2018 AMC 10A Problem 15

Let $O$ be the center of the big circle and $P$, $Q$ the centers of the small circles tangent at $A$ and $B$ respectively. Draw the segments joining the centers.

• Internal tangency at $A$: $O$, $P$, $A$ are collinear with $OP = 13 - 5 = 8$ and $OA = 13$. Likewise $OQ = 8$, $OB = 13$.
• External tangency of the small circles: $PQ = 5 + 5 = 10$.

Triangles $OPQ$ and $OAB$ share the angle at $O$, and $\dfrac{OA}{OP} = \dfrac{OB}{OQ} = \dfrac{13}{8}$, so they are similar with ratio $\tfrac{13}{8}$. Hence
$$AB = \frac{13}{8} \cdot PQ = \frac{13}{8} \cdot 10 = \frac{65}{4}.$$

Since $\gcd(65, 4) = 1$, $m + n = 65 + 4 = 69$.

The answer is $\boxed{\textbf{(D)}\ 69}$.

Tangent circles come with two free facts: the point of tangency is on the line of centers, and the distance between centers is $R + r$ (external) or $R - r$ (internal). Drawing all the center-to-center and center-to-tangency segments turns a circle problem into a triangle problem, and here the triangle $OAB$ is just a dilation of $OPQ$ from the common vertex $O$.

Coordinates: put $O = (0,0)$ and $P = (-5, -h)$, $Q = (5, -h)$ with $25 + h^2 = 64$, so $h = \sqrt{39}$. Then $A = \tfrac{13}{8} P$ and $B = \tfrac{13}{8} Q$, so $AB = \tfrac{13}{8} \cdot 10 = \tfrac{65}{4}$. The similarity does the same work without the square root.

• Taking the small centers to be 13 - 5 = 8 apart (instead of 10, the sum of their radii) or forgetting that A lies on line OP beyond P.
• Placing $A$ and $B$ at the same height as $P$ and $Q$ (so $AB = 10$ or $AB = 13$-ish guesses), or computing the height $\sqrt{39}$ by coordinates and making a radical error; the similarity ratio $\tfrac{13}{8}$ needs no square roots.