It is easy to find three triangulations, each consisting of four tetrahedra. Are there more?
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No, these are all. The edge graph of the octahedron has no $K_4$ subgraph, so you have to add a new edge to make a triangulation. The only possible places for a new edge are connecting opposite vertices. You can only add one such edge, as any two meet in their interior. So every triangulation of the octahedron (without new vertices) adds exactly one of the three edges $e$ between opposite vertices. The only $K_4$'s in that graph are the four tetrahedra arranged around $e$, so any traingulation using $e$ must use a subset of those tetrahedra. We need all of them to fill in the octahedron.
answered Jan 3, 2018 at 17:55
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$\begingroup$ "You can only add one such edge" -- Why can't you have 2 edges crossing each other? You first divide the octahedron into 2 square pyramids, then each pyramid into 2 tetrahedra. This would give a total of 3+2*3 = 9 triangulations. $\endgroup$– WoodCommented Jan 4, 2018 at 5:06
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2$\begingroup$ Because a triangulation of a polytope $P$, by definition, is a collection of simplices whose union is $P$ and where any two simplices intersect in a common face of both. (See, for example, the textbook of de Loera, Rambau and Santos.) $\endgroup$ Commented Jan 4, 2018 at 14:06
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$\begingroup$ Thanks! I just want to point out that your definition doesn't seem to agree with the one given on this question: mathoverflow.net/questions/45863/triangulations-of-polyhedra $\endgroup$– WoodCommented Jan 6, 2018 at 3:15