2006 AMC 10A Problem 16

Question

A circle of radius

1

is tangent to a circle of radius

2.

The sides of

\triangle ABC

are tangent to the circles as shown, and the sides

AB

and

AC

are congruent. What is the area of

\triangle ABC?

\dfrac{35}{2}

15\sqrt{2}

\dfrac{64}{3}

16\sqrt{2}

24

Let

O

and

O'

be the centers; the right triangles cut off along

AC

are similar

The centers lie

3

and

6

from

A,

and the base is

8

below

A

Let

O, O'

be the centers of the small and large circles, and let

D

be the point where the small circle touches

AC.

The right triangles cut off along

AC

are similar, so

\frac{AO}{1} = \frac{AO + 3}{2},

giving

AO = 3

and

AO' = 6.

The tangent length is

AD = \sqrt{AO^2 - 1^2}

= \sqrt{3^2 - 1^2}

= 2\sqrt2.

Let

F

be the midpoint of

BC

; then

AF = AO' + 2 = 8.

Since

\triangle ADO \sim \triangle AFC,

we get

\frac{FC}{1} = \frac{AF}{2\sqrt2} = \frac{8}{2\sqrt2} = 2\sqrt2.

Thus

BC = 4\sqrt2,

and the area is

\frac12 \cdot BC \cdot AF

= \frac12 \cdot 4\sqrt2 \cdot 8

= 16\sqrt2.

Thus, the correct answer is D.

Problem 16 in other years