2011 AMC 10B Problems/Problem 20

Problem

Rhombus $ABCD$ has side length $2$ and $\angle B = 120$ °. 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 $R$ ?

$\textbf{(A)}\ \frac{\sqrt{3}}{3} \qquad\textbf{(B)}\ \frac{\sqrt{3}}{3} \qquad\textbf{(C)}\ \frac{2\sqrt{3}}{3} \qquad\textbf{(D)}\ 1 + \frac{\sqrt{3}}{3} \qquad\textbf{(E)}\ 2$

Solution

Any point on the perpendicular bisector of two other points is equidistant from both of those points. This line will divide divide the rhombus into two sections. The region containing $B$ is also the region where the points will be closer to $B$ than the other vertex. The region $R$ we are looking for is the intersection of the sections containing R from the perpendicular bisectors of $AB, CB,$ and $DB.$

$[asy] unitsize(8mm); defaultpen(linewidth(0.8pt)+fontsize(10pt)); dotfactor=4; pair A=(4,0), B=(2,2sqrt(3)), C=(-2,2sqrt(3)), D=(0,0); fill((0,2sqrt(3))--B--(3,sqrt(3))--(2,(2sqrt(3))/3)--(0,(4sqrt(3))/3)--cycle,lightgray); draw(A--B--C--D--cycle); draw(D--(0,2sqrt(3))); draw(D--(3,sqrt(3))); draw(A--C); label("$A$",A,SE); label("$B$",B,NE); label("$C$",C,NW); label("$D$",D,SW); label("$2$",(B--C),N); [/asy]$

The region turns to be an irregular pentagon. We can make it easier to find the area of this by dividing it into a triangle and a trapezoid.

$[asy] unitsize(8mm); defaultpen(linewidth(0.8pt)+fontsize(10pt)); dotfactor=4; pair A=(0,2), B=(2sqrt(3),0), C=((4sqrt(3))/3,-2), D=((-4sqrt(3))/3,-2), E=(-2sqrt(3),0); pair O=(0,0); draw(A--B--C--D--E--cycle); draw(B--E); draw(A--O,gray); label("$A$",A,N); label("$B$",B); label("$C$",C,SE); label("$D$",D,SW); label("$E$",E,W); label("$O$",O,S); label("$1$",midpoint(E--A),NW); label("$1$",midpoint(B--A),NE); label("$\frac{1}{2}$",midpoint(A--O),E); label("$\frac{\sqrt{3}}{2}$",midpoint(B--O),S); label("$\frac{\sqrt{3}}{2}$",midpoint(E--O),S); [/asy]$

We know $AE=AB=1$ because they are each half of the sides of the rhombus. $\triangle EAO$ and $\triangle BAO$ are $30-60-90$ triangles because $AO$ is an angle bisector of the $120^\circ$ angle $EAB.$ We can solve for those triangles. The area of triangle $\triangle AEB$ is $\frac{1}{2}bh = \frac{1}{2} \cdot \sqrt{3} \cdot \frac{1}{2} = \frac{\sqrt{3}}{4}$

Another thing we can do is find the height of trapezoid $BCDE.$ Diagonal $BD$ back in the first diagram is equal to $2$ because it would form two equilateral triangles. Because the diagonals of a rhombus bisect each other, the distance from $A$ to $CD$ (in the second diagram) is $1.$ Subtract $AO$ from that, and we get the height of trapezoid $BCDE$ is $\frac{1}{2}.$

To find the area of the trapezoid, we still need to find the length of the other base, $CD.$

$[asy] unitsize(8mm); defaultpen(linewidth(0.8pt)+fontsize(9pt)); dotfactor=4; pair A=(0,2), B=(2sqrt(3),0), C=((4sqrt(3))/3,-2), D=((-4sqrt(3))/3,-2), E=(-2sqrt(3),0); pair O=(0,0), X=((4sqrt(3))/3,0); draw(A--B--C--D--E--cycle); draw(B--E); draw(X--C,gray); label("$A$",A,N); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); label("$E$",E,NW); label("$X$",X,NW); label("$\frac{1}{2}$",midpoint(X--C),W); label("$\sqrt{3}$",midpoint(B--E),N); label("$\frac{\sqrt{3}}{6}$",midpoint(X--B),N); [/asy]$

Since $AB \perp BC$ and $\angle ABE = 30^\circ, \triangle XBC$ is another $30-60-90\text{ triangle }$ and $XB = \frac{\sqrt{3}}{6}.$

$BCDE$ is an isosceles trapezoid, so $CD = EB - 2XB = \sqrt{3} - \frac{\sqrt{3}}{3} = \frac{2\sqrt{3}}{3}.$ The area of the trapezoid is $\frac{h}{2}(b_1+b_2) = \frac{1}{4} (\frac{2\sqrt{3}}{3} + \sqrt{3}) = \frac{1}{4} \cdot \frac{5\sqrt{3}}{3} = \frac{5\sqrt{3}}{12}$

Find the total area of the pentagon

$\frac{5\sqrt{3}}{12} + \frac{\sqrt{3}}{4} = \frac{5\sqrt{3}}{12} + \frac{3\sqrt{3}}{12} = \frac{8\sqrt{3}}{12} = \boxed{\mathrm{(C) \ } \frac{2\sqrt{3}}{3}}$

2011 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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2011 AMC 10B Problems/Problem 20

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