HMMT 二月 2004 · 冲刺赛 · 第 31 题

HMMT February 2004 — Guts Round — Problem 31

[10] P is a point inside triangle ABC , and lines AP, BP, CP intersect the opposite ◦ sides BC, CA, AB in points D, E, F , respectively. It is given that ∠ AP B = 90 , and that AC = BC and AB = BD . We also know that BF = 1, and that BC = 999. Find AF .

解析

P is a point inside triangle ABC , and lines AP, BP, CP intersect the opposite sides ◦ BC, CA, AB in points D, E, F , respectively. It is given that ∠ AP B = 90 , and that AC = BC and AB = BD . We also know that BF = 1, and that BC = 999. Find AF . Solution: 499 / 500 B F A P D E C Let AC = BC = s , AB = BD = t . Since BP is the altitude in isosceles triangle ABD , it bisects angle B . So, the Angle Bisector Theorem in triangle ABC given AE/EC = AB/BC = t/s . Meanwhile, CD/DB = ( s − t ) /t . Now Ceva’s theorem gives us ( ) ( ) AF AE CD s − t = · = F B EC DB s AB s − t 2 s − t st ⇒ = 1 + = ⇒ F B = . F B s s 2 s − t Now we know s = 999, but we need to find t given that st/ (2 s − t ) = F B = 1. So st = 2 s − t ⇒ t = 2 s/ ( s + 1), and then 2 AF s − t ( s − s ) / ( s + 1) s − 1 499 AF = F B · = 1 · = = = . F B s s s + 1 500 9