2021 AMC Fall 10A Problem 19

A rolling disk of radius $1$ covers every point within $1$ of the path of its center. So for each disk, first find where the center goes, then see what it covers.

Inside the square.

1. The center stays $1$ away from each wall, so it moves around a smaller square of side $s - 2$.
2. The disk covers a band along the walls, $2$ wide. If it filled that band completely, the area would be the big square minus the untouched middle square of side $s - 4$:
$$s^2 - (s - 4)^2 = s^2 - (s^2 - 8s + 16) = 8s - 16.$$
3. But the disk is round, so it cannot reach into the corners. At a corner, the closest the center gets is $1$ from each wall. The $1 \times 1$ corner square is then covered only by a quarter circle of radius $1$. The missed piece at each corner has area $1 - \frac{\pi}{4}$.
4. So
$$A = 8s - 16 - 4\left(1 - \frac{\pi}{4}\right) = 8s - 16 - 4 + \pi = 8s - 20 + \pi.$$

Outside the square.

1. The disk covers everything within $2$ of the square, outside it. Along each side that is a $s \times 2$ strip: $4 \times 2s = 8s$.
2. At each corner the disk swings around the corner point, covering a quarter circle of radius $2$, area $\frac14 \pi \times 2^2 = \pi$. Four corners give $4\pi$.
3. Outside area: $8s + 4\pi$.

Set up the equation.

8. The outside area is $2A$:
$$8s + 4\pi = 2(8s - 20 + \pi) = 16s - 40 + 2\pi.$$
9. Move terms: $40 + 4\pi - 2\pi = 16s - 8s$, so $8s = 40 + 2\pi$ and
$$s = 5 + \frac{2\pi}{8} = 5 + \frac{\pi}{4}.$$
10. So $a = 5$, $b = 1$, $c = 4$, and $a + b + c = 10$.

The answer is $\boxed{\textbf{(A)}\ 10}$.

To find what a rolling disk sweeps, trace its center and then "thicken" that path by the radius. Around an outside corner the thickened path bulges into a quarter circle. In an inside corner, the round disk cannot fill the corner and leaves a small gap: a square minus a quarter circle. Breaking both regions into rectangles and quarter circles gives two simple expressions in $s$.

Inside. The disk's center stays $1$ unit from the walls, tracing a square of side $s - 2$. The disk covers every point within $1$ of that path: the band of width $2$ along the walls, minus what it cannot reach in the corners. Near a corner, the nearest center position is the corner of the center-path, $1$ unit from each wall; the unit square in the corner is covered only by the quarter disk of radius $1$ around that point. So each corner has an unreachable sliver of area $1 - \tfrac{\pi}{4}$, and
$$A = s^2 - (s - 4)^2 - 4\left(1 - \tfrac{\pi}{4}\right) = 8s - 16 - 4 + \pi = 8s - 20 + \pi.$$

Outside. Rolling around the outside, the disk covers every point within $2$ of the square but outside it: four $s \times 2$ rectangles along the sides plus four quarter disks of radius $2$ at the corners (the disk pivots around each corner). Area:
$$4 \cdot 2s + 4 \cdot \tfrac{1}{4}\pi \cdot 2^2 = 8s + 4\pi.$$

Equation. The outside area is $2A$:
$$8s + 4\pi = 2(8s - 20 + \pi) = 16s - 40 + 2\pi \quad\Longrightarrow\quad 8s = 40 + 2\pi \quad\Longrightarrow\quad s = 5 + \frac{\pi}{4}.$$

So $a = 5$, $b = 1$, $c = 4$ and $a + b + c = 10$.

The answer is $\boxed{\textbf{(A)}\ 10}$.