Competition · AMC preparation · step 4 of 4

AMC 10 · 2020B · #20

Grade 8 geometry-3d
Archetype Compound Area by Decomposition / Difference └ Sub-archetype Circle-Polygon Hybrid
volume-rectangular-prismvolume-cylindersurface-areaminkowski-sum identify-subproblemsdimensional-analysis ↑ Prerequisites: volume-rectangular-prismsurface-area
📏 Long solution 💡 3 insights
Problem
B is a solid box with edge lengths 1, 3, 4. For r ≥ 0 let S(r) be the set of points within distance r of some point of B (the box 'inflated' by r). Its volume is a cubic polynomial in r: V(S(r)) = a r³ + b r² + c r + d. Compute bcad\frac{bc}{ad}.

Pick an answer.

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

AMC 10 2020 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Identify Subproblems

Tool #7 (Subproblems): the inflated box S(r) is the union of four very different pieces — the original box, slabs over each face, rounded cylinders over each edge, and spherical caps at each corner. Computing each piece's volume separately is far easier than tackling the whole shape at once. Tool #1 (Diagram) and Tool #17 (Visualize): a quick sketch shows the four types of pieces and confirms that the 12 edge-cylinders are quarter cylinders (one quarter circle per edge) and the 8 corner spheres are one-eighth spheres (one octant per corner).

1STEP 1

Take the box's own volume

Piece 1, the box itself: volume 1 · 3 · 4 = 12, the constant term, so d = 12.

V₁ = 1 · 3 · 4 = 12, d = 12
2STEP 2

Add the six face slabs

Piece 2, six face slabs: total volume is the box's surface area times r, and SA = 2(3 + 4 + 12) = 38, so c = 38.

SA(B) = 2(1 · 3 + 1 · 4 + 3 · 4) = 38, V₂ = 38 r, c = 38
3STEP 3

Add the twelve edge pieces

Piece 3, twelve edge quarter-cylinders: four per length make a full cylinder, so V₃ = π r²(1 + 3 + 4) = 8π r², giving b = 8π.

V₃ = π r² (1 + 3 + 4) = 8 π r², b = 8π
4STEP 4

Add the eight corner pieces

Piece 4, eight corner eighth-spheres: together one full sphere, V₄ = 43\frac{4}{3}π r³, so a = 43\frac{4}{3}π.

V₄ = 8 · 1/8 · 4/3π r³ = 4/3π r³, a = 4/3π
5STEP 5

Divide the coefficients

Read off a = 43\frac{4}{3}π, b = 8π, c = 38, d = 12; then bcad\frac{bc}{ad} = 8π⋅3843π⋅12\frac{8π·38}{\frac{4}{3}π·12} = 304π16π\frac{304π}{16π} = 19, the π cancels.

bc/ad = (8π · 38)/(4/3π · 12) = 304π/16π = 19 → (B)
Answer
19
The cancellation of π is the strongest reasonableness signal: b and a both carry one factor of π from the cylinders and sphere, while c and d are π-free. So bcad\frac{bc}{ad} is a clean rational number, matching the integer answer 19. The numbers themselves match the box's geometry — surface area 38, edge-sum 8, volume 12 — so the factorization checks out.
💡Key takeaway

This AMC 10 problem only needs Grade 8 cylinder-and-sphere volume formulas you already know — split the inflated box into four pieces (the box, slabs, edge-cylinders, corner-spheres). Read off a = 4π3\frac{4π}{3}, b = 8π, c = 38, d = 12, then bcad\frac{bc}{ad} = 304π16π\frac{304π}{16π} = 19. The answer is (B).

  • Take the box's own volume
  • Add the six face slabs
  • Add the twelve edge pieces
  • Add the eight corner pieces
  • Divide the coefficients

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