Answer:A triangle ABC with positive integer side lengths has perimeter 35, centroid G, and incenter I. If ∠GIC=90∘, what is the length of AB¯¯¯¯¯¯¯¯?
We use barycentric coordinates with reference triangle △ABC. Letting k=a+b+c, we have the displacement vectors GI→=(13−ak,13−bk,13−ck) and IC→=(ak,bk,ck−1). EFFT gives us
a2(4b+c3k−13−2bck2)+b2(4a+c3k−13−2ack2)+c2(a+b3k−2abk2)=0
. Multiplying by 3k2, writing k=a+b+c, and expanding out, the equation turns into
a2((a+b+c)(3b−a)−6bc)+b2((a+b+c)(3a−b)−6ac)+c2((a+b)(a+b+c)−6ab)=0.
Expanding further and factoring an a+b+c out, we have
(a+b+c)((a+b)(4ab−a2−b2+c2)−6abc)=0.
Notice that when c=a+b the term (a+b)(4ab−a2−b2+c2)−6abc) vanishes, so we factor further to obtain
(a+b+c)(c−a−b)(a2+b2+c(a+b)−4ab)=0⟹a2+b2+c(a+b)−4ab=0⟹(a+b)2+c(a+b)=6ab⟹35(35−c)=6ab.
From the triangle inequality, we have c≤17 and we also know 6|35−c; it's now easy to find a triple (a,b,c)=(10,14,11) satisfying the equation above, and the answer is 11.
Hope this helps......