Let (G, φ) be a 2-connected plane graph in which every vertex is incident to one
3-face, one 5-face, and two (opposite) 4-faces. Determine the number of faces in (G, φ).
So I know I need to figure out the number of vertices and edges to find the total number of faces using the formula. I am just stuck on how I can determine the number of vertices per face, take for example the 3-face?
Best Answer
Remark: The way you've stated the question I'm not totally convinced it is solvable; I will solve the problem under the additional assumption that each vertex is incident with one face of length 3, one face of length 5, two (opposite) faces of length 4, and no others (equivalently, I'm assuming the graph is 4-regular).
Suppose $G$ has order $n$. Since each vertex is incident with one face of length 3, one face of length 5, two (opposite) faces of length 4, we have that there're $n$ incidences of vertices with faces of length 3 and 5, while there are $2n$ incidences of vertices with faces of length 4. Of course, any face of length $l$ is incident with exactly $l$ vertices, so this is to say that there're $n/3$, $n/5$, and $2n/4 = n/2$ faces of length 3, 5, and 4 respectively, or that there're $n(1/2 + 1/3 + 1/5)$ = $(31/30)n$ faces in total (note that we are using our assumption that there're no other faces here).
It is elementary to see that $$\sum{l(F_i)} = 2|E(G)|$$ (Note again our use of the assumption that there're no other faces; to see the fact, simply apply the handshaking lemma to the dual graph). From this, using that $\sum{l(F_i)}$ = $4n$ (since $4(n/2) + 3(n/3) + 5(n/5) = 4n$), we have that $|E(G)|$ = $4n/2$ = $2n$ (alternatively, we may apply the handshaking lemma directly to see this, since we noted that our assumption is equivalent to $G$ being 4-regular). By Euler's formula, we may then compute $$n - 2n + (31/30)n = 2 \iff n = 60$$ (using simple algebra). Since we found above that there're exactly $(31/30)n$ faces in $G$, plugging in $n = 60$ yields 62 faces, completing the problem.
As a final remark, note that the graph we found does, in fact, exist: https://en.wikipedia.org/wiki/Rhombicosidodecahedron and matches exactly the (assumed) parameters of your question.