2006 AMC 12B Problem 15

Question

Circles with centers

O

and

P

have radii

2

and

4,

respectively, and are externally tangent. Points

A

and

B

are on the circle centered at

O,

and points

C

and

D

are on the circle centered at

P,

such that

AD

and

BC

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

AOBCPD?

18\sqrt{3}

24\sqrt{2}

36

24\sqrt{3}

32\sqrt{2}

The centers are

OP = 2 + 4 = 6

apart, and the radii to the tangent points are perpendicular to the tangent

AOPD

is a right trapezoid; find its slant side with the Pythagorean theorem, then double the area

The circles are externally tangent, so

OP = 2 + 4 = 6.

In quadrilateral

AOPD,

both

OA = 2

and

PD = 4

are perpendicular to the tangent line

AD,

making it a right trapezoid.

Drawing the line through

O

parallel to

AD

creates a right triangle with hypotenuse

OP = 6

and one leg

PD - OA = 2,

so

AD = \sqrt{6^2 - 2^2} = \sqrt{32} = 4\sqrt2.

The trapezoid

AOPD

has area

\frac{1}{2}(2 + 4)(4\sqrt2) = 12\sqrt2.

By symmetry the hexagon

AOBCPD

is made of two such trapezoids, so its area is

2 \cdot 12\sqrt2 = 24\sqrt2.

Thus, the correct answer is B.

Problem 15 in other years