Harvard-MIT Mathematics Tournament

Solution:

The desired region consists of a small square and four "circle segments," i.e. regions of a circle bounded by a chord and an arc. The side of this small square is just the chord of a unit circle that cuts off an angle of $30^{\circ}$, and the circle segments are bounded by that chord and the circle. Using the law of cosines (in an isosceles triangle with unit leg length and vertex angle $30^{\circ}$), we find that the square of the length of the chord is equal to $2-\sqrt{3}$. We can also compute the area of each circle segment, namely $\frac{\pi}{12}-\frac{1}{2}(1)(1) \sin 30^{\circ}=\frac{\pi}{12}-\frac{1}{4}$. Hence, the desired region has area $2-\sqrt{3}+4\left(\frac{\pi}{12}-\frac{1}{4}\right)=\frac{\pi}{3}+1-\sqrt{3}$.