10. $[\mathbf{8}]$ You are given two diameters $A B$ and $C D$ of circle $\Omega$ with radius 1 . A circle is drawn in one of the smaller sectors formed such that it is tangent to $A B$ at $E$, tangent to $C D$ at $F$, and tangent to $\Omega$ at $P$. Lines $P E$ and $P F$ intersect $\Omega$ again at $X$ and $Y$. What is the length of $X Y$, given that $A C=\frac{2}{3}$ ?
Answer: $\frac{4 \sqrt{2}}{3}$ Let $O$ denote the center of circle $\Omega$. We first prove that $O X \perp A B$ and $O Y \perp C D$. Consider the homothety about $P$ which maps the smaller circle to $\Omega$. This homothety takes $E$ to $X$ and also takes $A B$ to the line tangent to circle $\Omega$ parallel to $A B$. Therefore, $X$ is the midpoint of the arc $A B$, and so $O X \perp A B$. Similarly, $O Y \perp C D$.
Let $\theta=\angle A O C$. By the Law of Sines, we have $A C=2 \sin \frac{\theta}{2}$. Thus, $\sin \frac{\theta}{2}=\frac{1}{3}$, and $\cos \frac{\theta}{2}=\sqrt{1-\left(\frac{1}{3}\right)^{2}}=$ $\frac{2 \sqrt{2}}{3}$. Therefore,
$$
\begin{aligned}
X Y & =2 \sin \frac{\angle X O Y}{2} \\
& =2 \sin \left(90^{\circ}-\frac{\theta}{2}\right) \\
& =2 \cos \frac{\theta}{2} \\
& =\frac{4 \sqrt{2}}{3}
\end{aligned}
$$