2018 AMC 10A Problem 15

Circle problems become triangle problems once you connect the centers. Let $O$ be the center of the big circle, and $P$, $Q$ the centers of the small circles that touch the big one at $A$ and $B$.

1. Big circle and a small circle touching at $A$. When two circles touch, the touching point lies on the line through their centers. So $O$, $P$, $A$ are on one line. The distance $OA$ is the big radius, $13$, and $PA$ is the small radius, $5$, so $OP = 13 - 5 = 8$. In the same way, $O$, $Q$, $B$ are on one line with $OQ = 8$ and $OB = 13$.
2. The two small circles touch each other. Their centers are one radius plus the other apart: $PQ = 5 + 5 = 10$.
3. Compare two triangles. Triangle $OAB$ is triangle $OPQ$ stretched away from $O$: both $OA$ and $OB$ are $\tfrac{13}{8}$ times as long as $OP$ and $OQ$, and the angle at $O$ is the same. So the two triangles have the same shape, and every length in $OAB$ is $\tfrac{13}{8}$ times the matching length in $OPQ$.
4. Find $AB$.
$$AB = \frac{13}{8} \cdot PQ = \frac{13}{8} \cdot 10 = \frac{65}{4}.$$
5. Finish. $65$ and $4$ share no common factor, so $m + n = 65 + 4 = 69$.

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

Touching circles give two facts for free: the touching point is on the line of centers, and the distance between centers is the sum of radii (touching from outside) or the difference (one inside the other). Drawing those segments turns the picture into two triangles of the same shape sharing the corner $O$, so one scale factor gives the answer.

Coordinates. Put $O = (0,0)$ and the small centers at $P = (-5, -h)$ and $Q = (5, -h)$, so $PQ = 10$. Since $OP = 8$, we need $25 + h^2 = 64$, so $h = \sqrt{39}$. The point $A$ is on the ray from $O$ through $P$ at distance $13$, so $A = \tfrac{13}{8}P$; likewise $B = \tfrac{13}{8}Q$. Their $x$-coordinates are $\pm \tfrac{65}{8}$ and their $y$-coordinates match, so $AB = \tfrac{65}{4}$.