diff --git "a/HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl" "b/HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl"
--- "a/HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl"
+++ "b/HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl"
@@ -5,19 +5,19 @@
 {"year": "2025", "tier": "T4", "problem_label": "3", "problem_type": null, "exam": "HMMT", "problem": "Jacob rolls two fair six-sided dice. If the outcomes of these dice rolls are the same, he rolls a third fair six-sided die. Compute the probability that the sum of outcomes of all the dice he rolls is even.", "solution": "Answer: \\(\\frac{5}{12}\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n3. [5]", "solution_match": "\nProposed by: Rishabh Das  \n\n"}}
 {"year": "2025", "tier": "T4", "problem_label": "3", "problem_type": null, "exam": "HMMT", "problem": "Jacob rolls two fair six-sided dice. If the outcomes of these dice rolls are the same, he rolls a third fair six-sided die. Compute the probability that the sum of outcomes of all the dice he rolls is even.", "solution": "There's a \\(\\frac{1}{2} - \\frac{1}{6} = \\frac{1}{3}\\) probability that he rolls an even number without getting doubles: whatever the first roll is, there is a \\(\\frac{1}{2}\\) chance that the second roll is of opposite parity, and we subtract the \\(\\frac{1}{6}\\) chance that the second roll is the same.  \n\nThere's a \\(\\frac{1}{6} \\cdot \\frac{1}{2} = \\frac{1}{12}\\) probability that he gets doubles and then rolls an even number.  \n\nSumming \\(\\frac{1}{3}\\) and \\(\\frac{1}{12}\\) gets us \\(\\left[\\frac{5}{12}\\right]\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n3. [5]", "solution_match": "\nSolution: "}}
 {"year": "2025", "tier": "T4", "problem_label": "4", "problem_type": null, "exam": "HMMT", "problem": "Let \\(\\triangle ABC\\) be an equilateral triangle with side length 4. Across all points \\(P\\) inside triangle \\(\\triangle ABC\\) satisfying \\([PAB] + [PAC] = [PBC]\\) , compute the minimum possible length of \\(PA\\) .  \n\n(Here, \\([XYZ]\\) denotes the area of triangle \\(\\triangle XYZ\\) .)", "solution": "Answer: \\(\\sqrt{3}\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n4. [5]", "solution_match": "\nProposed by: Isabella Zhu  \n\n"}}
-{"year": "2025", "tier": "T4", "problem_label": "4", "problem_type": null, "exam": "HMMT", "problem": "Let \\(\\triangle ABC\\) be an equilateral triangle with side length 4. Across all points \\(P\\) inside triangle \\(\\triangle ABC\\) satisfying \\([PAB] + [PAC] = [PBC]\\) , compute the minimum possible length of \\(PA\\) .  \n\n(Here, \\([XYZ]\\) denotes the area of triangle \\(\\triangle XYZ\\) .)", "solution": 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 \n\nThe area condition implies \\([ABC] = 2[PBC]\\) . Hence, \\(P\\) lies on the \\(A\\) - midline of \\(\\triangle ABC\\) . Therefore, the minimum possible value of \\(PA\\) is the distance from \\(A\\) to this midline. This is achieved by taking \\(P\\) to be the foot of the perpendicular from \\(A\\) to the \\(A\\) - midline. This distance is half the altitude of \\(ABC\\) , which has side length 4, so the answer is \\(\\frac{1}{2} (2\\sqrt{3}) = \\boxed{\\sqrt{3}}\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n4. [5]", "solution_match": "\nSolution:\n\n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "4", "problem_type": null, "exam": "HMMT", "problem": "Let \\(\\triangle ABC\\) be an equilateral triangle with side length 4. Across all points \\(P\\) inside triangle \\(\\triangle ABC\\) satisfying \\([PAB] + [PAC] = [PBC]\\) , compute the minimum possible length of \\(PA\\) .  \n\n(Here, \\([XYZ]\\) denotes the area of triangle \\(\\triangle XYZ\\) .)", "solution": "![md5:69fadc0b114370291d5cfb809590a975](69fadc0b114370291d5cfb809590a975.jpeg)\n  \n\nThe area condition implies \\([ABC] = 2[PBC]\\) . Hence, \\(P\\) lies on the \\(A\\) - midline of \\(\\triangle ABC\\) . Therefore, the minimum possible value of \\(PA\\) is the distance from \\(A\\) to this midline. This is achieved by taking \\(P\\) to be the foot of the perpendicular from \\(A\\) to the \\(A\\) - midline. This distance is half the altitude of \\(ABC\\) , which has side length 4, so the answer is \\(\\frac{1}{2} (2\\sqrt{3}) = \\boxed{\\sqrt{3}}\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n4. [5]", "solution_match": "\nSolution:\n\n\n"}}
 {"year": "2025", "tier": "T4", "problem_label": "5", "problem_type": null, "exam": "HMMT", "problem": "Compute the largest possible radius of a circle contained in the region defined by \\(|x + |y||\\leq 1\\) in the coordinate plane.", "solution": "Answer: \\(\\boxed {2\\sqrt{2} - 2 = 2(\\sqrt{2} - 1)}\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n5. [6]", "solution_match": "\nProposed by: Rishabh Das  \n\n"}}
-{"year": "2025", "tier": "T4", "problem_label": "5", "problem_type": null, "exam": "HMMT", "problem": "Compute the largest possible radius of a circle contained in the region defined by \\(|x + |y||\\leq 1\\) in the coordinate plane.", "solution": 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 \n\nAfter drawing the graph, it's clear that the circle should pass through \\((- 1,0)\\) and be tangent to \\(y = x - 1\\) and \\(y = - x + 1\\) . Letting the radius of this circle be \\(r\\) , we have \\(r\\sqrt{2} +r = 2\\) , so \\(\\boxed {r = 2\\sqrt{2} - 2}\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n5. [6]", "solution_match": "\nSolution:  \n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "5", "problem_type": null, "exam": "HMMT", "problem": "Compute the largest possible radius of a circle contained in the region defined by \\(|x + |y||\\leq 1\\) in the coordinate plane.", "solution": "![md5:2694b339b7413b14bf3ffd174cfea391](2694b339b7413b14bf3ffd174cfea391.jpeg)\n  \n\nAfter drawing the graph, it's clear that the circle should pass through \\((- 1,0)\\) and be tangent to \\(y = x - 1\\) and \\(y = - x + 1\\) . Letting the radius of this circle be \\(r\\) , we have \\(r\\sqrt{2} +r = 2\\) , so \\(\\boxed {r = 2\\sqrt{2} - 2}\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n5. [6]", "solution_match": "\nSolution:  \n\n"}}
 {"year": "2025", "tier": "T4", "problem_label": "6", "problem_type": null, "exam": "HMMT", "problem": "Let \\(\\triangle ABC\\) be an equilateral triangle. Point \\(D\\) lies on segment \\(\\overline{BC}\\) such that \\(BD = 1\\) and \\(DC = 4\\) . Points \\(E\\) and \\(F\\) lie on rays \\(\\overrightarrow{AC}\\) and \\(\\overrightarrow{AB}\\) , respectively, such that \\(D\\) is the midpoint of \\(\\overline{EF}\\) . Compute \\(EF\\) .", "solution": "Answer: \\(\\boxed {2\\sqrt {13}}\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n6. [6]", "solution_match": "\nProposed by: Pitchayut Saengrungkonkka\n\n\n\n"}}
-{"year": "2025", "tier": "T4", "problem_label": "6", "problem_type": null, "exam": "HMMT", "problem": "Let \\(\\triangle ABC\\) be an equilateral triangle. Point \\(D\\) lies on segment \\(\\overline{BC}\\) such that \\(BD = 1\\) and \\(DC = 4\\) . Points \\(E\\) and \\(F\\) lie on rays \\(\\overrightarrow{AC}\\) and \\(\\overrightarrow{AB}\\) , respectively, such that \\(D\\) is the midpoint of \\(\\overline{EF}\\) . Compute \\(EF\\) .", "solution": 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 \n\nLet \\(C^{\\prime}\\) be the reflection of \\(C\\) over \\(D\\) . Then, \\(\\overline{{E C}}\\parallel \\overline{{C^{\\prime}F}}\\) since \\(E C F C^{\\prime}\\) is a parallelogram. Thus, \\(B F C^{\\prime}\\) is an equilateral triangle, so \\(B F = B C^{\\prime} = 3\\) and \\(\\angle F B D = 120^{\\circ}\\) . By Law of Cosines, we get \\(D F = \\sqrt{3^{2} + 3\\cdot1 + 1^{2}} = \\sqrt{13}\\) and \\(E F = \\boxed {2\\sqrt {13}}\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n6. [6]", "solution_match": "\nSolution:  \n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "6", "problem_type": null, "exam": "HMMT", "problem": "Let \\(\\triangle ABC\\) be an equilateral triangle. Point \\(D\\) lies on segment \\(\\overline{BC}\\) such that \\(BD = 1\\) and \\(DC = 4\\) . Points \\(E\\) and \\(F\\) lie on rays \\(\\overrightarrow{AC}\\) and \\(\\overrightarrow{AB}\\) , respectively, such that \\(D\\) is the midpoint of \\(\\overline{EF}\\) . Compute \\(EF\\) .", "solution": "![md5:932a7555b8a60afcb53d5e3fcf9e922c](932a7555b8a60afcb53d5e3fcf9e922c.jpeg)\n  \n\nLet \\(C^{\\prime}\\) be the reflection of \\(C\\) over \\(D\\) . Then, \\(\\overline{{E C}}\\parallel \\overline{{C^{\\prime}F}}\\) since \\(E C F C^{\\prime}\\) is a parallelogram. Thus, \\(B F C^{\\prime}\\) is an equilateral triangle, so \\(B F = B C^{\\prime} = 3\\) and \\(\\angle F B D = 120^{\\circ}\\) . By Law of Cosines, we get \\(D F = \\sqrt{3^{2} + 3\\cdot1 + 1^{2}} = \\sqrt{13}\\) and \\(E F = \\boxed {2\\sqrt {13}}\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n6. [6]", "solution_match": "\nSolution:  \n\n"}}
