Planar Graph and Hexagons

Question

I need to show that it is impossible to have a connected planar graph, all of whose faces are hexagons, with each vertex degree at least 3. I am thinking the eulers THM where if g is a connected planar graph, v-e+f=2 but do not know where to go from there. Thanks

Answer 1

Hint: use the handshaking lemma, then dualize and use the handshaking lemma again.

By the handshaking lemma we have $2E\geq 3V$. The dual graph has $2-V+E$ vertices, all of which have degree six. The handshaking lemma implies $2E=6(2-V+E)$ which simplifies to $2E=3V-6$. Therefore $3V-6\geq 3V$, which is the same as $-6\geq 0$.