Iranian Mathematical Olympiad

Let $O$ be the center of $\omega$, and $H'$ be the orthocenter of triangle $AXY$. Denote the intersection of $AH'$ and $OM$ as $D$. It is evident that quadrilateral $AHMD$ is a parallelogram, so $2OM = AH = MD$. This result shows that $D$ is a fixed point. If we prove $\frac{AH'}{AD}$ is constant, it follows that with the movement of $A$ on $\omega$, $H'$ moves on a circle such that $D$ is the homothety center between the two circles.

Lemma 1. In triangle $ABC$ with orthocenter $H$, we have $|BC \cot \angle A| = |AH|$.

Proof. If we let $O$ be the circumcenter of triangle $ABC$ and $M$ be the midpoint of $BC$, we know that $2OM = AH$. Additionally, $|R \cos \angle A| = OM$ where $R$ is the radius of the circumcircle of triangle $ABC$. Therefore, according to the Law of Sines,
$$AH = 2OM = |2R \cos \angle A| = |BC \cot A|$$
the result follows.

Since $AHMD$ is a parallelogram, we can write $MH = AD$. According to the above lemma, we also have $|XY \cot \angle A| = AH'$. Since $|\cot A|$ is constant, it suffices to show that $\frac{XY}{MH}$ is constant.

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In the following we will use directed angles modulo $180^\circ$ denoted as $\angle ABC$. Let $A'$ be the antipodal of $A$ with respect to $\omega$. As $BHCA'$ is a parallelogram, $M$ is the midpoint of $AH'$. Note that $\angle XHA' = \angle XBA' = 90^\circ$. So $A'BXH$ is cyclic which results in $\angle A'XH = \angle A'BH = \angle BAC$. Analogously, $\angle HYA' = \angle BAC$ which means $XA'Y$ is an isosceles triangle and $H$ is the midpoint of $XY$. At last we can write:
$$\frac{XY}{MH} = \frac{4XH}{HA'} = 4 \cot \angle BAC$$
which is constant and the claim is proven. ■