 {"year": "2025", "tier": "T4", "problem_label": "7", "problem_type": null, "exam": "HMMT", "problem": "The number  \n\n\\[\\frac{9^{9} - 8^{8}}{1001}\\]  \n\nis an integer. Compute the sum of its prime factors.", "solution": "Answer: 231", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n7. [6]", "solution_match": "\nProposed by: Derek Liu  \n\n"}}
 {"year": "2025", "tier": "T4", "problem_label": "7", "problem_type": null, "exam": "HMMT", "problem": "The number  \n\n\\[\\frac{9^{9} - 8^{8}}{1001}\\]  \n\nis an integer. Compute the sum of its prime factors.", "solution": "Observe  \n\n\\[{9^{9}-8^{8}=27^{6}-16^{6}}\\] \\[{=(27^{2}-16^{2})(27^{4}+27^{2}\\cdot16^{2}+16^{4})}\\] \\[{=(27-16)(27+16)(27^{2}-27\\cdot16+16^{2})(27^{2}+27\\cdot16+16^{2})}\\] \\[{=11\\cdot43\\cdot553\\cdot1417.}\\]  \n\nThe remaining factorizations are motivated by the fact that \\(1001 = 7\\cdot 11\\cdot 13\\) . We see that \\(553 = 7\\cdot 79\\) and \\(1417 = 13\\cdot 109\\) , so the answer is \\(43 + 79 + 109 = \\boxed {231}\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n7. [6]", "solution_match": "\nSolution: "}}
 {"year": "2025", "tier": "T4", "problem_label": "8", "problem_type": null, "exam": "HMMT", "problem": "A checkerboard is a rectangular grid of cells colored black and white such that the top-left corner is black and no two cells of the same color share an edge. Two checkerboards are distinct if and only if they have a different number of rows or columns. For example, a \\(20 \\times 25\\) checkerboard and a \\(25 \\times 20\\) checkerboard are considered distinct.  \n\nCompute the number of distinct checkerboards that have exactly 41 black cells.", "solution": "Answer: 9", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n8. [6]", "solution_match": "\nProposed by: Albert Wang  \n\n"}}
 {"year": "2025", "tier": "T4", "problem_label": "8", "problem_type": null, "exam": "HMMT", "problem": "A checkerboard is a rectangular grid of cells colored black and white such that the top-left corner is black and no two cells of the same color share an edge. Two checkerboards are distinct if and only if they have a different number of rows or columns. For example, a \\(20 \\times 25\\) checkerboard and a \\(25 \\times 20\\) checkerboard are considered distinct.  \n\nCompute the number of distinct checkerboards that have exactly 41 black cells.", "solution": "Since there is a black corner on the checkerboard, the number of white squares is at most the number of black squares. So, the board either has 40 or 41 white squares. Therefore, we want to compute the number of ordered pairs \\((r,c)\\) with a product of 81 or 82. Since \\(81 = 3^{4}\\) has 5 divisors and \\(82 = 41\\cdot 2\\) has 4 divisors, there are \\(\\boxed{9}\\) checkerboards with exactly 41 black cells.", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n8. [6]", "solution_match": "\nSolution: "}}
 {"year": "2025", "tier": "T4", "problem_label": "9", "problem_type": null, "exam": "HMMT", "problem": "Let \\(P\\) and \\(Q\\) be points selected uniformly and independently at random inside a regular hexagon \\(ABCDEF\\) . Compute the probability that segment \\(\\overline{PQ}\\) is entirely contained in at least one of the quadrilaterals \\(ABCD\\) , \\(BCDE\\) , \\(CDEF\\) , \\(DEFA\\) , \\(EFAB\\) , or \\(FABC\\) .", "solution": "Answer: \\(\\boxed{5}\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n9. [7]", "solution_match": "\nProposed by: Isabella Zhu  \n\n"}}
-{"year": "2025", "tier": "T4", "problem_label": "9", "problem_type": null, "exam": "HMMT", "problem": "Let \\(P\\) and \\(Q\\) be points selected uniformly and independently at random inside a regular hexagon \\(ABCDEF\\) . Compute the probability that segment \\(\\overline{PQ}\\) is entirely contained in at least one of the quadrilaterals \\(ABCD\\) , \\(BCDE\\) , \\(CDEF\\) , \\(DEFA\\) , \\(EFAB\\) , or \\(FABC\\) .", "solution": 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 \n\nLet \\(O\\) be the center of the hexagon. Without loss of generality, assume \\(P\\) is in \\(\\triangle ABO\\) . Then, segment \\(PQ\\) is entirely contained in one of the given quadrilaterals if and only if \\(Q\\) is not in \\(\\triangle DEO\\) . The probability that \\(Q\\) is in \\(\\triangle DEO\\) is \\(\\frac{|DEO|}{|ABCDEF|} = \\frac{1}{6}\\) , so the answer is \\(\\boxed{5}\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n9. [7]", "solution_match": "\nSolution:  \n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "9", "problem_type": null, "exam": "HMMT", "problem": "Let \\(P\\) and \\(Q\\) be points selected uniformly and independently at random inside a regular hexagon \\(ABCDEF\\) . Compute the probability that segment \\(\\overline{PQ}\\) is entirely contained in at least one of the quadrilaterals \\(ABCD\\) , \\(BCDE\\) , \\(CDEF\\) , \\(DEFA\\) , \\(EFAB\\) , or \\(FABC\\) .", "solution": "![md5:0d46ba45e4b43766265ed8aab827c8da](0d46ba45e4b43766265ed8aab827c8da.jpeg)\n  \n\nLet \\(O\\) be the center of the hexagon. Without loss of generality, assume \\(P\\) is in \\(\\triangle ABO\\) . Then, segment \\(PQ\\) is entirely contained in one of the given quadrilaterals if and only if \\(Q\\) is not in \\(\\triangle DEO\\) . The probability that \\(Q\\) is in \\(\\triangle DEO\\) is \\(\\frac{|DEO|}{|ABCDEF|} = \\frac{1}{6}\\) , so the answer is \\(\\boxed{5}\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n9. [7]", "solution_match": "\nSolution:  \n\n"}}
 {"year": "2025", "tier": "T4", "problem_label": "10", "problem_type": null, "exam": "HMMT", "problem": "A square of side length 1 is dissected into two congruent pentagons. Compute the least upper bound of the perimeter of one of these pentagons.", "solution": "Answer: \\(\\boxed{2 + 3\\sqrt{2}}\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n10. [7]", "solution_match": "\nProposed by: Isabella Zhu  \n\n"}}
-{"year": "2025", "tier": "T4", "problem_label": "10", "problem_type": null, "exam": "HMMT", "problem": "A square of side length 1 is dissected into two congruent pentagons. Compute the least upper bound of the perimeter of one of these pentagons.", "solution": "![](data:image/jpeg;base64,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 \n\nLet \\(P_{1}\\) and \\(P_{2}\\) be the two congruent pentagons. Let \\(p(P)\\) denote the perimeter of polygon \\(P\\) .\n\n\n\nWe give an upper bound for \\(p(P_{1}) + p(P_{2})\\) . Note that since a square has four sides, at least four sides of \\(P_{1}\\) and \\(P_{2}\\) combined lie on the sides of the square. These sides have total length at most 4, the perimeter of \\(ABCD\\) .  \n\nEach of the remaining sides has length at most \\(\\sqrt{2}\\) , since the longest possible length of a segment inside \\(ABCD\\) is \\(\\sqrt{2}\\) . There are at most 6 remaining sides, so  \n\n\\[p(P_{1}) + p(P_{2})\\leq 4 + 6\\sqrt{2}.\\]  \n\nSince \\(P_{1}\\) and \\(P_{2}\\) are congruent, this implies  \n\n\\[p(P_{1}) = p(P_{2})\\leq \\boxed {2 + 3\\sqrt{2}}.\\]  \n\nThis least upper bound can be achieved by placing \\(X\\) close to \\(C\\) and \\(Y\\) close to \\(A\\) , as seen in the diagram.", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n10. [7]", "solution_match": "\nSolution:  \n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "10", "problem_type": null, "exam": "HMMT", "problem": "A square of side length 1 is dissected into two congruent pentagons. Compute the least upper bound of the perimeter of one of these pentagons.", "solution": "![md5:a779607648888b3176dd8192def6d28b](a779607648888b3176dd8192def6d28b.jpeg)\n  \n\nLet \\(P_{1}\\) and \\(P_{2}\\) be the two congruent pentagons. Let \\(p(P)\\) denote the perimeter of polygon \\(P\\) .\n\n\n\nWe give an upper bound for \\(p(P_{1}) + p(P_{2})\\) . Note that since a square has four sides, at least four sides of \\(P_{1}\\) and \\(P_{2}\\) combined lie on the sides of the square. These sides have total length at most 4, the perimeter of \\(ABCD\\) .  \n\nEach of the remaining sides has length at most \\(\\sqrt{2}\\) , since the longest possible length of a segment inside \\(ABCD\\) is \\(\\sqrt{2}\\) . There are at most 6 remaining sides, so  \n\n\\[p(P_{1}) + p(P_{2})\\leq 4 + 6\\sqrt{2}.\\]  \n\nSince \\(P_{1}\\) and \\(P_{2}\\) are congruent, this implies  \n\n\\[p(P_{1}) = p(P_{2})\\leq \\boxed {2 + 3\\sqrt{2}}.\\]  \n\nThis least upper bound can be achieved by placing \\(X\\) close to \\(C\\) and \\(Y\\) close to \\(A\\) , as seen in the diagram.", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n10. [7]", "solution_match": "\nSolution:  \n\n"}}
 {"year": "2025", "tier": "T4", "problem_label": "11", "problem_type": null, "exam": "HMMT", "problem": "Let \\(f(n) = n^{2} + 100\\) . Compute the remainder when \\(f(f(\\dots f(f(1))\\dots))\\) is divided by \\(10^{4}\\) .", "solution": "Answer: 3101", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n11. [7]", "solution_match": "\nProposed by: Pitchayut Saengrungkonkha  \n\n"}}
