2023 AMC 10A Problem 18

Problem 18 of 25IntermediateGeometry Combinatorics

A rhombic dodecahedron is a solid with

12

congruent rhombus faces. At every vertex,

3

4

edges meet, depending on the vertex. How many vertices have exactly

3

edges meeting?

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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,

x = 8.

Therefore, the answer is D.

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