2009 OIM Problems/Problem 4

Problem

Let $ABC$ be a triangle with $AB \ne AC$ . Let $I$ be the incenter of $ABC$ and $P$ the other point of intersection of the exterior bisector of angle $A$ with the circumcircle of $ABC$ . The line $PI$ intersects for the second time the circumcircle of $ABC$ at point $J$ . Show that the circumcircles of triangles $JIB$ and $JIC$ are tangent to $IC$ and $IB$ , respectively.

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

Clearly $P$ is the midpoint of arc $BAC$ . Let $BI$ , $CI$ intersect the circumcircle of $ABC$ at $D$ , $E$ respectively. It is well known that $PDIE$ is a parallelogram. Therefore, $\angle ICJ=\angle ECJ=\angle EPJ=\angle BEJ$ , which implies BI tangent to the circumcircle of $JIC$ . Similarly, $CI$ is tangent to the circumcircle of $JIB$ .

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2009 OIM Problems/Problem 4

Problem

Solution

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