\) with respect to the incircle, by Brocard's theorem, we have \\(D^{\\prime}\\) , \\(X\\) , and \\(T^{\\prime}\\) are collinear. It is well- known that \\(A T \\parallel I M\\) . Therefore, \\(\\angle D P M = \\angle D D^{\\prime}X = \\angle D D^{\\prime}T^{\\prime} = \\angle D T T^{\\prime} = \\angle D I M\\) . Consequently, \\(I M D P\\) is cyclic, and \\(\\angle I P M = \\angle I D M = 90^{\\circ}\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n8. [50]", "solution_match": "\nSolution 2: "}}
-{"year": "2025", "tier": "T4", "problem_label": "8", "problem_type": null, "exam": "HMMT", "problem": "Let \\(\\triangle ABC\\) be a triangle with incenter \\(I\\) . The incircle of triangle \\(\\triangle ABC\\) touches \\(\\overline{BC}\\) at \\(D\\) . Let \\(M\\) be the midpoint of \\(\\overline{BC}\\) , and let line \\(AI\\) meet the circumcircle of triangle \\(\\triangle ABC\\) again at \\(L \\neq A\\) . Let \\(\\omega\\) be the circle centered at \\(L\\) tangent to \\(AB\\) and \\(AC\\) . If \\(\\omega\\) intersects segment \\(\\overline{AD}\\) at point \\(P\\) , prove that \\(\\angle IPM = 90^{\\circ}\\) .", "solution": "Let \\(\\omega\\) be tangent to \\(A B\\) and \\(A C\\) at \\(E\\) and \\(F\\) , respectively. Note that these are the feet of the altitudes from \\(L\\) to \\(A B\\) and \\(A C\\) , and \\(L\\) lies on the circumcircle of \\(\\triangle A B C\\) by Fact 5. As \\(M\\) is clearly the foot from \\(L\\) to \\(B C\\) , it follows that \\(E\\) , \\(F\\) , and \\(M\\) are collinear on the Simson Line of \\(L\\) with respect to \\(\\triangle A B C\\) .  \n\nLastly, we want \\(P\\) to be on the circle with diameter \\(I M\\) . This circle intersects \\(E F\\) again at the foot from \\(M\\) to \\(A I\\) , which is the midpoint of \\(E F\\) . Let this point be \\(M^{\\prime}\\) . Consider the homothety sending the incircle to \\(\\omega\\) . This clearly sends \\(D\\) to the second intersection of \\(A D\\) and \\(\\omega\\) , which is \\(P^{\\prime}\\) , and it sends \\(I\\) to \\(L\\) . Note that \\(A P\\cdot A P^{\\prime} = A E^{2} = A M\\cdot A L\\) , as the circle with diameter \\(L E\\) is tangent to \\(A E\\) . Thus, \\(P P^{\\prime}M^{\\prime}L\\) is cyclic. Since \\(I D \\parallel L P^{\\prime}\\) , \\(I\\) lies on \\(M^{\\prime}L\\) , and \\(D\\) lies on \\(P P^{\\prime}\\) . By Reim's, we also have \\(P D M^{\\prime}I\\) is cyclic. As \\(I M\\) is a diameter of \\((D M^{\\prime}I)\\) , we have \\(\\angle I P M = 90^{\\circ}\\) .\n\n\n![](data:image/jpeg;base64,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"metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n8. [50]", "solution_match": "\nSolution 3: "}}
