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

(Created page with "==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 <...")
 
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<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>
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==Solution==
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Let <math>O</math> be the center of the smaller circle, and let <math>r</math> be its radius.  Then <math>OT = OP = OQ = r</math> and <math>AO</math> = 2r<math>, since </math>\triangle AOP<math> and </math>\triangle AOQ<math> are </math>30-60-90<math> triangles.  So </math>AT = 3r<math>.  Since </math>\triangle AOP \sim \triangle ATB<math>, </math>\frac{AP}{AB} = \frac{AO}{AT} = \frac{2}{3}<math>.  Since </math>AB = 12<math>, </math>AP = 8<math> and thus </math>PQ = 8<math>. </math>\fbox{(C)}$.
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-j314andrews

Revision as of 02:38, 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

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

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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