2004 AMC 10A Problem 23

Question

Circles

A,

B,

and

C

are externally tangent to each other and internally tangent to circle

D.

Circles

B

and

C

are congruent. Circle

A

has radius

1

and passes through the center of

D.

What is the radius of circle

B?

\dfrac{2}{3}

\dfrac{\sqrt{3}}{2}

\dfrac{7}{8}

\dfrac{8}{9}

\dfrac{1 + \sqrt{3}}{3}

Answer: D

Concepts:tangent circles coordinate geometry Pythagorean Theorem

Difficulty rating: 1990

Small hint:

Since

A

passes through

D

’s center and is internally tangent, circle

D

has radius

2

Big hint:

Place the centers on coordinates; circle

B

of radius

r

has center at distance

1 + r

from

A

and

2 - r

from

D

Solution:

Because circle

A

passes through

D

’s center and is internally tangent to

D,

circle

D

has radius

2.

Place

D

’s center at the origin and

A

’s center at

(-1, 0).

Let circle

B

have radius

r

and center

(x, r),

using the symmetry of

B

and

C

about the horizontal axis. Tangency gives

\begin{aligned} (x + 1)^2 + r^2 &= (1 + r)^2, \\ x^2 + r^2 &= (2 - r)^2. \end{aligned}

Subtracting yields

x = 3r - 2.

Substituting into the second equation gives

9r^2 - 8r = 0,

so

r = \dfrac{8}{9}.

Thus, the correct answer is D.

2004 AMC 10A Problem 23

Answer: D

Difficulty rating: 1990

Small hint:

Big hint:

Solution:

Problem 23 in other years