, 2015

Geometry Difficulty 5.0 AIME, harder Prove it Romania

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.

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.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.