HMMT 二月 2026 · 冲刺赛 · 第 5 题

HMMT February 2026 — Guts Round — Problem 5

[6] Let ABC be a right triangle with ∠ ABC = 90 and AB < BC . Let M be the midpoint of AC . Let T ◦ be the unique point lying on the segment BC such that ∠ BM T = 90 . Given that AB = 5 and M T = 3 , compute CT .

解析

[6] Let ABC be a right triangle with ∠ ABC = 90 and AB < BC . Let M be the midpoint of AC . ◦ Let T be the unique point lying on the segment BC such that ∠ BM T = 90 . Given that AB = 5 and M T = 3 , compute CT . Proposed by: Jason Mao √ 7 7 11 √ Answer: = 11 11 Solution: A M C B T Since M is the midpoint of AC , AM = M B = M C . In particular, triangle M BC is isosceles, so ◦ ∠ M BT = ∠ ACB . We also have ∠ T M B = ∠ ABC = 90 , so △ ABC ∼ △ T M B . Let AM = M B = M C = 5 x . The similarity yields 3 T B = = ⇒ T B = 6 x. 5 10 x Pythagorean’s Theorem on △ M BT gives 3 2 2 2 √ (6 x ) = (5 x ) + 3 = ⇒ x = . 11 ©2026 HMMT AB Note that BC = M B · , so T M 5 7 7 √ CT = BC − T B = 5 x · − 6 x = x = . 3 3 11