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HMMT 二月 2003 · 几何 · 第 2 题

HMMT February 2003 — Geometry — Problem 2

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

英文原题

  1. As shown, U and C are points on the sides of triangle M N H such that M U = s ,
    U N = 6, N C = 20, CH = s , HM = 25. If triangle U N C and quadrilateral M U CHhave equal areas, what is s ?
    6N
    U20
    s c ms
    H25
解析

英文解析

  1. As shown, U and C are points on the sides of triangle M N H such that M U = s ,
    U N = 6, N C = 20, CH = s , HM = 25. If triangle U N C and quadrilateral M U CHhave equal areas, what is s ?
    6N
    U20
    s c ms
    H25
    Solution: 4
    Using brackets to denote areas, we have [ M CH ] = [ U N C ] + [ M U CH ] = 2[ U N C ]. Onthe other hand, triangles with equal altitudes have their areas in the same ratio astheir bases, so
    [ M N H ] [ M N H ] [ M N C ] N H M N s + 20 s + 6
    2 = = · = · = · .
    [ U N C ] [ M N C ] [ U N C ] N C U N 20 6
    Clearing the denominator gives ( s + 20)( s + 6) = 240, and solving the quadratic givess = 4 or − 30. Since s > 0, we must have s = 4. 1