HMMT-Nov 2010 team, Problem 6

$A B$ is a diameter of circle $O . X$ is a point on $A B$ such that $A X=3 B X$. Distinct circles $\omega_{1}$ and $\omega_{2}$ are tangent to $O$ at $T_{1}$ and $T_{2}$ and to $A B$ at $X$. The lines $T_{1} X$ and $T_{2} X$ intersect $O$ again at $S_{1}$ and $S_{2}$. What is the ratio $\frac{T_{1} T_{2}}{S_{1} S_{2}}$ ?

Solution: Since the problem only deals with ratios, we can assume that the radius of $O$ is 1 . As we have proven in Problem 5, points $S_{1}$ and $S_{2}$ are midpoints of arc $A B$. Since $A B$ is a diameter, $S_{1} S_{2}$ is also a diameter, and thus $S_{1} S_{2}=2$. Let $O_{1}, O_{2}$, and $P$ denote the center of circles $\omega_{1}, \omega_{2}$, and $O$. Since $\omega_{1}$ is tangent to $O$, we have $P O_{1}+O_{1} X=1$. But $O_{1} X \perp A B$. So $\triangle P O_{1} X$ is a right triangle, and $O_{1} X^{2}+X P^{2}=O_{1} P^{2}$. Thus, $O_{1} X^{2}+1 / 4=\left(1-O_{1} X\right)^{2}$, which means $O_{1} X=\frac{3}{8}$ and $O_{1} P=\frac{5}{8}$. Since $T_{1} T_{2} \| O_{1} O_{2}$, we have $T_{1} T_{2}=O_{1} O_{2} \cdot \frac{P T_{1}}{P O_{1}}=2 O_{1} X \cdot \frac{P T_{1}}{P O_{1}}=2\left(\frac{3}{8}\right) \frac{1}{5 / 8}=\frac{6}{5}$. Thus $\frac{T_{1} T_{2}}{S_{1} S_{2}}=\frac{6 / 5}{2}=\frac{3}{5}$. $\fbox{\frac{3}{5}}$.

Answer: \frac{3}{5}