HMMT February, 2025

Solution:

Let $O_{1}$, $O_{2}$, and $O$ be the centers of $\omega_{1}$, $\omega_{2}$, and the circumcircle of $\triangle AXY$, respectively.
We claim that triangle $O O_{1}O_{2}$ is isosceles with $O O_{1} = O O_{2}$, and thus in particular $O$ always lies on the perpendicular bisector of $O_{1}O_{2}$.

To this end, observe that $O O_{1} \perp A X$ and $O_{1}O_{2} \perp A B$, so $\angle O O_{1}O_{2} = \angle X A B$. Analogously, $\angle O O_{2}O_{1} = \angle Y A B$. So indeed $O O_{1}O_{2}$ is isosceles, and we are done.

Solution:

Let $A^{\prime}$ be the $A$-antipode in circle $(A X Y)$. It suffices to show that $A^{\prime}$ lies on a fixed line. We will show that this line is one that is parallel to $A B$.
Let $M$ be the second intersection of line $A B$ with circle $(A X Y)$, and let $N$ be the antipode of $M$ on this circle. Since $A M A^{\prime}N$ is a rectangle with $N$ lying on the line through $A$ perpendicular to $A B$, it suffices to show that $N$ is fixed (independent of $X$ and $Y$).
To this end, take an inversion at $A$ with arbitrary radius, denoting images with $\bullet \mapsto \bullet^{*}$.
Observe that $X^{*}$ and $Y^{*}$ lie on the fixed lines $\ell_{1} = \omega_{1}^{*}$ and $\ell_{2} = \omega_{2}^{*}$. Let $\ell$ be the line through $A$ perpendicular to $A B^{*}$, and suppose that $\ell_{1}$ and $\ell_{2}$ intersect $\ell$ at $P$ and $Q$, respectively.
Since $\angle X A B = \angle B A Y$, we have $\angle X^{*}A B^{*} = \angle B^{*}A Y^{*}$. Circles $\omega_{1}$, $\omega_{2}$, and $(A X Y)$ are mapped to lines $B^{*}X^{*}$, $B^{*}Y^{*}$, and $X^{*}Y^{*}$. As $A N \perp A B$, it follows that $N^{*}$ is the intersection of $X^{*}Y^{*}$ and $\ell$.
Finally, observe that $(N^{*},A;P,Q)\stackrel {B^{*}}{=}(N^{*},M^{*};X,Y)$ is a harmonic bundle, as $A M^{*}$ bisects $\angle X^{*}A Y^{*}$ and $\angle M^{*}A N^{*} = 90^{\circ}$. Since $A$, $P$, and $Q$ are fixed, so is $N^{*}$. Thus $N$ is fixed, and $O$ lies on the perpendicular bisector of $A N$, which is also fixed.