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Case 1: Each face is a triangle. Let \(f\) be the number of faces. There are then \(3f/2\) edges. Using Euler's formula we have \(v - 3f/2 + f = 2\) so \(v = 2 + f/2\text{.}\) Now each vertex has the same degree, say \(k\text{.}\) So the number of edges is also \(kv/2\text{.}\) Putting this together gives

\begin{equation*} e = \frac{3f}{2} = \frac{k(2+f/2)}{2}\text{,} \end{equation*}

which says

\begin{equation*} k = \frac{6f}{4+f}\text{.} \end{equation*}

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