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Harvard-MIT Mathematics Tournament

Geometry Difficulty 5.4 AIME, harder Prove it United States

Problem:

Points $A$ and $B$ lie on circle $\omega$ . Point $P$ lies on the extension of segment $A B$ past $B$ . Line $\ell$ passes through $P$ and is tangent to $\omega$ . The tangents to $\omega$ at points $A$ and $B$ intersect $\ell$ at points $D$ and $C$ respectively. Given that $A B=7$ , $B C=2$ , and $A D=3$ , compute $B P$ .

Solution

Solution:

Say that $\ell$ is tangent to $\omega$ at point $T$ . Observing equal tangents, write

$C D = C T + D T = B C + A D = 5.$

Let the tangents to $\omega$ at $A$ and $B$ intersect each other at $Q$ . Working from Menelaus applied to triangle $C D Q$ and line $A B$ gives
$\begin{aligned} -1 & = \frac{D A}{A Q} \cdot \frac{Q B}{B C} \cdot \frac{C P}{P D} \\ & = \frac{D A}{B C} \cdot \frac{C P}{P C + C D} \\ & = \frac{3}{2} \cdot \frac{C P}{P C + 5}, \end{aligned}$
from which $P C = 10$ . By power of a point, $P T^{2} = A P \cdot B P$ , or $12^{2} = B P \cdot (B P + 7)$ , from which $B P = 9$ .

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