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Annual Harvard-MIT Mathematics Tournament

Geometry Difficulty 5.0 AIME, harder Prove it United States

Problem:

Circle $\Omega$ has radius $13$ . Circle $\omega$ has radius $14$ and its center $P$ lies on the boundary of circle $\Omega$ . Points $A$ and $B$ lie on $\Omega$ such that chord $AB$ has length $24$ and is tangent to $\omega$ at point $T$ . Find $AT \cdot BT$ .

Solution

Solution:

Let $M$ be the midpoint of chord $AB$ ; then $AM = BM = 12$ and by the Pythagorean theorem on triangle $AMO$ we have $MO = 5$ .

Note that $\angle AOM = \angle AOB / 2 = \angle APB = \angle APT + \angle TPB$ , or $\tan(\angle AOM) = \tan(\angle APT + \angle TPB)$ . Applying the tangent addition formula,
$\begin{aligned} \frac{AM}{MO} &= \frac{\frac{AT}{TP} + \frac{BT}{TP}}{1 - \frac{AT}{TP} \cdot \frac{BT}{TP}} \\ &= \frac{AB \cdot TP}{TP^2 - AT \cdot BT} \end{aligned}$
from which
$AT \cdot BT = TP^2 - \frac{AB \cdot TP \cdot MO}{AM} = 14^2 - \frac{24 \cdot 14 \cdot 5}{12} = 56.$

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