HMMT 二月 2002 · 冲刺赛 · 第 52 题

英文解析

52. Let ABCD be a quadrilateral, and let E, F, G, H be the respective midpoints of
    AB, BC, CD, DA . If EG = 12 and F H = 15, what is the maximum possible area of
    ABCD ?
    Solution: The area of EF GH is EG · F H sin θ/ 2, where θ is the angle between EG and
    F H . This is at most 90. However, we claim the area of ABCD is twice that of EF GH . Tosee this, notice that EF = AC/ 2 = GH , F G = BD/ 2 = HE , so EF GH is a parallelogram.
    The half of this parallelogram lying inside triangle DAB has area ( BD/ 2)( h/ 2), where h isthe height from A to BD , and triangle DAB itself has area BD · h/ 2 = 2 · ( BD/ 2)( h/ 2). Asimilar computation holds in triangle BCD , proving the claim. Thus, the area of ABCD isat most 180 . And this maximum is attainable — just take a rectangle with AB = CD =
    15 , BC = DA = 12.