2011 AMC 10B Problem 20

Since $\angle B = 120^\circ$, the angles at $A$ and $C$ are $60^\circ$, so triangles $ABD$ and $CBD$ are both equilateral with side $2$ (in particular $BD = 2$). The rhombus is their union along $BD$.

A point is closer to $B$ than to $A$ exactly when it lies on $B$'s side of the perpendicular bisector of $AB$, and similarly for $C$ and $D$. Inside equilateral triangle $ABD$, the perpendicular bisectors of $AB$ and $BD$ meet at the center $O$ of the triangle, and the points closer to $B$ than to $A$ or $D$ form the kite with vertices $B$, the midpoint of $AB$, $O$, and the midpoint of $BD$. By the threefold symmetry of the equilateral triangle, this kite is one third of the triangle. (The bisector of $BC$ does not cut into it: every point of that kite is within distance $1$ of $B$ but is at least as far from $C$.)

Triangle $ABD$ has area $\frac{\sqrt{3}}{4} \cdot 2^2 = \sqrt{3}$, so the kite has area $\frac{\sqrt{3}}{3}$. The same happens in triangle $CBD$.

Total area of $R$: $2 \cdot \frac{\sqrt{3}}{3} = \frac{2\sqrt{3}}{3}$, so the answer is $\boxed{\textbf{(C)}\ \frac{2\sqrt{3}}{3}}$.

"Closer to $B$ than to the other vertices" is a Voronoi cell bounded by perpendicular bisectors. Decomposing the rhombus into equilateral triangles turns each half of the cell into a symmetric piece whose area is a fixed fraction of the triangle, so no messy vertex coordinates are needed.

Coordinates with $B = (0, 0)$, $A = (1, \sqrt{3})$, $C = (1, -\sqrt{3})$, $D = (2, 0)$: the region is $x \le 1$, $x + \sqrt{3}|y| \le 2$, $|y| \le \sqrt{3}x$, a pentagon with vertices $(0,0)$, $(\tfrac{1}{2}, \pm\tfrac{\sqrt{3}}{2})$, $(1, \pm\tfrac{1}{\sqrt{3}})$. The shoelace formula gives $\frac{2\sqrt{3}}{3}$.

• Thinking the region closer to $B$ is a quarter of the rhombus, or bounding it only by the bisector of $BD$ and forgetting the bisectors of $BA$ and $BC$.
• Computing $BD$ as the long diagonal $2\sqrt{3}$ instead of the short one of length $2$.