Six regular hexagons surround a regular hexagon of side length $1$ as shown. What is the area of $\triangle ABC$? $\textbf {(A) } 2\sqrt{3} \qquad$ $ \textbf {(B) } 3\sqrt{3} \qquad$ $ \textbf {(C) } 1+3\sqrt{2} \qquad$ $ \textbf {(D) } 2+2\sqrt{3} \qquad$ $ \textbf {(E) } 3+2\sqrt{3}$
Solution
Put the center of the middle hexagon at the origin. The centers of the outer hexagons are at distance $\sqrt3$ from it, and from the figure $A=\left(\frac{3\sqrt3}2,-\frac12\right)$, $B=\left(-\sqrt3,1\right)$, $C=(0,-2)$.
$\overrightarrow{CA}=\left(\frac{3\sqrt3}2,\frac32\right)$ and $\overrightarrow{CB}=\left(-\sqrt3,3\right)$, so $CA=3$, $CB=2\sqrt3$, and $\overrightarrow{CA}\cdot\overrightarrow{CB}=-\frac92+\frac92=0$. So $\triangle ABC$ has a right angle at $C$.
A square of side length $2$ has a regular octagon of side length $2$ constructed externally on each side. Let $A,B,C,D$ be the centers of the four octagons, listed in order around the square. What is the area of square $ABCD$?
$\textbf {(A) } 2\sqrt{3} \qquad$ $ \textbf {(B) } 3\sqrt{3} \qquad$ $ \textbf {(C) } 1+3\sqrt{2} \qquad$ $ \textbf {(D) } 2+2\sqrt{3} \qquad$ $ \textbf {(E) } 3+2\sqrt{3}$ 