+{"year": "2025", "tier": "T4", "problem_label": "8", "problem_type": null, "exam": "HMMT", "problem": "Let \\(\\triangle ABC\\) be a triangle with incenter \\(I\\) . The incircle of triangle \\(\\triangle ABC\\) touches \\(\\overline{BC}\\) at \\(D\\) . Let \\(M\\) be the midpoint of \\(\\overline{BC}\\) , and let line \\(AI\\) meet the circumcircle of triangle \\(\\triangle ABC\\) again at \\(L \\neq A\\) . Let \\(\\omega\\) be the circle centered at \\(L\\) tangent to \\(AB\\) and \\(AC\\) . If \\(\\omega\\) intersects segment \\(\\overline{AD}\\) at point \\(P\\) , prove that \\(\\angle IPM = 90^{\\circ}\\) .", "solution": "Solution 1: Let \\(X\\) and \\(Y\\) be the bottom and top point on \\(\\omega\\) (i.e., the tangents of \\(X\\) and \\(Y\\) to \\(\\omega\\) are parallel to \\(BC\\) , and \\(Y\\) and \\(A\\) lie on the same side of \\(BC\\) ). Note that \\(A\\) , \\(P\\) , \\(D\\) , and \\(X\\) are collinear by homothety between the incircle and \\(\\omega\\) . The key claim is the following.  \n\nClaim 1. Line \\(IY\\) is tangent to \\(\\omega\\) .  \n\nProof. Let the line through \\(I\\) parallel to \\(BC\\) meet \\(AB\\) and \\(AC\\) at \\(B^{\\prime}\\) and \\(C^{\\prime}\\) , respectively. Notice that \\(B^{\\prime}L\\) is the perpendicular bisector of \\(BI\\) , so \\(B^{\\prime}L\\) externally bisects \\(\\angle AB^{\\prime}C^{\\prime}\\) . Similarly, \\(C^{\\prime}L\\) externally bisects \\(\\angle AC^{\\prime}B^{\\prime}\\) . Hence, \\(L\\) is the excenter of \\(\\triangle AB^{\\prime}C^{\\prime}\\) , which means that \\(B^{\\prime}C^{\\prime}\\) is tangent to \\(\\omega\\) . \\(\\square\\)  \n\n![md5:9d25c968fd58984cde61b27542d89c49](9d25c968fd58984cde61b27542d89c49.jpeg)\n  \n\nNow, we note that \\(LY \\perp BC\\) , so \\(L\\) , \\(Y\\) , and \\(M\\) are collinear (on the perpendicular bisector of \\(BC\\) ). Since \\(\\angle YPX = 90^{\\circ}\\) and \\(\\angle YMD = 90^{\\circ}\\) , \\(PDMY\\) is cyclic. However, \\(IYMD\\) is a rectangle, so \\(IPDMY\\) is a cyclic pentagon. Hence, \\(\\angle IPM = \\angle IDM = 90^{\\circ}\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n8. [50]", "solution_match": "\nProposed by: Pitchayut Saengrungkongka  \n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "8", "problem_type": null, "exam": "HMMT", "problem": "Let \\(\\triangle ABC\\) be a triangle with incenter \\(I\\) . The incircle of triangle \\(\\triangle ABC\\) touches \\(\\overline{BC}\\) at \\(D\\) . Let \\(M\\) be the midpoint of \\(\\overline{BC}\\) , and let line \\(AI\\) meet the circumcircle of triangle \\(\\triangle ABC\\) again at \\(L \\neq A\\) . Let \\(\\omega\\) be the circle centered at \\(L\\) tangent to \\(AB\\) and \\(AC\\) . If \\(\\omega\\) intersects segment \\(\\overline{AD}\\) at point \\(P\\) , prove that \\(\\angle IPM = 90^{\\circ}\\) .", "solution": "Let the incircle touch \\(AC\\) and \\(AB\\) at \\(E\\) and \\(F\\) , respectively. Let \\(DI\\) intersect \\(EF\\) at \\(X\\) . Let \\(D'\\) be the other intersection of \\(AD\\) and the incircle.\n\n\n\nClaim 2. \\(P M \\parallel D^{\\prime}X\\) .  \n\nProof. Consider the homothety at \\(A\\) that sends \\(\\omega\\) to the incircle. It sends \\(L\\) to \\(I\\) and \\(P\\) to \\(D^{\\prime}\\) . Furthermore, it's well- known that \\(X\\) lies on \\(A M\\) . Because \\(I X \\parallel L M\\) , we also have that the homothety sends \\(M\\) to \\(X\\) . These facts imply that \\(D^{\\prime}X \\parallel P M\\) . \\(\\square\\)  \n\n![md5:3588a05bc805619ce4de2b7bfb81bb3c](3588a05bc805619ce4de2b7bfb81bb3c.jpeg)\n  \n\nLet \\(T\\) be the antipode of \\(D\\) on the incircle. Let \\(A T\\) intersect the incircle again at \\(T^{\\prime}\\) . Since \\(X\\) lies on the polar of \\(A\\) with respect to the incircle, by Brocard's theorem, we have \\(D^{\\prime}\\) , \\(X\\) , and \\(T^{\\prime}\\) are collinear. It is well- known that \\(A T \\parallel I M\\) . Therefore, \\(\\angle D P M = \\angle D D^{\\prime}X = \\angle D D^{\\prime}T^{\\prime} = \\angle D T T^{\\prime} = \\angle D I M\\) . Consequently, \\(I M D P\\) is cyclic, and \\(\\angle I P M = \\angle I D M = 90^{\\circ}\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n8. [50]", "solution_match": "\nSolution 2: "}}
+{"year": "2025", "tier": "T4", "problem_label": "8", "problem_type": null, "exam": "HMMT", "problem": "Let \\(\\triangle ABC\\) be a triangle with incenter \\(I\\) . The incircle of triangle \\(\\triangle ABC\\) touches \\(\\overline{BC}\\) at \\(D\\) . Let \\(M\\) be the midpoint of \\(\\overline{BC}\\) , and let line \\(AI\\) meet the circumcircle of triangle \\(\\triangle ABC\\) again at \\(L \\neq A\\) . Let \\(\\omega\\) be the circle centered at \\(L\\) tangent to \\(AB\\) and \\(AC\\) . If \\(\\omega\\) intersects segment \\(\\overline{AD}\\) at point \\(P\\) , prove that \\(\\angle IPM = 90^{\\circ}\\) .", "solution": "Let \\(\\omega\\) be tangent to \\(A B\\) and \\(A C\\) at \\(E\\) and \\(F\\) , respectively. Note that these are the feet of the altitudes from \\(L\\) to \\(A B\\) and \\(A C\\) , and \\(L\\) lies on the circumcircle of \\(\\triangle A B C\\) by Fact 5. As \\(M\\) is clearly the foot from \\(L\\) to \\(B C\\) , it follows that \\(E\\) , \\(F\\) , and \\(M\\) are collinear on the Simson Line of \\(L\\) with respect to \\(\\triangle A B C\\) .  \n\nLastly, we want \\(P\\) to be on the circle with diameter \\(I M\\) . This circle intersects \\(E F\\) again at the foot from \\(M\\) to \\(A I\\) , which is the midpoint of \\(E F\\) . Let this point be \\(M^{\\prime}\\) . Consider the homothety sending the incircle to \\(\\omega\\) . This clearly sends \\(D\\) to the second intersection of \\(A D\\) and \\(\\omega\\) , which is \\(P^{\\prime}\\) , and it sends \\(I\\) to \\(L\\) . Note that \\(A P\\cdot A P^{\\prime} = A E^{2} = A M\\cdot A L\\) , as the circle with diameter \\(L E\\) is tangent to \\(A E\\) . Thus, \\(P P^{\\prime}M^{\\prime}L\\) is cyclic. Since \\(I D \\parallel L P^{\\prime}\\) , \\(I\\) lies on \\(M^{\\prime}L\\) , and \\(D\\) lies on \\(P P^{\\prime}\\) . By Reim's, we also have \\(P D M^{\\prime}I\\) is cyclic. As \\(I M\\) is a diameter of \\((D M^{\\prime}I)\\) , we have \\(\\angle I P M = 90^{\\circ}\\) .\n\n\n![md5:bd8de87f9a899904f1b27f0c1c33d28a](bd8de87f9a899904f1b27f0c1c33d28a.jpeg)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n8. [50]", "solution_match": "\nSolution 3: "}}
