HMMT 二月 2026 · 冲刺赛 · 第 30 题

HMMT February 2026 — Guts Round — Problem 30

[16] Let ABC be a triangle with AB = 60 , AC = 67 , and BC = 69 . The incircle ω of triangle ABC ′ touches sides BC , CA , and AB at D , E , and F , respectively. Let D be the point diametrically opposite ′ ′ to D in ω . Let the common chord of the circumcircles of triangles BD F and CD E meet line BC at X . Compute BX . 2 π 2 π 3 2

解析

[16] Let ABC be a triangle with AB = 60 , AC = 67 , and BC = 69 . The incircle ω of triangle ′ ABC touches sides BC , CA , and AB at D , E , and F , respectively. Let D be the point diametrically ′ ′ opposite to D in ω . Let the common chord of the circumcircles of triangles BD F and CD E meet line BC at X . Compute BX . Proposed by: Pitchayut Saengrungkongka Answer: 45 Solution: A ′ D E F ′ I I B C X D P The crux of the problem lies in the following claim. ©2026 HMMT ′ ′ Claim 1. If I is the reflection of I across perpendicular bisector of BC , then I lies on the radical axis. ′ ′ Proof. Let P be second intersection of ⊙ ( BD F ) and ⊙ ( CD E ) . Note that ∠ A ′ ′ ′ ′ ◦ ∠ BP C = ∠ BP D + ∠ CP D = ∠ AF D + ∠ AED = 90 − , 2 so P ∈ ⊙ ( BIC ) . Finally, we note that ∠ B ′ ′ ′ ′ ∠ BP I = ∠ BCI = = ∠ AF D = ∠ BP D , 2 ′ ′ so I , D , P are collinear. Once we have this, the answer extraction is strikingly clean. We easily compute s = 98 , so BD = 31 ′ and CD = 38 . If D is the foot from I to BC , then BD = 38 , so DD = 7 . Finally, D is the 1 1 1 1 midpoint of DX , so XD = 7 , so BX = BD + D X = 38 + 7 = 45 . 1 1 1 2 π 2 π 3 2