Polyhedra containing hexagones only It is well-known that Euler's theorem gives raison d'être to polyhedra containing exactly 12 pentagons if they are connected by 3 in a vertex. The number of hexagons may be arbitrary (in fact >1). In the same configuration, a polyhedron built of hexagons only is impossible. On the other hand, we do know some beautiful pentagonal constructions of more than 12 pentagons connected in vertexes, say, by 5 and 3. My question is: has anybody proved that a purely hexagons-built polyhedron is impossible, no matter how many hexagons are met in a vertex?
My kind request is to send the expected answer also to my own address:
vlsot@math.bas.bg 
With best wishes,
Vladimir Sotirov (Bulgaria)
www.math.bas.bg/~vlsot 
 A: This is a pretty standard graph theory lemma but, if you haven't seen it before, you haven't seen it before:
Theorem There are no planar graphs where every face is a disc with $\geq 6$ sides and each vertex has degree $\geq 3$.
Proof Let $F_k$ be the number of faces with $k$ sides and $V_m$ the number of vertices of degree $m$. Writing $(V, E, F)$ for the total number of vertices, edges and faces, we have
$$E = \frac{1}{2} \sum_k k F_k = \frac{1}{2} \sum_m m V_m$$
and hence
$$E = \sum_k \frac{k}{6} F_k + \sum_m \frac{m}{3} V_m.$$
Plugging into Euler's $F+V-E=2$, we get that
$$\sum_k \frac{6-k}{6} F_k + \sum_m \frac{3-m}{3} V_m = 2.$$
So either there is a vertex of degree $<3$ or a face of size $<6$. $\square$
Remark If you allow vertices of degree $2$, such examples exist topologically: Take a tetrahedron and put a vertex in the middle of each edge.
Remark More generally, if $1/p+1/q \geq 1/2$, a similar argument shows that there is either a vertex of degree $\geq p$ or a face of degree $\geq 3$. The interesting cases are $(p,q) = (3,6)$, $(4,4)$ and $(6,3)$.
