2008 AMC 12A Problems/Problem 13

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(Redirected from 2008 AMC 10A Problems/Problem 16)

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?

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

Solution

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

2008 AMC 12A (Problems • 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

2008 AMC 10A (Problems • 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

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2008 AMC 12A Problems/Problem 13

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