 {"year": "2025", "tier": "T4", "problem_label": "11", "problem_type": null, "exam": "HMMT", "problem": "Let \\(f(n) = n^{2} + 100\\) . Compute the remainder when \\(f(f(\\dots f(f(1))\\dots))\\) is divided by \\(10^{4}\\) .", "solution": "We claim that \\(f^{k}(n)\\equiv 1 + 100(2^{n} - 1)\\bmod 10^{4}\\) . We can see this by induction, as  \n\n\\[f(1 + 100(2^{n} - 1)) = (1 + 100(2^{n} - 1))^{2} + 100\\] \\[\\qquad \\equiv 1 + 200(2^{n} - 1) + 100\\pmod {10^{4}}\\] \\[\\qquad \\equiv 1 + 100(2^{n + 1} - 1)\\pmod {10^{4}}.\\]  \n\nThus, it suffices to compute \\(2^{2025}\\bmod 100\\) . We note that \\(2^{2025}\\equiv 0\\) (mod 4) and by Euler's Totient theorem, \\(2^{2025}\\equiv 2^{5}\\) (mod 25), so \\(2^{2025}\\equiv 32\\) (mod 100). Hence, we can compute  \n\n\\[f^{2025}(1)\\equiv 1 + 100(31)\\equiv \\boxed {3101}\\bmod 10^{4}.\\]", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n11. [7]", "solution_match": "\nSolution: "}}
 {"year": "2025", "tier": "T4", "problem_label": "12", "problem_type": null, "exam": "HMMT", "problem": "Holden has a collection of polygons. He writes down a list containing the measure of each interior angle of each of his polygons. He writes down the list \\(30^{\\circ}\\) , \\(50^{\\circ}\\) , \\(60^{\\circ}\\) , \\(70^{\\circ}\\) , \\(90^{\\circ}\\) , \\(100^{\\circ}\\) , \\(120^{\\circ}\\) , \\(160^{\\circ}\\) , and \\(x^{\\circ}\\) , in some order. Compute \\(x\\) .", "solution": "Answer: 220", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n12. [7]", "solution_match": "\nProposed by: Rishabh Das  \n\n"}}
@@ -25,10 +25,10 @@
 {"year": "2025", "tier": "T4", "problem_label": "13", "problem_type": null, "exam": "HMMT", "problem": "A number is upwards if its digits in base 10 are nondecreasing when read from left to right. Compute the number of positive integers less than \\(10^{6}\\) that are both upwards and multiples of 11.", "solution": "Answer: 219", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n13. [9]", "solution_match": "\nProposed by: Srinivas Arun  \n\n"}}
 {"year": "2025", "tier": "T4", "problem_label": "13", "problem_type": null, "exam": "HMMT", "problem": "A number is upwards if its digits in base 10 are nondecreasing when read from left to right. Compute the number of positive integers less than \\(10^{6}\\) that are both upwards and multiples of 11.", "solution": "For a number \\(d_{5}d_{4}d_{3}d_{2}d_{1}d_{0}\\) (allowing leading 0s) to be upwards and a multiple of 11, we must have  \n\n\\[d_{5}\\leq d_{4}\\leq d_{3}\\leq d_{2}\\leq d_{1}\\leq d_{0},\\]  \n\n\\[d_{0} - d_{1} + d_{2} - d_{3} + d_{4} - d_{5}\\equiv 0\\pmod {11}.\\]  \n\nNote that \\(d_{0} - d_{1}\\) , \\(d_{2} - d_{3}\\) , and \\(d_{4} - d_{5}\\) are all nonnegative. Thus,  \n\n\\[0\\leq (d_{0} - d_{1}) + (d_{2} - d_{3}) + (d_{4} - d_{5})\\] \\[\\leq (d_{0} - d_{1}) + (d_{1} - d_{2}) + (d_{2} - d_{3}) + (d_{3} - d_{4}) + (d_{4} - d_{5})\\] \\[= d_{0} - d_{5}\\] \\[\\leq 9.\\]  \n\nTherefore,  \n\n\\[(d_{0} - d_{1}) + (d_{2} - d_{3}) + (d_{4} - d_{5}) = 0,\\]  \n\nwhich can only occur when \\(d_{0} = d_{1}\\) , \\(d_{2} = d_{3}\\) , and \\(d_{4} = d_{5}\\) , i.e. the number is of the form \\(aabccc\\) . We can easily verify that all numbers of the form \\(aabccc\\) for digits \\(a \\leq b \\leq c\\) satisfy our conditions, so we simply have to count them.  \n\nThere are \\(\\binom{12}{3} = 220\\) such triples of digits \\((a,b,c)\\) . However, one of these triples is \\((0,0,0)\\) , which corresponds to the number 0. Thus our answer is \\(220 - 1 = \\left\\lfloor \\frac{219}{2}\\right\\rfloor\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n13. [9]", "solution_match": "\nSolution: "}}
 {"year": "2025", "tier": "T4", "problem_label": "14", "problem_type": null, "exam": "HMMT", "problem": "A parallelogram \\(P\\) can be folded over a straight line so that the resulting shape is a regular pentagon with side length 1. Compute the perimeter of \\(P\\) .", "solution": "Answer: \\(\\boxed{5 + \\sqrt{5}}\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n14. [9]", "solution_match": "\nProposed by: Arul Kolla  \n\n"}}
-{"year": "2025", "tier": "T4", "problem_label": "14", "problem_type": null, "exam": "HMMT", "problem": "A parallelogram \\(P\\) can be folded over a straight line so that the resulting shape is a regular pentagon with side length 1. Compute the perimeter of \\(P\\) .", "solution": 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 \n\nIn regular pentagon \\(ABCDE\\) (labeled clockwise), reflect \\(ABDE\\) across \\(AB\\) to obtain \\(ABD'E'\\) . Then, \\(CDE'D'\\) is one such parallelogram \\(P\\) . The length of \\(CD'\\) is  \n\n\\[CB + BD = 1 + 2\\cos \\angle CBD = 1 + 2\\cos (\\pi /5) = 1 + \\frac{\\sqrt{5} + 1}{2} = \\frac{\\sqrt{5} + 3}{2}.\\]  \n\nHence, the perimeter of the desired parallelogram is \\(2\\left(1 + \\frac{\\sqrt{5} + 3}{2}\\right) = \\left\\lfloor \\frac{5 + \\sqrt{5}}{2}\\right\\rfloor\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n14. [9]", "solution_match": "\nSolution:  \n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "14", "problem_type": null, "exam": "HMMT", "problem": "A parallelogram \\(P\\) can be folded over a straight line so that the resulting shape is a regular pentagon with side length 1. Compute the perimeter of \\(P\\) .", "solution": "![md5:49af5caed6e82e485305f9f5f5390039](49af5caed6e82e485305f9f5f5390039.jpeg)\n  \n\nIn regular pentagon \\(ABCDE\\) (labeled clockwise), reflect \\(ABDE\\) across \\(AB\\) to obtain \\(ABD'E'\\) . Then, \\(CDE'D'\\) is one such parallelogram \\(P\\) . The length of \\(CD'\\) is  \n\n\\[CB + BD = 1 + 2\\cos \\angle CBD = 1 + 2\\cos (\\pi /5) = 1 + \\frac{\\sqrt{5} + 1}{2} = \\frac{\\sqrt{5} + 3}{2}.\\]  \n\nHence, the perimeter of the desired parallelogram is \\(2\\left(1 + \\frac{\\sqrt{5} + 3}{2}\\right) = \\left\\lfloor \\frac{5 + \\sqrt{5}}{2}\\right\\rfloor\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n14. [9]", "solution_match": "\nSolution:  \n\n"}}
 {"year": "2025", "tier": "T4", "problem_label": "15", "problem_type": null, "exam": "HMMT", "problem": "Right triangle \\(\\triangle DEF\\) with \\(\\angle D = 90^{\\circ}\\) and \\(\\angle F = 30^{\\circ}\\) is inscribed in equilateral triangle \\(\\triangle ABC\\) such that \\(D\\) , \\(E\\) , and \\(F\\) lie on segments \\(\\overline{BC}\\) , \\(\\overline{CA}\\) , and \\(\\overline{AB}\\) , respectively. Given that \\(BD = 7\\) and \\(DC = 4\\) , compute \\(DE\\) .", "solution": "Answer: \\(\\sqrt{13}\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n15. [9]", "solution_match": "\nProposed by: Pitchayut Saengrungkonka  \n\n"}}
-{"year": "2025", "tier": "T4", "problem_label": "15", "problem_type": null, "exam": "HMMT", "problem": "Right triangle \\(\\triangle DEF\\) with \\(\\angle D = 90^{\\circ}\\) and \\(\\angle F = 30^{\\circ}\\) is inscribed in equilateral triangle \\(\\triangle ABC\\) such that \\(D\\) , \\(E\\) , and \\(F\\) lie on segments \\(\\overline{BC}\\) , \\(\\overline{CA}\\) , and \\(\\overline{AB}\\) , respectively. Given that \\(BD = 7\\) and \\(DC = 4\\) , compute \\(DE\\) .", "solution": 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 \n\nFrom \\(\\angle E = 60^{\\circ}\\) , we get that \\(\\angle AEF = 120^{\\circ} - \\angle CED = \\angle CDE\\) . Therefore, \\(\\triangle AEF \\sim \\triangle CDE\\) . Since \\(EF:DE = 2:1\\) , the ratio of similarity must be \\(2:1\\) , so \\(AE = 2CD = 8\\) . Recall \\(ABC\\) has side length \\(7 + 4 = 11\\) , so \\(EC = 11 - 8 = 3\\) . Law of Cosines on \\(\\triangle CDE\\) gives \\(DE^2 = \\sqrt{3^2 + 4^2 - 3\\cdot 4} = \\sqrt{13}\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n15. [9]", "solution_match": "\nSolution 1:  \n\n"}}
-{"year": "2025", "tier": "T4", "problem_label": "15", "problem_type": null, "exam": "HMMT", "problem": "Right triangle \\(\\triangle DEF\\) with \\(\\angle D = 90^{\\circ}\\) and \\(\\angle F = 30^{\\circ}\\) is inscribed in equilateral triangle \\(\\triangle ABC\\) such that \\(D\\) , \\(E\\) , and \\(F\\) lie on segments \\(\\overline{BC}\\) , \\(\\overline{CA}\\) , and \\(\\overline{AB}\\) , respectively. Given that \\(BD = 7\\) and \\(DC = 4\\) , compute \\(DE\\) .", "solution": 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)\n  \n\nLet \\(\\odot (DEF)\\) meet \\(AC\\) again at point \\(X\\) . Then, \\(\\angle FXA = 180^{\\circ} - \\angle FXE = \\angle FDE = 90^{\\circ}\\) and \\(\\angle XDC = 180^{\\circ} - \\angle DCX - \\angle DXC = 120^{\\circ} - \\angle DXE = 120^{\\circ} - \\angle DFE = 90^{\\circ}\\) . It follows that \\(CX = 2CD = 8\\) , so \\(AX = 11 - CX = 3\\) , and \\(AF = 2AX = 6\\) . Thus, Law of Cosines on \\(\\triangle AEF\\) gives \\(EF = \\sqrt{8^2 + 6^2 - 8\\cdot 6} = 2\\sqrt{13}\\) , implying that \\(DE = \\sqrt{13}\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n15. [9]", "solution_match": "\nSolution 2:  \n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "15", "problem_type": null, "exam": "HMMT", "problem": "Right triangle \\(\\triangle DEF\\) with \\(\\angle D = 90^{\\circ}\\) and \\(\\angle F = 30^{\\circ}\\) is inscribed in equilateral