AMC 8 · 2003 · #15

Grade 5 geometry-3d

Archetype Small-Domain Constrained Search └ Sub-archetype Factor-Pair Enumeration

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

📏 Short solution 💡 2 insights 📊 Diagram

📘 View easy version →

Problem

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?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Create a Physical Representation

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.

1STEP 1

Set up axes so the views become projections: the front view collapses y and shows (x,z); the side view collapses x and shows (y,z).

front: (x,z) side: (y,z)

2STEP 2

Decode the views into grid squares: the front L lights (0,0),(0,1),(1,0); the side L lights (0,0),(1,0),(1,1).

front lit: {(0,0),(0,1),(1,0)}, side lit: {(0,0),(1,0),(1,1)}

3STEP 3

Line cubes along a viewing axis: 4 cubes at (0,0,0),(0,0,1),(1,0,0),(1,1,0) project to the front and side L, and every cube shares a face.

cubes: {(0,0,0),(0,0,1),(1,0,0),(1,1,0)} → front L+side L, #cubes=4

4STEP 4

Try 3 cubes: the shared top cube forces the two bases to different x and y, so one floats with no face-neighbor — connectivity fails.

3 cubes → floating corner → rule broken

5STEP 5

The 4-cube figure works and 3 cubes cannot, so the minimum is 4.

#cubes_min = 4 → (B)

Answer

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.

💡Key takeaway

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.

More like this