HMMT-Nov 2011 gen, Problem 9

Let $P$ and $Q$ be points on line $l$ with $P Q=12$. Two circles, $\omega$ and $\Omega$, are both tangent to $l$ at $P$ and are externally tangent to each other. A line through $Q$ intersects $\omega$ at $A$ and $B$, with $A$ closer to $Q$ than $B$, such that $A B=10$. Similarly, another line through $Q$ intersects $\Omega$ at $C$ and $D$, with $C$ closer to $Q$ than $D$, such that $C D=7$. Find the ratio $A D / B C$.

Solution: We first apply the Power of a Point theorem repeatedly. Note that $Q A \cdot Q B=Q P^{2}=$ $Q C \cdot Q D$. Substituting in our known values, we obtain $Q A(Q A+10)=12^{2}=Q C(Q C+7)$. Solving these quadratics, we get that $Q A=8$ and $Q C=9$. We can see that $\frac{A Q}{D Q}=\frac{C Q}{B Q}$ and that $\angle A Q D=\angle C Q B$, so $Q A D \sim Q C B$. (Alternatively, going back to the equality $Q A \cdot Q B=Q C \cdot Q D$, we realize that this is just a Power of a Point theorem on the quadrilateral $A B D C$, and so this quadrilateral is cyclic. This implies that $\angle A D Q=\angle A D C=$ $\angle A B C=\angle Q B C$.) Thus, $\frac{A D}{B C}=\frac{A Q}{Q C}=\frac{8}{9}$. $\fbox{\frac{8}{9}}$.

Answer: \frac{8}{9}