HMMT 二月 2014 · 冲刺赛 · 第 19 题

英文解析

19. [ 11 ] Let ABCD be a trapezoid with AB ‖ CD . The bisectors of ∠ CDA and ∠ DAB meet at E , thebisectors of ∠ ABC and ∠ BCD meet at F , the bisectors of ∠ BCD and ∠ CDA meet at G , and thebisectors of ∠ DAB and ∠ ABC meet at H . Quadrilaterals EABF and EDCF have areas 24 and 36,
    respectively, and triangle ABH has area 25. Find the area of triangle CDG .
    Answer: Let M, N be the midpoints of AD, BC respectively. Since AE and DE are bisectors 256
    of supplementary angles, the triangle AED is right with right angle E . Then EM is the median of 7
    a right triangle from the right angle, so triangles EM A and EM D are isosceles with vertex M . Butthen ∠ M EA = ∠ EAM = ∠ EAB , so EM ‖ AB . Similarly, F N ‖ BA . Thus, both E and F are onthe midline of this trapezoid. Let the length of EF be x . Triangle EF H has area 1 and is similar totriangle ABH , which has area 25, so AB = 5 x . Then, letting the heights of trapezoids EABF and
    6 xh
    EDCF be h (they are equal since EF is on the midline), the area of trapezoid EABF is = 24. So
    9 xh 2
    the area of trapezoid EDCF is 36 = . Thus DC = 8 x . Then, triangle GEF is similar to and has
    1 642
    times the area of triangle CDG . So the area of triangle CDG is times the area of quadrilateral
    64 63
    EDCF , or .256 7