2023 AMC 10A Problem 18

Below is the professionally curated solution for Problem 18 of the 2023 AMC 10A, 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.

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

Difficulty rating: 1660

18.

A rhombic dodecahedron is a solid with $12$ congruent rhombus faces. At every vertex, $3$ or $4$ edges meet, depending on the vertex. How many vertices have exactly $3$ edges meeting?

$5$

$6$

$7$

$8$

$9$

Solution:

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

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