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Difference between revisions of "1981 AHSME Problems/Problem 23"

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==Problem==
 
==Problem==
 
Equilateral <math>\triangle ABC</math> is inscribed in a circle. A second circle is tangent internally to the circumcircle at <math>T</math> and tangent to sides <math>AB</math> and <math>AC</math> at points <math>P</math> and <math>Q</math>. If side <math>BC</math> has length <math>12</math>, then segment <math>PQ</math> has length
 
Equilateral <math>\triangle ABC</math> is inscribed in a circle. A second circle is tangent internally to the circumcircle at <math>T</math> and tangent to sides <math>AB</math> and <math>AC</math> at points <math>P</math> and <math>Q</math>. If side <math>BC</math> has length <math>12</math>, then segment <math>PQ</math> 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>
 +
  
 
<math>\textbf{(A)}\ 6\qquad\textbf{(B)}\ 6\sqrt{3}\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 8\sqrt{3}\qquad\textbf{(E)}\ 9</math>
 
<math>\textbf{(A)}\ 6\qquad\textbf{(B)}\ 6\sqrt{3}\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 8\sqrt{3}\qquad\textbf{(E)}\ 9</math>

Revision as of 02:41, 26 June 2025

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

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