Two-Edge-Length Polyhedra

Which 3-dimensional convex polyhedra, all of whose faces are triangles, can be made with isosceles triangle faces? In an earlier research column (this column is essentially self-contained but you might want to look at the earlier column in Consortium 116), I raised the question of whether it might be true that all possible (finite) triangulations in the plane with at least four vertices (see Figure 1 for examples) might be "realizable" as 3-dimensional bounded convex polyhedra with faces that were all isosceles triangles, an isosceles triangle being one with two sides of one length and one side of another length. Here I don't allow a triangle described as isosceles to be equilateral, a triangle where all three of its edges have the same length.

In the earlier column I mentioned the issue of trying to better understand which triangulations could be realized as polyhedra with all faces congruent isosceles triangles. I knew of several infinite classes of such congruent isosceles triangle polyhedra, and I knew examples of triangulations where it was not possible to realize the triangulation with a convex polyhedron having isosceles triangle faces that were congruent to one another. Below, when I use the term polyhedron I am referring to a convex bounded polyhedron in 3-dimensional space where all of the faces of the polyhedron are convex polygons.

I am not thinking about a polyhedron gotten, say, by pasting two congruent cubes along a face and counting the vertices of the faces where the pasting occurs as "vertices". Thus, two 1 • 1 • 1 cubes pasted in such a way would be the polyhedron of the same combinatorial type as the cube, having 8 vertices, 6 faces and 12 edges, with edges of length 1 or 2, and is the "solid" we would think of as a 1 • 1 • 2 box. Below, the polyhedra I have in mind are the convex hull of their "extreme points."