Inflating 2D Triangulations to 3D Convex Polyhedra  Ben Marlin  

Theorem 1Proof: First consider the fact that the convex polyhedra P* splits into two pieces in a natural way: all the faces of P* you would see while standing over it and looking down form the top P*t, and all the faces you would see while standing under it and looking up form the bottom P*b. Since P is convex, the edges of P' are contained in both P*t and P*b. If not, then with out loss of generality assume an edge of P' is not in P*t. Then it must be covered by some face of P*t, but then the projection of this face onto the xyplane would have to contain edges outside the polygon P which can't be. Now assume we have some edge that is in both P*t and P*b and is not in P'. This can only happen if the polyhedra P* is pinched at some point, but then P* could not have been obtained from the convex hull of P'. Thus the spatial polygon P' exactly splits P* into the two pieces P*t and P*b. Now, if we separate the two pieces of P* along the boundary P' and project them back onto the horizontal plane we note first that the boundary of each is exactly the polygon P as we should expect. Also, the edges that form the triangulated convex hull of the top and bottom of P* project onto diagonals of P since they connect vertices of P'. Forming two diagonal triangularizations Tt and Tb of P. If the projection of either failed to produce a proper triangularization there would be two possible causes:
Lastly we note that Tt and Tb can not share any diagonals. Since as noted above the only edges P*t and P*b have in common are the edges that come from P' and these are all edges of P so they can not be diagonals. So we see that by starting with any fixed, convex polygon P in the horizontal plane, assigning each vertex some vertical variation, and then taking the convex hull of the resulting point set of vertices we obtain a convex polyhedron P*. The projection of this convex polyhedron back onto the horizontal plane yields two proper triangulations of P that must be distinct. 