2011 AMC 10B Problem 20

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

Use perpendicular bisectors to locate points closer to

B

The desired region is the central pentagon cut out by two congruent small triangles

The points closer to

B

than to another vertex are bounded by the perpendicular bisectors of

BA

,

BC

, and

BD

. The bisector of

BD

is diagonal

AC

, so the desired region lies in

\triangle ABC

.

Triangle

ABC

has area

\dfrac12\cdot2\cdot2\sin120^\circ=\sqrt3

. The perpendicular bisectors of

BA

and

BC

cut off two congruent

30

-

60

-

90

triangles, each with area

\dfrac{\sqrt3}{6}

.

Therefore the desired area is

\sqrt3-2\cdot\dfrac{\sqrt3}{6}=\dfrac{2\sqrt3}{3}

.

Thus, C is the correct answer.

Problem 20 in other years