Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.2 AIME, harder Find the answer

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 $A D / B C$ .

Your answer

A number or a short expression. Spacing and $ signs are ignored.

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, $\frac{A D}{B C}=\frac{A Q}{Q C}=\frac{8}{9}$ .

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Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.