HMMT 二月 2019 · 冲刺赛 · 第 17 题

HMMT February 2019 — Guts Round — Problem 17

[ 9 ] Let ABC be a triangle with AB = 3, BC = 4, and CA = 5. Let A , A be points on side BC , 1 2 B , B be points on side CA , and C , C be points on side AB . Suppose that there exists a point P 1 2 1 2 such that P A A , P B B , and P C C are congruent equilateral triangles. Find the area of convex 1 2 1 2 1 2 hexagon A A B B C C . 1 2 1 2 1 2

解析

[ 9 ] Let ABC be a triangle with AB = 3, BC = 4, and CA = 5. Let A , A be points on side BC , 1 2 B , B be points on side CA , and C , C be points on side AB . Suppose that there exists a point P 1 2 1 2 such that P A A , P B B , and P C C are congruent equilateral triangles. Find the area of convex 1 2 1 2 1 2 hexagon A A B B C C . 1 2 1 2 1 2 Proposed by: Michael Ren √ 12+22 3 Answer: 15 Since P is the shared vertex between the three equilateral triangles, we note that P is the incenter of ABC since it is equidistant to all three sides. Since the area is 6 and the semiperimeter is also 6, we can calculate the inradius, i.e. the altitude, as 1, which in turn implies that the side length of the 2 √ equilateral triangle is . Furthermore, since the incenter is the intersection of angle bisectors, it is 3 easy to see that AB = AC , BC = BA , and CA = CB . Using the fact that the altitudes from 2 1 2 1 2 1 P to AB and CB form a square with the sides, we use the side lengths of the equilateral triangle to 1 1 1 √ √ √ compute that AB = AC = 2 − , BA = BC = 1 − , and CB = CA = 3 − . We have that 2 1 1 2 1 2 3 3 3 the area of the hexagon is therefore ( ) √ ( ) ( ) ( ) 2 2 2 1 1 4 1 1 1 1 3 12 + 22 3 √ √ √ 6 − 2 − · + 1 − + 3 − · = . 2 5 2 2 5 15 3 3 3