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 meet?
$\textbf{(A) }5\qquad\textbf{(B) }6\qquad\textbf{(C) }7\qquad\textbf{(D) }8\qquad\textbf{(E) }9$
Note Euler's formula where $\text{Vertices}+\text{Faces}-\text{Edges}=2$. There are $12$ faces and the number of edges is $24$ because there are 12 faces each with four edges and each edge is shared by two faces. Now we know that there are $14$ vertices on the figure. Now note that the sum of the degrees of all the points is twice the number of edges. Let $x=$ the amount of vertices with $3$ edges. Now we know $\frac{3x+4(14-x)}{2}=24$. Solving this system of equations gives $x = 8$ so the answer is $\boxed{\textbf{(D) }8}$.
~aiden22gao ~zgahzlkw (LaTeX) ~ESAOPS (Simplified)