2008 AMC 12A Problems/Problem 13

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.

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2008 AMC 12A Problems/Problem 13

Contents

Problem

Solution 1

Solution 2

See also