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Difference between revisions of "2008 AMC 12A Problems/Problem 13"

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Let <math>P</math> be the center of the small circle with radius <math>r</math>, and let <math>Q</math> be the point where the small circle is [[tangent]] to <math>OA</math>, and finally, let <math>C</math> be the point where the small circle is tangent to the big circle with radius <math>R</math>.  Then <math>PQO</math> is a right triangle, and a 30-60-90 triangle at that.  So, <math>OP = 2PQ</math>.  Since <math>OP = OC - PC = OC - r = R - r</math>, we have <math>R - r = 2PQ</math>, or <math>R - r = 2r</math>, or <math>\frac {1}{3} = \frac {r}{R}</math>.  Then the ratio of areas will be <math>\frac {1}{3}</math> squared, or <math>\frac {1}{9}</math>.
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Let <math>P</math> be the center of the small circle with radius <math>r</math>, and let <math>Q</math> be the point where the small circle is [[tangent]] to <math>OA</math>, and finally, let <math>C</math> be the point where the small circle is tangent to the big circle with radius <math>R</math>.  Then <math>PQO</math> is a right triangle, and a 30-60-90 triangle at that.  So, <math>OP = 2PQ</math>.  Since <math>OP = OC - PC = OC - r = R - r</math>, we have <math>R - r = 2PQ</math>, or <math>R - r = 2r</math>, or <math>\frac {1}{3} = \frac {r}{R}</math>.  Then the ratio of areas will be <math>\frac {1}{3}</math> squared, or <math>\frac {1}{9} \Rightarrow \mathbf{(B)}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 00:06, 23 February 2008

Problem

Points $A$ and $B$ lie on a circle centered at $O$, and $\angle AOB = 60^\circ$. A second circle is internally tangent to the first and tangent to both $\overline{OA}$ and $\overline{OB}$. What is the ratio of the area of the smaller circle to that of the larger circle?

$\textbf{(A)}\ \frac {1}{16} \qquad \textbf{(B)}\ \frac {1}{9} \qquad \textbf{(C)}\ \frac {1}{8} \qquad \textbf{(D)}\ \frac {1}{6} \qquad \textbf{(E)}\ \frac {1}{4}$

Solution

[asy] size(300); defaultpen(0.8); pair O=(0,0), A=(3,0), B=(3/2,3/2*3^.5), C=(3^.5,1), D=(3^.5,0), F=(1.5*3^.5,1.5); picture p = new picture; draw(p,Circle(O,0.2)); clip(p,O--C--A--cycle); add(p); draw(Circle(O,3)); dot(A); dot(B); dot(C); dot(O); draw(A--O--B); draw(O--C--D); draw(C--F); draw(D-(0.2,0)--D-(0.2,-0.2)--D-(0,-0.2)); draw(Circle(C,1)); label("\(30^{\circ}\)",(.53,.1),O); label("\(r\)",(C+D)/2,E); label("\(2r\)",(O+C)/2,SE); label("\(O\)",O,SW); label("\(r\)",(C+F)/2,SE); label("\(R\)",(O+A)/2-(0,0.3),S); label("\(P\)",C,NW); label("\(Q\)",D,SE); [/asy]

Let $P$ be the center of the small circle with radius $r$, and let $Q$ be the point where the small circle is tangent to $OA$, and finally, let $C$ be the point where the small circle is tangent to the big circle with radius $R$. Then $PQO$ is a right triangle, and a 30-60-90 triangle at that. So, $OP = 2PQ$. Since $OP = OC - PC = OC - r = R - r$, we have $R - r = 2PQ$, or $R - r = 2r$, or $\frac {1}{3} = \frac {r}{R}$. Then the ratio of areas will be $\frac {1}{3}$ squared, or $\frac {1}{9} \Rightarrow \mathbf{(B)}$.

See also

2008 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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
All AMC 12 Problems and Solutions
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