2010 AMC 10B Problem 16

Let $r = \frac{\sqrt3}{3}$ and $O$ the common center. Each side of the square is at distance $\frac12$ from $O$; since $\frac12 < r < \frac{\sqrt2}{2}$, the circle pokes out through each side but does not reach the corners. The desired region is four congruent circular segments.

Consider one side and let $P, Q$ be the points where the circle crosses it. In right triangle from $O$ to the midpoint of the side to $P$, the leg along the perpendicular is $\frac12$ and the hypotenuse is $r$, so
$$\cos\angle = \frac{1/2}{\sqrt3/3} = \frac{\sqrt3}{2},$$
an angle of $30^\circ$. Hence $\angle POQ = 60^\circ$, and triangle $OPQ$ is equilateral with side $r$.

Segment area $=$ sector $-$ triangle:
$$\frac{60}{360}\pi r^2 - \frac{\sqrt3}{4} r^2 = \frac{\pi}{6}\cdot\frac13 - \frac{\sqrt3}{4}\cdot\frac13 = \frac{\pi}{18} - \frac{\sqrt3}{12}.$$

Four segments: $4\left(\frac{\pi}{18} - \frac{\sqrt3}{12}\right) = \frac{2\pi}{9} - \frac{\sqrt3}{3}$.

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

Overlap of a circle and a polygon is always a sum of circular segments, each equal to a sector minus an isosceles triangle. The radius $\frac{\sqrt3}{3}$ was chosen so that the half-angle is a $30$-$60$-$90$ angle; whenever a distance-to-radius ratio is $\frac{\sqrt3}{2}$, $\frac{\sqrt2}{2}$ or $\frac12$, expect a special angle.

Estimate: the circle's area is $\frac{\pi}{3} \approx 1.047$, only slightly more than the square's $1$, and most of the circle lies inside. The answer must be small but positive: (B) $\approx 0.698 - 0.577 = 0.12$ fits, while (C) $\approx 0.17$ and (D) $0.25$ are plausible only if you compute; (A) $\approx 0.047$ ignores the square's corners sticking out.

• Subtracting the whole square from the circle (the circle does not contain the square), or using a 30-degree angle for the segment.
• Computing the triangle inside the sector with base $r$ instead of using the equilateral triangle of side $r$, or forgetting to multiply the segment by $4$.