Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.0 AIME, harder Find the answer

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 $A B$ has length 24 and is tangent to $\omega$ at point $T$ . Find $A T \cdot B T$ .

Your answer

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

Solution

Let $M$ be the midpoint of chord $A B$ ; then $A M=B M=12$ and Pythagoras on triangle $A M O$ gives $M O=5$ . Note that $\angle A O M=\angle A O B / 2=\angle A P B=\angle A P T+\angle T P B$ or $\tan (\angle A O M)=\tan (\angle A P T+\angle T P B)$ . Applying the tangent addition formula, $\frac{A M}{M O} =\frac{\frac{A T}{T P}+\frac{B T}{T P}}{1-\frac{A T}{T P} \cdot \frac{B T}{T P}} =\frac{A B \cdot T P}{T P^{2}-A T \cdot B T}$ from which $A T \cdot B T=T P^{2}-A B \cdot T P \cdot M O / A M=14^{2}-24 \cdot 14 \cdot 5 / 12=56$ .

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