PUMaC 2013 · 几何（A 组） · 第 2 题

PUMaC 2013 — Geometry (Division A) — Problem 2

[ 3 ] An equilateral triangle is given. A point lies on the incircle of the triangle. If the smallest two distances from the point to the sides of the triangle is 1 and 4, the sidelength of this √ a b equilateral triangle can be expressed as where ( a, c ) = 1 and b is not divisible by the c square of an integer greater than 1. Find a + b + c .

解析

[ 3 ] An equilateral triangle is given. A point lies on the incircle of the triangle. If the smallest two distances from the point to the sides of the triangle is 1 and 4, the sidelength of this p a b equilateral triangle can be expressed as where ( a, c ) = 1 and b is not divisible by the c square of an integer greater than 1. Find a + b + c . Solution Let the triangle be ABC , and let the point on incircle be P . Let the feet of the two shorter pedals be X and Y , and assume P is closest to A among A, B, C . Then AXP Y p p is a cyclic quadrilateral so \ XP Y = 120 . We have from this that XY = 21 ) AP = 2 7. p 14 28 3 Setting up an equation on r , the inradius, we get r = , and thus the sidelength is . 3 3