# 2011 AMC 10B Problems/Problem 20

The following problem is from both the 2011 AMC 12B #16 and 2011 AMC 10B #20, so both problems redirect to this page.

## 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 $R$?

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

## Solution

Suppose that $P$ is a point in the rhombus $ABCD$ and let $\ell_{BC}$ be the perpendicular bisector of $\overline{BC}$. Then $PB < PC$ if and only if $P$ is on the same side of $\ell_{BC}$ as $B$. The line $\ell_{BC}$ divides the plane into two half-planes; let $S_{BC}$ be the half-plane containing $B$. Let us define similarly $\ell_{BD},S_{BD}$ and $\ell_{BA},S_{BA}$. Then $R$ is equal to $ABCD \cap S_{BC} \cap S_{BD} \cap S_{BA}$. The region turns out to be an irregular pentagon. We can make it easier to find the area of this region by dividing it into four triangles:

$[asy] unitsize(8mm); defaultpen(linewidth(0.8pt)+fontsize(10pt)); dotfactor=4; pair A=(4,0), B=(2,2sqrt(3)), C=(-2,2sqrt(3)), D=(0,0), E=(B+C)/2, F=(B+C+D)/3, G=(A+C)/2, H=(A+B+D)/3, I=(A+B)/2; fill((0,2sqrt(3))--B--(3,sqrt(3))--(2,(2sqrt(3))/3)--(0,(4sqrt(3))/3)--cycle,gray); draw(A--B--C--D--cycle); draw(D--(0,2sqrt(3))); draw(D--(3,sqrt(3))); draw(A--C); draw(F--B--H); draw(B--G); label("A",A,SE);label("B",B,NE);label("C",C,NW);label("D",D,SW); label("E",E,N);label("F",F,SW);label("G",G,SW);label("H",H,S);label("I",I,NE); label("2",(D--C),SW); [/asy]$ Since $\triangle BCD$ and $\triangle BAD$ are equilateral, $\ell_{BC}$ contains $D$, $\ell_{BD}$ contains $A$ and $C$, and $\ell_{BA}$ contains $D$. Then $\triangle BEF \cong \triangle BGF \cong \triangle BGH \cong \triangle BIH$ with $BE = 1$ and $EF = \frac{1}{\sqrt{3}}$ so $[BEF] = \frac{1}{2}\cdot 1 \cdot \frac{\sqrt{3}}{3}$. Multiply this by 4 and it turns out that the pentagon has area $\boxed{(C)\frac{2\sqrt{3}}{3}}$.

## Solution 2

We follow the steps shown above until we draw pentagon $BIHFE$. We know that rhombus $ABCD$ can be divided into equilateral triangles $\triangle ABD$ and $\triangle CBD$. Using the $30-60-90$ special right triangle rules, we find the height of the equilateral triangles (and the height of the rhombus) to be $\sqrt{3}$. Therefore, the area of $ABCD$ is $2\sqrt{3}$. We now have to take off the areas $\triangle CDA$, $\triangle CEF$, and $\triangle AIH$ to get the desired shape. $\triangle CDA$ is just half of $ABCD$ $(\sqrt {3})$ and $\triangle AIH$ and $\triangle CEF$ are each $\frac{\sqrt {3}}{6}$, for a total area of $2\sqrt {3}-\sqrt {3}-\frac{\sqrt{3}}{6}-\frac{\sqrt{3}}{6}=\boxed{(C)\frac{2\sqrt{3}}{3}}$.

## Solution 3

We split rhombus $ABCD$ into two equilateral triangles, $ABD$ and $BCD$. In triangle $ABD$, the probability that a randomly selected point is closer to $B$ than either other point is $\frac{1}{3}$ (why?). Similarly, in triangle $BCD$, the same principle applies. Thus, the area of the region closer to $B$ than $A$, $C$, or $D$ is $\frac{1}{3} [ABD] + \frac{1}{3} [BCD]$. Since $ABD$ and $BCD$ are congruent, we have $\frac{1}{3} [ABD] + \frac{1}{3} [BCD] = \frac{2}{3} [ABD] = \frac{2}{3} \cdot \frac{s^2\sqrt{3}}{4} = \frac{2}{3} \cdot \frac{(2)^2\sqrt{3}}{4} = \boxed{\frac{2\sqrt3}{3} = C}$, and we are done.

## Solution 4

$[asy] unitsize(8mm); defaultpen(linewidth(0.8pt)+fontsize(10pt)); dotfactor=4; pair A=(4,0), B=(2,2sqrt(3)), C=(-2,2sqrt(3)), D=(0,0), E=(B+C)/2, F=(B+C+D)/3, G=(A+B+D)/3, H=(A+B)/2; fill((0,2sqrt(3))--B--(3,sqrt(3))--(2,(2sqrt(3))/3)--(0,(4sqrt(3))/3)--cycle,gray); draw(A--B--C--D--cycle); draw(D--(0,2sqrt(3))); draw(D--(3,sqrt(3))); draw(A--C); draw(F--B--H); draw(B--G); label("A",A,SE);label("B",B,NE);label("C",C,NW);label("D",D,SW); label("E",E,N);label("F",F,SW);label("G",G,S);label("H",H,E); label("2",(D--C),SW); [/asy]$ Since $H$ and $E$ are halfway between $AB$ and $CB$, respectively, we know that $\overline{BH}=\overline{BE}=1$. By symmetry, $\Delta BFG$ is equilateral, so $\angle FBG=60^\circ\implies\angle EBF=\angle HBG=30^\circ$ and therefore $\Delta EBF$ and $\Delta HBG$ are 30-60-90 right triangles. Thus, $[\Delta EBF]=[\Delta BFG]=\dfrac1{2\sqrt3}$. We know that $\overline{FB}=\overline{GB}=\dfrac2{\sqrt3}$, so therefore $[\Delta BFG]=\dfrac{\sqrt3}4\left(\dfrac2{\sqrt3}\right)^2=\dfrac1{\sqrt3}$. Summing these three regions, we get $\dfrac1{2\sqrt3}+\dfrac1{2\sqrt3}+\dfrac1{\sqrt3}=\boxed{\textbf{(C)}~\dfrac{2\sqrt3}3}$. ~ Technodoggo, Asymptote diagram modified from Solution 1

~IceMatrix