triangle \\(\\triangle ABC\\) such that \\(D\\) , \\(E\\) , and \\(F\\) lie on segments \\(\\overline{BC}\\) , \\(\\overline{CA}\\) , and \\(\\overline{AB}\\) , respectively. Given that \\(BD = 7\\) and \\(DC = 4\\) , compute \\(DE\\) .", "solution": "![md5:ddd982a2b8a39ca0e846cc3a28985b3e](ddd982a2b8a39ca0e846cc3a28985b3e.jpeg)\n  \n\nFrom \\(\\angle E = 60^{\\circ}\\) , we get that \\(\\angle AEF = 120^{\\circ} - \\angle CED = \\angle CDE\\) . Therefore, \\(\\triangle AEF \\sim \\triangle CDE\\) . Since \\(EF:DE = 2:1\\) , the ratio of similarity must be \\(2:1\\) , so \\(AE = 2CD = 8\\) . Recall \\(ABC\\) has side length \\(7 + 4 = 11\\) , so \\(EC = 11 - 8 = 3\\) . Law of Cosines on \\(\\triangle CDE\\) gives \\(DE^2 = \\sqrt{3^2 + 4^2 - 3\\cdot 4} = \\sqrt{13}\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n15. [9]", "solution_match": "\nSolution 1:  \n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "15", "problem_type": null, "exam": "HMMT", "problem": "Right triangle \\(\\triangle DEF\\) with \\(\\angle D = 90^{\\circ}\\) and \\(\\angle F = 30^{\\circ}\\) is inscribed in equilateral triangle \\(\\triangle ABC\\) such that \\(D\\) , \\(E\\) , and \\(F\\) lie on segments \\(\\overline{BC}\\) , \\(\\overline{CA}\\) , and \\(\\overline{AB}\\) , respectively. Given that \\(BD = 7\\) and \\(DC = 4\\) , compute \\(DE\\) .", "solution": "![md5:6bdde14db96ad51963d491cef9d82b2a](6bdde14db96ad51963d491cef9d82b2a.jpeg)\n  \n\nLet \\(\\odot (DEF)\\) meet \\(AC\\) again at point \\(X\\) . Then, \\(\\angle FXA = 180^{\\circ} - \\angle FXE = \\angle FDE = 90^{\\circ}\\) and \\(\\angle XDC = 180^{\\circ} - \\angle DCX - \\angle DXC = 120^{\\circ} - \\angle DXE = 120^{\\circ} - \\angle DFE = 90^{\\circ}\\) . It follows that \\(CX = 2CD = 8\\) , so \\(AX = 11 - CX = 3\\) , and \\(AF = 2AX = 6\\) . Thus, Law of Cosines on \\(\\triangle AEF\\) gives \\(EF = \\sqrt{8^2 + 6^2 - 8\\cdot 6} = 2\\sqrt{13}\\) , implying that \\(DE = \\sqrt{13}\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n15. [9]", "solution_match": "\nSolution 2:  \n\n"}}
 {"year": "2025", "tier": "T4", "problem_label": "16", "problem_type": null, "exam": "HMMT", "problem": "The Cantor set is defined as the set of real numbers \\(x\\) such that \\(0 \\leq x < 1\\) and the digit 1 does not appear in the base-3 expansion of \\(x\\) . Two numbers are uniformly and independently selected at random from the Cantor set. Compute the expected value of their absolute difference.  \n\n(Formally, one can pick a number \\(x\\) uniformly at random from the Cantor set by first picking a real number \\(y\\) uniformly at random from the interval \\([0,1)\\) , writing it out in binary, reading its digits as if they were in base- 3, and setting \\(x\\) to 2 times the result.)", "solution": "Answer: \\(\\boxed{\\frac{2}{5}}\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n16. [9]", "solution_match": "\nProposed by: Derek Liu  \n\n"}}
 {"year": "2025", "tier": "T4", "problem_label": "16", "problem_type": null, "exam": "HMMT", "problem": "The Cantor set is defined as the set of real numbers \\(x\\) such that \\(0 \\leq x < 1\\) and the digit 1 does not appear in the base-3 expansion of \\(x\\) . Two numbers are uniformly and independently selected at random from the Cantor set. Compute the expected value of their absolute difference.  \n\n(Formally, one can pick a number \\(x\\) uniformly at random from the Cantor set by first picking a real number \\(y\\) uniformly at random from the interval \\([0,1)\\) , writing it out in binary, reading its digits as if they were in base- 3, and setting \\(x\\) to 2 times the result.)", "solution": "Let \\(d\\) be the expected value of the absolute difference. Observe that the Cantor set is made up of two smaller copies of itself, each scaled down by a factor of 3. There is a \\(\\frac{1}{2}\\) chance that the two selected numbers are in the same copy, in which case the expected value of their absolute difference is \\(\\frac{1}{3} d\\) . Otherwise, we can write them as \\(\\frac{2 + x}{3}\\) and \\(\\frac{y}{3}\\) for independently and uniformly randomly selected \\(x\\) and \\(y\\) in the Cantor set. Their difference is \\(\\frac{2 + (x - y)}{3}\\) , which by symmetry has expected value \\(\\frac{2}{3}\\) . Thus  \n\n\\[d = \\frac{1}{2}\\cdot \\frac{1}{3} d + \\frac{1}{2}\\cdot \\frac{2}{3}\\Rightarrow d = \\left[\\frac{2}{5}\\right].\\]", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n16. [9]", "solution_match": "\nSolution: "}}
 {"year": "2025", "tier": "T4", "problem_label": "17", "problem_type": null, "exam": "HMMT", "problem": "Let \\(f\\) be a quadratic polynomial with real coefficients, and let \\(g_{1}\\) , \\(g_{2}\\) , \\(g_{3}\\) , ... be a geometric progression of real numbers. Define \\(a_{n} = f(n) + g_{n}\\) . Given that \\(a_{1}\\) , \\(a_{2}\\) , \\(a_{3}\\) , \\(a_{4}\\) , and \\(a_{5}\\) are equal to 1, 2, 3, 14, and 16, respectively, compute \\(\\frac{g_{2}}{g_{1}}\\) .", "solution": "Answer: \\(\\boxed{\\frac{19}{10}}\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n17. [11]", "solution_match": "\nProposed by: Pitchayut Saengrungkonkha  \n\n"}}
@@ -44,13 +44,13 @@
 {"year": "2025", "tier": "T4", "problem_label": "22", "problem_type": null, "exam": "HMMT", "problem": "Let \\(a\\) , \\(b\\) , and \\(c\\) be real numbers such that \\(a^2(b + c) = 1\\) , \\(b^2(c + a) = 2\\) , and \\(c^2(a + b) = 5\\) . Given that there are three possible values for \\(abc\\) , compute the minimum possible value of \\(abc\\) .", "solution": "Answer: \\(\\frac{- 5 - \\sqrt{5}}{2}\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n22. [12]", "solution_match": "\nProposed by: Pitchayut Saengrungkonkha  \n\n"}}
 {"year": "2025", "tier": "T4", "problem_label": "22", "problem_type": null, "exam": "HMMT", "problem": "Let \\(a\\) , \\(b\\) , and \\(c\\) be real numbers such that \\(a^2(b + c) = 1\\) , \\(b^2(c + a) = 2\\) , and \\(c^2(a + b) = 5\\) . Given that there are three possible values for \\(abc\\) , compute the minimum possible value of \\(abc\\) .", "solution": "Let \\(x = abc\\) . Multiplying all equations together and simplifying gives  \n\n\\[(abc)^2 (a + b)(b + c)(c + a) = 10,\\]  \n\n\\[(abc)^2 \\left(a^2(b + c) + b^2(c + a) + c^2(a + b) + 2abc\\right) = 10,\\]  \n\n\\[x^2 (1 + 2 + 5 + 2x) = 10,\\]  \n\n\\[x^2 (x + 4) = 5.\\]  \n\nThe resulting cubic factors as \\((x - 1)(x^2 + 5x + 5) = 0\\) . Therefore, the smallest possible value of \\(abc\\) is \\(\\frac{- 5 - \\sqrt{5^2 - 4\\cdot 5}}{2} = \\frac{- 5 - \\sqrt{5}}{2}\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n22. [12]", "solution_match": "\nSolution: "}}
 {"year": "2025", "tier": "T4", "problem_label": "23", "problem_type": null, "exam": "HMMT", "problem": "Regular hexagon \\(ABCDEF\\) has side length 2. Circle \\(\\omega\\) lies inside the hexagon and is tangent to segments \\(\\overline{AB}\\) and \\(\\overline{AF}\\) . There exist two perpendicular lines tangent to \\(\\omega\\) that pass through \\(C\\) and \\(E\\) , respectively. Given that these two lines do not intersect on line \\(AD\\) , compute the radius of \\(\\omega\\) .", "solution": "Answer: \\(\\frac{3\\sqrt{3} - 3}{2} = \\frac{3}{2} (\\sqrt{3} - 1)\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n23. [12]", "solution_match": "\nProposed by: Karthik Venkata Vedula  \n\n"}}
-{"year": "2025", "tier": "T4", "problem_label": "23", "problem_type": null, "exam": "HMMT", "problem": "Regular hexagon \\(ABCDEF\\) has side length 2. Circle \\(\\omega\\) lies inside the hexagon and is tangent to segments \\(\\overline{AB}\\) and \\(\\overline{AF}\\) . There exist two perpendicular lines tangent to \\(\\omega\\) that pass through \\(C\\) and \\(E\\) , respectively. Given that these two lines do not intersect on line \\(AD\\) , compute the radius of \\(\\omega\\) .", "solution": 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 \n\nLet \\(O\\) be the center of \\(\\omega\\) , and let the two tangent lines intersect at \\(P\\) . Note that \\(O\\) lies on the external angle bisector of \\(\\angle CPE\\) because the tangents are symmetric about line \\(PO\\) . Additionally, \\(O\\) lies on the perpendicular bisector of \\(CE\\) by symmetry. By Fact 5, \\(COPE\\) is cyclic and \\(\\angle COE = 90^{\\circ}\\) . To finish, observe that \\(\\angle COD = 45^{\\circ}\\) . Dropping the altitude \\(CH\\) down to \\(AD\\) gives \\(OH = CH = \\sqrt{3}\\) . So, \\(AO = AH - OH = 3 - \\sqrt{3}\\) . The desired answer is then \\(\\frac{\\sqrt{3}}{2} \\cdot AO = \\left[\\frac{3\\sqrt{3} - 3}{2}\\right]\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n23. [12]", "solution_match": "\nSolution 1:  \n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "23", "problem_type": null, "exam": "HMMT", "problem": "Regular hexagon \\(ABCDEF\\) has side length 2. Circle \\(\\omega\\) lies inside the hexagon and is tangent to segments \\(\\overline{AB}\\) and \\(\\overline{AF}\\) . There exist two perpendicular lines tangent to \\(\\omega\\) that pass through \\(C\\) and \\(E\\) , respectively. Given that these two lines do not intersect on line \\(AD\\) , compute the radius of \\(\\omega\\) .", "solution": "![md5:dcb968f470d9995223369f2d83efd4f0](dcb968f470d9995223369f2d83efd4f0.jpeg)\n  \n\nLet \\(O\\) be the center of \\(\\omega\\) , and let the two tangent lines intersect at \\(P\\) . Note that \\(O\\) lies on the external angle bisector of \\(\\angle CPE\\) because the tangents are symmetric about line \\(PO\\) . Additionally, \\(O\\) lies on the perpendicular bisector of \\(CE\\) by symmetry. By Fact 5, \\(COPE\\) is cyclic and \\(\\angle COE = 90^{\\circ}\\) . To finish, observe that \\(\\angle COD = 45^{\\circ}\\) . Dropping the altitude \\(CH\\) down to \\(AD\\) gives \\(OH = CH = \\sqrt{3}\\) . So, \\(AO = AH - OH = 3 - \\sqrt{3}\\) . The desired answer is then \\(\\frac{\\sqrt{3}}{2} \\cdot AO = \\left[\\frac{3\\sqrt{3} - 3}{2}\\right]\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n23. [12]", "solution_match": "\nSolution 1:  \n\n"}}
