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2020 AMC 10B Problem 20

2020 AMC 10B Problem 20AMC 10 Step by Step

Name: Date: September 13, 2026

Solid Geometry · difficulty 4 of 5 · about 6 min · position P16-20

Let $B$ be a right rectangular prism (box) with edges lengths $1,$ $3,$ and $4$ , together with its interior. For real $r\geq0$ , let $S(r)$ be the set of points in $3$ -dimensional space that lie within a distance $r$ of some point in $B$ . The volume of $S(r)$ can be expressed as $ar^{3} + br^{2} + cr +d$ , where $a,$ $b,$ $c,$ and $d$ are positive real numbers. What is $\frac{bc}{ad}?$

A)
$6$
B)
$19$
C)
$24$
D)
$26$
E)
$38$

Answer

Key insight

The r-neighborhood of a box is the box, plus slabs on the faces, quarter-cylinders along the edges, and eighth-spheres at the corners; each contributes one power of r.

Solution

Sort the points of $S(r)$ by which part of the box is nearest to them.

The box itself: volume $1 \cdot 3 \cdot 4 = 12$ . So $d = 12$ .
Points nearest a face: a slab of thickness $r$ on each face, total volume (surface area) $\times\, r$ . Surface area $= 2(1\cdot3 + 3\cdot4 + 4\cdot1) = 38$ . So $c = 38$ .
Points nearest an edge: along each edge a quarter-cylinder of radius $r$ (the slabs on the two adjacent faces meet at a right angle, leaving a $90^\circ$ wedge). Total edge length $= 4(1 + 3 + 4) = 32$ , so the volume is $\tfrac14 \pi r^2 \cdot 32 = 8\pi r^2$ . So $b = 8\pi$ .
Points nearest a vertex: an eighth of a sphere at each of the $8$ corners, together one full sphere: $\tfrac43 \pi r^3$ . So $a = \tfrac{4\pi}{3}$ .

Then
$\frac{bc}{ad} = \frac{8\pi \cdot 38}{\frac{4\pi}{3}\cdot 12} = \frac{304\pi}{16\pi} = 19.$

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

Why this works

"All points within $r$ of a convex solid" decomposes by dimension: the original volume (constant), face slabs (linear in $r$ ), edge wedges (quadratic), and corner spheres (cubic). The wedge angles at a box are $90^\circ$ , so the edge pieces are quarter-cylinders and the corner pieces are eighth-spheres; those fractions reassemble into one cylinder of length $\tfrac14\cdot$ (edge total) and exactly one sphere.

The trap

Using full cylinders along the edges or full spheres at the corners, or forgetting that the edge terms use quarter-cylinders (factor 1/4).

Common mistakes

Using full cylinders along the edges or full spheres at the corners, or forgetting that the edge terms use quarter-cylinders (factor 1/4).
Miscounting the surface area (e.g. $1\cdot3 + 3\cdot4 + 4\cdot1 = 19$ without doubling) or the edge total (there are $12$ edges, four of each length).

Techniques

Cut the figure into known shapes (triangles, rectangles, sectors)