Another easy way to visualize this is to place two square pyramids next to each other on the sand, sharing one square edge. Then join their apices with another edge. It’s easy to see that (1) you’ve just created a regular tetrahedron, and (2) the new faces you’ve just created are coplanar with the already existing square pyramid faces.

]]>Thanks! I understood the idea, but was having great trouble lining up the planes in my mind. Your construction clarified the situation very nicely. ]]>

Then dissect this construction by a plane containing the 4 vertices of the octahedron. What’s left is a square-based pyramid with 2 tetrahedrons attached to it. ]]>

Your 9,8,8,8,7,3 example is the sort of thing I was looking for, based on the assumption that the 8s would form a triangle (although in your example I think the 3 is just large enough that we *can* form a triangle in this case — something like 2.01 is small enough, though, and would surely break the situation when the 8s all have a common vertex, too). The last case (or the last 3 cases, as you note), are trickier.

Perhaps it’s easier to classify workable situations if we also use some tool like Ptolemy’s Inequality: http://mathworld.wolfram.com/PtolemyInequality.html — in a tetrahedron ABCD, we have AB * CD < AC * BD + AD * BC. Unfortunately, this isn’t even sensitive enough to find the 4s and 7 example; I wonder if it can find any example that doesn’t have a violation of the triangle inequality?

]]>I wonder if 4,4,4,4,4,7 is the smallest integral counterexample. Certainly this is the best for five congruent and one large. Four congruent and two congruent can’t produce a counterexample. I am not sure how to check something on the order of 9, 8, 8, 8, 7, 3. It seems that it would need to fail three cases – equilateral triangle, 8’s emanating from a single vertex, and 8’s lined end to end, but not closed. And that third case looks like 3 cases.

]]>@Austin, for a more conceptual explanation: consider a cube. One can choose four vertices so that the segments joining them are face diagonals of the cube, i.e., so that they form a regular tetrahedron. Cutting this tetrahedron out, we are left with four pyramids, each with three isosceles right triangular faces and one equilateral triangular face. Note that two such pyramids share an edge with the tetrahedron, and the three angles sum to make the flat side of the cube. Now, glue these four pyramids together to form the square pyramid in the question; when you attach the tetrahedron, the same angles are reconstructed.

Alternative viewpoint: chop space up with the planes x + y + z = 2n, x + y – z = 2n, x – y + z = 2n and -x + y + z = 2n for all integers n. The regions they determine are regular tetrahedra and regular octahedra.

]]>