Iranian Mathematical Olympiad

Let $\omega$ be the circumcircle of triangle $AXY$. Since the reflection of line $BC$ with respect to $XY$ is tangent to $\omega$, the reflection of $\omega$ with respect to $XY$ is tangent to line $BC$ at a point that is denoted by $T$.

Let $P$ be the second intersection point of circumcircles of triangles $BXT$ and $CYT$ (other than $T$). It is claimed that circumcircles of triangles $BOC$ and $AXY$ are tangent at $P$.

First, note that since $T$ lies on the reflection of $\omega$ with respect to $XY$, $\angle XTY = \angle XAY = \angle BAC$. Therefore,
$$\begin{align*} \angle BPC &= \angle BPT + \angle CPT = \angle BXT + \angle CYT \\ &= 360^\circ - \angle AXT - \angle AYT = \angle XAY + \angle XTY \\ &= 2\angle BAC = \angle BOC. \end{align*}$$
This implies that $P$ lies on the circumcircle of triangle $BOC$. On the other hand,
$$180^\circ - \angle BAC = \angle XBT + \angle YCT = \angle XPY.$$
Therefore, $P$ lies on the circumcircle of triangle $AXY$. Now it suffices to show that these two circles have the same tangent line at $P$, or equivalently, $\angle BPX = \angle BCP + \angle XAP$. Since $\angle BCP = \angle TYP$ and $\angle XAP = \angle XYP$, $\angle BCP + \angle XAP = \angle TYP + \angle XYP = \angle XYT$. However, according to the assumptions, $\angle XYT = \angle XTB = \angle BPX$, which completes the proof.