HMMT 二月 2021 · 团队赛 · 第 9 题
HMMT February 2021 — Team Round — Problem 9
题目详情
- [90] Let scalene triangle ABC have circumcenter O and incenter I . Its incircle ω is tangent to sides BC , CA , and AB at D , E , and F , respectively. Let P be the foot of the altitude from D to EF , and let line DP intersect ω again at Q 6 = D . The line OI intersects the altitude from A to BC at T . Given that OI ‖ BC , show that P Q = P T .
解析
- [90] Let scalene triangle ABC have circumcenter O and incenter I . Its incircle ω is tangent to sides BC , CA , and AB at D , E , and F , respectively. Let P be the foot of the altitude from D to EF , and let line DP intersect ω again at Q 6 = D . The line OI intersects the altitude from A to BC at T . Given that OI ‖ BC , show that P Q = P T . Proposed by: Carl Schildkraut, Milan Haiman Solution: Let H be the orthocenter of ∆ DEF . We first claim that O, I, H are collinear. We present two proofs. Proof 1. Invert about ω . Circle ( ABC ) inverts to a circle with center on OI , but A, B, C invert to the midpoints of EF, F D, DE , respectively, so the nine-point center of ∆ DEF is on OI . As this center is the midpoint of IH , we get that H, I, O are collinear. Proof 2. Let Q , Q , Q be the second intersections of the D − , E − , and F − altitudes, respectively, A B C in ∆ DEF with ω . We claim ∆ Q Q Q is homothetic with ∆ ABC . Indeed, as Q is the reflection A B C B of H over DF and Q is the reflection of H over DE , DQ = DQ , so the perpendicular bisector of C B C Q Q is line ID . As ID ⊥ BC , Q Q || BC , whence the homothety follows. This homothety takes B C B C the incircle to the circumcircle, so it is centered on line OI . However, it also takes the incenter H of Q Q Q to the incenter I of ABC , so it is centered on line IH . So, O, I, H are collinear. A B C As P is the midpoint of QH , it suffices to show that P is on the circle with diameter QH , or that ◦ ∠ QT H = 90 . As AT ⊥ T H = IO , it suffices to show that Q is on line AT . We also present two proofs of this. ′ ′ Proof 1. Let D be the antipode of D , and let AD intersect BC at X . As X is the A -extouch point, the midpoint M of DX is also the midpoint of BC . We have ′ OM ID DD = = DX M X DX 2 ′ ◦ ′ ′ ′ and ∠ OM X = ∠ D DX = 90 , so D , O, X are collinear, so D is on line AO . As QD || EF , AQ and ′ AD are isogonal in ∠ BAC , so AQ and AO are isogonal, which means Q is on the A -altitude, finishing the proof. Proof 2. Let Γ denote the circumcircle of ∆ ABC , and let M be the midpoint of arc BC on Γ not containing A . ′ Lemma. The intersection T of M D and the A -altitude to BC is on the line through I parallel to BC . ′ ′ ′ Proof. Let D = M A ∩ BC . As ∠ D BM = ∠ CBM = ∠ CAM = ∠ M AB , 4 D BM ∼ 4 BAM , and 2 2 ′ M I = M B = M D · M A. ′ Since AT ‖ ID , we have ′ M T M A M I = = , ′ M D M I M D ′ ′ so IT || DD , finishing the proof. By the above lemma, T is on M D . Consider a homothety centered at T that takes D to M . It takes ω to a circle centered on line IT that is tangent to Γ at M ; since O is on line IT this circle must be Γ itself. So, T is the exsimilicenter of Γ and ω . By Proof 2 above, T is the center of the homothety which sends Q Q Q to ABC , so T , Q = Q , and A are collinear, finishing the proof. A B C A