Problem 20 · 2024 AMC 8
Hard
Geometry & Measurement
spatial-reasoningcareful-counting
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Answer: D — 3 equilateral triangles.
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Hint 1 of 2
A cube only offers three distances between vertices: edge (1), face diagonal (√2), space diagonal (√3). For all three sides to match, which single distance can do it?
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Hint 2 of 2
Only face diagonals (√2) work. So find P's face-diagonal neighbors, then count how many TRIANGLES they form — that's a "choose 2" once you know how many neighbors there are.
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Approach: use face-diagonal length as the only valid side
- List the only three possible vertex-to-vertex distances in a cube: edge = 1, face diagonal = √2, space diagonal = √3. An equilateral triangle needs all three sides equal, and edges or space diagonals can't close up into a triangle — only face diagonals (√2) can.
- From P, the vertices a face-diagonal away are the three that sit across each of P's three faces: R, T, V (one per face).
- Now check pairs: any two of {R, T, V} are themselves a face-diagonal apart, so each pair plus P is an equilateral triangle. With 3 such neighbors, the number of triangles is "choose 2 of 3" = 3: {P,R,T}, {P,R,V}, {P,T,V}.
- 3 triangles. This transfers: in 3-D distance problems, first list the few possible lengths, decide which can build the shape, then it becomes a small counting ("choose 2") problem.
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