Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.1 National olympiad Prove it Belarus

Let $O$ be the circumcenter of an acute-angled triangle $ABC$ . Let $AH$ be the altitude of this triangle, $M$ , $N$ , $P$ , $Q$ be the midpoints of the segments $AB$ , $AC$ , $BH$ , $CH$ , respectively.

Let $\omega_1$ and $\omega_2$ be the circumcircles of the triangles $AMN$ and $POQ$ .
Prove that one of the intersection points of $\omega_1$ and $\omega_2$ belongs to the altitude $AH$ .

Solution

Let $M$ , $N$ and $R$ be the midpoints of the sides $AB$ , $AC$ , $BC$ , respectively. Let $X$ denote the foot of the perpendicular from $O$ on $AH$ . We claim that $X$ is the point of intersection of $\omega_1$ and $\omega_2$ .

Since $O$ is the center of the circumcircle of the triangle $ABC$ , we have $OR \perp BC$ , $ON \perp AC$ and $OM \perp AB$ . So $O$ , $M$ , $A$ , $N$ and $X$ lie on the same circle $\omega_1$ , and $OA$ is the diameter of $\omega_1$ .

Show that $PR = HQ$ . Indeed,
$PR = BR - BP = \frac{1}{2}(BC - BH) = \frac{1}{2}HC = HQ.$
From $PR = HQ$ and $OX \parallel BC$ it follows that $POXQ$ is an isosceles trapezium. Therefore $X$ belongs to $\omega_2$ .

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