Talk:Tetrahedron packing

Aristotle
Quote"Aristotle claimed that tetrahedra could fill space completely."Aristotle was right, but I believe he also claimed that regular octahedra could not fill space completely, in which claim he was mistaken.

Quote"Tetrahedra do not tile space."This is also mistaken.

If you accept the definition of a tiling in Wikipaedia viz."More formally, a tessellation or tiling is a cover of the Euclidean plane by a countable number of closed sets, called tiles, such that the tiles intersect only on their boundaries."(interpreting plane as space), then space can be tiled either by regular tetrahedra or by regular octahedra.

A regular octahedron can be dissected into 6 regular octahedra and 8 regular tetrahedra, each with an edge length of half the original. Similarly A regular tetrahedron can be dissected into a regular octahedron and 4 regular tetrahedra, each with an edge length of half the original.

So to construct a tiling of space by regular tetrahedra you start with a tiling by regular octahedra and tetrahedra of unit edge. You then dissect all the octahedra as above. You then dissect the newly formed octahedra, then the octahedra so formed and so on. The set of all tetrahedra that remain at all stages is then a tiling of space.

To tile with octahedra you start with the same tiling by regular octahedra and tetrahedra, but dissect the tetrahedra at each stage instead.

There are no tilings in either case if a positive lower bound on the size of the tiles is mandated, which you can see by looking at the solid angles at the edges of the tiles. In this case you have to be able to add a set of angles to $$4\pi$$ at a point on the edge of any tile, but you can't. Martin Rattigan (talk) 02:21, 30 April 2015 (UTC)


 * The restriction to *congruent* tetrahedra is implied, but even if it is not, your construction, clever as it is, is not, I believe, a tessellation under the definition you gave. Yes, it consists of a countable number of closed sets that intersect only on their boundary. However, it does not cover the Euclidean space. It leaves out a set of zero measure of points at the limit of nested sequences of octahedra. Some such limit points (a countable set) will be at the boundary of tetrahedra, and will be covered that way, but that still leaves out an uncountable number of uncovered points. Eigenbra (talk) 19:26, 30 April 2015 (UTC)

Yes, you're right. I didn't see that. Back to the drawing board. Thanks for the response. Martin Rattigan (talk) 22:24, 30 April 2015 (UTC)