3. A circle $\omega$ with center at point $O$ is circumscribed around triangle $ABC$. Circle $\omega_{1}$ touches the line $AB$ at point $A$ and passes through point $C$, while circle $\omega_{2}$ touches the line $AC$ at point $A$ and passes through point $B$. A line through point $A$ intersects circle $\omega_{1}$ again at point $X$ and circle $\omega_{2}$ again at point $Y$. Point $M$ is the midpoint of segment $XY$. Find the angle $OMX$.
Answer: $90^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_627b343ad954e1e3a31bg-33.jpg?height=497&width=691&top_left_y=1970&top_left_x=702)

Solution 1. Let $\alpha=\angle B A C$. The angle between the tangent $A B$ and the chord $A C$ of circle $\omega_{1}$ is equal to the inscribed angle that subtends $A C$, i.e., $\angle A X C=\angle B A C=\alpha$. Similarly, we get $\angle A Y B=\alpha$. Extend quadrilateral $C X Y B$ to form triangle $X Y Z$ (see figure). Note that

$$
\angle B O C=2 \angle B A C=2 \alpha=\angle Z X Y+\angle Z Y X=180^{\circ}-\angle X Z Y
$$

which means quadrilateral $O B Z C$ is cyclic. Then

$$
\angle O Z B=\angle O C B=\angle O B C=\angle O Z C
$$

Therefore, ray $Z O$ is the angle bisector of isosceles triangle $X Z Y$, which is also the median and altitude. Thus, point $M$ lies on ray $Z O$ and $\angle O M X=90^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_627b343ad954e1e3a31bg-34.jpg?height=500&width=691&top_left_y=818&top_left_x=702)

Solution 2. Let line $X Y$ intersect circle $\omega$ again at point $K$. Since line $A C$ is tangent to circle $\omega_{2}$,

$$
\angle B Y K=\angle B Y A=\angle B A C
$$

Angle $Y B A$ subtends arc $Y A$ of circle $\omega_{2}$, not containing point $B$. Therefore, it is equal to the angle between chord $Y A$ and the tangent to $\omega_{2}$ at point $A$, i.e., angle $X A C$. Considering the cyclic quadrilateral $A B C K$, we get

$$
\angle Y B A=\angle X A C=\angle K A C=\angle K B C
$$

which implies $\angle Y B K=\angle A B C$. Therefore, triangles $Y B K$ and $A B C$ are similar by two angles. Then

$$
\frac{Y K}{A C}=\frac{B K}{B C} \Longleftrightarrow Y K=\frac{A C \cdot B K}{B C}
$$

Similarly, it can be verified that triangles $C A X$ and $C B K$ are similar and $\frac{A X}{B K}=\frac{A C}{B C}$. Thus,

$$
A X=\frac{A C \cdot B K}{B C}=Y K
$$

From this, it follows that point $M$ is the midpoint of segment $A K$. Note that point $O$ is the circumcenter of triangle $A K B$ and, therefore, lies on the perpendicular bisector of side $A K$. Hence, $\angle O M X=\angle O M K=90^{\circ}$.