 {"year": "2025", "tier": "T4", "problem_label": "9", "problem_type": null, "exam": "HMMT", "problem": "Let \\(\\mathbb{Z}\\) be the set of integers. Determine, with proof, all primes \\(p\\) for which there exists a function \\(f\\colon \\mathbb{Z}\\to \\mathbb{Z}\\) such that for any integer \\(x\\) ,  \n\n\\(f(x + p) = f(x)\\) and \\(p\\) divides \\(f(x + f(x)) - x\\) .", "solution": "Answer: \\(\\boxed{p = 5}\\) and all primes \\(p\\equiv \\pm 1\\) (mod 5)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n9. [60]", "solution_match": "\nProposed by: Marin Hristov Hristov  \n\n"}}
 {"year": "2025", "tier": "T4", "problem_label": "9", "problem_type": null, "exam": "HMMT", "problem": "Let \\(\\mathbb{Z}\\) be the set of integers. Determine, with proof, all primes \\(p\\) for which there exists a function \\(f\\colon \\mathbb{Z}\\to \\mathbb{Z}\\) such that for any integer \\(x\\) ,  \n\n\\(f(x + p) = f(x)\\) and \\(p\\) divides \\(f(x + f(x)) - x\\) .", "solution": "We work in \\(\\mathbb{F}_{p}\\) , treating \\(f\\) as a map from \\(\\mathbb{F}_{p}\\) to itself. Clearly, \\(p = 2\\) doesn't work. For \\(p > 2\\) such that 5 is a quadratic residue mod \\(p\\) , as well as \\(p = 5\\) itself, there exists some \\(\\alpha\\) such that \\((2\\alpha +1)^{2}\\equiv 5\\) (mod \\(p\\) ). Taking \\(f(x) = \\alpha x\\) then works because  \n\n\\[f(x + f(x)) - x = (\\alpha^{2} + \\alpha -1)x = \\frac{1}{4}\\left((2\\alpha +1)^{2} - 5\\right)x\\equiv 0\\pmod {p}.\\]  \n\nTo prove no other primes satisfy the conditions in the problem statement, note that \\(f\\) is surjective, as for any \\(x\\) , \\(f(x + f(x)) = x\\) . As \\(\\mathbb{F}_{p}\\) is finite, \\(f\\) is bijective. Plugging in \\(x = f(y)\\) yields  \n\n\\[f(f(y) + f(f(y))) = f(y)\\implies f(y) + f(f(y)) = y.\\]  \n\nSince \\(f\\) is bijective, there exists \\(z\\in \\mathbb{F}_{p}\\) such that \\(f(z) = 0\\) , then \\(z = f(z + f(z)) = f(z) = 0\\) . Therefore, \\(f(0) = 0\\) . This is the only fixed point, as any fixed point \\(d\\) would satisfy \\(d = f(d) + f(f(d)) = 2d\\) ,\n\n\n\nwhich is impossible if \\(d \\neq 0\\) . Hence the remaining residues form nontrivial cycles \\(y\\) , \\(f(y)\\) , \\(f(f(y))\\) , etc. If the cycle containing \\(y\\) is of length \\(n\\) , then  \n\n\\[\\begin{array}{r l} & {f^{-1}(y) = y + f(y),}\\\\ & {f^{-2}(y) = f^{-1}(y) + y = 2y + f(y),}\\\\ & {\\qquad \\vdots}\\\\ & {f^{-(n - 1)}(y) = f(y) = F_{n}y + F_{n - 1}f(y),}\\\\ & {\\qquad y = F_{n + 1}y + F_{n}f(y),} \\end{array} \\quad (by induction)\\]  \n\nwhere \\(F_{k}\\) is the \\(k\\) - th Fibonacci number. As \\(y \\neq 0\\) and \\(f(y) \\neq 0\\) , the last two equations tell us  \n\n\\[F_{n}^{2} \\equiv \\left(\\frac{(1 - F_{n - 1})f(y)}{y}\\right) \\left(\\frac{(1 - F_{n + 1})y}{f(y)}\\right) \\equiv (F_{n + 1} - 1)(F_{n - 1} - 1) \\pmod {p}.\\]  \n\nLet \\(A = F_{n + 1} - 1\\) and \\(B = F_{n - 1} - 1\\) for brevity. The last equation becomes  \n\n\\[(A - B)^{2} \\equiv AB \\equiv \\frac{1}{4} ((A + B)^{2} - (A - B)^{2}) \\pmod {p} \\Longrightarrow (A + B)^{2} \\equiv 5(A - B)^{2} \\pmod {p}.