2010 AMC 10B Problem 16

First work out what the region looks like. Call the common center $O$ and the radius $r = \frac{\sqrt3}{3} \approx 0.577$.

1. Where does the circle go? The center is $\frac12 = 0.5$ away from each side of the square, and $\frac{\sqrt2}{2} \approx 0.707$ away from each corner. The radius $0.577$ is between these. So the circle pokes out through the middle of every side, but the square's corners stick out past the circle.
2. So the region inside the circle but outside the square is four equal caps, one on each side. Find the area of one cap.
3. One cap. Take the top side. Let $M$ be its midpoint, and let $P$ and $Q$ be the two points where the circle crosses it. Triangle $OMP$ has a right angle at $M$, with $OM = \frac12$ and $OP = r = \frac{\sqrt3}{3}$. By the Pythagorean theorem,
$$MP^2 = \frac13 - \frac14 = \frac{1}{12}, \qquad MP = \frac{1}{2\sqrt3} = \frac{\sqrt3}{6}.$$
4. Notice that $OP = \frac{\sqrt3}{3}$ is exactly twice $MP = \frac{\sqrt3}{6}$. A right triangle whose long side is twice a short side is a $30$-$60$-$90$ triangle, and the $30^\circ$ angle sits opposite the short side $MP$, at $O$. So $\angle MOP = 30^\circ$, and by symmetry $\angle POQ = 60^\circ$.
5. The cap is the $60^\circ$ slice of the circle (the sector $OPQ$) with triangle $OPQ$ taken away.
- The slice is $\frac{60}{360} = \frac16$ of the circle: $\frac16 \cdot \pi r^2 = \frac16 \cdot \frac{\pi}{3} = \frac{\pi}{18}$.
- Triangle $OPQ$ has base $PQ = 2 \cdot \frac{\sqrt3}{6} = \frac{\sqrt3}{3}$ and height $OM = \frac12$, so its area is $\frac12 \cdot \frac{\sqrt3}{3} \cdot \frac12 = \frac{\sqrt3}{12}$.
- One cap: $\frac{\pi}{18} - \frac{\sqrt3}{12}$.
6. Four caps:
$$4\left(\frac{\pi}{18} - \frac{\sqrt3}{12}\right) = \frac{2\pi}{9} - \frac{\sqrt3}{3}.$$

As a check, this is about $0.698 - 0.577 = 0.12$, a small positive area, which fits four thin caps.

The answer is $\boxed{\textbf{(B)}\ \dfrac{2\pi}{9}-\dfrac{\sqrt{3}}{3}}$.

When a circle and a polygon overlap, the pieces of circle cut off by the sides are caps, and each cap is a slice of the circle minus a triangle. The radius here was picked so that the slice angle comes out to a nice $60^\circ$. Always start by comparing the radius with the distance to a side and to a corner, so you know what shape you are dealing with.