PUMaC 2016 · 团队赛 · 第 7 题
PUMaC 2016 — Team Round — Problem 7
题目详情
- (5) In triangle ABC , let S be on BC and T be on AC so that AS ⊥ BC and BT ⊥ AC , and let AS and BT intersect at H . Let O be the center of the circumcircle of 4 AHT , P be the center of the circumcircle of 4 BHS , and G be the other point of intersection (besides H ) of the two circles. Let GH and OP intersect at X . If AB = 14, BH = 6, and HA = 11, then m XO − XP can be written in simplest form as . Find m + n . n
解析
- Since AHT and BHS are right triangles, O is the midpoint of AH and P is the midpoint of BH . This means that ∠ BGH and ∠ AGH are right angles, so G is the foot of the altitude 1 from C to AB . Note also that 4 HOP ∼ 4 HAB with factor . We thus have 2 2 2 2 2 2 2 2 2 XO − XP XO − XP ( HO − HX ) − ( HP − HX ) XO − XP = = = XO − XP 7 7 2 2 2 2 HO − HP 11 − 6 85 = = = , 7 28 28 so our answer is 85 + 28 = 113 . Problem written by Eric Neyman. 1 109