A few days ago I asked this question. It turned out to be a known unsolved problem in mathematics: The Kelvin problem.
Now I'd like to slightly change the question:

What is the optimal way to divide 3-space into pieces of equal volume with the least total surface area, whereby only isohedral polyhedra are allowed?

I hope, this will be easier to solve than the original Kelvin problem.


1 Answer 1


There aren't many isohedra (all faces the same) that are also space-filling polyhedra. Two of them are the cube and the rhombic dodecahedron. A third is a space-filling tetrahedron, called "Sommerville 1" or "Baumgartner T", with vertex set $(001, 021, 110, 112)$.

From the limited selection, the rhombic dodecahedron is optimal.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .