HMMT 二月 2004 · 冲刺赛 · 第 31 题
HMMT February 2004 — Guts Round — Problem 31
题目详情
英文原题
- [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 , andthat AC = BC and AB = BD . We also know that BF = 1, and that BC = 999.
Find AF .
解析
英文解析
- 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
PAFB
Let AC = BC = s , AB = BD = t . Since BP is the altitude in isosceles triangle cED
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 theoremgives 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. Sost = 2 s − t ⇒ t = 2 s/ ( s + 1), and then
AF s − t ( s − s ) / ( s + 1) s − 1 4992
AF = F B · = 1 · = = = .
F B s s s + 1 500 9