AMC 8 · 2003 · #15

Easy mode Grade 5

Problem

Imagine a shape built out of unit cubes (cubes of side $1$ ). Every cube in the shape must touch at least one other cube along a full face.

We are told the shape's front view and its side view (shown in the figure). Both views are made of unit squares.

What is the smallest number of cubes that can make a shape with this front view and this side view?

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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