AMC 8 · 2023 · #17

Grade 6 geometry-3d
spatial-visualizationpolyhedron-netsface-adjacency physical-representationcasework ↑ Prerequisites: spatial-visualizationpolyhedron-nets
📏 Medium solution 💡 3 insights 📊 Diagram
Problem
A regular octahedron has eight equilateral-triangle faces, with four faces meeting at each vertex. The flat net shows seven numbered faces (1 through 7) plus one face labeled Q. When the net is folded into the octahedron drawn on the right, face Q becomes a slanted upward-pointing triangle on the front of the solid. Which numbered face ends up sharing the right-hand edge of Q (the edge marked "?" in the 3D picture)?

Pick an answer.

(A)
1
(B)
2
(C)
3
(D)
4
(E)
5

AMC 8 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Create a Physical Representation

This is a classic "fold a net into a solid" puzzle, exactly what Tool #10 (Create a Physical Representation) is for: copy the net onto paper, cut it out, and fold — the face that lands on Q's right edge appears in your hands. If no paper is available, walk through the same moves with Tool #17 (Visualize Spatial Relationships). Tool #2 (Make a Systematic List) keeps the search organized — face Q has three edges, two of which are already glued (to 6 and 7), so we only need to chase one free edge around the net's perimeter. Tool #3 (Eliminate Possibilities) shows that several answer choices (2, 3, 5) are already locked into other faces' neighborhoods, leaving only 1 as a feasible match.

1STEP 1

Copy and fold the net; face Q keeps net-neighbors 6 and 7, leaving only its right edge — the one the question asks about — still loose.

2STEP 2

Trace the net's outer boundary; the eight faces with a free edge are Q, 6, 4, 1, 2, 3, 5, 7, and Q's loose edge must seal to one of them.

perimeter faces (with one free edge each): Q, 6, 4, 1, 2, 3, 5, 7
3STEP 3

Faces 2, 3, 5 sit on the opposite cap (only vertex-touching Q) and 4 seals to 6, not Q — eliminating all but 1 across Q's free edge.

eliminated: 2, 3, 4, 5 → remaining candidate: 1
4STEP 4

As the cap folds, corners Q, 6, 4, 1 meet at the top vertex; 6 lands left of Q, 4 across, and 1 on Q's right.

cap vertex: Q ⇔ 6 ⇔ 4 ⇔ 1 ⇔ Q
5STEP 5

Face 1 is Q's right-hand neighbor, so the answer is choice (A).

right neighbor of Q = 1 → (A)
Answer
1
Sanity-check the geometry. A regular octahedron has 6 vertices and at each vertex exactly 4 faces meet, so the four faces Q, 6, 4, 1 sharing a vertex is consistent with the solid's structure. Each face of Q should have exactly one neighbor across each of its three edges — in the net it already has 6 and 7, so it needs exactly one more, found here to be 1. Faces 2, 3, 5 together with 7 form the opposite cap of the octahedron (the four faces meeting at the bottom vertex), so they cannot touch Q along an edge — exactly matching the elimination step.
💡Key takeaway

This AMC 8 problem only needs the Grade 6 idea of "a flat net is a 3D figure waiting to be folded" you already know!