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HMMT 二月 2012 · 几何 · 第 1 题

HMMT February 2012 — Geometry — Problem 1

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

英文原题

Let ABC be a triangle with ∠A= 90◦,AB= 1, and AC= 2. Let ℓbe a line through Aperpendicular toBC, and let the perpendicular bisectors of ABandACmeet ℓatEandF, respectively. Find the length of segment EF.

解析

英文解析

  1. ABC is an isosceles triangle with AB = 2 and ∡ ABC = 90 . D is the midpoint of BC and E is on
    AC such that the area of AEDB is twice the area of ECD . Find the length of DE .
    √
    Let F be the foot of the perpendicular from E to BC . We have [ AEDB ] + [ EDC ] =17
    Answer:
    2 1 43
    [ ABC ] = 2 ⇒ [ EDC ] = . Since we also have [ EDC ] = ( EF )( DC ) , we get EF = F C = . So
    3 2 3
    √
    1 17
    F D = , and ED = by the Pythagorean Theorem.
    3 3
    °