# 2018 AMC 10A Problems/Problem 15

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## Problem

Two circles of radius $5$ are externally tangent to each other and are internally tangent to a circle of radius $13$ at points $A$ and $B$, as shown in the diagram. The distance $AB$ can be written in the form $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$? $[asy] draw(circle((0,0),13)); draw(circle((5,-6.2),5)); draw(circle((-5,-6.2),5)); label("B", (9.5,-9.5), S); label("A", (-9.5,-9.5), S); [/asy]$ $\textbf{(A) } 21 \qquad \textbf{(B) } 29 \qquad \textbf{(C) } 58 \qquad \textbf{(D) } 69 \qquad \textbf{(E) } 93$

## Solution $[asy] draw(circle((0,0),13)); draw(circle((5,-6.25),5)); draw(circle((-5,-6.25),5)); label("A", (-8.125,-10.15), S); label("B", (8.125,-10.15), S); draw((0,0)--(-8.125,-10.15)); draw((0,0)--(8.125,-10.15)); draw((-5,-6.25)--(5,-6.25)); draw((-8.125,-10.15)--(8.125,-10.15)); label("X", (0,0), N); label("Y", (-5,-6.25),NW); label("Z", (5,-6.25),NE); [/asy]$

Let the center of the surrounding circle be $X$. The circle that is tangent at point $A$ will have point $Y$ as the center. Similarly, the circle that is tangent at point $B$ will have point $Z$ as the center. Connect $AB$, $YZ$, $XA$, and $XB$. Now observe that $\triangle XYZ$ is similar to $\triangle XAB$ by SAS.

Writing out the ratios, we get $$\frac{XY}{XA}=\frac{YZ}{AB} \Rightarrow \frac{13-5}{13}=\frac{5+5}{AB} \Rightarrow \frac{8}{13}=\frac{10}{AB} \Rightarrow AB=\frac{65}{4}.$$ Therefore, our answer is $65+4= \boxed{\textbf{(D) } 69}$.

- Whiz

## Video Solution 2 by OmegaLearn

~ pi_is_3.14

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