# Most regular way to triangulate $\mathbb{R}^3$?

By "regular", I'm going by a property of Delaunay Triangulation, which is to maximize the minimum angle. In $\mathbb{R}^2$, tiling with equilateral triangles gives you a minimum angle of 60 degrees. Any other triangulation would give you a smaller minimum angle, so in this sense, triangulating $\mathbb{R}^2$ with equilateral triangles is the most regular way.

What's the best way to triangulate $\mathbb{R}^3$ so that the minimum solid angle of the tetrahedra is maximized?

I assume that the dual tessellation of closely-packed spheres would give a good triangulation since that works in $\mathbb{R}^2$. You get some regular tetrahedra and regular octahedra which can be divided into 4 congruent tetrahedra which contain the minimum solid angle. Is there a way to check if this is maximized?

• An estimate (rather trivial) would be the solid angle at the corner of a pyramid with a square base, with height 1/2 the side of the base: Six of these form a cube. (This ought not be the best.) We need to further split the pyramid into two tetrahedra (making the solid angle even smaller). – Mirko Mar 26 '14 at 15:41
• I would check what happens by using close-packing of spheres and taking dual tessellations. en.wikipedia.org/wiki/Close-packing_of_equal_spheres – user126154 Mar 26 '14 at 16:17
• Though I understand you do not require that all tetrahedra of the triangulation be congruent, your question seems closely related to the still unsolved problem of classifying the tetrahedra that tile $\mathbb{R}^3$. For a survey of known results, see Egon Schulte's article arxiv.org/pdf/1005.3836.pdf – Wlodek Kuperberg Mar 26 '14 at 22:36