 {"year": "2025", "tier": "T4", "problem_label": "23", "problem_type": null, "exam": "HMMT", "problem": "Regular hexagon \\(ABCDEF\\) has side length 2. Circle \\(\\omega\\) lies inside the hexagon and is tangent to segments \\(\\overline{AB}\\) and \\(\\overline{AF}\\) . There exist two perpendicular lines tangent to \\(\\omega\\) that pass through \\(C\\) and \\(E\\) , respectively. Given that these two lines do not intersect on line \\(AD\\) , compute the radius of \\(\\omega\\) .", "solution": "Another way to get \\(\\angle COE = 90^{\\circ}\\) is as follows.  \n\nLet \\(\\omega\\) meet the tangents from \\(C\\) and \\(E\\) at \\(Q\\) and \\(R\\) , respectively. Observe \\(OC = OE\\) (as \\(O\\) lies on the perpendicular bisector of \\(CE\\) ) and \\(OQ = OR\\) , so \\(\\triangle OCQ \\stackrel{\\triangle}{=} \\triangle OER\\) . Then \\(\\angle COE = \\angle QOR = 90^{\\circ}\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n23. [12]", "solution_match": "\nSolution 2: "}}
 {"year": "2025", "tier": "T4", "problem_label": "24", "problem_type": null, "exam": "HMMT", "problem": "For any integer \\(x\\) , let  \n\n\\[f(x) = 100! \\left(1 + x + \\frac{x^{2}}{2!} + \\frac{x^{3}}{3!} + \\dots + \\frac{x^{100}}{100!}\\right).\\]  \n\nA positive integer \\(a\\) is chosen such that \\(f(a) - 20\\) is divisible by \\(101^{2}\\) . Compute the remainder when \\(f(a + 101)\\) is divided by \\(101^{2}\\) .", "solution": "Answer: [1939]", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n24. [12]", "solution_match": "\nProposed by: Pitchayut Saengrungkonkha  \n\n"}}
 {"year": "2025", "tier": "T4", "problem_label": "24", "problem_type": null, "exam": "HMMT", "problem": "For any integer \\(x\\) , let  \n\n\\[f(x) = 100! \\left(1 + x + \\frac{x^{2}}{2!} + \\frac{x^{3}}{3!} + \\dots + \\frac{x^{100}}{100!}\\right).\\]  \n\nA positive integer \\(a\\) is chosen such that \\(f(a) - 20\\) is divisible by \\(101^{2}\\) . Compute the remainder when \\(f(a + 101)\\) is divided by \\(101^{2}\\) .", "solution": "By the binomial theorem,  \n\n\\[(a + 101)^{n} \\equiv a^{n} + \\binom{n}{1} a^{n - 1}101 = a^{n} + 101na^{n - 1} \\pmod {101^{2}}.\\]\n\n\n\nUsing this gives (all congruences are modulo \\(101^{2}\\) )  \n\n\\[f(a + 101) = 100!\\sum_{n = 0}^{100}\\frac{(a + 101)^{n}}{n!}\\] \\[\\equiv 100!\\sum_{n = 0}^{100}\\left(\\frac{a^{n}}{n!} +\\frac{101n a^{n - 1}}{n!}\\right)\\] \\[\\equiv f(a) + 100!\\cdot 101\\sum_{n = 1}^{100}\\frac{a^{n - 1}}{(n - 1)!}\\] \\[\\equiv f(a) + 101f(a) - 100!\\cdot 101\\frac{a^{100}}{100!}\\] \\[\\equiv f(a) + 101(f(a) - 1)\\] \\[\\equiv 20 + 101(20 - 1) = \\boxed {1939}\\quad (\\mathrm{mod}101^{2}).\\]", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n24. [12]", "solution_match": "\nSolution 1: "}}
 {"year": "2025", "tier": "T4", "problem_label": "24", "problem_type": null, "exam": "HMMT", "problem": "For any integer \\(x\\) , let  \n\n\\[f(x) = 100! \\left(1 + x + \\frac{x^{2}}{2!} + \\frac{x^{3}}{3!} + \\dots + \\frac{x^{100}}{100!}\\right).\\]  \n\nA positive integer \\(a\\) is chosen such that \\(f(a) - 20\\) is divisible by \\(101^{2}\\) . Compute the remainder when \\(f(a + 101)\\) is divided by \\(101^{2}\\) .", "solution": "The above solution can be viewed as a consequence of Hensel's lemma as follows. Because 101 is prime, for any integer \\(x\\) not divisible by 101, we have that  \n\n\\[f^{\\prime}(x) = 100!\\left(1 + x + \\frac{x^{2}}{2!} +\\dots +\\frac{x^{99}}{99!}\\right) = f(x) - x^{100}\\equiv f(x) - 1\\pmod {101}.\\]  \n\nClearly \\(101 \\nmid a\\) . Hence, by Hensel's lemma, we get that  \n\n\\[f(a + 101) \\equiv f(a) + 101f^{\\prime}(a) \\equiv 20 + 101 \\cdot 19 \\equiv \\boxed {1939} \\pmod {101^{2}}.\\]  \n\nRemark. All possible \\(a\\) 's are \\(a \\equiv 1012, 6670\\) , and 9885 (mod \\(101^{2}\\) ). They all lead to the same answer.", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n24. [12]", "solution_match": "\nSolution 2: "}}
 {"year": "2025", "tier": "T4", "problem_label": "25", "problem_type": null, "exam": "HMMT", "problem": "Let \\(A B C D\\) be a trapezoid such that \\(A B \\parallel C D\\) , \\(A D = 13\\) , \\(B C = 15\\) , \\(A B = 20\\) , and \\(C D = 34\\) . Point \\(X\\) lies inside the trapezoid such that \\(\\angle X A B = 2\\angle X B A\\) and \\(\\angle X D C = 2\\angle X C D\\) . Compute \\(X D - X A\\) .", "solution": "Answer: 4", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n25. [14]", "solution_match": "\nProposed by: Pitchayut Saengrungkonkha  \n\n"}}
-{"year": "2025", "tier": "T4", "problem_label": "25", "problem_type": null, "exam": "HMMT", "problem": "Let \\(A B C D\\) be a trapezoid such that \\(A B \\parallel C D\\) , \\(A D = 13\\) , \\(B C = 15\\) , \\(A B = 20\\) , and \\(C D = 34\\) . Point \\(X\\) lies inside the trapezoid such that \\(\\angle X A B = 2\\angle X B A\\) and \\(\\angle X D C = 2\\angle X C D\\) . Compute \\(X D - X A\\) .", "solution": "![](data:image/jpeg;base64,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)\n  \n\nConstruct point \\(P\\) on \\(A B\\) such that \\(X A = X P\\) and point \\(Q\\) on \\(C D\\) such that \\(X D = X Q\\) . The angle condition gives \\(Q C = X Q = X D\\) and \\(P B = X P = X A\\) . Moreover, \\(A D Q P\\) is an isosceles trapezoid.\n\n\n\nLet \\(S\\) be the projection of \\(A\\) onto \\(CD\\) , and let \\(T\\) be on \\(CD\\) such that \\(AT \\parallel BC\\) . Then \\(ADT\\) is a 13- 14- 15 triangle, so \\(DS = 5\\) . Therefore, \\(QD - PA = 10\\) . Finally, we get  \n\n\\[XD - XA = QC - PB = (34 - QD) - (20 - PA) = 14 - 10 = \\boxed{4}.\\]", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n25. [14]", "solution_match": "\nSolution:  \n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "25", "problem_type": null, "exam": "HMMT", "problem": "Let \\(A B C D\\) be a trapezoid such that \\(A B \\parallel C D\\) , \\(A D = 13\\) , \\(B C = 15\\) , \\(A B = 20\\) , and \\(C D = 34\\) . Point \\(X\\) lies inside the trapezoid such that \\(\\angle X A B = 2\\angle X B A\\) and \\(\\angle X D C = 2\\angle X C D\\) . Compute \\(X D - X A\\) .", "solution": "![md5:7a6b1d63c0cdadfc836209379f9acbdc](7a6b1d63c0cdadfc836209379f9acbdc.jpeg)\n  \n\nConstruct point \\(P\\) on \\(A B\\) such that \\(X A = X P\\) and point \\(Q\\) on \\(C D\\) such that \\(X D = X Q\\) . The angle condition gives \\(Q C = X Q = X D\\) and \\(P B = X P = X A\\) . Moreover, \\(A D Q P\\) is an isosceles trapezoid.\n\n\n\nLet \\(S\\) be the projection of \\(A\\) onto \\(CD\\) , and let \\(T\\) be on \\(CD\\) such that \\(AT \\parallel BC\\) . Then \\(ADT\\) is a 13- 14- 15 triangle, so \\(DS = 5\\) . Therefore, \\(QD - PA = 10\\) . Finally, we get  \n\n\\[XD - XA = QC - PB = (34 - QD) - (20 - PA) = 14 - 10 = \\boxed{4}.\\]", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n25. [14]", "solution_match": "\nSolution:  \n\n"}}
 {"year": "2025", "tier": "T4", "problem_label": "26", "problem_type": null, "exam": "HMMT", "problem": "Isabella has a bag with 20 blue diamonds and 25 purple diamonds. She repeats the following process 44 times: she removes a diamond from the bag uniformly at random, then puts one blue diamond and one purple diamond into the bag. Compute the expected number of blue diamonds in the bag after all 44 repetitions.", "solution": "Answer: \\(\\boxed{\\frac{173}{4}}\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n26. [14]", "solution_match": "\nProposed by: Henrick Rabinovitz, Srinivas Arun  \n\n"}}
