HMMT 二月 2019 · 冲刺赛 · 第 26 题

HMMT February 2019 — Guts Round — Problem 26

[ 15 ] Let ABC be a triangle with AB = 13 , BC = 14 , CA = 15. Let I , I , I be the A, B, C excenters A B C of this triangle, and let O be the circumcenter of the triangle. Let γ , γ , γ be the corresponding A B C excircles and ω be the circumcircle. X is one of the intersections between γ and ω . Likewise, Y is an A intersection of γ and ω , and Z is an intersection of γ and ω . Compute B C cos ∠ OXI + cos ∠ OY I + cos ∠ OZI . A B C 2 2 2

解析

[ 15 ] Let ABC be a triangle with AB = 13 , BC = 14 , CA = 15. Let I , I , I be the A, B, C excenters A B C of this triangle, and let O be the circumcenter of the triangle. Let γ , γ , γ be the corresponding A B C excircles and ω be the circumcircle. X is one of the intersections between γ and ω . Likewise, Y is an A intersection of γ and ω , and Z is an intersection of γ and ω . Compute B C cos ∠ OXI + cos ∠ OY I + cos ∠ OZI . A B C Proposed by: Andrew Gu 49 Answer: − 65 √ Let r , r , r be the exradii. Using OX = R, XI = r , OI = R ( R + 2 r ) (Euler’s theorem for A B C A A A A excircles), and the Law of Cosines, we obtain 2 2 R + r − R ( R + 2 r ) r A A A cos ∠ OXI = = − 1 . A 2 Rr 2 R A r + r + r A B C Therefore it suffices to compute − 3 . Since 2 R ( ) 1 1 1 1 8 abc abc r + r + r − r = 2 K + + − = 2 K = = 4 R A B C 2 − a + b + c a − b + c a + b − c a + b + c (4 K ) K r 65 where K = [ ABC ], this desired quantity the same as − 1. For this triangle, r = 4 and R = , so 2 R 8 4 49 the answer is − 1 = − . 65 / 4 65 2 2 2