Once I asked the following question.
Question 1 (solved): Do any two faces share with at most one edge in a 3-connected plane graph?
As anilch reminded, the dual graph of a 3-vertex-connected (or,3-edge -connected) planar graph is simple. So the question is solved.
The following broader question might be interesting.
Question 2: Do any two faces share with at most two vertices in a 3-vertex-connected plane graph?
If ''3" is not guaranteed, what if we choose "4" or "5"? Disprove it or prove it.
I'm looking for some counterexamples, but so far I haven't found any.
Without finding a counterexample, I thought the following claim would be true:
Claim: Let $G$ be a $3$-vertex-connected graph. Then any two faces $f_1$ and $f_2$ have at most two vertices in common. Further if $f_1$ and $f_2$ share two vertices $v_1,v_2$, then the edge $v_1v_2$ is the unique edge shared by $f_1$ and $f_2$.
A Outline of proof: If there are two faces $f_1$ and $f_2$ such that they have three or more vertices in common. By properly choosing two shared vertices $x,y$, we can construct a closed curve that is crosses through these two vertices and falls in $f_1$ and $f_2$. By the closed curve, we can see clearly that $x, y$ is a 2-cut set, a contradiction.