 {"year": "2025", "tier": "T4", "problem_label": "26", "problem_type": null, "exam": "HMMT", "problem": "Isabella has a bag with 20 blue diamonds and 25 purple diamonds. She repeats the following process 44 times: she removes a diamond from the bag uniformly at random, then puts one blue diamond and one purple diamond into the bag. Compute the expected number of blue diamonds in the bag after all 44 repetitions.", "solution": "Let \\(a = 20\\) and \\(b = 25\\) be the initial numbers of blue and purple diamonds, respectively, and let \\(c = 44\\) be the number of times Isabella performs the operation. Suppose that at some point, the bag contains \\(x\\) blue diamonds and \\(y\\) purple diamonds, for \\(x + y = z\\) total diamonds. After one step, the bag will have \\(z + 1\\) diamonds. The expected change in the number of blue diamonds in this step is \\((- x / z) + 1 = y / z\\) , and likewise this quantity for purple diamonds is \\(x / z\\) . Thus, the expected change in the difference between the number of blue and purple diamonds is \\((y - x) / z\\) . Since this difference was initially \\(x - y\\) , the expected value of this difference is multiplied by \\((z - 1) / z\\) at each step (regardless of \\(x - y\\) ). Since \\(z\\) starts at \\(a + b\\) and ends at \\(a + b + c\\) , the expected difference after \\(c\\) operations is  \n\n\\[(a - b)\\cdot \\prod_{z = a + b}^{a + b + c - 1}\\frac{z - 1}{z} = \\frac{(a + b - 1)(a - b)}{(a + b + c - 1)},\\]  \n\nand as the total number of diamonds is \\(a + b + c\\) , the expected number of blue diamonds at the end is  \n\n\\[\\frac{a + b + c}{2} +\\frac{(a + b - 1)(a - b)}{2(a + b + c - 1)}.\\]  \n\nPlugging in \\(a = 20\\) , \\(b = 25\\) , and \\(c = 44\\) gives us the answer, \\(\\boxed{\\frac{173}{4}}\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n26. [14]", "solution_match": "\nSolution: "}}
 {"year": "2025", "tier": "T4", "problem_label": "27", "problem_type": null, "exam": "HMMT", "problem": "Compute the number of ordered pairs \\((m, n)\\) of odd positive integers both less than 80 such that  \n\n\\[\\gcd (4^{m} + 2^{m} + 1,4^{n} + 2^{n} + 1) > 1.\\]", "solution": "Answer: \\(\\boxed{820}\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n27. [14]", "solution_match": "\nProposed by: Pitchayut Saengrungkonkha  \n\n"}}
@@ -59,13 +59,13 @@
 {"year": "2025", "tier": "T4", "problem_label": "28", "problem_type": null, "exam": "HMMT", "problem": "Let \\(f\\) be a function from nonnegative integers to nonnegative integers such that \\(f(0) = 0\\) and  \n\n\\[f(m) = f\\left(\\left\\lfloor \\frac{m}{2}\\right\\rfloor\\right) + \\left\\lceil \\frac{m}{2}\\right\\rceil^{2}\\]  \n\nfor all positive integers \\(m\\) . Compute  \n\n\\[\\frac{f(1)}{1\\cdot2} +\\frac{f(2)}{2\\cdot3} +\\frac{f(3)}{3\\cdot4} +\\dots +\\frac{f(31)}{31\\cdot32}.\\]  \n\n(Here, \\(\\lfloor z\\rfloor\\) is the greatest integer less than or equal to \\(z\\) , and \\(\\lceil z\\rceil\\) is the least positive integer greater than or equal to \\(z\\) .)", "solution": "For all positive integers \\(n\\) , let \\(\\omega (n) = f(n) - f(n - 1)\\) . We claim that \\(\\omega (n)\\) is the largest odd divisor of \\(n\\) for all \\(n > 0\\) . Indeed, for all positive integers \\(k\\) , we have  \n\n\\[\\omega (2k) = f(2k) - f(2k - 1) = f(k) + k^{2} - (f(k - 1) + k^{2}) = f(k) - f(k - 1) = \\omega (k)\\]  \n\nand  \n\n\\[\\omega (2k + 1) = f(2k + 1) - f(2k) = f(k) + (k + 1)^{2} - (f(k) + k^{2}) = 2k + 1.\\]  \n\nInducting on the positive integers implies that \\(\\omega (n)\\) is indeed the largest odd divisor of \\(n\\) .  \n\nWe can now rewrite the sum as  \n\n\\[\\sum_{n = 1}^{31}\\frac{f(n)}{n(n + 1)} = \\sum_{n = 1}^{31}\\left(\\frac{f(n)}{n} -\\frac{f(n)}{n + 1}\\right) = \\left(\\sum_{n = 1}^{30}\\frac{f(n) - f(n - 1)}{n}\\right) - \\frac{f(31)}{32}.\\]  \n\nNote that by using the original recursive definition, we can compute  \n\n\\[f(31) = 16^{2} + 8^{2} + 4^{2} + 2^{2} + 1^{2} = 341.\\]  \n\nMoreover, we also see that \\(\\frac{f(n) - f(n - 1)}{n} = \\frac{\\omega(n)}{n} = 2^{-\\nu_{2}(n)}\\) , where \\(2^{\\nu_{2}(n)}\\) is the largest power of 2 dividing \\(n\\) . Thus, our desired sum is  \n\n\\[\\left(\\sum_{n = 1}^{31}2^{-\\nu_{2}(n)}\\right) - \\frac{341}{32} = 16\\cdot 2^{-0} + 8\\cdot 2^{-1} + 4\\cdot 2^{-2} + 2\\cdot 2^{-3} + 1\\cdot 2^{-4} - \\frac{341}{32} = \\left\\lfloor \\frac{341}{32}\\right\\rfloor\\]", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n28. [14]", "solution_match": "\nSolution 1: "}}
 {"year": "2025", "tier": "T4", "problem_label": "28", "problem_type": null, "exam": "HMMT", "problem": "Let \\(f\\) be a function from nonnegative integers to nonnegative integers such that \\(f(0) = 0\\) and  \n\n\\[f(m) = f\\left(\\left\\lfloor \\frac{m}{2}\\right\\rfloor\\right) + \\left\\lceil \\frac{m}{2}\\right\\rceil^{2}\\]  \n\nfor all positive integers \\(m\\) . Compute  \n\n\\[\\frac{f(1)}{1\\cdot2} +\\frac{f(2)}{2\\cdot3} +\\frac{f(3)}{3\\cdot4} +\\dots +\\frac{f(31)}{31\\cdot32}.\\]  \n\n(Here, \\(\\lfloor z\\rfloor\\) is the greatest integer less than or equal to \\(z\\) , and \\(\\lceil z\\rceil\\) is the least positive integer greater than or equal to \\(z\\) .)", "solution": "From the original recursion, for all positive integers \\(n\\) , we have  \n\n\\[\\frac{f(2n)}{2n(2n + 1)} = \\frac{f(n) + n^{2}}{2n(2n + 1)} = \\frac{f(n)}{2n(2n + 1)} +\\frac{n}{2(2n + 1)}\\]\n\n\n\nand  \n\n\\[\\frac{f(2n + 1)}{(2n + 1)(2n + 2)} = \\frac{f(n) + (n + 1)^{2}}{(2n + 1)(2n + 2)} = \\frac{f(n)}{(2n + 1)(2n + 2)} +\\frac{n + 1}{2(2n + 1)}.\\]  \n\nAdding these two equations gives  \n\n\\[\\frac{f(2n)}{2n(2n + 1)} +\\frac{f(2n + 1)}{(2n + 1)(2n + 2)} = \\frac{f(n)}{2n(n + 1)} +\\frac{1}{2}.\\]  \n\nThus, if \\(T(n) = \\sum_{k = 1}^{n} \\frac{f(k)}{k(k + 1)}\\) , we have  \n\n\\[T(2n + 1) = \\sum_{k = 1}^{2n + 1}\\frac{f(k)}{k(k + 1)}\\] \\[\\qquad = \\frac{f(1)}{1(2)} +\\sum_{m = 1}^{n}\\left(\\frac{f(2m)}{2m(2m + 1)} +\\frac{f(2m + 1)}{(2m + 1)(2m + 2)}\\right)\\] \\[\\qquad = \\frac{1}{2} +\\sum_{m = 1}^{n}\\left(\\frac{f(m)}{2m(m + 1)} +\\frac{1}{2}\\right)\\] \\[\\qquad = \\frac{n + 1}{2} +\\frac{1}{2} T(n).\\]  \n\nStarting from \\(T(1) = 1\\) , we can compute \\(T(3) = \\frac{5}{4}\\) , \\(T(7) = \\frac{21}{8}\\) , \\(T(15) = \\frac{85}{16}\\) , and finally, \\(T(31) = \\left\\lfloor \\frac{341}{32} \\right\\rfloor\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n28. [14]", "solution_match": "\nSolution 2: "}}
 {"year": "2025", "tier": "T4", "problem_label": "29", "problem_type": null, "exam": "HMMT", "problem": "Points \\(A\\) and \\(B\\) lie on circle \\(\\omega\\) with center \\(O\\) . Let \\(X\\) be a point inside \\(\\omega\\) . Suppose that \\(XO = 2\\sqrt{2}\\) , \\(XA = 1\\) , \\(XB = 3\\) , and \\(\\angle AXB = 90^{\\circ}\\) . Points \\(Y\\) and \\(Z\\) are on \\(\\omega\\) such that \\(Y \\neq A\\) and triangles \\(\\triangle AXB\\) and \\(\\triangle YXZ\\) are similar with the same orientation. Compute \\(XY\\) .", "solution": "Answer: \\(\\frac{11}{5}\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n29. [16]", "solution_match": "\nProposed by: Ethan Liu  \n\n"}}
-{"year": "2025", "tier": "T4", "problem_label": "29", "problem_type": null, "exam": "HMMT", "problem": "Points \\(A\\) and \\(B\\) lie on circle \\(\\omega\\) with center \\(O\\) . Let \\(X\\) be a point inside \\(\\omega\\) . Suppose that \\(XO = 2\\sqrt{2}\\) , \\(XA = 1\\) , \\(XB = 3\\) , and \\(\\angle AXB = 90^{\\circ}\\) . Points \\(Y\\) and \\(Z\\) are on \\(\\omega\\) such that \\(Y \\neq A\\) and triangles \\(\\triangle AXB\\) and \\(\\triangle YXZ\\) are similar with the same orientation. Compute \\(XY\\) .", "solution": "Consider a rotation about \\(X\\) by \\(90^{\\circ}\\) followed by a homothety with ratio \\(\\frac{1}{3}\\) that sends \\(B\\) to \\(A\\) . This sends \\(\\omega\\) to \\(\\omega^{\\prime}\\) with radius \\(\\frac{1}{3}\\) of the radius of \\(\\omega\\) and center \\(O^{\\prime}\\) . Since \\(A\\) is the image of \\(B\\) under this rotation, we know \\(A\\) lies on both circles; the same argument shows \\(Y\\) must lie on both circles. Thus, \\(Y\\) is the reflection of \\(A\\) over \\(OO^{\\prime}\\) . In particular, this means that \\(XY = AX^{\\prime}\\) , where \\(X^{\\prime}\\) is the reflection of \\(X\\) over \\(OO^{\\prime}\\) .  \n\nLet \\(M\\) be the midpoint of \\(AB\\) . Note that because \\(\\triangle OXO^{\\prime} \\sim \\triangle BXA\\) , we also have \\(\\triangle XOX^{\\prime} \\sim \\triangle XMA\\) , as they are both isosceles and \\(\\angle XOX^{\\prime} = 2\\angle XOO^{\\prime} = 2\\angle XBA = \\angle XMA\\) . This implies that \\(\\triangle XOM \\sim \\triangle XX^{\\prime}A\\) . Thus, we know that \\(AX^{\\prime} = OM \\cdot \\frac{XA}{XM} = \\frac{2}{\\sqrt{10}} OM\\) . It remains to compute \\(OM\\) ; noting that the distance between \\(O\\) and the foot from \\(X\\) to \\(AB\\) is \\(\\frac{2}{5}\\sqrt{10}\\) , and that the altitude of \\(\\triangle AXY\\) has length \\(\\frac{3}{10}\\sqrt{10}\\) , we get that the distance from \\(O\\) to \\(AB\\) is  \n\n\\[\\frac{3}{10}\\sqrt{10} +\\sqrt{\\left(2\\sqrt{2}\\right)^{2} - \\left(\\frac{2}{5}\\sqrt{10}\\right)^{2}} = \\frac{11}{10}\\sqrt{10}\\]  \n\nby the Pythagorean theorem, which means that \\(XY = AX^{\\prime} = \\left\\lfloor \\frac{11}{5} \\right\\rfloor\\) .\n\n\n![](data:image/jpeg;base64,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"metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n29. [16]", "solution_match": "\nSolution 1: "}}
