Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.5 AIME Prove it Taiwan

Given that acute triangle $\triangle ABC$ is not an isosceles triangle, points $O$ and $I$ are the circumcenter and incenter of $\triangle ABC$ respectively. The incircle of $\triangle ABC$ is tangent to the three sides $BC$ , $CA$ , $AB$ at points $D$ , $E$ , $F$ respectively. If line $AI$ intersects $OD$ at point $P$ , $BI$ intersects $OE$ at point $Q$ , and $CI$ intersects $OF$ at point $R$ , and $M$ is the circumcenter of $\triangle PQR$ . Prove that: $I$ , $M$ , $O$ are collinear.

Solution

(i) Let $R$ , $r$ denote the circumradius and inradius of $\triangle ABC$ respectively. First we prove that $OP : PD = R : r$ .

The proof is as follows. Extend $AP$ to meet the circumcircle of $\triangle ABC$ at $A'$ . Since $AI$ bisects $\angle BAC$ , $A'$ is the midpoint of arc $BA'C$ , so $OA'$ is perpendicular to $BC$ . Also, since $BC$ is tangent to the incircle at $D$ , $ID$ is perpendicular to $BC$ , hence $ID$ is parallel to $OA'$ . Thus $\triangle IPD \sim \triangle A'PO$ , and therefore $OP : PD = OA' : ID = R : r$ . This proves the claim.

Next we prove the original proposition. By (i), $OP : PD = OQ : QE = OR : RF = R : r$ , so $\triangle PQR$ is the image of $\triangle DEF$ under a homothety centered at $O$ with ratio $OP : OD = R : (R+r)$ .

Therefore, the center of the homothety $O$ , the circumcenter $I$ of $\triangle DEF$ , and the circumcenter $M$ of $\triangle PQR$ are collinear. This completes the proof!

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from zh; metadata (topic, difficulty) added by this project.