2023 AMC 10A Problem 18

Attempt Problem 18 of the 2023 AMC 10A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2023 AMC 10A solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

18.

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?
55
66
77
88
99
Answer: D
Concepts:Euler’s Polyhedron Formulapolyhedrongraph theorydouble counting
Small hint:
The solid has 1212 faces and 2424 edges; use Euler’s formula to get the number of vertices
Big hint:
If xx vertices have degree 33 and the rest degree 4,4, the sum of all degrees is 2⋅242 \cdot 24
Solution:
Each rhombus has 44 edges, and every edge is shared by 22 faces, so E=12⋅42=24.E = \frac{12 \cdot 4}{2} = 24. With F=12,F = 12, Euler’s formula gives V=2−F+E=14.V = 2 - F + E = 14. Suppose xx vertices have 33 edges and the other 14−x14 - x have 4.4. The degrees sum to twice the edge count: 3x+4(14−x)=2E=48,3x + 4(14 - x) = 2E = 48, so x=8.x = 8. Therefore, the answer is D.
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