AMC 8 · 2003 · #15
Grade 5 geometry-3dProblem
A figure is constructed from unit cubes. Each cube shares at least one face with another cube. What is the minimum number of cubes needed to build a figure with the front and side views shown?
Pick an answer.
AMC 8 2003 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 3D figure is built from unit cubes so that every cube shares at least one full face with another cube. Its front view is an L-shape ($2$ unit squares stacked on the left, $1$ unit square to the right of the bottom one), and its side view is the mirror L-shape ($1$ unit square on the left, $2$ unit squares stacked on the right). What is the smallest number of unit cubes that can produce both views?
Givens: Building blocks are unit cubes; every cube must touch at least one other cube on a full face; Front view (looking along one horizontal axis) shows an L-shape of $3$ unit squares: positions $(0,0)$, $(0,1)$, $(1,0)$ in (column, row); Side view (looking along the perpendicular horizontal axis) shows the mirror L-shape of $3$ unit squares; Answer choices: (A) $3$, (B) $4$, (C) $5$, (D) $6$, (E) $7$
Unknowns: The minimum number of unit cubes that can produce both the front and the side view
Understand
Restated: A 3D figure is built from unit cubes so that every cube shares at least one full face with another cube. Its front view is an L-shape ($2$ unit squares stacked on the left, $1$ unit square to the right of the bottom one), and its side view is the mirror L-shape ($1$ unit square on the left, $2$ unit squares stacked on the right). What is the smallest number of unit cubes that can produce both views?
Givens: Building blocks are unit cubes; every cube must touch at least one other cube on a full face; Front view (looking along one horizontal axis) shows an L-shape of $3$ unit squares: positions $(0,0)$, $(0,1)$, $(1,0)$ in (column, row); Side view (looking along the perpendicular horizontal axis) shows the mirror L-shape of $3$ unit squares; Answer choices: (A) $3$, (B) $4$, (C) $5$, (D) $6$, (E) $7$
Plan
Primary tool: #10 Create a Physical Representation
Secondary: #17 Visualize Spatial Relationships, #3 Eliminate Possibilities
The problem hands us two 2D views and asks for a 3D figure — exactly the trigger for Tool #10 (Create a Physical Representation): place real or imagined unit cubes on a grid until both shadows match. Tool #17 (Visualize Spatial Relationships) then helps us share cubes between the two views by lining them up along a single axis so one cube counts in both projections. Tool #3 (Eliminate Possibilities) is the multiple-choice safety net: the smallest choice is $3$, and the front view alone has $3$ squares — checking whether $3$ cubes can satisfy both views and the connectivity rule lets us rule it out and lock in $4$.
Execute — Answer: B
5.G.A.1 Step 1 - Set up axes so the views correspond to projections.
- Let $x$ point to the right (along the front view), $y$ point away (depth, along the side view), and $z$ point up.
- The front view collapses the $y$-axis: a cube at $(x,y,z)$ shows up at column $x$, row $z$ of the front picture.
- The side view collapses the $x$-axis: the same cube shows up at column $y$, row $z$ of the side picture.
💡 Reading a 3D location as two 2D pictures is the Grade 5 coordinate-axes idea, just extended to a third axis.
5.G.A.1 Step 2 - Decode the two views into grid squares.
- The front L-shape lights up the three positions $(x,z) = (0,0), (0,1), (1,0)$.
- The mirror-L side view lights up $(y,z) = (0,0), (1,0), (1,1)$.
- So $z = 1$ is needed only at $x = 0$ in the front, and only at $y = 1$ in the side.
💡 Listing the lit squares of each view as ordered pairs turns the picture into a checklist we can match.
5.G.A.1 Step 3 - Build a figure that minimizes cubes by lining cubes up along a viewing axis.
- Place $4$ unit cubes at the positions $(x,y,z) = (0,0,0), (0,1,0), (1,1,0), (1,1,1)$.
- Check the front view: the four cubes project to front squares $(0,0), (0,0), (1,0), (1,1)$ — but we need $(0,1)$, not $(1,1)$.
- Reposition: try $(0,0,0), (0,0,1), (1,0,0), (1,1,0)$ instead.
- Front projections: $(0,0), (0,1), (1,0), (1,0)$.
- The distinct front squares are $(0,0), (0,1), (1,0)$ — the L-shape, perfect.
- Side projections: $(0,0), (0,1), (0,0), (1,0)$.
