HMMT 二月 2017 · 几何 · 第 9 题

HMMT February 2017 — Geometry — Problem 9

Let ABC be a triangle, and let BCDE, CAF G, ABHI be squares that do not overlap the triangle with centers X, Y, Z respectively. Given that AX = 6, BY = 7, and CZ = 8, find the area of triangle XY Z .

解析

Let ABC be a triangle, and let BCDE, CAF G, ABHI be squares that do not overlap the triangle with centers X, Y, Z respectively. Given that AX = 6, BY = 7, and CZ = 8, find the area of triangle XY Z . Proposed by: Sam Korsky √ 21 15 Answer: 4 By the degenerate case of Von Aubel’s Theorem we have that Y Z = AX = 6 and ZX = BY = 7 and √ 21 15 XY = CZ = 8 so it suffices to find the area of a 6 − 7 − 8 triangle which is given by . 4 To prove that AX = Y Z , note that by LoC we get 2 2 b c 2 Y X = + + bc sin ∠ A 2 2 and 2 a 2 2 AX = b + − ab (cos ∠ C − sin ∠ C ) 2 2 a 2 = c + − ac (cos ∠ B − sin ∠ B ) 2 2 2 b + c + a ( b sin ∠ C + c sin ∠ B ) = 2 2 2 b c = + + ah 2 2 where h is the length of the A -altitude of triangle ABC . In these calculations we used the well-known fact that b cos ∠ C + c cos ∠ B = a which can be easily seen by drawing in the A -altitude. Then since bc sin ∠ A and ah both equal twice the area of triangle ABC , we are done.