1981 AHSME Problems/Problem 23

Problem

Equilateral $\triangle ABC$ is inscribed in a circle. A second circle is tangent internally to the circumcircle at $T$ and tangent to sides $AB$ and $AC$ at points $P$ and $Q$ . If side $BC$ has length $12$ , then segment $PQ$ has length

$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 6\sqrt{3}\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 8\sqrt{3}\qquad\textbf{(E)}\ 9$

Solution

Let $O$ be the center of the smaller circle, and let $r$ be its radius. Then $OT = OP = OQ = r$ and $AO$ = 2r $, since$ \triangle AOP $and$ \triangle AOQ $are$ 30-60-90 $triangles. So$ AT = 3r $. Since$ \triangle AOP \sim \triangle ATB $,$ \frac{AP}{AB} = \frac{AO}{AT} = \frac{2}{3} $. Since$ AB = 12 $,$ AP = 8 $and thus$ PQ = 8 $.$ \fbox{(C)}$.

-j314andrews

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1981 AHSME Problems/Problem 23

Problem

Solution