2013 AMC 10A Problem 20

Put the center at $O$ and let the starting square have sides $x = \pm\frac12$, $y = \pm\frac12$. Its vertices lie on the circle of radius $\frac{\sqrt2}{2}$; the rotated copies' vertices stay on that circle. The swept region has $90^\circ$ rotational symmetry, so compute the part in the angular range $0^\circ$ to $90^\circ$ (measured from the positive $x$-axis) and multiply by $4$.

Directions $45^\circ$ to $90^\circ$. The starting vertex $V = \left(\frac12, \frac12\right)$ at $45^\circ$ travels along the circle to $90^\circ$, so every point of this $45^\circ$ sector of radius $\frac{\sqrt2}{2}$ is swept: area $\frac{45}{360}\pi\cdot\frac12 = \frac{\pi}{16}$.

Directions $0^\circ$ to $45^\circ$. No vertex ever points here. In direction $\theta$ the boundary of a square is at distance $\frac{1}{2\cos\alpha}$, where $\alpha \le 45^\circ$ is the angle to the nearest side's perpendicular, and this is largest at the two extreme positions. So the region here is bounded by the side $x = \frac12$ of the starting square (directions $22.5^\circ$ to $45^\circ$) and the corresponding side of the final square (directions $0^\circ$ to $22.5^\circ$). These two sides meet at $Q = \left(\frac12, \frac12\tan 22.5^\circ\right) = \left(\frac12, \frac{\sqrt2 - 1}{2}\right)$. The region is the kite $O U Q V$ with $U = \left(\frac{\sqrt2}{2}, 0\right)$, which is two copies of triangle $OQV$:
$$2 \cdot \frac12 \cdot \frac12 \cdot \left(\frac12 - \frac{\sqrt2 - 1}{2}\right) = \frac{2 - \sqrt2}{4}.$$

Per quarter: $\frac{\pi}{16} + \frac{2 - \sqrt2}{4}$. Times $4$:
$$\frac{\pi}{4} + 2 - \sqrt2.$$

The answer is $\boxed{\textbf{(C)}\ 2 - \sqrt2 + \frac{\pi}{4}}$.

A region swept by a rotating convex shape is bounded partly by arcs traced by vertices and partly by edges of the extreme positions. Sorting the directions into "a vertex passes here" (full sector) and "no vertex passes here" (polygonal, governed by the first and last positions) turns the problem into sectors plus triangles. Symmetry cuts the work to one eighth of the figure.

Start from the union of the first and last squares, an eight-pointed star made of $8$ kites like $OUQV$: area $8 \cdot \frac{2 - \sqrt2}{4} = 4 - 2\sqrt2$. Each of the four notches between neighboring star points is filled by a vertex arc, adding a $45^\circ$ sector minus the kite already counted: $4\left(\frac{\pi}{16} - \frac{2 - \sqrt2}{4}\right) = \frac{\pi}{4} - 2 + \sqrt2$. Total $2 - \sqrt2 + \frac{\pi}{4}$. As a check, this is about $1.37$, between the star ($\approx 1.17$) and the full disk ($\approx 1.57$).

• Answering the whole circumscribed disk (pi/2) or the union of the first and last squares, instead of the disk near the vertices and the squares elsewhere.
• Using sectors of radius $\frac12$ (the inradius) instead of the circumradius $\frac{\sqrt2}{2}$ for the arcs traced by the vertices.