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 (ProblemsAnswer KeyResources)
Preceded by
Problem 49
Followed by
Last Question
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