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AMC 10 · 2020 · #20

Grade 8 geometry-3d
Archetype Compound Area by Decomposition / Difference
volume-rectangular-prismvolume-cylindersurface-areaminkowski-sum identify-subproblemsdimensional-analysis ↑ Prerequisites: volume-rectangular-prismsurface-area
📏 Long solution 💡 3 insights

Problem

Let BB be a right rectangular prism (box) with edges lengths 1,1, 3,3, and 44, together with its interior. For real r≥0r\geq0, let S(r)S(r) be the set of points in 33-dimensional space that lie within a distance rr of some point in BB. The volume of S(r)S(r) can be expressed as ar3+br2+cr+dar^{3} + br^{2} + cr +d, where a,a, b,b, c,c, and dd are positive real numbers. What is bcad?\frac{bc}{ad}?

(A) 6(B) 19(C) 24(D) 26(E) 38\textbf{(A) } 6 \qquad\textbf{(B) } 19 \qquad\textbf{(C) } 24 \qquad\textbf{(D) } 26 \qquad\textbf{(E) } 38

Pick an answer.

(A)
6
(B)
19
(C)
24
(D)
26
(E)
38
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Toolkit + CCSS Solution

Understand

Restated: $B$ is a solid box with edge lengths $1$, $3$, $4$. For $r \ge 0$ 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 $r$: $V(S(r)) = a r^3 + b r^2 + c r + d$. Compute $\dfrac{bc}{ad}$.

Givens: $B$ is a right rectangular prism with edge lengths $1, 3, 4$; $S(r) = \{P : \text{distance}(P, B) \le r\}$; $V(S(r)) = a r^3 + b r^2 + c r + d$ with positive $a, b, c, d$; Choices: (A) $6$, (B) $19$, (C) $24$, (D) $26$, (E) $38$

Unknowns: $\dfrac{bc}{ad}$

Understand

Restated: $B$ is a solid box with edge lengths $1$, $3$, $4$. For $r \ge 0$ 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 $r$: $V(S(r)) = a r^3 + b r^2 + c r + d$. Compute $\dfrac{bc}{ad}$.

Givens: $B$ is a right rectangular prism with edge lengths $1, 3, 4$; $S(r) = \{P : \text{distance}(P, B) \le r\}$; $V(S(r)) = a r^3 + b r^2 + c r + d$ with positive $a, b, c, d$; Choices: (A) $6$, (B) $19$, (C) $24$, (D) $26$, (E) $38$

Plan

Primary tool: #7 Identify Subproblems

Secondary: #1 Draw a Diagram, #17 Visualize Spatial Relationships

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).

Execute — Answer: B

#7 Identify Subproblems 5.MD.C.5 Step 1
  • Piece 1 — the box itself.
  • Volume $V_1 = 1 \cdot 3 \cdot 4 = 12$.
  • This is the constant term, so $d = 12$.
$$V_1 = 1 \cdot 3 \cdot 4 = 12,\quad d = 12$$

💡 The box contributes the constant: no $r$ dependence.

#7 Identify Subproblems 6.G.A.4 Step 2
  • Piece 2 — six rectangular slabs over the six faces.
  • Each face $f$ contributes a slab of base area $f$ and thickness $r$, so the total volume is the surface area of the box times $r$.
  • Surface area $= 2(1 \cdot 3 + 1 \cdot 4 + 3 \cdot 4) = 2(3 + 4 + 12) = 38$.
  • So $V_2 = 38 r$, giving $c = 38$.
$$\text{SA}(B) = 2(1 \cdot 3 + 1 \cdot 4 + 3 \cdot 4) = 38,\quad V_2 = 38 r,\quad c = 38$$

💡 Push each face outward by $r$ — total volume is surface area times $r$.

#17 Visualize Spatial Relationships 8.G.C.9 Step 3
  • Piece 3 — twelve quarter-cylinders along the twelve edges.
  • The box has $4$ edges of length $1$, $4$ edges of length $3$, $4$ edges of length $4$.
  • Around each edge of length $\ell$, $S(r)$ adds a quarter-cylinder of radius $r$ and length $\ell$, volume $\tfrac{\pi r^2 \ell}{4}$.
  • Four such pieces along parallel edges combine into a full cylinder (the four quarters of the disk around the edge).
  • Total: $V_3 = \pi r^2 (1 + 3 + 4) = 8 \pi r^2$, so $b = 8 \pi$.
$$V_3 = \pi r^2 (1 + 3 + 4) = 8 \pi r^2,\quad b = 8\pi$$

💡 Four quarter-cylinders along the four parallel edges sum to one full cylinder.

