# 1953 AHSME Problems/Problem 50

## Problem

One of the sides of a triangle is divided into segments of $6$ and $8$ units by the point of tangency of the inscribed circle. If the radius of the circle is $4$, then the length of the shortest side is $\textbf{(A) \ } 12 \mathrm{\ units} \qquad \textbf{(B) \ } 13 \mathrm{\ units} \qquad \textbf{(C) \ } 14 \mathrm{\ units} \qquad \textbf{(D) \ } 15 \mathrm{\ units} \qquad \textbf{(E) \ } 16 \mathrm{\ units}$

## Solution

Let the triangle have side lengths $14, 6+x,$ and $8+x$. The area of this triangle can be computed two ways. We have $A = rs$, and $A = \sqrt{s(s-a)(s-b)(s-c)}$, where $s = 14+x$ is the semiperimeter. Therefore, $4(14+x)=\sqrt{(14+x)(x)(8)(6)}$. Solving gives $x = 7$ as the only valid solution. This triangle has sides $13,14$ and $15$, so the shortest side is $\boxed{\textbf{(B) \ } 13 \mathrm{\ units}}$.

## See Also

 1953 AHSC (Problems • Answer Key • Resources) Preceded byProblem 49 Followed byLast Question 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 All AHSME Problems and Solutions

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