2004 AMC 12A Problem 24

Question

A plane contains points

A

and

B

with

AB = 1.

Let

S

be the union of all disks of radius

1

in the plane that cover

\overline{AB}.

What is the area of

S?

2\pi + \sqrt{3}

\dfrac{8\pi}{3}

3\pi - \dfrac{\sqrt{3}}{2}

\dfrac{10\pi}{3} - \sqrt{3}

4\pi - 2\sqrt{3}

Answer: C

Concepts:area decomposition sector circle

Difficulty rating: 2350

Small hint:

A radius-

1

disk covers

\overline{AB}

exactly when its center lies within

1

of both

A

and

B

Big hint:

Find that lens-shaped region

R

of centers, then

S

is all points within

1

of

R

Solution:

A radius-

1

disk covers segment

\overline{AB}

exactly when its center is within

1

of both

A

and

B.

That region

R

is the lens where the two unit circles centered at

A

and

B

overlap.

Each unit circle passes through the other’s center, so the lens is bounded by two

120^\circ

arcs. Two

120^\circ

sectors of area

\tfrac{\pi}{3}

overlap in two equilateral triangles of total area

\tfrac{\sqrt3}{2},

giving

R

area

\tfrac{2\pi}{3} - \tfrac{\sqrt3}{2}.

The set

S

consists of all points within

1

of

R.

Beyond

R

itself, this adds two

60^\circ

sectors of radius

1

(each area

\tfrac{\pi}{6}

) and two

120^\circ

annuli of outer radius

2

and inner radius

1