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Spring Mathematical Tournament

Geometry Difficulty 6.1 National Olympiad Prove it Bulgaria

(Stoyan Boev) The incircle of an acute $\triangle ABC$ touches the sides $AB$ , $BC$ and $CA$ at points $P$ , $Q$ and $R$ , respectively. The orthocenter $H$ of $\triangle ABC$ lies on the segment $QR$ .
a) Prove that $PH \perp QR$ .
b) Let $I$ and $O$ be the incenter and circumcenter of $\triangle ABC$ , and $N$ the common point of $AB$ and the excircle to this side. Prove that the points $I$ , $O$ and $N$ are collinear.

Solution

a) Since $\angle RAH = \angle QBH$ and
$\angle ARH = 180^{\circ} - \angle CRQ = 180^{\circ} - \angle CQR = \angle BQH,$
then $\triangle ARH \sim \triangle QBH$ . Hence
$\frac{AH}{BH} = \frac{AR}{BQ} = \frac{AP}{BP}$
and $HP$ is the bisector of $\angle AHB$ . Then
$\angle RHP = \angle RHA + \angle AHP = \angle QHB + \angle BHP = \angle QHP$
which implies that $PH \perp RQ$ .

b) If $AC = AB$ , then $N \equiv P \equiv M$ and the points $I, O$ and $N$ lie on the bisector of $AB$ .
Let now $AC \neq AB$ . Since $PH \perp RQ$ and $CI \perp RQ$ , it follows that $HP \parallel CI$ . On the other hand, $CH \parallel IP$ and hence $CHPI$ is a parallelogram. Then $CH = IP$ . If $M$ is the midpoint of $AB$ , then $CH = 2OM = 2R \cos \gamma$ and so $IP = 2OM$ . Since $AP = BN$ , then $M$ is the midpoint of $PN$ . Therefore $O$ is the midpoint of $IN$ .

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