HMMT 十一月 2020 · 团队赛 · 第 8 题

HMMT November 2020 — Team Round — Problem 8

[50] Altitudes BE and CF of acute triangle ABC intersect at H . Suppose that the altitudes of triangle a 2 EHF concur on line BC . If AB = 3 and AC = 4 , then BC = , where a and b are relatively prime b positive integers. Compute 100 a + b .

解析

[50] Altitudes BE and CF of acute triangle ABC intersect at H . Suppose that the altitudes of triangle 2 a EHF concur on line BC . If AB = 3 and AC = 4 , then BC = , where a and b are relatively prime b positive integers. Compute 100 a + b . Proposed by: Jeffrey Lu Answer: 33725 Solution: A E F H B C P Let P be the orthocenter of 4 EHF . Then EH ⊥ F P and EH ⊥ AC , so F P is parallel to AC . Similarly, EP is parallel to AB . Using similar triangles gives BP CP AE AF AB cos A AC cos A 1 = + = + = + , BC BC AC AB AC AB 12 2 2 2 12 337 so cos A = . Then by the law of cosines, BC = 3 + 4 − 2(3)(4)( ) = . 25 25 25