AMC 8 · 1999 · #20

Grade 6 geometry-3d

Archetype Casework by Position / Vertex Type

spatial-visualizationsystematic-enumeration physical-representationcasework ↑ Prerequisites: spatial-visualization

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

Figure 1 is called a "stack map." The numbers tell how many cubes are stacked in each position. Fig. 2 shows these cubes, and Fig. 3 shows the view of the stacked cubes as seen from the front.

Which of the following is the front view for the stack map in Fig. 4?

Pick an answer.

(A)

Front view with column heights 2, 3, 4 plus extra blocks (heights 4 left-stack, 2 middle, 3 right) — irregular L

(B)

Front view with column heights 2, 3, 4 (left to right) — staircase rising to the right

(C)

Front view with column heights 4, 3 only (two columns)

(D)

Front view with column heights 2, 2, 4 (left to right)

(E)

Front view with column heights 1, 4, 4 (left to right)

View mode:

Toolkit + CCSS Solution

Understand

Restated: A "stack map" is a top-down grid where each cell's number tells how many unit cubes are stacked at that position. The stack map in Figure 4 has two rows (back and front) and three columns (left, middle, right): back row reads $2,\ 2,\ 4$ and front row reads $1,\ 3,\ 1$. Which of the five pictures (A)-(E) shows the FRONT view of this 3D structure?

Givens: Stack map (top-down) for Figure 4: back row = $(2, 2, 4)$, front row = $(1, 3, 1)$, with columns labeled left/middle/right; Each entry is the number of unit cubes stacked at that ground-plan cell; The front view looks straight at the structure along the back-to-front axis, so it collapses the two depth rows into one silhouette; Answer choices (A)-(E) are five different front-view silhouettes drawn from unit squares

Unknowns: Which choice (A)-(E) is the correct front view of the stacked-cube structure

Understand

Plan

Primary tool: #17 Visualize Spatial Relationships

Secondary: #10 Create a Physical Representation, #3 Eliminate Possibilities

The problem hands us a 2D top-down map and asks what the 3D structure looks like from the front — the canonical trigger for Tool #17 (Visualize Spatial Relationships). The key spatial insight is that the front view of a column is whichever stack in that column is tallest, because the taller one hides the shorter one. Tool #10 (Create a Physical Representation) is the safety net: if the mental projection is shaky, build the stack from real cubes (or sketch the silhouette column by column on paper) and look at it from the front. Tool #3 (Eliminate Possibilities) is the multiple-choice cross-check: once we know the three column heights, only one of the five pictures has exactly those heights in that order.

Execute — Answer: B

#17 Visualize Spatial Relationships 5.G.A.1 Step 1

Set up axes so the projection rule is clear.
Let the left-to-right direction be the $x$-axis, back-to-front be the $y$-axis (depth), and up be the $z$-axis.
The front view is what you see when you look along $y$ toward the viewer: it collapses depth, so two stacks in the same column (same $x$, different $y$) overlap into one silhouette whose height equals the tallest of the two.

$$\text{front-view height at column } x = \max_y(\text{stack height at }(x,y))$$

💡 Reading a top-down grid as $(x, y)$ positions and projecting onto the $(x, z)$ plane is the Grade 5 coordinate-axes idea, extended to a third axis.

#17 Visualize Spatial Relationships 5.G.A.1 Step 2

Read off the stack map for Figure 4.
Write the back row and front row aligned by column so the pairs are easy to compare.

$$\text{column heights } \to \begin{array}{c|c|c|c} & \text{left} & \text{middle} & \text{right} \\ \hline \text{back} & 2 & 2 & 4 \\ \text{front} & 1 & 3 & 1 \end{array}$$

💡 Laying the two rows on top of each other column by column turns the map into three pairs of heights — exactly the input to the $\max$ rule.

#10 Create a Physical Representation 6.SP.A.3 Step 3

Apply the $\max$ rule to each column.
The taller stack wins because it hides the shorter one from the viewer.

$$\max(2, 1)=2,\quad \max(2, 3)=3,\quad \max(4, 1)=4$$

💡 If you actually stack physical cubes on a table and crouch to look from the front, the only height you can see for each column is the tallest one — the rule writes itself.

#3 Eliminate Possibilities 6.SP.A.3 Step 4

Compare the front-view heights $(2, 3, 4)$ to the five choices and eliminate.
The required silhouette is a clean staircase rising from height $2$ on the left to $3$ in the middle to $4$ on the right.

$$(2, 3, 4) \;\Rightarrow\; \text{rule out (A)},(C),(D),(E);\;\; \text{(B) matches} \;\Rightarrow\; \textbf{(B)}$$

💡 Only (B) is the staircase $2$-$3$-$4$ left to right. (C) has only two columns, and (A), (D), (E) have at least one column whose height does not match — each can be crossed off in one glance.

[1] #17 5.G.A.1 Set up axes so the projection rule is clear. Let the left-to-right direction be

[2] #17 5.G.A.1 Read off the stack map for Figure 4. Write the back row and front row aligned by

[3] #10 6.SP.A.3 Apply the $\max$ rule to each column. The taller stack wins because it hides the

[4] #3 6.SP.A.3 Compare the front-view heights $(2, 3, 4)$ to the five choices and eliminate. Th

Review

Reasonableness: Sanity check by walking through what each cube looks like from the front. The right column has a $4$-tall stack at the back and a $1$-tall stack at the front, so the front silhouette must reach up to height $4$ — choices (A), (D), and (B) are the only ones with a $4$-tall column on the right; (E) has $4$ on the right too. Now check the middle: the tallest stack there is $3$ (the $3$ in the front row), so the middle silhouette is $3$ tall — that eliminates (D) (middle is $2$) and (E) (middle is $4$). The left column's tallest stack is $2$, so the silhouette there is $2$ tall — that eliminates (A) (left is $4$). (B) survives with heights $2, 3, 4$ left to right. Also: the total visible front-face area is $2+3+4 = 9$ unit squares, which is exactly what (B) shows — internal consistency.

Alternative: Tool #10 (Create a Physical Representation): grab $13$ blocks ($2+2+4+1+3+1$) and build the structure on a table following the stack map — two cubes in the back-left cell, one cube in the front-left cell, and so on. Then crouch down and look straight at it from the front. You will literally see a staircase $2, 3, 4$ unit squares tall, matching (B). This is the most foolproof check for any 3D visualization question.

CCSS standards used (min grade 6)

5.G.A.1 Use a pair of perpendicular number lines, called axes, to define a coordinate system (Reading the stack map as $(x, y)$ ground-plan positions with stack height $z$, and identifying the front view as the projection onto the $(x, z)$ plane.)
6.SP.A.3 Recognize that a measure of center for a numerical data set summarizes its values with a single number (Applying the $\max$ summary (a measure that pulls out the tallest stack in each column) to compress each column's two heights into the single visible silhouette height.)

⭐ From the front, the tallest stack in each column hides the shorter one — take the max of each column and the silhouette $2, 3, 4$ is the answer (B).

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 6)

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