3. A circle $\omega$ is circumscribed around triangle $A B C$. Circle $\omega_{1}$ touches the line $A B$ at point $A$ and passes through point $C$, while circle $\omega_{2}$ touches the line $A C$ at point $A$ and passes through point $B$. A tangent is drawn from point $A$ to circle $\omega$, which intersects circle $\omega_{1}$ again at point $X$ and intersects circle $\omega_{2}$ again at point $Y$. Find the ratio $\frac{A X}{X Y}$.
Answer: $\frac{1}{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_627b343ad954e1e3a31bg-28.jpg?height=492&width=799&top_left_y=1967&top_left_x=640)

Solution 1. Let $O$ be the center of $\omega, \alpha=\angle B A C$. The angle between the tangent $A B$ and the chord $A C$ of the circle $\omega_{1}$ is equal to the inscribed angle that subtends $A C$, i.e., $\angle A X C=\alpha$. Similarly, we obtain $\angle A Y B=\alpha$. Extend the quadrilateral $C X Y B$ to the triangle $X Y Z$ (see the 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 that the 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, the ray $Z O$ is the bisector of the isosceles triangle $X Z Y$, which is also the median and altitude. The radius $O A$ of the circle $\omega$ is perpendicular to its tangent $X Y$ and, therefore, parallel to $Z O$. Thus, the point $A$ lies on the ray $Z O$, from which $A X=A Y$ and $\frac{A X}{X Y}=\frac{1}{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_627b343ad954e1e3a31bg-29.jpg?height=511&width=688&top_left_y=884&top_left_x=707)

Solution 2. The line $X Y$ is tangent to the circle $\omega$, hence

$$
\angle B A Y=\angle B C A \quad \text { and } \quad \angle C A X=\angle C B A .
$$

Since the line $A B$ is tangent to the circle $\omega_{1}$, and the line $A C$ is tangent to the circle $\omega_{2}$, we also have

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

Thus, triangles $A B C, X A C$ and $Y B A$ are similar by two angles. Therefore,

$$
\frac{A X}{A C}=\frac{B A}{B C} \quad \text { and } \quad \frac{A Y}{A B}=\frac{C A}{C B}
$$

Thus,

$$
A X=\frac{A B \cdot A C}{B C}=A Y \quad \text { and } \quad \frac{A X}{X Y}=\frac{A X}{A X+A Y}=\frac{1}{2} .
$$

Solution 3. Let the radii of the circles $\omega, \omega_{1}$ and $\omega_{2}$ be $r, r_{1}, r_{2}$ respectively. Also, let

$$
\alpha=\angle B A C, \quad \beta=\angle A B C, \quad \gamma=\angle A C B
$$

The angle between the tangent $A B$ and the chord $A C$ of the 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 obtain

$$
\angle A Y B=\alpha, \quad \angle X A C=\beta, \quad \angle Y A B=\gamma
$$

Then triangles $A X C$ and $A Y B$ are similar to $A B C$ by two angles, from which $\angle A C X=\gamma$ and $\angle A B Y=\beta$. By the Law of Sines,

$$
A C=2 r_{1} \sin \alpha=2 r \sin \beta \Longleftrightarrow \frac{\sin \alpha}{\sin \beta}=\frac{r}{r_{1}} \quad \text { and } \quad A B=2 r_{2} \sin \alpha=2 r \sin \gamma \Longleftrightarrow \frac{\sin \gamma}{\sin \alpha}=\frac{r_{2}}{r}
$$

Therefore,

$$
\frac{A X}{A Y}=\frac{r_{1} \sin \gamma}{r_{2} \sin \beta}=\frac{r_{1}}{r_{2}} \cdot \frac{\sin \gamma}{\sin \alpha} \cdot \frac{\sin \alpha}{\sin \beta}=\frac{r_{1}}{r_{2}} \cdot \frac{r}{r_{1}} \cdot \frac{r_{2}}{r}=1
$$

from which the answer follows.