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

(Duplicate, standardized answer choices, added AMC 10 box, style)
(Solution - Minor edits)
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== Solution ==
 
== Solution ==
−
<center><asy>
+
<asy>size(200);
−
size(300);
+
defaultpen(fontsize(10));
−
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);
 
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;  
 
picture p = new picture;  
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draw(D-(0.2,0)--D-(0.2,-0.2)--D-(0,-0.2));
 
draw(D-(0.2,0)--D-(0.2,-0.2)--D-(0,-0.2));
 
draw(Circle(C,1));
 
draw(Circle(C,1));
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label("\(30^{\circ}\)",(.53,.1),O);
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label("\(30^{\circ}\)",(0.65,0.15),O);
 
label("\(r\)",(C+D)/2,E);
 
label("\(r\)",(C+D)/2,E);
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label("\(2r\)",(O+C)/2,SE);
+
label("\(2r\)",(O+C)/2,NNE);
 
label("\(O\)",O,SW);
 
label("\(O\)",O,SW);
 
label("\(r\)",(C+F)/2,SE);
 
label("\(r\)",(C+F)/2,SE);
 
label("\(R\)",(O+A)/2-(0,0.3),S);
 
label("\(R\)",(O+A)/2-(0,0.3),S);
 
label("\(P\)",C,NW);
 
label("\(P\)",C,NW);
−
label("\(Q\)",D,SE);
+
label("\(Q\)",D,SE);</asy>
−
</asy></center>
 
  
−
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>.
+
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>. Also, 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 <math>30-60-90</math> triangle at that. Therefore, <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 B</math>.
  
 
== See also ==
 
== See also ==

Revision as of 01:28, 26 April 2008

The following problem is from both the 2008 AMC 12A #13 and 2008 AMC 10A #16, so both problems redirect to this page.

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?

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

Solution

[asy]size(200); defaultpen(fontsize(10)); 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("\(30^{\circ}\)",(0.65,0.15),O); label("\(r\)",(C+D)/2,E); label("\(2r\)",(O+C)/2,NNE); 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 $OA$. Also, 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. Therefore, $OP=2PQ$.

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 $\frac{1}{9}\Rightarrow B$.

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
2008 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
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 10 Problems and Solutions
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