AMC 8 · 2009 · #25
Grade 6 geometry-3dProblem
A one-cubic-foot cube is cut into four pieces by three cuts parallel to the top face of the cube. The first cut is foot from the top face. The second cut is foot below the first cut, and the third cut is foot below the second cut. From the top to the bottom the pieces are labeled A, B, C, and D. The pieces are then glued together end to end as shown in the second diagram. What is the total surface area of this solid in square feet?
Pick an answer.
Toolkit + CCSS Solution
Understand
Restated: A $1 \times 1 \times 1$ ft cube is sliced by three horizontal cuts into four flat slabs A, B, C, D with heights $\tfrac{1}{2}$, $\tfrac{1}{3}$, $\tfrac{1}{17}$, and the leftover. The four slabs are then placed side by side on the ground in the order D, A, B, C. Find the total surface area, in square feet, of this new solid.
Givens: Original cube: $1 \times 1 \times 1$ ft, volume $1$ ft$^3$; Cuts are parallel to the top face, so each slab has a $1 \times 1$ ft base; Heights from top: $h_A = \tfrac{1}{2}$, $h_B = \tfrac{1}{3}$, $h_C = \tfrac{1}{17}$ ft; $h_D = 1 - (h_A + h_B + h_C)$ since the four heights must sum to $1$ ft; Final arrangement: slabs sit on the floor in the order D, A, B, C, each touching its neighbor on a $1$-ft-wide face; Answer choices: (A) $6$, (B) $7$, (C) $\tfrac{419}{51}$, (D) $\tfrac{158}{17}$, (E) $11$
Unknowns: Total surface area of the new solid in square feet
Understand
Restated: A $1 \times 1 \times 1$ ft cube is sliced by three horizontal cuts into four flat slabs A, B, C, D with heights $\tfrac{1}{2}$, $\tfrac{1}{3}$, $\tfrac{1}{17}$, and the leftover. The four slabs are then placed side by side on the ground in the order D, A, B, C. Find the total surface area, in square feet, of this new solid.
Givens: Original cube: $1 \times 1 \times 1$ ft, volume $1$ ft$^3$; Cuts are parallel to the top face, so each slab has a $1 \times 1$ ft base; Heights from top: $h_A = \tfrac{1}{2}$, $h_B = \tfrac{1}{3}$, $h_C = \tfrac{1}{17}$ ft; $h_D = 1 - (h_A + h_B + h_C)$ since the four heights must sum to $1$ ft; Final arrangement: slabs sit on the floor in the order D, A, B, C, each touching its neighbor on a $1$-ft-wide face; Answer choices: (A) $6$, (B) $7$, (C) $\tfrac{419}{51}$, (D) $\tfrac{158}{17}$, (E) $11$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #1 Draw a Diagram, #11 Look for Symmetry / Invariants
The new solid is a row of four short rectangular blocks of different heights. Instead of counting faces one by one, Tool #7 (Identify Subproblems) lets us split the total surface area into five tidy groups — top, bottom, front, back, and the staircase of side faces — and add. Tool #1 (Draw a Diagram) is the picture that makes those groups obvious. The big payoff comes from Tool #11 (Symmetry / Invariants): the front and back are each one rectangle whose total width is $4$ ft and whose heights sum to exactly $1$ ft, so each has area $1$ ft$^2$ without ever needing the messy value of $h_D = \tfrac{11}{102}$.
Execute — Answer: E
5.NF.A.1 Step 1 - Use the invariant that the four heights add to the original cube height of $1$ ft to find $h_D$.
- We do not even need this value for the front/back faces, but we will use it for the side faces.
💡 Adding fractions with unlike denominators ($2$, $3$, $17$) over a common denominator $102$ is Grade 5 fraction addition.
6.G.A.4 Step 2 - Split the surface of the new solid into five subproblems: (1) the top, (2) the bottom, (3) the front, (4) the back, and (5) the side step pattern at the ends and between slabs.
- We will compute each group and add.
💡 Grouping the faces of a 3D figure to compute surface area is exactly the Grade 6 surface-area-from-nets idea.
3.MD.C.7 Step 3 - Top and bottom.
- Each slab keeps its $1 \times 1$ ft top and its $1 \times 1$ ft bottom from the original cube.
