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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 <...")
 
 
(3 intermediate revisions by the same user not shown)
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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>
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defaultpen(linewidth(.8pt));
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pair B = origin;
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pair A = dir(60);
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pair C = dir(0);
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pair circ = circumcenter(A,B,C);
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pair P = intersectionpoint(circ--(circ + (-1,0)),A--B);
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pair Q = intersectionpoint(circ--(circ + (1,0)),A--C);
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label("$A$",A,N);
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label("$B$",B,SW);
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label("$C$",C,SE);
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label("$P$",P,NW);
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label("$Q$",Q,NE);
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label("$T$",(0.5,-0.3),S);
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draw(A--B--C--cycle);
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draw(circumcircle(A,B,C));
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draw(P--Q);
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draw(Circle((0.5,0.09),0.385));
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</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>
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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 = 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>.
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-j314andrews
 +
 +
==See also==
 +
 +
{{AHSME box|year=1981|num-b=22|num-a=24}}
 +
{{MAA Notice}}

Latest revision as of 14:29, 28 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

See also

1981 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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