2013 AMC 10A Problem 20

Put the center of the square at $O = (0, 0)$, with the starting square's sides on $x = \pm\tfrac12$ and $y = \pm\tfrac12$. Turn it $45^\circ$ counterclockwise.

1. How far the corners are from the center. A corner like $V = \left(\tfrac12, \tfrac12\right)$ is half a diagonal away: $\sqrt{\tfrac14 + \tfrac14} = \tfrac{\sqrt2}{2}$. As the square turns, each corner moves along the circle of radius $\tfrac{\sqrt2}{2}$, through one eighth of a full turn.
2. Cut the picture into 8 slices. Draw lines from $O$ at $0^\circ, 45^\circ, 90^\circ, \ldots$ This makes $8$ slices of $45^\circ$. Because the picture looks the same after a quarter turn, the slices come in two kinds, $4$ of each. Work out one of each kind, then multiply by $4$.
3. Slice from $45^\circ$ to $90^\circ$: a corner passes through. The corner $V$ starts at $45^\circ$ and ends at $90^\circ$. At every moment the whole segment from $O$ to that corner is inside the square. So every point of this slice within distance $\tfrac{\sqrt2}{2}$ of $O$ gets covered, and nothing farther out does. The region is a pie slice, $\tfrac18$ of the circle:
$$\frac18 \times \pi \times \left(\frac{\sqrt2}{2}\right)^2 = \frac18 \times \pi \times \frac12 = \frac{\pi}{16}.$$
4. Slice from $0^\circ$ to $45^\circ$: no corner passes through. Here the square only ever shows one side to this slice. That side turns steadily, and in each direction it sticks out farthest either at the start or at the end of the turn. So the region here is just what the starting square or the final square covers:
- the starting square's side $x = \tfrac12$, coming down from $V = \left(\tfrac12, \tfrac12\right)$;
- the final square's side from its corner $U = \left(\tfrac{\sqrt2}{2}, 0\right)$ to its corner $\left(0, \tfrac{\sqrt2}{2}\right)$, which is the line $x + y = \tfrac{\sqrt2}{2}$.
5. Where those two sides cross. Put $x = \tfrac12$ into $x + y = \tfrac{\sqrt2}{2}$: $y = \tfrac{\sqrt2 - 1}{2}$. Call this point $Q = \left(\tfrac12, \tfrac{\sqrt2 - 1}{2}\right)$.
6. Area of this slice. The region is the four-sided shape $O\,U\,Q\,V$, made of two mirror-image triangles, $OQV$ and $OQU$. In triangle $OQV$, the side $QV$ is upright on the line $x = \tfrac12$ with length
$$\frac12 - \frac{\sqrt2 - 1}{2} = \frac{2 - \sqrt2}{2},$$
and $O$ is $\tfrac12$ away from that line. So triangle $OQV$ has area $\tfrac12 \times \tfrac{2 - \sqrt2}{2} \times \tfrac12 = \tfrac{2 - \sqrt2}{8}$, and the whole shape has twice that, $\tfrac{2 - \sqrt2}{4}$.
7. Add up all 8 slices.
$$4 \times \frac{\pi}{16} + 4 \times \frac{2 - \sqrt2}{4} = \frac{\pi}{4} + 2 - \sqrt2.$$

Sanity check. This is about $1.37$. It should be more than the square's area $1$ and less than the full circle through the corners, $\tfrac{\pi}{2} \approx 1.57$. It is.

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

When a shape turns, the region it sweeps has two kinds of edge: curved arcs traced by the corners, and straight sides from the first and last positions. Sorting the directions into "a corner passes here" (a pie slice) and "no corner passes here" (only the first and last positions matter) turns a strange shape into pie slices plus triangles. Symmetry means you only need to work out two slices.

Start from the star made by the first and last squares together. It is $8$ copies of the shape $OUQV$, with area $8 \times \tfrac{2 - \sqrt2}{4} = 4 - 2\sqrt2$. The moving corners then fill in $4$ of the star's slices up to the circle: in each, a pie slice of area $\tfrac{\pi}{16}$ replaces a star piece of area $\tfrac{2 - \sqrt2}{4}$. So the total is
$$4 - 2\sqrt2 + 4\left(\frac{\pi}{16} - \frac{2 - \sqrt2}{4}\right) = 2 - \sqrt2 + \frac{\pi}{4}.$$

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