- With four slabs in a row, the top is four squares and so is the bottom.
💡 Area of a rectangle as length $\times$ width, then adding congruent pieces, is Grade 3 area work.
6.G.A.4 Step 4 - Front and back.
- Looking at the row of slabs from the front, you see four rectangles each $1$ ft wide, with heights $h_D, h_A, h_B, h_C$ that add to $1$ ft.
- The total front area is therefore $1$ ft $\times 1$ ft — the messy individual heights cancel.
- The back is identical.
💡 Spotting that the heights are forced to sum to $1$ is an invariant move that bypasses computing each height.
6.G.A.4 Step 5 - Side faces.
- The arrangement is D, A, B, C, so the exposed vertical pieces perpendicular to the row are: the left end (height $h_D$), three step rectangles between neighbors (heights $|h_A - h_D|$, $|h_B - h_A|$, $|h_C - h_B|$), and the right end (height $h_C$).
- Each is $1$ ft deep.
💡 A side-view diagram makes the staircase of exposed rectangles easy to list without missing any.
5.NF.A.1 Step 6 - Convert each side rectangle to the common denominator $102$ and add.
- The numerators are $11, 40, 17, 28, 6$, which sum to $102$.
💡 Adding several fractions with the same denominator $102$ is straightforward Grade 5 fraction arithmetic.
3.MD.C.7 Step 7 Add the five subtotals to get the total surface area.
💡 Combining the area subtotals is the final addition step in a Grade 3 area task.
5.NF.A.1 Use the invariant that the four heights add to the original cube height of $1$ f 6.G.A.4 Split the surface of the new solid into five subproblems: (1) the top, (2) the b 3.MD.C.7 Top and bottom. Each slab keeps its $1 \times 1$ ft top and its $1 \times 1$ ft 6.G.A.4 Front and back. Looking at the row of slabs from the front, you see four rectang 6.G.A.4 Side faces. The arrangement is D, A, B, C, so the exposed vertical pieces perpen 5.NF.A.1 Convert each side rectangle to the common denominator $102$ and add. The numerat 3.MD.C.7 Add the five subtotals to get the total surface area. Review
Reasonableness: The original cube has surface area $6$ ft$^2$. Cutting and rearranging never removes material, only exposes hidden faces, so the new surface area must be at least $6$ — answer (A) is therefore suspicious. The rearrangement exposes three internal horizontal cuts ($3$ extra top faces + $3$ extra bottom faces = $6$ extra ft$^2$) and hides three side strips of total area $1$ where neighbors touch, while exposing roughly the same staircase area on the side. The net change works out to $+5$, giving $6 + 5 = 11$ ft$^2$. This matches (E) and rules out the other choices.
Alternative: Tool #9 (Solve an Easier Problem): replace the awkward heights by simpler ones — say two slabs of heights $\tfrac{1}{2}$ and $\tfrac{1}{2}$ — and check that top $+$ bottom $+$ front $+$ back $+$ sides gives $2 + 2 + 1 + 1 + 1 = 7$ when the row has $2$ slabs (and $4 + 4 + 1 + 1 + 1 = 11$ when the row has $4$ slabs whose heights sum to $1$). The structure shows that the answer depends only on the number of slabs and the cube being a unit cube, not on the specific heights — so the fraction $\tfrac{1}{17}$ is a distraction, confirming the answer is the clean integer $11$.
CCSS standards used (min grade 6)
3.MD.C.7Relate area to multiplication and addition operations (Computing each rectangular face area as length $\times$ width and adding congruent face areas for the top and bottom totals.)5.NF.A.1Add and subtract fractions with unlike denominators (Finding $h_D = 1 - (\tfrac{1}{2} + \tfrac{1}{3} + \tfrac{1}{17}) = \tfrac{11}{102}$ and summing the side-face fractions to $\tfrac{102}{102} = 1$.)6.G.A.4Represent three-dimensional figures using nets and find surface area (Recognising the row of slabs as a 3D figure whose surface area is the sum of its face groups (top, bottom, front, back, sides).)
⭐ This AMC 8 problem only needs Grade 6 surface-area-from-nets reasoning you already know!
⭐ This AMC 8 problem only needs Grade 6 surface-area-from-nets reasoning you already know!
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