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

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(Solution 2)
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== Solution 2 ==
 
== Solution 2 ==
The incenter of an equilateral triangle is the same as the centroid of an equilateral triangle. This means the center of the inscribed circle is also the centroid. From properties of median lengths, the radius of the large circle is 3 times the radius of the small circle. <math>\frac{1}{3}=\frac{1}{9}</math>. The answer is B.
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The incenter of an equilateral triangle is the same as the centroid of an equilateral triangle. This means the center of the inscribed circle is also the centroid. From properties of median lengths, the radius of the large circle is 3 times the radius of the small circle. <math>(\frac{1}{3})^2=\frac{1}{9}</math>. The answer is B.
 
 
Then the ratio of areas will be <math>\frac{1}{3}</math> squared, or <math>\frac{1}{9}\Rightarrow \boxed{\text{B}}</math>.
 
  
 
== See also ==
 
== See also ==

Revision as of 01:49, 24 December 2020

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 1

[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 $\frac{1}{3}=\frac{r}{R}$.

Solution 2

The incenter of an equilateral triangle is the same as the centroid of an equilateral triangle. This means the center of the inscribed circle is also the centroid. From properties of median lengths, the radius of the large circle is 3 times the radius of the small circle. $(\frac{1}{3})^2=\frac{1}{9}$. The answer is B.

See also

2008 AMC 12A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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