2021 AMC Fall 10A Problem 19

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 $\boxed{\textbf{(A)}\ 10}$.

A rolling disk sweeps the set of points within its radius of the center's path, so first describe the path, then thicken it. Outside a convex corner the thickened path bulges into a quarter circle; inside a concave (interior) corner the disk cannot fill the corner and leaves a square-minus-quarter-circle gap. Decomposing both regions into rectangles and quarter circles gives two clean expressions in $s$ and $\pi$.

• Treating the inside sweep as a full frame of width 2, s^2 - (s-4)^2, and forgetting the four unreachable corner slivers of area 1 - pi/4 each.
• Giving the outside region square corners ($8s + 16$) instead of quarter circles, or using radius $1$ quarter circles ($8s + \pi$) for the outer corners.