2011 AMC 12B Problem 16

Question

Rhombus

ABCD

has side length

2

and

\angle B=120^\circ.

Region

R

consists of all points inside the rhombus that are closer to vertex

B

than any of the other three vertices. What is the area of

R?

\dfrac{\sqrt{3}}{3}

\dfrac{\sqrt{3}}{2}

\dfrac{2\sqrt{3}}{3}

1+\dfrac{\sqrt{3}}{3}

2

Answer: C

Concepts:perpendicular bisector rhombus area

Difficulty rating: 1850

Small hint:

The boundary of

R

lies along the perpendicular bisectors of the sides through

B

Big hint:

R

is a pentagon made of four congruent

30

-

60

-

90

triangles, each with a leg of length

1

Solution:

Let

E

and

H

be the midpoints of

AB

and

BC.

The perpendicular bisector of

AB

through

E

meets diagonal

AC

at

F,

and the perpendicular bisector of

BC

through

H

meets

AC

at

G.

The region

R

is the pentagon

BEFGH.

Triangle

AFE

is a

30

-

60

-

90^\circ

triangle with

AE=1,

so its area is

\dfrac12\cdot1\cdot\dfrac{1}{\sqrt3}=\dfrac{\sqrt3}{6}.

Triangles

BFE

and

BGH

are congruent to it, and

\triangle FBG

is equilateral, splitting into two more copies.

Hence

R

consists of four congruent triangles, giving area

4\cdot\dfrac{\sqrt3}{6}=\dfrac{2\sqrt3}{3}.

Thus, the correct answer is C.