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

$[asy] defaultpen(linewidth(.8pt)); pair B = origin; pair A = dir(60); pair C = dir(0); pair circ = circumcenter(A,B,C); pair P = intersectionpoint(circ--(circ + (-1,0)),A--B); pair Q = intersectionpoint(circ--(circ + (1,0)),A--C); label("$A$",A,N); label("$B$",B,SW); label("$C$",C,SE); label("$P$",P,NW); label("$Q$",Q,NE); label("$T$",(0.5,-0.3),S); draw(A--B--C--cycle); draw(circumcircle(A,B,C)); draw(P--Q); draw(Circle((0.5,0.09),0.385)); [/asy]$

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

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

1981 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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1981 AHSME Problems/Problem 23

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