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

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

@@ Line 6: / Line 6: @@
 ==Solution==
-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)}$.
+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 = 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)}</math>.
 -j314andrews

Page

Toolbox

Search

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

Revision as of 02:38, 26 June 2025

Problem

Solution