# Difference between revisions of "2017 AMC 8 Problems/Problem 22"

## Problem 22

In the right triangle $ABC$, $AC=12$, $BC=5$, and angle $C$ is a right angle. A semicircle is inscribed in the triangle as shown. What is the radius of the semicircle? $[asy] draw((0,0)--(12,0)--(12,5)--(0,0)); draw(arc((8.67,0),(12,0),(5.33,0))); label("A", (0,0), W); label("C", (12,0), E); label("B", (12,5), NE); label("12", (6, 0), S); label("5", (12, 2.5), E);[/asy]$

$\textbf{(A) }\frac{7}{6}\qquad\textbf{(B) }\frac{13}{5}\qquad\textbf{(C) }\frac{59}{18}\qquad\textbf{(D) }\frac{10}{3}\qquad\textbf{(E) }\frac{60}{13}$

## Solution

We can reflect triangle $ABC$ on line $AC.$ This forms the triangle $AB'C$ and a circle out of the semicircle. Let us call the center of the circle $O.$

We can see that Circle $O$ is the incircle of $AB'C.$ We can use the formula for finding the radius of the incircle to solve this problem. The area of $AB'C$ is $12\times5 = 60.$ The semiperimeter is $5+13 = 18.$ Simplifying $\dfrac{60}{18} = \dfrac{10}{3}.$ Our answer is therefore $\boxed{\textbf{(D)}\ \frac{10}{3}}.$

## Solution

We immediately see that $AB=13$, and we label the center of the semicircle $O$. Drawing radius $OD$ with length $x$ such that $OD$ is tangent to $AB$, we immediately see that $ODB\cong OCB$ because of HL congruence, so $BD=5$ and $DA=8$. By similar triangles $ODA$ and $BCA$, we see that $\frac{8}{12}=\frac{x}{5}\implies 12x=40\implies x=\frac{10}{3}\implies\boxed{\textbf{D}}$.