AMC 8 · 2014 · #19
Grade 6 geometry-3dProblem
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?
Pick an answer.
AMC 8 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
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
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) $\tfrac{1}{3}$
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
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.
💡 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.
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 $0$.
💡 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.
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.
💡 A systematic list of the slot types — ordered by exposure $0,1,2,3$ — guarantees no slot is double-counted or missed.
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.
💡 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.
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.
💡 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.
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 $\gcd(5, 54) = 1$.
💡 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.
6.G.A.4 Compute the total surface area of the $3$-inch cube. A cube has $6$ congruent sq 6.G.A.4 Classify the $27$ slots by how many of the unit cube's faces would be exposed on 4.OA.A.3 Check the count of slots by listing the types from least exposed to most exposed 6.G.A.4 Place the $6$ white cubes greedily into the most-hidden slots (the ones at the t 6.G.A.4 Compute the white surface area from the placement. The center white cube shows $ 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.1Understand 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.3Solve 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.4Represent 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.
⭐ 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.
More like this
Same archetype — closest grade level first.
