Let $I$ be the incenter of the triangle $ABC$. Let $A_1,A_2$ be two distinct points on the line $BC$, let $B_1,B_2$ be two distinct points on the line $CA$, and let $C_1,C_2$ be two distinct points on the line $BA$ such that $AI = A_1I = A_2I$ and $BI = B_1I = B_2I$ and $CI = C_1I = C_2I$. Prove that $A_1A_2+B_1B_2+C_1C_2 = p$ where $p$ denotes the perimeter of $ABC.$ Pierre.
Problem
Source: Me
Tags: geometry, incenter, perimeter, trigonometry
Sailor
21.12.2004 19:51
We'll use the well-known fact: In any triangle ABC we have that the projection of O(circumcenter of ABC)to BC is equal to RcosA. Using for triangle A1AA2 we have r=AIcosA1AA2 hence cosA1AA2=2*sin <sup>2</sup>(A/2)-1. Using cosine theorem in A1AA2: A1A2=2(p-a). Adding similar conclusions we get the desired result.
Valiowk
22.12.2004 16:57
Let $D$ and $E$ be the feet of the perpendiculars from $I$ to $AB$ and $BC$ respectively. Then it's obvious that $\triangle ADI \equiv \triangle A_1EI$, so $AD = A_1E$. Similarly for sides $CA$ and $AB$.
kennyworm
29.12.2004 17:13
i thought it was quite a cute cutting and pasting of the triangle. kennyworm
kueh
21.01.2005 23:03
when i solved it, it didn't occur to me. but now its obvious (= i just used pythagoras without noticing the congruent triangles.
Valiowk
22.01.2005 15:41
3 consecutive posts by the Singapore team! Wow!