2008 AMC 10A Problem 16

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 small circle’s center lies on the bisector of

\angle AOB

The center is

2r

from

O

by a

30

-

60

-

90

triangle, and also

R - r

from

O

Let the radii be

r

and

R.

The small circle’s center

E

lies on the bisector of

\angle AOB,

so

OE

makes a

30^\circ

angle with

OA.

The perpendicular from

E

to

OA

has length

r,

and in the resulting

30

-

60

-

90

triangle

OE = 2r.

Since

OE = R - r,

we get

2r = R - r,

so

R = 3r

and the area ratio is

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

Thus, the correct answer is B.

Problem 16 in other years