# 2017 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 8

## Problem

$[asy] pair C=(0,0),A=(0,12),B=(5,0); draw(C--A--B--C,dot); MP("A",A,N);MP("C",C,SW);MP("B",B,E); real r=10/3; pair X=(0,r),Y=X+r*(12/13,5/13); draw(X--Y,dot); MP("X",X,W);MP("Y",Y,NE); draw(circle(X,r)); MP("12",(0,8),W);MP("5",(2.5,0),S); [/asy]$

Let $ABC$ be a right triangle with right angle at $C$. Suppose $AC = 12$ and $BC = 5$ and $CX$ is the diameter of a semicircle, where $X$ lies on $AC$ and the semicircle is tangent to side $AB$. Find the radius of the semicircle.

## Solution

Method 1:

We can compute the area in two ways: $\triangle{ABC} = \frac{1}{2}\cdot 5\cdot 12 = 30$ or $\triangle{ABC}=\triangle{ABX} +\triangle{CBX} \frac{1}{2}\cdot{r}\cdot{13}+\frac{1}{2}\cdot{r}\cdot{5} = 9r$. Setting the two areas equal we obtain $r = \frac{10}{3}$.

Method 2: Place the point C on the origin of the xy plane, $B$ at $(5,0)$ and $A$ at $(12,0)$. Point $X$ lies at cartesian coordinate $(0,r)$. The line AB has formula $y=\frac{12}{5}\cdot(5-x)$. The vector $\vec{XY}$ has coordinates $r(12/13,5/13)$ since it has length $r$ in the unit direction $(12,5)/13$ which is orthogonal to the line AB.

Then point Y has coordinates $X + \vec{XY}=(0,r)+r(12/13,5/13)$ and lies on the line $y=\frac{12}{5}\cdot(5-x)$. Substituting for these equations gives $r = \frac{10}{3}$.