Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.0 AIME, harder Prove it United States

Problem:

Circle $\Omega$ has radius $5$ . Points $A$ and $B$ lie on $\Omega$ such that chord $AB$ has length $6$ . A unit circle $\omega$ is tangent to chord $AB$ at point $T$ . Given that $\omega$ is also internally tangent to $\Omega$ , find $AT \cdot BT$ .

Solution

Solution:

Let $M$ be the midpoint of chord $AB$ and let $O$ be the center of $\Omega$ . Since $AM = BM = 3$ , Pythagoras on triangle $AMO$ gives $OM = 4$ .

Now let $\omega$ be centered at $P$ and say that $\omega$ and $\Omega$ are tangent at $Q$ . Because the diameter of $\omega$ exceeds $1$ , points $P$ and $Q$ lie on the same side of $AB$ . By tangency, $O$ , $P$ , and $Q$ are collinear, so that $OP = OQ - PQ = 4$ .

Let $H$ be the orthogonal projection of $P$ onto $OM$ ; then $OH = OM - HM = OM - PT = 3$ . Pythagoras on $OHP$ gives $HP^{2} = 7$ .

Finally,
$AT \cdot BT = AM^{2} - MT^{2} = AM^{2} - HP^{2} = 9 - 7 = 2$

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