AMC 8 · 2003 · #15

Grade 5 geometry-3d

Archetype Small-Domain Constrained Search

spatial-visualizationpolyhedron-netssystematic-enumeration physical-representationcaseworksystematic-enumeration ↑ Prerequisites: spatial-visualization

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Problem

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.

(A)

(B)

(C)

(D)

(E)

View mode:

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

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

#17 Visualize Spatial Relationships 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.

$$\text{front: } (x,z) \quad\text{side: } (y,z)$$

💡 Reading a 3D location as two 2D pictures is the Grade 5 coordinate-axes idea, just extended to a third axis.

#10 Create a Physical Representation 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.

$$\text{front lit: } \{(0,0),(0,1),(1,0)\},\quad \text{side lit: } \{(0,0),(1,0),(1,1)\}$$

💡 Listing the lit squares of each view as ordered pairs turns the picture into a checklist we can match.

#10 Create a Physical Representation 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.

$$\text{cubes: } \{(0,0,0),(0,0,1),(1,0,0),(1,1,0)\} \;\Rightarrow\; \text{front L}+\text{side L},\;\;\#\text{cubes}=4$$

💡 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.

#3 Eliminate Possibilities 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.

$$3 \text{ cubes} \;\Rightarrow\; \text{floating corner}\;\Rightarrow\; \text{rule broken}$$

💡 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.

#3 Eliminate Possibilities 5.G.A.1 Step 5

Conclude.
The configuration of $4$ cubes works and $3$ cubes cannot, so the minimum is $4$.

$$\#\text{cubes}_{\min} = 4 \;\Rightarrow\; \textbf{(B)}$$

💡 From the answer choices, $4$ is the only one consistent with both the construction and the elimination.

[1] #17 5.G.A.1 Set up axes so the views correspond to projections. Let $x$ point to the right (

[2] #10 5.G.A.1 Decode the two views into grid squares. The front L-shape lights up the three po

[3] #10 5.G.A.1 Build a figure that minimizes cubes by lining cubes up along a viewing axis. Pla

[4] #3 5.G.A.1 Eliminate $3$ cubes. Each view contains $3$ lit squares, so $3$ cubes would have

[5] #3 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.1 Use 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.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 5)

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