Show that there is no finite plane graph whose every face is a hexagon and each two neighboring faces have exactly one edge in common.
So this is one of the questions I am asked to solve for my discrete math 2 class. I believe I have to solve this using the concept of dual planar graph but I might be wrong
For contradiction suppose there exists a graph $G$ which is planar, and whose every face is a hexagon and each two neighboring faces have exactly one edge in common.
The dual of $G$ (call it $G'$) is a planar graph.
Every face of $G$ is adjacent to exactly 6 other faces by construction, so every vertex of $G'$ has degree 6.
There is a theorem that for all planar graphs with 3 or more vertices, $e\leq3v-6$ where $e$ is the number of edges and $v$ is the number of vertices. By handshake lemma, $G'$ has $3v$ edges since all its vertices have degree 6. This is a contradiction so $G$ cannot exist.