# Difference between revisions of "1964 AHSME Problems/Problem 13"

## Problem 13

A circle is inscribed in a triangle with side lengths $8, 13$, and $17$. Let the segments of the side of length $8$, made by a point of tangency, be $r$ and $s$, with $r. What is the ratio $r:s$?

$\textbf{(A)}\ 1:3 \qquad \textbf{(B)}\ 2:5 \qquad \textbf{(C)}\ 1:2 \qquad \textbf{(D)}\ 2:3 \qquad \textbf{(E)}\ 3:4$

## Solution 1

Label our triangle $ABC$ where that $AB=17$, $AC=13$, and $BC=8$. Let $L$, $J$, and $K$ be the tangency points of $BC$, $AC$, and $AB$ respectively. Let $AK=x$, which implies $AJ=x$. Thus $KB=BL=17-x$ and $JC=LC=13-x$.

Since $BL+LC=(17-x)+(13-x)=8$, $x=11$. Thus $r:s=CL:BL=13-x:17-x=2:6=1:3$, hence our answer is $\fbox{A}$.

## Solution 2

Let the three sides of $8, 13, 17$ be split into two lengths each, for a total of six segments. Because the vertices of the triangle form two congruent external tangents to the triangle each, there are really only three unique lengths, and two of these three lengths each make up a side. Thus, we can make a system of three equations in there variables: $a+b = 8, b+c=13, c+a=17$.

Adding all three equations and dividing by $2$ gives $a+b+c = 19$. Subtracting each equation from that gives $(a, b, c) = (6,2,11)$

Thus, $a+b$ makes up the shortest side of $8$. The ratio is $2:6$, or $1:3$, which is answer $\fbox{A}$.