AMC 8 · 1999 · #20

Grade 6 geometry-3d

Archetype Casework by Position / Vertex Type └ Sub-archetype Grid / Solid Cell Position

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

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

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?

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)

AMC 8 1999 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Visualize Spatial Relationships

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.

1STEP 1

Looking from the front collapses each column's two stacks into one silhouette, whose height is the taller of the two.

front-view height at column x = max_y(stack height at (x,y))

2STEP 2

Pair the back row over the front row column by column, so each column shows its two competing heights.

column heights → & left & middle & right ; back & 2 & 2 & 4 ; front & 1 & 3 & 1

3STEP 3

Take the maximum of each column — the taller stack hides the shorter — giving front-view heights 2, 3, 4.

max(2, 1)=2, max(2, 3)=3, max(4, 1)=4

4STEP 4

Match heights 2, 3, 4 against the five pictures: only (B) is that left-to-right staircase; every other choice misses a column.

(2, 3, 4) → rule out (A),(C),(D),(E); (B) matches → (B)

Answer

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

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.

💡Key takeaway

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).

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