11.5. If $\Varangle B A Y=\alpha$ and $\Varangle A B Y=\beta$ then $\Varangle B Y X=\beta+\alpha$. Furthermore $\Varangle A C B=$ $\Varangle A Y B=180^{\circ}-\alpha-\beta$, implying $\Varangle B A C=$ $\Varangle A B C=\frac{\beta+\alpha}{2}$. Thus $\Varangle A Y C=\frac{\beta+\alpha}{2}$ and $\Varangle Y C X=\Varangle Y A C=\frac{\beta+\alpha}{2}-\alpha=\frac{\beta-\alpha}{2}$. The Sine theorem for $\triangle B Y X$ and $\triangle C Y X$ gives

![](https://cdn.mathpix.com/cropped/2024_06_03_f8ccd71b4db367e7c2a9g-194.jpg?height=367&width=476&top_left_y=637&top_left_x=1141)

$$
\frac{X Y}{X B}=\frac{\sin \alpha}{\sin (\alpha+\beta)}, \quad \frac{X Y}{X C}=\frac{\sin \frac{\beta-\alpha}{2}}{\sin \frac{\beta+\alpha}{2}}
$$

and since $X B=X C$ we have $\frac{\sin \alpha}{\sin (\alpha+\beta)}=\frac{\sin \frac{\beta-\alpha}{2}}{\sin \frac{\beta+\alpha}{2}}$. Hence

$$
\sin \alpha=2 \cos \frac{\beta+\alpha}{2} \sin \frac{\beta-\alpha}{2}=\sin \beta-\sin \alpha
$$

and therefore $\frac{A Y}{B Y}=\frac{\sin \beta}{\sin \alpha}=2$.
Problem 11.5. The tangent lines to the circumcircle $k$ of an isosceles $\triangle A B C$, $A C=B C$, at the points $B$ and $C$ meet at point $X$. If $A X$ meets $k$ at point $Y$, find the ratio $\frac{A Y}{B Y}$.

Emil Kolev