2020 AMC 10B Problem 20

Warm-up in two dimensions. Take a flat rectangle and all points within $r$ of it. You get the rectangle, plus a strip of width $r$ along each side, plus a quarter-circle of radius $r$ at each corner. The four quarter-circles together make one full circle. The same idea works for the box.

Sort the points of $S(r)$ by which part of the box they are closest to.

1. The box itself. Volume $1 \times 3 \times 4 = 12$. This has no $r$, so $d = 12$.
2. Next to a face. Over each face sits a flat slab of thickness $r$. Its volume is (area of face) $\times\, r$. The six faces have total area
$$2(1 \times 3 + 3 \times 4 + 4 \times 1) = 2(3 + 12 + 4) = 38,$$
so the slabs have volume $38r$, and $c = 38$.
3. Next to an edge. Along each edge, the slabs on the two faces meeting there leave a gap shaped like a quarter of a cylinder (the faces meet at a right angle, which is a quarter turn). The box has $4$ edges of each length $1$, $3$ and $4$, so the total edge length is $4(1 + 3 + 4) = 32$. The quarter-cylinders add up to
$$\frac14 \times \pi r^2 \times 32 = 8\pi r^2,$$
so $b = 8\pi$.
4. Next to a corner. At each of the $8$ corners sits one eighth of a ball of radius $r$. Eight eighths make one whole ball:
$$\frac43 \pi r^3,$$
so $a = \frac{4\pi}{3}$.
5. Compute.
$$\frac{bc}{ad} = \frac{8\pi \times 38}{\frac{4\pi}{3} \times 12} = \frac{304\pi}{16\pi} = 19.$$

The answer is $\boxed{\textbf{(B)}\ 19}$.

Points near a box split up by what they are nearest to: the inside, a face, an edge, or a corner. Faces give slabs (volume grows like $r$), edges give pieces of cylinders (like $r^2$), and corners give pieces of a ball (like $r^3$). All the angles of a box are right angles, so the edge pieces are quarter-cylinders and the eight corner pieces fit together into exactly one ball.