PUMaC 2014 · 几何（B 组） · 第 1 题

PUMaC 2014 — Geometry (Division B) — Problem 1

[ 3 ] Triangle ABC has lengths AB = 20 , AC = 14 , BC = 22. The median from B intersects AC at M and the angle bisector from C intersects AB at N and the median from B at P . Let p [ AM P N ] = for positive integers p, q coprime. Note that [ ABC ] denotes the area of triangle q [ ABC ] ABC . Find p + q

解析

[ 3 ] Triangle ABC has lengths AB = 20 , AC = 14 , BC = 22. The median from B intersects AC at M and the angle bisector from C intersects AB at N and the median from B at P . Let p [ AM P N ] = for positive integers p, q coprime. Note that [ ABC ] denotes the area of triangle q [ ABC ] ABC . Find p + q Solution: [ BM C ] M C 1 [ BN C ] BN 11 We have that = = since BM is a median, and = = from the [ ABC ] AC 2 [ ABC ] AB 18 angle bisector theorem. Now, we find the area of BP C . [ BP C ] BP We see = . Using mass points, A is assigned a mass of 11 meaning that the mass at [ BM C ] BM BP 22 [ BN P M C ] 1 11 22 1 191 M is 22 and the mass at B is 7. Therefore, = . So = + · = . BM 29 [ ABC ] 2 18 29 2 261 [ AM P N ] 191 70 Then, = 1 = . Hence p + q = 331 [ ABC ] 261 261