HMMT 二月 2002 · 冲刺赛 · 第 52 题

HMMT February 2002 — Guts Round — Problem 52

[ ± 8] 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 ?

解析

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 . To see 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 is the height from A to BD , and triangle DAB itself has area BD · h/ 2 = 2 · ( BD/ 2)( h/ 2). A similar computation holds in triangle BCD , proving the claim. Thus, the area of ABCD is at most 180 . And this maximum is attainable — just take a rectangle with AB = CD = 15 , BC = DA = 12.