Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.5 AIME, harder Prove it United States

Problem:

Let $P$ and $Q$ be points on line $l$ with $PQ = 12$ . 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 $AB = 10$ . Similarly, another line through $Q$ intersects $\Omega$ at $C$ and $D$ , with $C$ closer to $Q$ than $D$ , such that $CD = 7$ . Find the ratio $AD / BC$ .

Solution

Solution:

Answer: $\frac{8}{9}$

We first apply the Power of a Point theorem repeatedly. Note that $QA \cdot QB = QP^{2} = QC \cdot QD$ . Substituting in our known values, we obtain $QA(QA + 10) = 12^{2} = QC(QC + 7)$ . Solving these quadratics, we get that $QA = 8$ and $QC = 9$ .

We can see that $\frac{AQ}{DQ} = \frac{CQ}{BQ}$ and that $\angle AQD = \angle CQB$ , so $QAD \sim QCB$ . (Alternatively, going back to the equality $QA \cdot QB = QC \cdot QD$ , we realize that this is just a Power of a Point theorem on the quadrilateral $ABDC$ , and so this quadrilateral is cyclic. This implies that $\angle ADQ = \angle ADC = \angle ABC = \angle QBC$ .) Thus, $\frac{AD}{BC} = \frac{AQ}{QC} = \frac{8}{9}$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.