# What is the best way to cut a donut up into tetrahedra?

A donut is a solid $3$-dimensional object known, in fancier mathematical terms, as a solid torus. We are interested in chopping a donut into tetrahedra, solids built with $4$ triangular faces. Perhaps it would be better to see that we are interested in building a solid torus out of tetrahedra. So take a bunch of tetrahedra and glue them together faces-to-faces; we are doing topology here, so it is OK to bend and deform them, but anytime you glue two tetrahedra together, you should glue one triangular face of one tetrahedron to one triangular face of the other one. It is even OK to glue a tetrahedron to itself, you should just glue two different faces to each other. (This, obviously, takes a lot of bending and deforming, but that's OK.) Now, notice that after you've done this, the surface of the donut will itself be divided up into some number of triangles.

Here is the question: Suppose you start with a particular way of dividing the surface of the donut into exactly two triangles. (One way of doing this is illustrated in the picture here.) Then there are many different ways that you might build the donut from tetrahedra so as to give you that particular "triangulation" of the surface. Of all these different ways, which one has the least number of tetrahedra?

Something to start with is to try to think of how many different ways there are to divide the surface of a donut into exactly two triangles. For each one of these ways, we expect a different answer to the question above. I'll try to say some more about this problem in a later post. Thanks to Stephan Tillmann for telling me about this problem. There is a particular conjecture about the answer to this problem, but I'll leave that unstated for now - that's another story.