2008 AMC 12A Problem 13

Question

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

OA

and

OB.

What is the ratio of the area of the smaller circle to that of the larger circle?

\dfrac{1}{16}

\dfrac{1}{9}

\dfrac{1}{8}

\dfrac{1}{6}

\dfrac{1}{4}

The center of the small circle lies on the bisector of

\angle AOB

Dropping a radius to

OA

makes a

30

-

60

-

90

triangle in which

OE = 2r

Let

r

and

R

be the radii of the smaller and larger circles, and let

E

be the center of the smaller circle. By symmetry

E

lies on the bisector of

\angle AOB,

so

OE

makes a

30^\circ

angle with

OA.

Dropping the radius

ED

perpendicular to

OA

gives a

30

-

60

-

90

triangle with

OE = 2 \cdot ED = 2r.

Since the circles are internally tangent,

OE = R - r.

Then

R - r = 2r,

so

R = 3r

and

\tfrac{r}{R} = \tfrac{1}{3}.

The ratio of areas is

\left(\dfrac{1}{3}\right)^2 = \dfrac{1}{9}.

Thus, B is the correct answer.

Problem 13 in other years