2014 AMC 10B Problem 13

The whole picture is built around the center of the middle hexagon, so measure everything from there. Call that center $O$.

1. Put in coordinates. Place $O$ at $(0, 0)$ with the middle hexagon's corners at $(1, 0)$, $\left(\frac12, \frac{\sqrt3}{2}\right)$, $\left(-\frac12, \frac{\sqrt3}{2}\right)$, $(-1, 0)$, $\left(-\frac12, -\frac{\sqrt3}{2}\right)$, $\left(\frac12, -\frac{\sqrt3}{2}\right)$. (A regular hexagon of side $1$ is six equilateral triangles of side $1$ meeting at the center, so each corner is $1$ from $O$.)
2. Where are $A$, $B$, $C$? Each of them is a corner where two outer hexagons meet, sitting straight out from a corner of the middle hexagon. Two outer hexagons share an edge of length $1$ that starts at a corner of the middle hexagon and points straight away from $O$. So each of $A$, $B$, $C$ is $1 + 1 = 2$ from $O$, in the same direction as a corner of the middle hexagon.
3. Using every other corner, as in the figure: $A = (-2, 0)$, $B = (1, \sqrt3)$, $C = (1, -\sqrt3)$. (Each is twice a corner of the middle hexagon.)
4. Area. $B$ and $C$ lie on the vertical line $x = 1$, so take $BC$ as the base: its length is $\sqrt3 - (-\sqrt3) = 2\sqrt3$.
5. The height is the horizontal distance from $A$ to that line: from $x = -2$ to $x = 1$ is $3$.
6. Area $= \frac12 \times 2\sqrt3 \times 3 = 3\sqrt3$.

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

In a picture made of regular hexagons, the center of the middle one is the natural place to measure from, and every length is a whole number or a whole number times $\sqrt3$. Once you know how far each corner of the triangle is from the center, the triangle is fixed. Choosing a base that lies on a vertical line makes the height easy to read off.

Skip coordinates. The three corners $A$, $B$, $C$ are each $2$ from $O$ and spaced evenly around it, $120^\circ$ apart, so $ABC$ is equilateral. Drop a perpendicular from $O$ to side $AB$: it makes a $30$-$60$-$90$ triangle with hypotenuse $OA = 2$, so half of $AB$ is $\sqrt3$ and $AB = 2\sqrt3$. The area of an equilateral triangle with side $s$ is $\frac{\sqrt3}{4}s^2 = \frac{\sqrt3}{4} \times 12 = 3\sqrt3$.

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