-{"year": "2025", "tier": "T4", "problem_label": "29", "problem_type": null, "exam": "HMMT", "problem": "Points \\(A\\) and \\(B\\) lie on circle \\(\\omega\\) with center \\(O\\) . Let \\(X\\) be a point inside \\(\\omega\\) . Suppose that \\(XO = 2\\sqrt{2}\\) , \\(XA = 1\\) , \\(XB = 3\\) , and \\(\\angle AXB = 90^{\\circ}\\) . Points \\(Y\\) and \\(Z\\) are on \\(\\omega\\) such that \\(Y \\neq A\\) and triangles \\(\\triangle AXB\\) and \\(\\triangle YXZ\\) are similar with the same orientation. Compute \\(XY\\) .", "solution": 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 \n\nLet \\(M\\) be the midpoint of \\(AB\\) . We will find \\(MO\\) first.  \n\nLet the internal bisector of \\(\\angle A X B\\) intersect \\(\\odot (A X B)\\) at \\(P\\) . From Ptolemy, \\(X P = 2\\sqrt{2} = X O\\) . Let \\(X^{\\prime}\\) be the foot of altitude from \\(X\\) to \\(M O\\) . Observe that \\(O\\) is the reflection of \\(P\\) across \\(X X^{\\prime}\\) . By the\n\n\n\narea of \\(\\triangle A X B\\) , we have \\(X^{\\prime}M = \\frac{3}{\\sqrt{10}}\\) . Therefore,  \n\n\\[M O = O X^{\\prime} + X^{\\prime}M = X^{\\prime}P + X^{\\prime}M = 2X^{\\prime}M + M P = \\frac{11\\sqrt{10}}{10}.\\]  \n\nBy the spiral similarity \\(\\triangle X A B\\mapsto \\triangle X Y Z\\) , we have that \\(A Y\\perp B Z\\) and \\(\\angle A O B + \\angle Y O Z = 90^{\\circ}\\) Therefore, \\(\\triangle X A M\\sim \\triangle X Y N\\) where \\(N\\) is the midpoint of \\(Y Z\\) . Thus, \\(Y Z = \\frac{11\\sqrt{10}}{5}\\) and \\(X Y = \\left[\\frac{11}{5}\\right]\\)  \n\nNote: if \\(\\triangle X A B\\widetilde{\\sim}\\triangle X Y Z\\) , just consider another \\(\\triangle X Y^{\\prime}Z^{\\prime}\\) caused by reflecting \\(\\triangle X Y Z\\) across \\(X O\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n29. [16]", "solution_match": "\nSolution 2:  \n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "29", "problem_type": null, "exam": "HMMT", "problem": "Points \\(A\\) and \\(B\\) lie on circle \\(\\omega\\) with center \\(O\\) . Let \\(X\\) be a point inside \\(\\omega\\) . Suppose that \\(XO = 2\\sqrt{2}\\) , \\(XA = 1\\) , \\(XB = 3\\) , and \\(\\angle AXB = 90^{\\circ}\\) . Points \\(Y\\) and \\(Z\\) are on \\(\\omega\\) such that \\(Y \\neq A\\) and triangles \\(\\triangle AXB\\) and \\(\\triangle YXZ\\) are similar with the same orientation. Compute \\(XY\\) .", "solution": "Consider a rotation about \\(X\\) by \\(90^{\\circ}\\) followed by a homothety with ratio \\(\\frac{1}{3}\\) that sends \\(B\\) to \\(A\\) . This sends \\(\\omega\\) to \\(\\omega^{\\prime}\\) with radius \\(\\frac{1}{3}\\) of the radius of \\(\\omega\\) and center \\(O^{\\prime}\\) . Since \\(A\\) is the image of \\(B\\) under this rotation, we know \\(A\\) lies on both circles; the same argument shows \\(Y\\) must lie on both circles. Thus, \\(Y\\) is the reflection of \\(A\\) over \\(OO^{\\prime}\\) . In particular, this means that \\(XY = AX^{\\prime}\\) , where \\(X^{\\prime}\\) is the reflection of \\(X\\) over \\(OO^{\\prime}\\) .  \n\nLet \\(M\\) be the midpoint of \\(AB\\) . Note that because \\(\\triangle OXO^{\\prime} \\sim \\triangle BXA\\) , we also have \\(\\triangle XOX^{\\prime} \\sim \\triangle XMA\\) , as they are both isosceles and \\(\\angle XOX^{\\prime} = 2\\angle XOO^{\\prime} = 2\\angle XBA = \\angle XMA\\) . This implies that \\(\\triangle XOM \\sim \\triangle XX^{\\prime}A\\) . Thus, we know that \\(AX^{\\prime} = OM \\cdot \\frac{XA}{XM} = \\frac{2}{\\sqrt{10}} OM\\) . It remains to compute \\(OM\\) ; noting that the distance between \\(O\\) and the foot from \\(X\\) to \\(AB\\) is \\(\\frac{2}{5}\\sqrt{10}\\) , and that the altitude of \\(\\triangle AXY\\) has length \\(\\frac{3}{10}\\sqrt{10}\\) , we get that the distance from \\(O\\) to \\(AB\\) is  \n\n\\[\\frac{3}{10}\\sqrt{10} +\\sqrt{\\left(2\\sqrt{2}\\right)^{2} - \\left(\\frac{2}{5}\\sqrt{10}\\right)^{2}} = \\frac{11}{10}\\sqrt{10}\\]  \n\nby the Pythagorean theorem, which means that \\(XY = AX^{\\prime} = \\left\\lfloor \\frac{11}{5} \\right\\rfloor\\) .\n\n\n![md5:9f73282b07dbc89124eba6f1161ff910](9f73282b07dbc89124eba6f1161ff910.jpeg)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n29. [16]", "solution_match": "\nSolution 1: "}}
+{"year": "2025", "tier": "T4", "problem_label": "29", "problem_type": null, "exam": "HMMT", "problem": "Points \\(A\\) and \\(B\\) lie on circle \\(\\omega\\) with center \\(O\\) . Let \\(X\\) be a point inside \\(\\omega\\) . Suppose that \\(XO = 2\\sqrt{2}\\) , \\(XA = 1\\) , \\(XB = 3\\) , and \\(\\angle AXB = 90^{\\circ}\\) . Points \\(Y\\) and \\(Z\\) are on \\(\\omega\\) such that \\(Y \\neq A\\) and triangles \\(\\triangle AXB\\) and \\(\\triangle YXZ\\) are similar with the same orientation. Compute \\(XY\\) .", "solution": "![md5:79e94bdb5b607f1c5bb5ad0a8ff2bf6a](79e94bdb5b607f1c5bb5ad0a8ff2bf6a.jpeg)\n  \n\nLet \\(M\\) be the midpoint of \\(AB\\) . We will find \\(MO\\) first.  \n\nLet the internal bisector of \\(\\angle A X B\\) intersect \\(\\odot (A X B)\\) at \\(P\\) . From Ptolemy, \\(X P = 2\\sqrt{2} = X O\\) . Let \\(X^{\\prime}\\) be the foot of altitude from \\(X\\) to \\(M O\\) . Observe that \\(O\\) is the reflection of \\(P\\) across \\(X X^{\\prime}\\) . By the\n\n\n\narea of \\(\\triangle A X B\\) , we have \\(X^{\\prime}M = \\frac{3}{\\sqrt{10}}\\) . Therefore,  \n\n\\[M O = O X^{\\prime} + X^{\\prime}M = X^{\\prime}P + X^{\\prime}M = 2X^{\\prime}M + M P = \\frac{11\\sqrt{10}}{10}.\\]  \n\nBy the spiral similarity \\(\\triangle X A B\\mapsto \\triangle X Y Z\\) , we have that \\(A Y\\perp B Z\\) and \\(\\angle A O B + \\angle Y O Z = 90^{\\circ}\\) Therefore, \\(\\triangle X A M\\sim \\triangle X Y N\\) where \\(N\\) is the midpoint of \\(Y Z\\) . Thus, \\(Y Z = \\frac{11\\sqrt{10}}{5}\\) and \\(X Y = \\left[\\frac{11}{5}\\right]\\)  \n\nNote: if \\(\\triangle X A B\\widetilde{\\sim}\\triangle X Y Z\\) , just consider another \\(\\triangle X Y^{\\prime}Z^{\\prime}\\) caused by reflecting \\(\\triangle X Y Z\\) across \\(X O\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n29. [16]", "solution_match": "\nSolution 2:  \n\n"}}
 {"year": "2025", "tier": "T4", "problem_label": "30", "problem_type": null, "exam": "HMMT", "problem": "Let \\(a\\) , \\(b\\) , and \\(c\\) be real numbers satisfying the system of equations  \n\n\\[a\\sqrt{1 + b^{2}} +b\\sqrt{1 + a^{2}} = \\frac{3}{4},\\] \\[b\\sqrt{1 + c^{2}} +c\\sqrt{1 + b^{2}} = \\frac{5}{12},\\mathrm{and}\\] \\[c\\sqrt{1 + a^{2}} +a\\sqrt{1 + c^{2}} = \\frac{21}{20}.\\]  \n\nCompute \\(a\\) .", "solution": "Answer: \\(\\frac{7}{2\\sqrt{30}}\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n30. [16]", "solution_match": "\nProposed by: Pitchayut Saengrungkonga  \n\n"}}
 {"year": "2025", "tier": "T4", "problem_label": "30", "problem_type": null, "exam": "HMMT", "problem": "Let \\(a\\) , \\(b\\) , and \\(c\\) be real numbers satisfying the system of equations  \n\n\\[a\\sqrt{1 + b^{2}} +b\\sqrt{1 + a^{2}} = \\frac{3}{4},\\] \\[b\\sqrt{1 + c^{2}} +c\\sqrt{1 + b^{2}} = \\frac{5}{12},\\mathrm{and}\\] \\[c\\sqrt{1 + a^{2}} +a\\sqrt{1 + c^{2}} = \\frac{21}{20}.\\]  \n\nCompute \\(a\\) .", "solution": "Recall that the functions \\(\\sinh (x) = \\frac{e^{x} - e^{- x}}{2}\\) and \\(\\cosh (x) = \\frac{e^{- x} + e^{- x}}{2}\\) satisfy the relation  \n\n\\[\\sinh (x + y) = \\sinh (x)\\cosh (y) + \\cosh (x)\\sinh (y) = \\sinh (x)\\sqrt{1 + \\sinh (y)^{2}} +\\sinh (y)\\sqrt{1 + \\sinh (x)^{2}}.\\]  \n\nSince \\(\\sinh\\) is surjective, we can perform the substitution \\(a = \\sinh (x)\\) , \\(b = \\sinh (y)\\) , and \\(c = \\sinh (z)\\) , which turns the equations into  \n\n\\[\\sinh (x + y) = \\frac{2 - \\frac{1}{2}}{2},\\] \\[\\sinh (y + z) = \\frac{\\frac{3}{2} - \\frac{2}{3}}{2},\\] \\[\\sinh (z + x) = \\frac{\\frac{5}{2} - \\frac{2}{5}}{2}.\\]  \n\nThus, \\(x + y = \\log (2)\\) , \\(y + z = \\log (3 / 2)\\) , and \\(z + x = \\log (5 / 2)\\) . Solving these equations gives \\(x = \\log (\\sqrt{10 / 3})\\) , so \\(a = \\frac{1}{2}\\left(\\sqrt{\\frac{10}{3}} - \\sqrt{\\frac{3}{10}}\\right) = \\left[\\frac{7}{2\\sqrt{30}}\\right]\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n30. [16]", "solution_match": "\nSolution 1: "}}
 {"year": "2025", "tier": "T4", "problem_label": "30", "problem_type": null, "exam": "HMMT", "problem": "Let \\(a\\) , \\(b\\) , and \\(c\\) be real numbers satisfying the system of equations  \n\n\\[a\\sqrt{1 + b^{2}} +b\\sqrt{1 + a^{2}} = \\frac{3}{4},\\] \\[b\\sqrt{1 + c^{2}} +c\\sqrt{1 + b^{2}} = \\frac{5}{12},\\mathrm{and}\\] \\[c\\sqrt{1 + a^{2}} +a\\sqrt{1 + c^{2}} = \\frac{21}{20}.\\]  \n\nCompute \\(a\\) .", "solution": "We can find positive real numbers \\(x\\) , \\(y\\) , and \\(z\\) such that \\(a = \\frac{x^{2} - 1}{2x}\\) , \\(b = \\frac{y^{2} - 1}{2y}\\) , and \\(c = \\frac{z^{2} - 1}{2z}\\) . Then, the first equation becomes  \n\n\\[\\frac{x^{2} - 1}{2x} \\cdot \\frac{y^{2} + 1}{2y} + \\frac{y^{2} - 1}{2y} \\cdot \\frac{x^{2} + 1}{2x} = \\frac{3}{4},\\]  \n\nwhich simplifies to  \n\n\\[x y - \\frac{1}{x y} = \\frac{3}{2},\\]  \n\nfrom which it follows that \\(x y = 2\\) . Similarly, \\(y z - \\frac{1}{y z} = \\frac{5}{6}\\) and \\(z x - \\frac{1}{z x} = \\frac{21}{10}\\) , so \\(y z = \\frac{3}{2}\\) and \\(z x = \\frac{5}{2}\\) . Thus \\(x = \\sqrt{(2 \\cdot \\frac{5}{2}) / (\\frac{3}{2})} = \\sqrt{\\frac{10}{3}}\\) , and \\(a = \\frac{(10 / 3) - 1}{2\\sqrt{10 / 3}} = \\left[\\frac{7}{2\\sqrt{30}}\\right]\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n30. [16]", "solution_match": "\nSolution 2: "}}
