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Conjecture: The optimal way to divide 3-space into pieces of equal volume with the least total surface area is the rhombic dodecahedral honeycomb.

Reasoning: "(The rhombic dodecahedral honeycomb) is the Voronoi diagram of the face-centered cubic sphere-packing, which is the densest possible packing of equal spheres in ordinary space." (Wikipedia) This resembles very closely how the regular hexagonal grid emerges from densest circle packing. (Also in real honeycombs, where the rhombic dodecahedron appears on one side of cells.)(see wiki:Honeycomb)
Furthermore the kissing number in 2-space is 6 and in 3-space it's 12. The "flat side" (edge->surface) should be where the objects meet. Therefore we expect a kind of dodecahedron.

Question: Is my conjecture right? If yes, how can I proof it?

References:
https://en.wikipedia.org/wiki/Honeycomb_conjecture
https://en.wikipedia.org/wiki/Rhombic_dodecahedral_honeycomb

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    $\begingroup$ Following your links leads to another article, the Weaire-Phelan structure and its polyhedral approximation. Although this example disproves the earlier conjecture of Kelvin, it suggests that the problem of "best" tiling in 3D is open. $\endgroup$
    – hardmath
    Oct 23, 2015 at 15:41
  • $\begingroup$ Thank you, now I remember reading that article again. In the references of it, I found that the answer to the original question is 'no'. As you can see, I edited the question now. I had "hoped" that a regular symmetric polyhedron would be the solution... $\endgroup$
    – MrFrety
    Oct 23, 2015 at 16:33
  • $\begingroup$ It is a reasonably accessible example of symmetry breaking. Contrast the 2D agreement between circle packing and hexagonal tiling. $\endgroup$
    – hardmath
    Oct 23, 2015 at 17:13
  • $\begingroup$ With your edit it is unclear what your question is now asking. If the original question is now answered and you want to ask a different but related one, it might be clearer to Readers if you post that answer below and then post the new Question including a link back to this one for context. $\endgroup$
    – hardmath
    Oct 23, 2015 at 17:19
  • $\begingroup$ @hardmath Ok... $\endgroup$
    – MrFrety
    Oct 27, 2015 at 15:58

1 Answer 1

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The optimal way to divide 3-space into pieces of equal volume with the least total surface area is the rhombic dodecahedral honeycomb.

Is my conjecture right?

No.
It has been described in the paper "A counter-example to Kelvin's conjecture on minimal surfaces" by D. Weaire & R. Phelan (link) that both the Kelvin tetrakaidecahedron and their own structure are closer to a solution to the problem which corresponds to this question. It is called the Kelvin problem and is still an open problem in mathematics.

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