HMMT 二月 2014 · 冲刺赛 · 第 14 题

HMMT February 2014 — Guts Round — Problem 14

[ 8 ] Let ABCD be a trapezoid with AB k CD and \ D = 90 . Suppose that there is a point E on CD such that AE = BE and that triangles AED and CEB are similar, but not congruent. Given that CD BC = 2014, find . AB AD

解析

[ 8 ] Let ABCD be a trapezoid with AB ‖ CD and ∠ D = 90 . Suppose that there is a point E on CD such that AE = BE and that triangles AED and CEB are similar, but not congruent. Given that CD BC = 2014, find . AB AD √ Answer: 4027 Let M be the midpoint of AB . Let AM = M B = ED = a , M E = AD = b , and AE = BE = c . Since 4 BEC ∼ 4 DEA , but 4 BEC is not congruent to 4 DAE , we must have 4 BEC ∼ 4 DEA . Thus, BC/BE = AD/DE = b/a , so BC = bc/a , and CE/EB = AE/ED = c/a , 2 c 2 2

a 2 c 1 c a so EC = c /a . We are given that CD/AB = = + = 2014 ⇒ = 4027. Thus, 2 2 2 a 2 a 2 a √ bc/a BC/AD = = c/a = 4027. b