HMMT 二月 2011 · ALGGEO 赛 · 第 10 题

英文解析

10. Let ABCD be a square of side length 13. Let E and F be points on rays AB and AD , respectively,
    so that the area of square ABCD equals the area of triangle AEF . If EF intersects BC at X and
    BX = 6, determine DF .
    √
    Answer: 13
    Y Cx f
    7 y d
    6X
    A B E
    Algebra & Geometry Individual Test
    First Solution
    Let Y be the point of intersection of lines EF and CD . Note that [ ABCD ] = [ AEF ] implies that
    [ BEX ] + [ DY F ] = [ CY X ]. Since 4 BEX ∼ 4 CY X ∼ 4 DY F , there exists some constant r such
    2 2 2 2 2 2
    that [ BEX ] = r · BX , [ Y DF ] = r · CX , and [ CY X ] = r · DF . Hence BX + DF = CX , so
    √ √
    √
    2 2
    DF = CX − BX = 49 − 36 = 13.
    Second Solution
    Let x = DF and y = Y D . Since 4 BXE ∼ 4 CXY ∼ 4 DF Y , we have
    BE CY DY y = = = .
    BX CX DF x
    6 y 13 − y y
    Using BX = 6, XC = 7 and CY = 13 − y we get BE = and = . Solving this last equationx 7 x
    13 xfor y gives y = . Now [ ABCD ] = [ AEF ] givesx +7
    ( )
    1 1 6 y
    169 = AE · AF = 13 + (13 + x ) .
    2 2 x
    78 y
    169 = 6 y + 13 x +
    6 x 78 x
    13 = + x +
    x + 7 x + 7
    0 = x − 13 .2
    √
    Thus x = 13. 2