2012 AMC 10A Problem 18

Question

The closed curve in the figure is made up of

9

congruent circular arcs each of length

\dfrac{2\pi}{3},

where each of the centers of the corresponding circles is among the vertices of a regular hexagon of side

2.

What is the area enclosed by the curve?

2\pi+6

2\pi+4\sqrt{3}

3\pi+4

2\pi+3\sqrt{3}+2

\pi+6\sqrt{3}

Answer: E

Concepts:sector regular polygon area decomposition

Difficulty rating: 1930

Small hint:

Each arc is a

120^\circ

arc of a unit circle

Big hint:

Rearrange the sector pieces around the regular hexagon

Solution:

Each arc has radius

1

and length

\frac{2\pi}{3},

so the corresponding circular sectors all have the same area. As the diagram shows, the sector pieces may be rearranged: everything added to the regular hexagon combines to exactly one unit circle.

The regular hexagon is made of six equilateral triangles of side

2,

so its area is

6\left(\dfrac{\sqrt3}{4}\cdot2^2\right)=6\sqrt3.

Adding the unit circle gives a total enclosed area of

\pi+6\sqrt3.

Thus, E is the correct answer.