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 $R?$

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

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

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

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

$2$

Solution:

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.

2011 AMC 10B Problem 20

Solution:

Problem 20 in Other Years