AMC 8 · 2014 · #19

Grade 6 geometry-3d

Archetype Casework by Position / Vertex Type

surface-areaspatial-visualizationoptimization-counting caseworkoptimization-counting ↑ Prerequisites: surface-areaspatial-visualization

📏 Medium solution 💡 4 insights

Problem

A cube with $3$ -inch edges is to be constructed from $27$ smaller cubes with $1$ -inch edges. Twenty-one of the cubes are colored red and $6$ are colored white. If the $3$ -inch cube is constructed to have the smallest possible white surface area showing, what fraction of the surface area is white?

$\textbf{(A) }\frac{5}{54}\qquad\textbf{(B) }\frac{1}{9}\qquad\textbf{(C) }\frac{5}{27}\qquad\textbf{(D) }\frac{2}{9}\qquad\textbf{(E) }\frac{1}{3}$

Pick an answer.

(A)

$frac{5}{54}$

(B)

$frac{1}{9}$

(C)

$frac{5}{27}$

(D)

$frac{2}{9}$

(E)

$frac{1}{3}$

View mode:

Toolkit + CCSS Solution

Understand

Restated: A $3$-inch cube is built from twenty-seven $1$-inch unit cubes: $21$ red and $6$ white. Place the white cubes so that the white area showing on the outside is as small as possible. What fraction of the $3$-inch cube's surface area is white?

Givens: Large cube: $3$ inches per edge, built from $27$ unit cubes; $21$ unit cubes are red, $6$ are white; We choose the placement of the white cubes to minimize the white area visible on the surface; Answer choices: (A) $\tfrac{5}{54}$, (B) $\tfrac{1}{9}$, (C) $\tfrac{5}{27}$, (D) $\tfrac{2}{9}$, (E) $\tfrac{1}{3}$

Unknowns: The fraction $\dfrac{\text{minimum white surface area}}{\text{total surface area of the }3\text{-inch cube}}$

Understand

Plan

Primary tool: #10 Create a Physical Representation

Secondary: #7 Identify Subproblems, #2 Make a Systematic List

This is a 3D placement problem, exactly the trigger for Tool #10 — picture (or build with $27$ blocks) a $3 \times 3 \times 3$ cube and notice that not every slot exposes the same number of faces. Tool #7 (Identify Subproblems) splits the task into three clean pieces: (i) find the total surface area, (ii) classify the $27$ slots by how many faces they expose, (iii) place the $6$ white cubes greedily in the most-hidden slots. Tool #2 (Systematic List) handles step (ii): list the slot types in order from least exposed to most exposed and count each.

Execute — Answer: A

#7 Identify Subproblems 6.G.A.4 Step 1

Compute the total surface area of the $3$-inch cube.
A cube has $6$ congruent square faces, each of side $3$ inches, so each face is $3 \times 3 = 9$ square inches.

$$\text{Total surface area} = 6 \times 9 = 54 \text{ sq in}$$

💡 Splitting the cube's exterior into $6$ square faces is the Grade 6 surface-area move — count the faces, find one face's area, multiply.

#10 Create a Physical Representation 6.G.A.4 Step 2

Classify the $27$ slots by how many of the unit cube's faces would be exposed on the outside.
Picture (or build) the $3 \times 3 \times 3$ cube and look at each slot type: a corner slot touches $3$ outer faces, an edge slot (between two corners) touches $2$, a face-center slot touches $1$, and the very middle slot touches $0$.

$$\text{corners: }8,\ \text{edges: }12,\ \text{face-centers: }6,\ \text{center: }1$$

💡 Physically handling (or imagining) the cube makes the four slot types obvious — corners stick out in $3$ directions, edges in $2$, face-centers in $1$, the center in none.

#2 Make a Systematic List 4.OA.A.3 Step 3

Check the count of slots by listing the types from least exposed to most exposed.
This is also a quick sanity check that we have all $27$ unit cubes.

$1 + 6 + 12 + 8 = 27$ \checkmark

💡 A systematic list of the slot types — ordered by exposure $0,1,2,3$ — guarantees no slot is double-counted or missed.