- The distinct side squares are $(0,0), (0,1), (1,0)$.
- Both views match the required L-shapes (front L, side L; the mirror direction is just a labeling choice of which horizontal axis is "side").
- Connectivity: the cube at $(0,0,1)$ sits on top of $(0,0,0)$ (shared face), $(0,0,0)$ shares a face with $(1,0,0)$, and $(1,0,0)$ shares a face with $(1,1,0)$ — every cube touches another on a full face.
💡 Sharing one $z=1$ cube between the two views (it projects to both top squares) is the minimum-cube trick: stack along the axis the views collapse.
5.G.A.1 Step 4 - Eliminate $3$ cubes.
- Each view contains $3$ lit squares, so $3$ cubes would have to produce both views with no overlap wasted — each cube must contribute one distinct front square and one distinct side square.
- But the front view needs a cube at $z = 1$ (top of the L), and the side view also needs a cube at $z = 1$.
- If $3$ cubes are forced to cover both, the top of the front L and the top of the side L would have to be the same cube, sitting at some $(x_0, y_0, 1)$.
- The two bottom cubes would then sit at $z = 0$, at positions $(x_0, y_0', 0)$ and $(x_0', y_0, 0)$ to hit the remaining front and side squares.
- That bottom pair shares no face with each other (different $x$ and different $y$), and the top cube hangs above just one of them, leaving the other floating with no face-neighbor.
- The figure violates the connectivity rule, so $3$ cubes are not enough.
💡 Even when $3$ cubes could in principle cast the right shadows, the "every cube must touch another" rule forces an extra cube — so $4$ is the true minimum.
5.G.A.1 Step 5 - Conclude.
- The configuration of $4$ cubes works and $3$ cubes cannot, so the minimum is $4$.
💡 From the answer choices, $4$ is the only one consistent with both the construction and the elimination.
5.G.A.1 Set up axes so the views correspond to projections. Let $x$ point to the right ( 5.G.A.1 Decode the two views into grid squares. The front L-shape lights up the three po 5.G.A.1 Build a figure that minimizes cubes by lining cubes up along a viewing axis. Pla 5.G.A.1 Eliminate $3$ cubes. Each view contains $3$ lit squares, so $3$ cubes would have 5.G.A.1 Conclude. The configuration of $4$ cubes works and $3$ cubes cannot, so the mini Review
Reasonableness: Counts make sense: each view has $3$ unit squares, so a lower bound is $3$ cubes (one per shadow if every cube hit a different shadow in both views). The connectivity rule pushes the minimum up by $1$, so $4$ is exactly the kind of "$3 + 1$" answer to expect, matching choice (B). The construction $\{(0,0,0),(0,0,1),(1,0,0),(1,1,0)\}$ has every cube touching another on a full face, and explicitly produces the L-shape in both projections — so $4$ is achievable. Choices (C)$5$, (D)$6$, (E)$7$ are not minimal because the explicit $4$-cube model already works, and (A)$3$ is impossible by the connectivity argument.
Alternative: Tool #17 (Visualize Spatial Relationships) without grabbing physical blocks: imagine the front L. The bottom row of the side L lives along the same $z = 0$ row, so place $2$ cubes along the $y$-axis at $z = 0$ to cover the front $(1,0)$ and side $(1,0)$ squares with one cube each. Add a cube at $(0,0,0)$ to also light front $(0,0)$ and side $(0,0)$. Finally stack one more cube on top of $(0,0,0)$ to light both the front $(0,1)$ and the side $(0,1)$ at once. That is the same $4$-cube figure, derived by mentally tracking which projection each cube contributes to.
CCSS standards used (min grade 5)
5.G.A.1Use a pair of perpendicular number lines, called axes, to define a coordinate system (Reading the front view as the $(x,z)$ projection and the side view as the $(y,z)$ projection of unit cubes located in 3D, then placing each cube at a chosen $(x,y,z)$ so both projections match the required L-shapes.)
⭐ Two flat views, one 3D answer: stack cubes along the line the views collapse, then add one more cube so nothing floats — Grade 5 coordinate thinking pins this AMC 8 problem to $4$ cubes.
⭐ Two flat views, one 3D answer: stack cubes along the line the views collapse, then add one more cube so nothing floats — Grade 5 coordinate thinking pins this AMC 8 problem to $4$ cubes.
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