Once you have pre-specified some simplices $S$ that must be included in your triangulation of the convex polytope $P$, what remains is the problem of triangulating a nonconvex region: $P \setminus S$. There are nonconvex polyhedra (in dimension 3) that cannot be triangulated. I believe one could make such an example from the [Schönhardt polyhedron][1], by insisting on the inclusion of three tetrahedra ($S$) external to that polyhedron as part of a triangulation of the convex hull ($P$) of two twisted triangles in parallel planes, so that $P \setminus S$ is the un-tetrahedralizable Schönhardt polyhedron (see below). And it is an NP-complete problem to decide if a given nonconvex polyhedron can be triangulated, a 1992 result of Ruppert and Seidel. ![alt text][2]<br /> <sub>([Image from Wikipedia][1])</sub> If you want to nevertheless hope that your region can be triangulated, you might explore [geometric bistellar flips][3] to underlie an approach. [1]: http://en.wikipedia.org/wiki/Sch%C3%B6nhardt_polyhedron [2]: https://upload.wikimedia.org/wikipedia/commons/thumb/f/f3/Sch%C3%B6nhardt_polyhedron.svg/220px-Sch%C3%B6nhardt_polyhedron.svg.png [3]: http://web.archive.org/web/20181114131513/https://www.voronoi.com/wiki/index.php?title=Bistellar_flips