# Are polyhedra of equal volume equidecomposable?

(written by lawrence krubner, however indented passages are often quotes). You can contact lawrence at: lawrence@krubner.com

Just as any pair of polygons of equal area can be decomposed and reassembled into the same square, can any pair of polyhedra of equal volume be decomposed and reassembled into the same cube? Are polyhedra of equal volume equidecomposable? Are they equicomplementable?

By the end of the nineteenth century there were several examples of equalvolume polyhedra that were both equidecomposable and equicomplementable, but there was no general solution. One simple example is prisms with the same height and equal area bases, stemming from the two-dimensional polygon result. In 1844 Gerling showed (and then Bricard proved again in 1896) that two mirror-image polyhedra are equidecomposable by cutting them up into congruent mirror-image pieces that can then be rotated into each other. There were also some specific tetrahedra equidecomposable with a cube, shown in 1896 by M.J.M. Hill.

We can reduce the problem about polyhedra in general to a problem about tetrahedra.

1.4 Definition III: Tetrahedron

A tetrahedron is a polyhedron with four triangular faces, six edges, and four vertices. In many current math textbooks the faces are required to be congruent. We are not going to require that any of the faces be congruent; our definition is closer to what many current math textbooks call a triangular pyramid.

Just as any polygon can be cut up into triangles, any polyhedron can be cut up into tetrahedra. First, we can cut any polyhedron into a finite number of convex polyhedra. These can each then be cut up into a finite number of pyramids with polygonal bases. Because any polygon can be cut up into a finite number of triangles, each of these polygonal pyramids can be cut up into a finite number of triangular pyramids. This means that if we can prove or disprove the Bolyai-Gerwien theorem in the third dimension for tetrahedra, then we have also proved or disproved it more generally for polyhedra. Below is an example of a division of a polyhedron into triangular pyramids, based off of figures 13.12 and 13.13 in Rajwade’s 2001 Convex Polyhedra.

The volume of a tetrahedron is ⅓ the area of the base multiplied by the height; as described by Euclid, any two tetrahedra with bases of equal areas and with equal heights will also have the same volume. This definition of volume is reminiscent of the parallel result in two dimensions: the volume of a triangle is ½ the length of the base multiplied by the height.

Unlike the area of a triangle, however, the volume of a tetrahedron and therefore the volume of a polyhedron is found through calculus, by dividing the three-dimensional polyhedron into infinitesimally thin two-dimensional cross sections and adding up their areas.

If the Bolyai-Gerwien theorem can be expanded into the third dimension, we can define the volume of any three dimensional polyhedron the same way we define the area of a two dimensional polygon, by breaking it up into discrete building blocks—tetrahedra in three dimensions and triangles in two—and reassembling the pieces into a cube or square. This would be an elementary solution, with no infinities (or calculus) required.

This problem was posed by C.F. Gauss. In a letter in 1844, Gauss expressed that he wanted to see a proof that used finitely rather than infinitely many pieces. By Hilbert’s time it was not yet solved.

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March 11, 2018 10:49 am

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