#17 Visualize Spatial Relationships 8.G.C.9 Step 4
  • Piece 4 — eight one-eighth spheres at the eight corners.
  • Together they make one full sphere of radius $r$: $V_4 = \tfrac{4}{3} \pi r^3$, so $a = \tfrac{4}{3} \pi$.
$$V_4 = 8 \cdot \tfrac{1}{8} \cdot \tfrac{4}{3}\pi r^3 = \tfrac{4}{3}\pi r^3,\quad a = \tfrac{4}{3}\pi$$

💡 Eight eighth-spheres at the corners fit together to make one whole sphere.

#7 Identify Subproblems 6.EE.A.2 Step 5
  • Read off the coefficients: $a = \tfrac{4}{3}\pi$, $b = 8\pi$, $c = 38$, $d = 12$.
  • Compute $\dfrac{bc}{ad} = \dfrac{8\pi \cdot 38}{\tfrac{4}{3}\pi \cdot 12} = \dfrac{304 \pi}{16 \pi} = 19$.
  • The $\pi$ cancels — clean integer answer.
$$\dfrac{bc}{ad} = \dfrac{8\pi \cdot 38}{\tfrac{4}{3}\pi \cdot 12} = \dfrac{304\pi}{16\pi} = 19 \;\Rightarrow\; \textbf{(B)}$$

💡 Plug the four coefficients into $bc/(ad)$ and the $\pi$ cancels.

[1] #7 5.MD.C.5 Piece 1 — the box itself. Volume $V_1 = 1 \cdot 3 \cdot 4 = 12$. This is the con
[2] #7 6.G.A.4 Piece 2 — six rectangular slabs over the six faces. Each face $f$ contributes a
[3] #17 8.G.C.9 Piece 3 — twelve quarter-cylinders along the twelve edges. The box has $4$ edges
[4] #17 8.G.C.9 Piece 4 — eight one-eighth spheres at the eight corners. Together they make one
[5] #7 6.EE.A.2 Read off the coefficients: $a = \tfrac{4}{3}\pi$, $b = 8\pi$, $c = 38$, $d = 12$

Review

Reasonableness: The cancellation of $\pi$ is the strongest reasonableness signal: $b$ and $a$ both carry one factor of $\pi$ from the cylinders and sphere, while $c$ and $d$ are $\pi$-free. So $bc/(ad)$ 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.

Alternative: Tool #14 (Finite Differences) / formula recall: for any convex body in $\mathbb{R}^3$, $V(S(r)) = V + (\text{SA}) r + (\text{mean width factor}) \pi r^2 + \tfrac{4}{3}\pi r^3$, which for a box becomes $V + (\text{SA}) r + \pi (\text{edge sum}/4) r^2 \cdot ?$ — read the standard Steiner-formula version directly. Both routes give the same coefficients $(a, b, c, d) = (\tfrac{4\pi}{3}, 8\pi, 38, 12)$ and the same answer $19$.

CCSS standards used (min grade 8)

  • 5.MD.C.5 Relate volume to the operations of multiplication and addition (Computing the box's volume $1 \cdot 3 \cdot 4 = 12$.)
  • 6.EE.A.2 Write, read, and evaluate expressions in which letters stand for numbers (Identifying $a, b, c, d$ as coefficients of $r^3, r^2, r^1, r^0$ and evaluating $\dfrac{bc}{ad}$.)
  • 6.G.A.4 Represent three-dimensional figures using nets and find surface area (Computing the box's surface area $2(1 \cdot 3 + 1 \cdot 4 + 3 \cdot 4) = 38$ for the slab piece.)
  • 8.G.C.9 Know the formulas for volumes of cones, cylinders, and spheres (Volume of the cylinder pieces ($\pi r^2 \ell$) and the sphere piece ($\tfrac{4}{3}\pi r^3$) that the corners and edges contribute.)

⭐ 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 $a = \tfrac{4\pi}{3}$, $b = 8\pi$, $c = 38$, $d = 12$, then $\dfrac{bc}{ad} = \dfrac{304\pi}{16\pi} = 19$. The answer is $\textbf{(B)}$.

⭐ 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 $a = \tfrac{4\pi}{3}$, $b = 8\pi$, $c = 38$, $d = 12$, then $\dfrac{bc}{ad} = \dfrac{304\pi}{16\pi} = 19$. The answer is $\textbf{(B)}$.

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