Jim.Buddenhagen@launchpad.unc.edu (Jim Buddenhagen) writes: >Start with a regular tetrahedron circumscribed by a sphere. Subdivide >each triangle into 4 triangles by connecting edge midpoints, and project >to the sphere (radially), to obtain a new polyhedron with all vertices >on the sphere and with 16 triangular faces. > >Repeat this procedure with the new polyhedron to get a third polyhedron. > >True or false: this third polyhedron is convex. > >(I found the answer surprising, what about you?) >--
A simpler example of this phenomenon can be gotten by taking a regular octahedron inscribed in the sphere and stellating the faces (that is, adding a vertex in the middle of each face, connected to the vertices of each face), and then pushing the new vertices out to the sphere. This will not be convex, no matter where you put the new vertices. What's more, there is no convex polyhedron inscribed in the sphere combinatorially equivalent to the stellated icosahedron (which only has 24 faces, but is not the smallest example).
These kinds of questions go back a long way (to Descartes), and seem to be finally well understood, through some work of mine on hyperbolic geometry; see the research announcement in the most recent (Oct 92) Bulletin of the AMS. I can send you the actual preprints/reprints, if you are interested.