9. Let $A C$ be a diameter of a circle $\omega$ of radius 1 , and let $D$ be the point on $A C$ such that $C D=1 / 5$. Let $B$ be the point on $\omega$ such that $D B$ is perpendicular to $A C$, and let $E$ be the midpoint of $D B$. The line tangent to $\omega$ at $B$ intersects line $C E$ at the point $X$. Compute $A X$.
Solution:
3
We first show that $A X$ is perpendicular to $A C$. Let the tangent to $\omega$ at $A$ intersect $C B$ at $Z$ and $C E$ at $X^{\prime}$. Since $Z A$ is parallel to $B D$ and $B E=E D, Z X^{\prime}=X^{\prime} A$. Therefore, $X^{\prime}$ is the midpoint of the hypotenuse of the right triangle $A B Z$, so it is also its circumcenter. Thus $X^{\prime} A=X^{\prime} B$, and since $X^{\prime} A$ is tangent to $\omega$ and $B$ lies on $\omega$, we must have that $X^{\prime} B$ is tangent to $\omega$, so $X=X^{\prime}$.
Let $O$ be the center of $\omega$. Then $O D=\frac{4}{5}$, so $B D=\frac{3}{5}$ and $D E=\frac{3}{10}$. Then $A X=$ $D E \cdot \frac{A C}{D C}=\frac{3}{10} \cdot \frac{2}{1 / 5}=3$.