2023 AMC 10A Problems/Problem 17

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?

$\textbf{(A) }5\qquad\textbf{(B) }6\qquad\textbf{(C) }7\qquad\textbf{(D) }8\qquad\textbf{(E) }9$

Solution

We see there are $\frac{12 \cdot 4}{2}$ edges. We have by Euler's Polyhedral Formula, $V-E+F=2$ meaning $V-24+12=2$ or $V=14$ . Let there be $a$ vertices that have $3$ edges meeting and $b$ vertices that have $4$ edges meeting. Hence, $a+b=14$ $3a+4b=48$ We find $b=6$ and $a=8$ , hence the answer is $8$ .

~SirAppel

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2023 AMC 10A Problems/Problem 17

Solution