Problem 17 · 2023 AMC 8
Hard
Geometry & Measurement
spatial-reasoningfolding
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Answer: A — Face 1.
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Hint 1 of 2
Don't try to fold the whole net in your head. Use the structure: an octahedron is two pyramids glued base-to-base, so its 8 faces split into a top 4 and a bottom 4, and exactly 4 faces meet at each tip.
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Hint 2 of 2
Hunt for 4 faces that all share one corner in the net — those become one pyramid (one hemisphere). Faces 2, 3, 4, 5 do that, so the other four {Q, 6, 7, 1} are Q's hemisphere; the answer must be among them.
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Approach: split the eight faces into two hemispheres of four
- Instead of mentally folding, use the octahedron's structure: it's two square pyramids base-to-base, so the 8 faces split into a top set of 4 and a bottom set of 4, and exactly 4 faces ring each apex.
- In the net, faces 2, 3, 4, 5 all meet at a single vertex — that vertex becomes one apex, making {2, 3, 4, 5} one hemisphere. Q can't be in that group, so by elimination Q's hemisphere is {Q, 6, 7, 1}.
- Within that ring of four around the top apex, the fold seats face 1 immediately to the right of Q. This transfers: for any solid-from-net puzzle, first sort faces into the groups that share a vertex — it slashes the possibilities before you fold anything.
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