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Taiwan IMO Selection Camp

Geometry Difficulty 4.5 AIME Prove it Taiwan

In triangle $ABC$ , $BC > AB$ . Let $L$ be the internal angle bisector of $\angle ABC$ . From $A, C$ draw perpendiculars to $L$ , and let the feet of these perpendiculars be $P, Q$ respectively. Let $M, N$ be the midpoints of side $AC$ and side $BC$ respectively. Let the circumcenter of triangle $PQM$ be $O$ , and let the second intersection point of this circle with $AC$ be $H$ . Prove that $O, M, N, H$ are concyclic.

Solution

Extend $AP$ to meet $BC$ at point $D$ ; then $P$ is the midpoint of $AD$ . Since $M$ is the midpoint of $AC$ , $PM$ is parallel to $CD$ (i.e., $BC$ ), and
$\angle QPM = \angle QBC = \frac{1}{2} \angle ABC.$
By the same reasoning, $\angle MQP = \frac{1}{2}\angle ABC$ , so $PM = QM$ .

$\angle QHC = \angle QHM = \angle QPM,$
that is, $\angle QHC = \angle QBC$ . Hence $Q,H,B,C$ are concyclic, and
$\angle BHC = \angle BQC = 90^\circ,$
so we get $HN = \frac{1}{2}BC = NQ$ .
Since $OH = OQ$ , $ON$ is the perpendicular bisector of segment $HQ$ . Also, since $\angle MQP = \frac{1}{2}\angle ABC$ and $N$ is the midpoint of $BC$ , we obtain that $Q,M,N$ are three collinear points. Therefore
$\angle NHO = \angle NQO = \angle MQO = \angle OMQ,$
which shows that $O,M,N,H$ are concyclic. 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.