 {"year": "2025", "tier": "T4", "problem_label": "31", "problem_type": null, "exam": "HMMT", "problem": "There exists a unique circle that is both tangent to the parabola \\(y = x^{2}\\) at two points and tangent to the curve \\(x = \\sqrt{\\frac{y^{3}}{1 - y}}\\) . Compute the radius of this circle.", "solution": "Answer: \\(\\frac{\\sqrt{5}}{2}\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n31. [16]", "solution_match": "\nProposed by: Karthik Venkata Vedula  \n\n"}}
-{"year": "2025", "tier": "T4", "problem_label": "31", "problem_type": null, "exam": "HMMT", "problem": "There exists a unique circle that is both tangent to the parabola \\(y = x^{2}\\) at two points and tangent to the curve \\(x = \\sqrt{\\frac{y^{3}}{1 - y}}\\) . Compute the radius of this circle.", "solution": 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 \n\nWe can square both sides of the second curve to get \\(x^{2} = \\frac{y^{3}}{1 - y}\\) , which further rearranges to  \n\n\\[\\frac{x^{2}}{(x^{2} + y^{2})^{2}} = \\frac{y}{x^{2} + y^{2}}.\\]  \n\nThis relation implies that curves \\(y = x^{2}\\) and \\(x^{2} = \\frac{y^{3}}{1 - y}\\) map to each other under inversion about the unit circle \\(x^{2} + y^{2} = 1\\) . Therefore, the unique circle we seek must be invariant under inversion about \\(x^{2} + y^{2} = 1\\) .  \n\nSince the circle is tangent to \\(y = x^{2}\\) , we know that the circle is of the form  \n\n\\[x^{2} + (y - y_{0})^{2} = r^{2}.\\]  \n\nWe know that the length of the tangent from \\((0,0)\\) to this circle is 1. Since the distance from \\((0,0)\\) to the center of the circle is \\(y_{0}\\) , using the Pythagorean Theorem gives \\(r^{2} + 1 = y_{0}^{2}\\) . Because the parabola \\(y = x^{2}\\) is tangent to this circle at two distinct points, the equation \\(x^{2} + (x^{2} - y_{0})^{2} = r^{2} = y_{0}^{2} - 1\\) must have two double roots. Therefore,  \n\n\\[x^{2} + (x^{2} - y_{0})^{2} - (y_{0}^{2} - 1) = x^{4} - (2y_{0} - 1)x^{2} + 1\\]  \n\nmust be a perfect square, so \\(2y_{0} - 1 = 2 \\implies y_{0} = \\frac{3}{2}\\) . This means \\(r^{2} = y_{0}^{2} - 1 = \\frac{5}{4}\\) , so the radius of the circle is \\(\\left[\\frac{\\sqrt{5}}{2}\\right]\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n31. [16]", "solution_match": "\nSolution:  \n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "31", "problem_type": null, "exam": "HMMT", "problem": "There exists a unique circle that is both tangent to the parabola \\(y = x^{2}\\) at two points and tangent to the curve \\(x = \\sqrt{\\frac{y^{3}}{1 - y}}\\) . Compute the radius of this circle.", "solution": "![md5:8ed7b13c95d2c464ecbb46a164de476c](8ed7b13c95d2c464ecbb46a164de476c.jpeg)\n  \n\nWe can square both sides of the second curve to get \\(x^{2} = \\frac{y^{3}}{1 - y}\\) , which further rearranges to  \n\n\\[\\frac{x^{2}}{(x^{2} + y^{2})^{2}} = \\frac{y}{x^{2} + y^{2}}.\\]  \n\nThis relation implies that curves \\(y = x^{2}\\) and \\(x^{2} = \\frac{y^{3}}{1 - y}\\) map to each other under inversion about the unit circle \\(x^{2} + y^{2} = 1\\) . Therefore, the unique circle we seek must be invariant under inversion about \\(x^{2} + y^{2} = 1\\) .  \n\nSince the circle is tangent to \\(y = x^{2}\\) , we know that the circle is of the form  \n\n\\[x^{2} + (y - y_{0})^{2} = r^{2}.\\]  \n\nWe know that the length of the tangent from \\((0,0)\\) to this circle is 1. Since the distance from \\((0,0)\\) to the center of the circle is \\(y_{0}\\) , using the Pythagorean Theorem gives \\(r^{2} + 1 = y_{0}^{2}\\) . Because the parabola \\(y = x^{2}\\) is tangent to this circle at two distinct points, the equation \\(x^{2} + (x^{2} - y_{0})^{2} = r^{2} = y_{0}^{2} - 1\\) must have two double roots. Therefore,  \n\n\\[x^{2} + (x^{2} - y_{0})^{2} - (y_{0}^{2} - 1) = x^{4} - (2y_{0} - 1)x^{2} + 1\\]  \n\nmust be a perfect square, so \\(2y_{0} - 1 = 2 \\implies y_{0} = \\frac{3}{2}\\) . This means \\(r^{2} = y_{0}^{2} - 1 = \\frac{5}{4}\\) , so the radius of the circle is \\(\\left[\\frac{\\sqrt{5}}{2}\\right]\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n31. [16]", "solution_match": "\nSolution:  \n\n"}}
 {"year": "2025", "tier": "T4", "problem_label": "32", "problem_type": null, "exam": "HMMT", "problem": "In the coordinate plane, a closed lattice loop of length \\(2n\\) is a sequence of lattice points \\(P_{0}, P_{1}, P_{2}, \\ldots , P_{2n}\\) such that \\(P_{0}\\) and \\(P_{2n}\\) are both the origin and \\(P_{i}P_{i + 1} = 1\\) for each \\(i\\) . A closed lattice loop of length 2026 is chosen uniformly at random from all such loops. Let \\(k\\) be the maximum integer such that the line \\(\\ell\\) with equation \\(x + y = k\\) passes through at least one point of the loop. Compute the expected number of indices \\(i\\) such that \\(0 \\leq i \\leq 2025\\) and \\(P_{i}\\) lies on \\(\\ell\\) .\n\n\n\n(A lattice point is a point with integer coordinates.)", "solution": "Answer: \\(\\frac{1013}{507}\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n32. [16]", "solution_match": "\nProposed by: Carlos Rodriguez, Jordan Lefkowitz  \n\n"}}
 {"year": "2025", "tier": "T4", "problem_label": "32", "problem_type": null, "exam": "HMMT", "problem": "In the coordinate plane, a closed lattice loop of length \\(2n\\) is a sequence of lattice points \\(P_{0}, P_{1}, P_{2}, \\ldots , P_{2n}\\) such that \\(P_{0}\\) and \\(P_{2n}\\) are both the origin and \\(P_{i}P_{i + 1} = 1\\) for each \\(i\\) . A closed lattice loop of length 2026 is chosen uniformly at random from all such loops. Let \\(k\\) be the maximum integer such that the line \\(\\ell\\) with equation \\(x + y = k\\) passes through at least one point of the loop. Compute the expected number of indices \\(i\\) such that \\(0 \\leq i \\leq 2025\\) and \\(P_{i}\\) lies on \\(\\ell\\) .\n\n\n\n(A lattice point is a point with integer coordinates.)", "solution": "We claim that if 2026 is replaced with \\(2n\\) , the answer is \\(\\frac{2n}{n + 1}\\) .  \n\nWrite the path as a sequence of \\(U\\) , \\(D\\) , \\(L\\) , and \\(R\\) moves. The possible sequences that can result are precisely those with an equal number of \\(U\\) 's and \\(D\\) 's, and an equal number of \\(R\\) 's and \\(L\\) 's. We first project this sequence onto a single dimension by converting each \\(U\\) and \\(R\\) to a 1, and each \\(D\\) and \\(L\\) to a \\(- 1\\) . The resulting sequence will have an equal number of 1's and \\(- 1\\) 's.  \n\nWe claim that every such sequence of \\(n\\) 1's and \\(n - 1\\) 's corresponds to the same number of closed lattice loops. Indeed, given such a sequence, a corresponding lattice loop can be made by replacing all 1's with \\(U\\) 's and \\(R\\) 's, and all \\(- 1\\) 's with \\(D\\) 's and \\(L\\) 's, so that there are an equal number of \\(U\\) 's and \\(D\\) 's. Nothing in this replacement process is order- dependent, so the number of ways to create this loop does not depend on the initial sequence.  \n\nWe can see that each of the 1's move the path towards \\(\\ell\\) and the \\(- 1\\) 's move the path away. Hence, we can think of the path as a one- dimensional walk, starting and ending in the same place, where the points on \\(\\ell\\) correspond exactly to the maxima of this one- dimensional walk.  \n\nUsing Catalan numbers, the probability that any given point is at the maxima is  \n\n\\[\\frac{\\frac{1}{n + 1}\\binom{2n}{n}}{\\binom{2n}{n}} = \\frac{1}{n + 1},\\]  \n\nand thus by linearity, the expected number of \\(i \\in [0, 2025]\\) such that \\(P_{i}\\) is on \\(\\ell\\) is \\(\\frac{2n}{n + 1}\\) . Plugging in \\(n = 1013\\) gives a final answer of \\(\\left[\\frac{1013}{507}\\right]\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n32. [16]", "solution_match": "\nSolution: "}}
 {"year": "2025", "tier": "T4", "problem_label": "33", "problem_type": null, "exam": "HMMT", "problem": "Estimate the total number of pages that teams submitted to the Team Round this year. (All pages associated to at least one problem number count as submitted pages, even blank cover sheets for a problem.)  \n\nSubmit a positive integer \\(E\\) . If the correct answer is \\(A\\) , you will receive \\(\\max \\left(0, \\left[20 \\left(1 - \\left(\\frac{|E - A|}{100}\\right)^{2 / 3}\\right)\\right]\\right)\\) points.", "solution": "Answer: \\(\\boxed{1003}\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-guts-solutions.jsonl", "problem_match": "\n33. [20]", "solution_match": "\nProposed by: Derek Liu  \n\n"}}