# How can you prove that every polyhedron can be dissected into tetrahedrons?

I have always loosely accepted the fact that any polyhedron can be dissected into tetrahedrons just like any polygon can be dissected into triangles. But how can I prove precisely that any polyhedron no matter their homeomorphic difference can be dissected into tetrahedrons?

Many kinds of proof are welcome especially proofs in a geometric sense rather than with formulas! I'd be especially thankful to receive explanations on how the proof works for both convex and non-convex polyhedrons.

A triangulation without gaps and overlaps of any convex polygonal face of a polyhedron is obtained by picking some point inside that face and connecting each vertex to that very point.

An according simplexial decomposition without gaps and overlaps of a convex polyhedron then is obtained by picking some point inside that solid and connecting any vertex as well as each chosen face center to that chosen body center. (Note that nothing is assumed about those "centers" except being some internal points each.)

When considering non-convex polygons or polyhedra, then just subdivide those into convex ones and the above statement could be applied individually, and therefore to those cases as well.

Btw., the same argument thus can be applied for any higher dimensional polytope as well, just by means of dimensional recursion.

--- rk

First, note that there is a terminology issue which may prove troublesome when navigating the literature: generally, a "triangulation" (as well as the weaker notion of "tiling") is required to not add vertices. If we adopt this meaning, then there are non-triangulable polyhedra; this was first shown by Lennes, and Braxton Carrigan's thesis seems a good expository source.

However, you don't appear to mind additional vertices. When we drop that requirement, and calling the resulting property "decomposability," it turns out that every polyhedron is decomposable. This is mentioned frequently in the computational geometry literature, e.g. in the beginning of this paper; however, that literature is (understandably!) more interested in "efficient" decompositions rather than the mere existence of decompositions, and I haven't yet found a source for this weaker fact.

(Actually, the linked paper only talks about decompositions into convex polyhedra. But this is enough, since every convex polyhedron is decomposable (indeed, *triangulable): first triangulate each face as usual. Now pick a vertex $$v$$ and connect each other vertex to $$v$$. Every triangle in a face of the polyhedron now generates a tetrahedron (or contains $$v$$, in which case ignore it) - given by its three vertices together with the vertex $$v$$ - and these tetrahedra give a triangulation of the whole polyhedron. So it's enough to show just that arbitrary polyhedra can be decomposed into convex polyhedra.)

• I now intuitively recognize that all a triangle on a face connected to a vertex on another face forms a tetrahedron and therefore convex polyhedrons can be triangulated. I'm wondering if there is perhaps a more strict proof explaining that no gaps can exist between tetrahedrons in this process? (In other words that no portion of the polyhedra is left out in the triangulation process) Oct 31, 2018 at 15:41
• @RoxaneMin Yes, I didn't prove that claim (or the claim about convex decomposability). If I remember correctly, it's not hard but a bit tedious. I don't immediately know a good source for this, but I'll look around when I have more time. Oct 31, 2018 at 15:58
• @RoxaneMin Incidentally, it would be reasonable if you unaccepted my answer until I can find a good source; you did ask for a proof after all. And unaccepting my answer might make someone else come through with a better one ... Oct 31, 2018 at 15:59