HMMT 二月 2013 · 冲刺赛 · 第 23 题

英文解析

23. [ 14 ] Let ABCD be a parallelogram with AB = 8, AD = 11, and ∠ BAD = 60 . Let X be on segment
    CD with CX/XD = 1 / 3 and Y be on segment AD with AY /Y D = 1 / 2. Let Z be on segment ABsuch that AX , BY , and DZ are concurrent. Determine the area of triangle XY Z .
    √
    19 3
    Answer: Let AX and BD meet at P . We have DP/P B = DX/AB = 3 / 4. Now, applying
    Ceva’s Theorem in triangle ABD , we see that 2
    AZ DP AY 3 1 3 = · = · = .
    ZB P B Y D 4 2 8
    Now,
    [ AY Z ] [ AY Z ] 1 1 3 1 = = · · = ,
    [ ABCD ] 2[ ABD ] 2 3 11 22
    and similarly
    [ DY X ] 1 2 3 1 = · · = .
    [ ABCD ] 2 3 4 4
    Also,
    ( )
    [ XCBZ ] 1 1 8 43 = + = .
    [ ABCD ] 2 4 11 88
    The area of XY Z is the rest of the fraction of the area of ABCD not covered by the three abovepolygons, which by a straightforward calculation 19 / 88 the area of ABCD , so our answer is
    √
    19 19 3
    °
    8 · 11 · sin 60 · = .
    88 2