PUMaC 2015 · 团队赛 · 第 9 题

PUMaC 2015 — Team Round — Problem 9

[ 6 ] Triangle ABC has AB = 5 , BC = 4 , CA = 6. Points D and E are on sides AB and AC , respectively, such that AD = AE = BC . Let CD and BE intersect at F and let AF and DE √ a b intersect at G . The length of F G can be expressed in the form in simplified form. What c is a + b + c ?

解析

[ 6 ] Triangle ABC has AB = 5 , BC = 4 , CA = 6. Points D and E are on sides AB and AC , respectively, such that AD = AE = BC . Let CD and BE intersect at F and let AF and DE √ a b intersect at G . The length of F G can be expressed in the form in simplified form. What c is a + b + c ? Solution: We use mass points: we assign a mass of 1 to A , a mass of 4 to B , and a mass of 2 1 to C . Then, the mass of D , E , and F are 5, 3, and 7, respectively. Then, F G = AG . 7 3 2 2 2 From the law of cosines, we have that 4 = 5 + 6 − 2 · 5 · 6 · cos A ⇔ cos A = . Let P be 4 the foot of the perpendicular from D to AE , so AP = 3 and by the Pythagorian theorem, √ √ √ DG 3 3 2 5 2 DE = 2 2. Then, since = , we have DG = and EG = . EG 5 4 4 By Stewart’s theorem on 4 ADE , we have: √ √ √ √ √ √ 5 2 3 2 3 2 5 2 2 2 2 4 · + 4 · = · · 2 2 + AG · 2 2 4 4 4 4 √ √ 226 1 226 So AG = and F G = , thus a + b + c = 255 . 4 28 Author: Bill Huang