\\]  \n\nAs 5 is not a quadratic residue, this implies \\(F_{n + 1} \\equiv F_{n - 1} \\equiv 1 \\pmod {p}\\) . Hence, if \\(d\\) is the smallest positive integer such that \\(F_{d} \\equiv 0 \\pmod {p}\\) and \\(F_{d + 1} \\equiv 1 \\pmod {p}\\) , then the Fibonacci sequence is periodic modulo \\(p\\) with period \\(d\\) , so \\(d \\mid n\\) . The sum of all cycle lengths (excluding the fixed point 0) is \\(p - 1\\) , so \\(d \\mid p - 1\\) . The following well- known lemma will give us a contradiction.  \n\nLemma 1. If 5 is not a quadratic residue modulo a prime \\(p\\) , then \\(p \\nmid F_{p - 1}\\) .  \n\nProof 1. Recall Binet's formula,  \n\n\\[F_{p - 1} = \\frac{1}{\\sqrt{5}} \\left(\\left(\\frac{1 + \\sqrt{5}}{2}\\right)^{p - 1} - \\left(\\frac{1 - \\sqrt{5}}{2}\\right)^{p - 1}\\right).\\]  \n\nMultiplying both sides by \\(2^{p - 1}\\) and expanding via the binomial theorem, we have  \n\n\\[2^{p - 1}F_{p - 1} = 2\\sum_{k = 0}^{\\frac{p - 3}{2}}5^{k}\\binom{p - 1}{2k + 1}.\\]  \n\nHowever, \\(\\binom{p- 1}{2k+1} \\equiv(- 1)^{2k+1} \\equiv- 1 \\pmod{p}\\) for all \\(k\\) , so  \n\n\\[p \\mid F_{p - 1} \\quad \\text{if and only if} \\quad p \\left| \\sum_{k = 0}^{\\frac{p - 3}{2}} 5^{k} = \\frac{5^{\\frac{p - 1}{2}} - 1}{5 - 1} \\right.\\]  \n\nTherefore \\(p \\mid F_{p - 1}\\) if and only if \\(5^{\\frac{p - 1}{2}} \\equiv 1 \\pmod {p}\\) , which doesn't hold if 5 is not a quadratic residue modulo \\(p\\) , as desired. \\(\\square\\)  \n\nProof 2. Work in \\(\\mathbb{F}_{p^{2}} = \\mathbb{F}_{p}[\\sqrt{5}]\\) . Since 5 is not a quadratic residue, we get that \\((\\sqrt{5})^{p} = - \\sqrt{5}\\) . Using the fact that \\((a + b)^{p} = a^{p} + b^{p}\\) (because all other terms have coefficient divisible by \\(p\\) ), we get that  \n\n\\[\\left(\\frac{1 + \\sqrt{5}}{2}\\right)^{p} = \\frac{1 - \\sqrt{5}}{2} \\implies \\left(\\frac{1 + \\sqrt{5}}{2}\\right)^{p - 1} = \\left(\\frac{1 - \\sqrt{5}}{2}\\right)^{2} = \\frac{3 - \\sqrt{5}}{2}.\\]\n\n\n\nSimilarly, \\(\\left(\\frac{1 - \\sqrt{5}}{2}\\right)^{p - 1} = \\frac{3 + \\sqrt{5}}{2}\\) . Hence, by Binet's formula,  \n\n\\[F_{p - 1} = \\frac{1}{\\sqrt{5}}\\left(\\left(\\frac{1 + \\sqrt{5}}{2}\\right)^{p - 1} - \\left(\\frac{1 - \\sqrt{5}}{2}\\right)^{p - 1}\\right)\\] \\[= \\frac{1}{\\sqrt{5}}\\left(\\frac{3 - \\sqrt{5}}{2} -\\frac{3 + \\sqrt{5}}{2}\\right) = -1,\\]  \n\nso it is not divisible by \\(p\\) .  \n\nHence, \\(F_{p - 1} \\neq 0\\) (mod \\(p\\) ) if 5 is not a quadratic residue modulo \\(p\\) , which contradicts \\(d \\mid p - 1\\) above. This completes the solution.", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n9. [60]", "solution_match": "\nSolution: "}}
 {"year": "2025", "tier": "T4", "problem_label": "10", "problem_type": null, "exam": "HMMT", "problem": "Determine, with proof, all possible values of \\(\\gcd (a^{2} + b^{2} + c^{2}, abc)\\) across all triples of positive integers \\((a, b, c)\\) .", "solution": "Answer: All positive integers \\(n\\) such that \\(\\nu_{p}(n) \\neq 1\\) for all prime \\(p \\equiv 3\\) (mod 4)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-team-solutions.jsonl", "problem_match": "\n10. [60]", "solution_match": "\nProposed by: Henrik Rabinovitz  \n\n"}}