PUMaC 2016 · 几何（A 组） · 第 4 题

PUMaC 2016 — Geometry (Division A) — Problem 4

Let 4 ABC be a triangle with integer side lengths such that BC = 2016. Let G be the centroid of 4 ABC and I be the incenter of 4 ABC . If the area of 4 BGC equals the area of 4 BIC , find the largest possible length of AB .

解析

The length of the altitude from I to BC , i.e. the inradius r , equals the length of the altitude from G to BC , which is one-third the height of the triangle, which we will call h . We thus have ah rp hp [ ABC ] = = = , 2 2 6 where a = BC = 2016 and p is the perimeter of the triangle, so p = 3 a , and indeed if p = 3 a then the conditions of the problem are satisfied. This means that A must lie on the ellipse with foci B and C such that AB + AC = 2 a , implying a major axis of length 2 a . Thus, 3 a AB < = 3024. Since AB is an integer, AB is at most 3023 . 2 Problem written by Eric Neyman.