2006 AMC 10B Problem 24

Question

Circles with centers at

O

and

P

have radii

2

and

4,

respectively, and are externally tangent. Points

A

and

B

on the circle with center

O

and points

C

and

D

on the circle with center

P

are such that

\overline{AD}

and

\overline{BC}

are common external tangents to the circles. What is the area of the concave hexagon

AOBCPD?

18\sqrt{3}

24\sqrt{2}

36

24\sqrt{3}

32\sqrt{2}

Answer: B

Concepts:tangent circles trapezoid right triangle

Difficulty rating: 2010

Small hint:

Draw

OF\parallel AD

meeting

PD

at

F

to form a rectangle and a right triangle

Big hint:

Then

DF=2,

FP=2,

and

OF=\sqrt{OP^2-FP^2}

with

OP=6

Solution:

The hexagon is symmetric about

\overline{OP},

so its area is twice that of trapezoid

AOPD.

Draw

OF\parallel AD

with

F

on

\overline{PD}.

Then

AOFD

is a rectangle, so

DF=OA=2

and

FP=PD-DF=4-2=2.

Since the circles are externally tangent,

OP=2+4=6,

so in right triangle

OFP,

OF=\sqrt{36-4}=4\sqrt2.

Trapezoid

AOPD

has parallel sides

OA=2

and

PD=4

with height

OF=4\sqrt2,

giving area

\tfrac12(2+4)(4\sqrt2)=12\sqrt2.

The hexagon area is

2\cdot12\sqrt2=24\sqrt2.

Thus, the correct answer is B.