HMMT 二月 2023 · 团队赛 · 第 9 题

HMMT February 2023 — Team Round — Problem 9

[75] Let ABC be a triangle with AB < AC . The incircle of triangle ABC is tangent to side BC at D and intersects the perpendicular bisector of segment BC at distinct points X and Y . Lines AX 2 and AY intersect line BC at P and Q , respectively. Prove that, if DP · DQ = ( AC − AB ) , then AB + AC = 3 BC .

解析

[75] Let ABC be a triangle with AB < AC . The incircle of triangle ABC is tangent to side BC at D and intersects the perpendicular bisector of segment BC at distinct points X and Y . Lines AX 2 and AY intersect line BC at P and Q , respectively. Prove that, if DP · DQ = ( AC − AB ) , then AB + AC = 3 BC . Proposed by: Luke Robitaille ′ Solution: Let E be the extouch point on BC , let I be the incenter, and D the reflection of D over 2 I . Note that DE = AC − AB , so DP · DQ = DE . Now let F be the reflection of E across D . The length condition implies ( E, F ; P, Q ) is a harmonic bundle. We also know XY is the perpendicular ′ bisector of DE , so the midpoint M of D E lies on XY . But then IM ‖ BC , so IM ⊥ XY , and M is ′ the midpoint of XY . Since A, D , E are collinear, this means AE bisects XY . Now consider projecting ( E, F ; P, Q ) onto XY . P and Q are taken to X and Y , while E is taken to the midpoint of XY . Thus, F is taken to the point at infinity, so AF ⊥ BC . Now since D is the midpoint of EF , we see that 1 a + b + c ′ AF = 2 DD , or h = 2 r , where h is the height from A and r is the inradius. But ah = r , so a a a 2 2 a + b + c a = , or 3 a = b + c , as desired. 2