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2023 AMC 10A Problem 18

Problem 18 of 25IntermediateGeometryCounting & Probability

A rhombic dodecahedron is a solid with 1212 congruent rhombus faces. At every vertex, 33 or 44 edges meet, depending on the vertex. How many vertices have exactly 33 edges meeting?

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Solution

Each rhombus has 44 edges, and every edge is shared by 22 faces, so E=1242=24.E = \frac{12 \cdot 4}{2} = 24. With F=12,F = 12, Euler’s formula gives V=2F+E=14.V = 2 - F + E = 14. Suppose xx vertices have 33 edges and the other 14x14 - x have 4.4. The degrees sum to twice the edge count: 3x+4(14x)=2E=48,3x + 4(14 - x) = 2E = 48, so x=8.x = 8. Therefore, the answer is D.

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Concepts: Euler’s Polyhedron Formula · polyhedron · graph theory · double counting

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.