Four circles of radius $1$ are each tangent to two sides of a square and externally tangent to a circle of radius $2$, as shown. What is the area of the square? $\text{(A)}\ 32 \qquad$ $ \text{(B)}\ 22 + 12\sqrt {2}\qquad$ $ \text{(C)}\ 16 + 16\sqrt {3}\qquad$ $ \text{(D)}\ 48 \qquad$ $ \text{(E)}\ 36 + 16\sqrt {2}$
Solution
By symmetry the centers of the small circles lie on the diagonals of the square. Along a diagonal: from a corner to the nearest small center is $\sqrt2$ (the center is $1$ from both sides), then $1+2$ to the center of the big circle, then $2+1$ to the opposite small center, then $\sqrt2$ to the corner. So the diagonal is $\sqrt2+1+4+1+\sqrt2=6+2\sqrt2$.
So the area of the square is $\frac12(6+2\sqrt2)^2=22+12\sqrt2$.
$\text{(A)}\ 32 \qquad$ $ \text{(B)}\ 22 + 12\sqrt {2}\qquad$ $ \text{(C)}\ 16 + 16\sqrt {3}\qquad$ $ \text{(D)}\ 48 \qquad$ $ \text{(E)}\ 36 + 16\sqrt {2}$ 
