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HMMT 十一月 2025 · THM 赛 · 第 8 题

HMMT November 2025 — THM Round — Problem 8

专题
Discrete Math / 离散数学
难度
L3
来源
HMMT

题目详情

  1. Let M ARS be a trapezoid with M A parallel to RS and side lengths M A = 11, AR = 17, RS = 22, and SM = 16. Point X lies on side M A such that the common chord of the circumcircles of triangles M XS and AXR bisects segment RS . Compute M X .
解析
  1. Let M ARS be a trapezoid with M A parallel to RS and side lengths M A = 11, AR = 17, RS = 22, and SM = 16. Point X lies on side M A such that the common chord of the circumcircles of triangles M XS and AXR bisects segment RS . Compute M X . Proposed by: Pitchayut Saengrungkongka 17 Answer: = 8 . 5 2 © 2025 HMMT Solution 1: Let Y be the second intersection of the two circumcircles. Let N = XY ∩ SR , P = M S ∩ AR , and construct Q on the line through P parallel to SR so that P QRS is an isosceles trapezoid, which is cyclic. Then we have that Y lies on this circle because ∠ SY R = ∠ SY X + ∠ XY R = ∠ P M X + ∠ P AX ◦ = 180 − ∠ SP R, so P SY RQ is cyclic. We also have ∠ QY R = ∠ QP R = ∠ P AX = ∠ XY R so Q , X , and Y are collinear. Q P A M X N R S Y Note that since △ P AM ∼ △ P RS and RS = 2 AM , we have QR = P S = 2 M S = 32 and QS = P R = 2 AR = 34. By Ptolemy’s on P QRS , we obtain 2 2 P R · QS − P S · QR 34 − 32 P Q = = = 6 . RS 22 Finally, P Q + SN 6 + 11 17 M X = = = . 2 2 2 Solution 2: Let the circumcircles of △ M XS and △ AXR intersect line SR again at Y and Z , respectively. Let N be the midpoint of SR . Because N lies on the radical axis of the two circles, we have SN · N Y = RN · N Z , so Y N = ZN . As a result, SY = RZ . © 2025 HMMT A M X S Z N Y R Let F and F be the feet of altitudes from M and A to SR , respectively. By Pythagoras’ theorem, M A q q 2 2 2 2 2 2 2 2 16 − SF = 17 − RF = ⇒ RF − SF = 17 − 16 = 33 . A M M A Since SF + RF = SR − M A = 11, we get that RF − SF = 33 / 11 = 3, so SF = 4 M A A M M and RF = 7. Because quadrilaterals M XSY and AXZR are both isosceles trapezoids, we have A SY = M X + 2 · SF = M X + 8 and RZ = AX + 2 · RF = AX + 14. Therefore, M X + 8 = AX + 14. M A Since M X + AX = M A = 11, we obtain that M X = 8 . 5 .