#7 Identify Subproblems 6.G.A.4 Step 4

Place the $6$ white cubes greedily into the most-hidden slots (the ones at the top of the list).
The $1$ center slot exposes $0$ faces, so put $1$ white cube there.
The next-best slots are the $6$ face-centers, each exposing only $1$ face — there are exactly enough of these for the remaining $5$ white cubes.

$$1 \text{ center} + 5 \text{ face-centers} = 6 \text{ white cubes placed}$$

💡 To minimize total exposed white, hide as many white cubes as possible and let the rest each show only $1$ face — this is the greedy subproblem of minimizing a sum of exposures.

#7 Identify Subproblems 6.G.A.4 Step 5

Compute the white surface area from the placement.
The center white cube shows $0$ square inches.
Each of the $5$ face-center white cubes shows one $1 \times 1 = 1$ square inch face.

$$\text{White area} = 1 \times 0 + 5 \times 1 = 5 \text{ sq in}$$

💡 Adding up the exposed face count from each slot turns the placement into a number — exactly the Grade 6 surface-area calculation, just applied to the white portion only.

#7 Identify Subproblems 3.NF.A.1 Step 6

Form the requested fraction: white surface area over total surface area.
The fraction $\tfrac{5}{54}$ is already in lowest terms because $\gcd(5, 54) = 1$.

$$\dfrac{\text{white}}{\text{total}} = \dfrac{5}{54} \;\Rightarrow\; \textbf{(A)}$$

💡 Comparing a part to the whole as a fraction is the Grade 3 fraction concept — $5$ of the $54$ unit squares on the surface are white.

[1] #7 6.G.A.4 Compute the total surface area of the $3$-inch cube. A cube has $6$ congruent sq

[2] #10 6.G.A.4 Classify the $27$ slots by how many of the unit cube's faces would be exposed on

[3] #2 4.OA.A.3 Check the count of slots by listing the types from least exposed to most exposed

[4] #7 6.G.A.4 Place the $6$ white cubes greedily into the most-hidden slots (the ones at the t

[5] #7 6.G.A.4 Compute the white surface area from the placement. The center white cube shows $

[6] #7 3.NF.A.1 Form the requested fraction: white surface area over total surface area. The fra

Review

Reasonableness: Sanity check the placement: any other placement is worse. If we instead put a white cube on an edge slot, it would show $2$ faces, raising the white area by at least $1$ square inch above $5$; on a corner slot it would show $3$ faces. So $5$ sq in is genuinely the minimum, and $\tfrac{5}{54} \approx 9.3\%$ — small, as expected when we hide white cubes inside. Choice (A) $\tfrac{5}{54}$ matches.

Alternative: Tool #16 (Change Focus / Complement) gives a fast cross-check on the answer choices. The $54$ surface unit squares are split between white and red; the answer choices are $\tfrac{5}{54}, \tfrac{1}{9} = \tfrac{6}{54}, \tfrac{5}{27} = \tfrac{10}{54}, \tfrac{2}{9} = \tfrac{12}{54}, \tfrac{1}{3} = \tfrac{18}{54}$. The smallest white count consistent with hiding one cube and exposing one face each on five others is $5$, which matches $\tfrac{5}{54}$ — and none of the other choices correspond to that minimum.

CCSS standards used (min grade 6)

3.NF.A.1 Understand a fraction $a/b$ as the quantity formed by $a$ parts of size $1/b$ (Expressing the final answer as the fraction $\tfrac{5}{54}$ of the surface — $5$ white unit squares out of $54$ total.)
4.OA.A.3 Solve multistep word problems with whole numbers, using the four operations (Counting the slots by type and checking $8 + 12 + 6 + 1 = 27$ to confirm every unit cube is accounted for.)
6.G.A.4 Represent three-dimensional figures using nets, and use the nets to find the surface area (Finding the total surface area ($6 \times 9 = 54$ sq in) and the white surface area ($5$ sq in) by counting outer unit-square faces on the $3 \times 3 \times 3$ cube.)

⭐ Hide one white cube in the very middle, and the other five in face-centers so each shows just one square — that is the smallest white surface possible.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: A

Review

CCSS standards used (min grade 6)

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