Simple connected plane graph G and its dual graph G*; if G is isomorphic to G*, then G is not bipartite? Let $G$ be a simple connected plane graph where $v>2$, and $G^*$ is its dual graph.
Prove that if $G$ is isomorphic to $G^*$, then $G$ is not bipartite.  
I know that $G$'s number of faces is equal to its number of vertices, but is this fact any use for this question? Would I also have to use the Euler's formula, $v-e+f=2$?  
Thanks in advance.
 A: In this problem, I shall assume that $G$ is a simple graph.  There are counterexamples if $G$ is not simple.  Indeed, there is a counterexample for any given number of vertices $v\geq 2$ (consider any star graph and replace every edge by a pair of parallel edges).
Hints:


*

*If $G \cong G^*$, how many edges does $G$ have as a function of the number of faces?

*What is the sum of the degrees of all faces?

*There is a face of degree exactly $3$.


Solution:
Suppose that $G\cong G^*$. If we can show that there is a face of $G$ of degree $3$, then this face is bounded by three edges forming a $3$-cycle of $G$.  Since a bipartite graph cannot have an odd cycle, $G$ is not bipartite.

Now, to show that $G$ has a face of degree $3$, we recall the Euler characteristic $v-e+f=2$, where $v$ is the number of vertices of $G$, $e$ the number of edges, and $f$ the number of faces.  Since $G$ is a simple graph, a face of $G$ must have degree at least $3$ (otherwise $G$ has a loop or parallel edges).  Since $G\cong G^*$, $v=f$.  That is, $e=2f-2$.  Now the sum of the degrees of $f$ is $2e=4f-4$.  Hence, there exists a face of degree at most $\frac{4f-4}{f}=4-\frac{4}{f}<4$.  That is, $G$ has a face of degree at most $3$.  This face must then have degree $3$.

For every $v\geq 4$, there is a simple planar graph $G$ on $v$ vertices such that $G$ is self-dual.  The wheel graphs are such examples.
