AMC 10 · 2023 · #18

Grade 8 geometry-2d
eulers-polyhedron-formulaface-adjacencypolyhedron-netsspatial-visualization identify-subproblemsconvert-to-algebradouble-counting ↑ Prerequisites: spatial-visualizationlinear-equations-two-var
📏 Short solution 💡 3 insights
Problem
A rhombic dodecahedron is a solid with 12 congruent rhombus faces. At each vertex either 3 or 4 edges meet. How many vertices have exactly 3 edges meeting?

Pick an answer.

(A)
5
(B)
6
(C)
7
(D)
8
(E)
9

AMC 10 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Identify Subproblems

The path is a clean three-step climb: (a) edges E from face/edge double-counting, (b) total vertices V from Euler's formula, (c) split V into degree-3 and degree-4 vertices using the handshake total 3x + 4y = 2E. Tool #7 (Identify Subproblems) names each rung. The final step is a 2 × 2 linear system in x and y, which is Tool #13 (Convert to Algebra). No clever symmetry argument is needed — every quantity falls out of two counting identities.

1STEP 1

Each rhombus face has 4 edges and every edge is shared by two faces, so E = 12 · 42\frac{4}{2} = 24.

E = (12⋅4)2\frac{(12 · 4)}{2} = 24
2STEP 2

Euler's V - E + F = 2 with F = 12 and E = 24 gives V = 2 + 24 - 12 = 14.

V = 2 + E - F = 2 + 24 - 12 = 14
3STEP 3

Let x, y be the degree-3 and degree-4 vertex counts: x + y = 14, and the handshake sum gives 3x + 4y = 48.

x + y = 14 ; 3x + 4y = 48
4STEP 4

Eliminate y: 4 · (x + y) - (3x + 4y) = 56 - 48, so x = 8 (and y = 6).

x = 8, y = 6 → (D) 8
Answer
8
Cross-check the handshake: 3 · 8 + 4 · 6 = 24 + 24 = 48 = 2 · 24 — matches 2E exactly. Cross-check Euler: V - E + F = 14 - 24 + 12 = 2 — matches. Cross-check totals: x + y = 8 + 6 = 14 = V. Geometrically, the rhombic dodecahedron is known to have 8 degree-3 vertices (the "acute" corners, sitting at the cube's 8 corners) and 6 degree-4 vertices (sitting at the cube's 6 face centers), confirming (D) 8.
💡Key takeaway

Two counting identities crack this solid: each edge sits on two faces (so E = (12⋅4)2\frac{(12 · 4)}{2} = 24), and Euler's V - E + F = 2 gives V = 14. Splitting the 14 vertices into degree-3 and degree-4 piles using "sum of degrees = 2E" yields a 2 × 2 system whose answer is (D) 8 degree-3 vertices.