The Bellows Conjecture // 37
exactly the same. Now, a polynomial equation may have many solutions, but the volume clearly changes continuously as the polyhedron flexes. The only way to change from one solution of the equation to a different one is to make a jump, and that’s not continuous. Therefore the volume cannot change. All very well, but does such a formula exist? There is one case where it definitely does: a classical formula for the volume of a tetrahedron in terms of its sides. Now, any polyhedron can be built up from tetrahedra, so the volume of the polyhedron is the sum of the volumes of its tetrahedral pieces. However, that’s not good enough. The resulting formula involves all the edges of all the pieces, many of which are ‘diagonal’ lines that cut across from one corner of the polyhedron to another. These are not edges of the polyhedron, and, for all we know, their lengths may change as the polyhedron flexes. Somehow the formula has to be tinkered with to get rid of these unwanted edges. A heroic calculation led to the amazing conclusion that such a formula does exist for an octahedron – a solid with eight triangular faces. It involves the 16th power of the volume, not the square. By 1996, Sabitov had found a way to do the same for any polyhedron, but it was very complicated, which may have been why the great mathematicians of earlier times had missed it. In 1997, however, Connelly, Sabitov and Walz found a far simpler approach, and the bellows conjecture became a theorem.
Same edges, different volumes.
I’d better point out that the existence of this formula does not imply that the volume of a polyhedron is uniquely determined by the lengths of its edges. A house with a roof has a smaller volume if you turn the roof upside down. These are two