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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: [email protected]

With best wishes, Vladimir Sotirov (Bulgaria)

www.math.bas.bg/~vlsot

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    $\begingroup$ Do you mean an honest polyhedron in a Euclidean space or just a polyhedral complex in the topological sense? In the latter case, they do exist! $\endgroup$
    – Alex Degtyarev
    Commented Mar 27, 2015 at 20:59
  • $\begingroup$ I do not insist that it consisted of "normal" plane hexagons. The hexagons might be decomposed in triangles and respectively, the “general” form would be a sphere. In fact, I am interested if there exists a Buckminster-style construction avoiding pentagonal points. In such a construction more than 3 “curbed” hexagons would be joined at some points (eventually, 6k triangles). Which is your topological example? Vladimir $\endgroup$
    – Vladimir Sotirov
    Commented Mar 27, 2015 at 21:32
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    $\begingroup$ There are lots of hexagonal decompositions of a torus (starting from a single hexagon glued to a torus) if you do not insist on genus $0$. One can also imagine a decomposition of $S^2$: two hexagons around the poles, connected by by $6$ quadrilaterals, which can be regarded as hexagons by inserting an extra vertex inside each ``vertical'' edge. But, of course, if you restrict the genus of the surface, there are bounds on the valencies of the vertices given by the Euler characteristic. $\endgroup$
    – Alex Degtyarev
    Commented Mar 27, 2015 at 21:38

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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 $(6,3)$.

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  • $\begingroup$ I am very sorry for my question. Probably it was appreciated as naive and therefore as an off-topic. It is a fact that my domain - modal logics - is far of geometry and topology. The polyhedra are of interest for me as beautiful constructions and I am very thankful to Alex Degtyarev and David Speyer for their kind replies and competent answers. $\endgroup$
    – Vladimir Sotirov
    Commented Mar 29, 2015 at 21:32

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