HMMT 二月 2014 · 冲刺赛 · 第 14 题
HMMT February 2014 — Guts Round — Problem 14
题目详情
英文原题
- [ 8 ] Let ABCD be a trapezoid with AB k CD and \ D = 90 . Suppose that there is a point E on CDsuch that AE = BE and that triangles AED and CEB are similar, but not congruent. Given that
CD BC = 2014, find .
AB AD
解析
英文解析
- [ 8 ] Let ABCD be a trapezoid with AB ‖ CD and ∠ D = 90 . Suppose that there is a point E on CDsuch 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 ,
c 2
2 2 + a
2 c 1 caso 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