2014 AMC 10B Problem 13

Let $O$ be the center of the middle hexagon. Draw $OA$, $OB$, $OC$.

Look at vertex $B$. The ray from $O$ through a vertex $V$ of the central hexagon reaches $V$ after distance $1$ (a regular hexagon's circumradius equals its side). Beyond $V$, two outer hexagons meet along a common edge that lies on the same ray by symmetry, and $B$ is the far end of that edge. So $OB = 1 + 1 = 2$. Likewise $OA = OC = 2$.

The three rays are spaced $120^\circ$ apart, so $\triangle ABC$ is equilateral with circumradius $2$. Dropping the perpendicular from $O$ to $AB$ gives a $30$-$60$-$90$ triangle with hypotenuse $2$, so half of $AB$ is $\sqrt3$ and $AB = 2\sqrt3$.

The area is
$$\frac{\sqrt3}{4}\,(2\sqrt3)^2 = \frac{\sqrt3}{4} \cdot 12 = 3\sqrt3.$$

The answer is $\boxed{\textbf{(B)}\ 3\sqrt{3}}$.

In a hexagonal tiling all the natural lengths are multiples of the side or of $\sqrt3$ times it, and the center of the middle hexagon is the organizing point. Measuring each vertex from $O$ converts a picture question into a circumradius question, and an equilateral triangle with circumradius $R$ has side $R\sqrt3$.

Place $O$ at the origin with the central hexagon's vertices at $(\pm1, 0)$, $(\pm\frac12, \pm\frac{\sqrt3}{2})$. Then $A = (-2, 0)$, $B = (1, \sqrt3)$, $C = (1, -\sqrt3)$; $BC = 2\sqrt3$ is vertical and $A$ is $3$ units from that line, so the area is $\frac12 \cdot 2\sqrt3 \cdot 3 = 3\sqrt3$.

• Estimating the side from the picture as 3 or 4 unit lengths instead of computing it from the 120-degree angles at the center.
• Using $OB = 2$ as the side length of the triangle rather than its circumradius, giving $\sqrt3$.