Track / Stage 5 / 31 of 400 #631 of 1964

Problem 631

AIME late

Geometry Difficulty 5.0 Prove it Danube Mathematical Competition · Romania · 2015

Let $ABCD$ be a cyclic quadrangle, let the diagonals $AC$ and $BD$ cross at $O$ , and let $I$ and $J$ be the incentres of the triangles $ABC$ and $ABD$ , respectively. The line $IJ$ crosses the segments $OA$ and $OB$ at $M$ and $N$ , respectively. Prove that the triangle $OMN$ is isosceles.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

We show that $\angle OMN \equiv \angle ONM$ . To this end, let the line $IJ$ cross the segments $AD$ and $BC$ at $P$ and $Q$ , respectively, and consider the position of $J$ relative to the line $AC$ , to write $\angle$ OMN $\equiv \angle$ AJP $\pm \angle$ JAM $\equiv \angle$ AJP $\pm (\angle$ JAD $\mp \angle$ CAD) $\equiv \angle$ AJP + $\angle$ JAD - $\angle$ CAD $\equiv \angle$ AJP + $\angle$ BAJ - $\angle$ CAD. Similarly, $\angle ONM \equiv \angle BIQ + \angle ABI - \angle CBD$ . Since the quadrangle $ABCD$ is cyclic, $\angle CAD \equiv \angle CBD$ and $\angle ACB \equiv \angle ADB$ . The latter implies that $\angle AIB \equiv \angle AJB$ , so the quadrangle $ABIJ$ is cyclic. Consequently, $\angle AJP \equiv \angle ABI$ and $\angle BAJ \equiv \angle BIQ$ , and the conclusion follows.

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