2023 AMC 10A Problem 18

Count edges first. Each of the $12$ rhombi has $4$ edges, and every edge belongs to exactly $2$ faces, so
$$E = \frac{12 \cdot 4}{2} = 24.$$

Euler's formula for a convex polyhedron, $V - E + F = 2$, gives $V = 2 + 24 - 12 = 14$.

Let $a$ vertices have $3$ edges and $b$ vertices have $4$ edges. Then $a + b = 14$, and since every edge has two endpoints, the degrees sum to $2E$:
$$3a + 4b = 48.$$
Subtracting $3(a + b) = 42$ gives $b = 6$, so $a = 8$.

The answer is $\boxed{\textbf{(D)}\ 8}$.

Polyhedron counting problems run on three tools: faces times sides over two gives edges, Euler's formula gives vertices, and the degree sum gives a second equation. You never need to picture the solid. This same template handles soccer balls (truncated icosahedra) and other mixed-degree solids.

Avoid Euler entirely with angles. Each rhombus has two acute and two obtuse angles, $24$ of each in total. In the rhombic dodecahedron the four-edge vertices are surrounded by four acute angles and the three-edge vertices by three obtuse angles (four obtuse angles could never fit around a convex corner). Hence $4b = 24$ gives $b = 6$ and $3a = 24$ gives $a = 8$.

• Counting edges as 12 times 4 = 48 without halving for the two faces each edge borders.
• Misremembering Euler's formula as $V + E - F = 2$ or $V - E + F = 0$.