Switzerland — Second Round, 2022

Solution:
Let $P$ be the intersection of $O X$ with the tangent through $Y$, and $Q$ any point on the tangent through $X$ such that $A$ and $Q$ are not on the same side of $O X$. Note that $\angle O A Y=\angle O Y A$, as $O Y=O A$. Also, by the tangent chord theorem, $\angle O A Y=\angle O Y P$. Similarly, $\angle O A X=\angle O X A$, as $O X=O A$ and by the tangent chord theorem, we have $\angle O A X=\angle O X Q$. We will now use the fact that the tangents through $X$ and $Y$ are parallel, so that $\angle O X Q=\angle O P Y$. We can now conclude, as the sum of the interior angles in the quadrilateral $X A Y P$ is $360^{\circ}$. Hence,
$$\begin{aligned} 360^{\circ} & =\angle O A Y+\angle O Y A+\angle O Y P+\angle Y P X+\angle P X A+\angle X A O \\ & =3 \cdot(\angle X A O+\angle O A Y) \\ & =3 \cdot \angle X A Y \end{aligned}$$
proving that $\angle X A Y=120^{\circ}$.

Solution:
As in Solution 1, we introduce $P$ and $Q$ and prove that $\angle O A Y=\angle O Y P$ and $\angle O A X=\angle O X Q$ by the tangent chord theorem. Now let $R$ be a point on the parallel to the tangents through $O$, such that $R$ and $A$ are on the same side as $X O$. As the lines are parallel, we have that $\angle O X Q=\angle X O R$ and $\angle O Y P=\angle Y O R$. This proves that
$$\begin{aligned} \angle X A Y & =\angle X A O+\angle O A Y \\ & =\angle O X Q+\angle O Y P \\ & =\angle X O R+\angle Y O R \\ & =\angle X O Y \end{aligned}$$
To conclude, we introduce a point $S$ on the $\operatorname{arc} X Y$ that does not contain $A$. Then, by the inscribed angle theorem, and using that $X A Y S$ is cyclic, we have
$$\begin{aligned} 180^{\circ} & =\angle X A Y+\angle X S Y \\ & =\angle X A Y+\frac{1}{2} \cdot \angle X O Y \\ & =\frac{3}{2} \cdot \angle X A Y \end{aligned}$$
proving that $\angle X A Y=120^{\circ}$.