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6.AP.2. FOUR PIECES
These dissections usually also work with a tetrahedron of spheres and hence these are related to ball pyramid puzzles, 6.AZ.

The first version I had in mind dissects each of the two pieces of 6.AP.1 giving four congruent rhombic pyramids. Alternatively, imagine a tetrahedron bisected by two of its midplanes, where a midplane goes halfway between a pair of opposite edges. This puzzle has been available in various versions since at least the 1970s, including one from Stokes Publishing Co., 1292 Reamwood Avenue, Sunnyvale, California, 94089, USA., but I have no idea of the original source. The same pieces are part of a more complex dissection of a cube, PolyPackPuzzle, which was produced by Stokes in 1996. (I bought mine from Key Curriculum Press.)

In 1997, Bill Ritchie, of Binary Arts, sent a quadrisection of the tetrahedron that they are producing. Each piece is a hexahedron. The easiest way to describe it is to consider the tetrahedron as a pile of spheres with four on an edge and hence 20 altogether. Consider a planar triangle of six of these spheres with three on an edge and remove one vertex sphere to produce a trapezium (or trapezoid) shape. Four of these assemble to make the tetrahedron. Writing this has made me realise that Ray Bathke has made and sold these 5-sphere pieces as Pyramid 4 for a few years. However, the solid pieces used by Binary Arts are distinctly more deceptive.

Len Gordon produced another quadrisection of the 20 sphere tetrahedron 0 0

using the planar shape at the right. This was c1980?? 0 0 0  
David Singmaster. Sums of squares and pyramidal numbers. MG 66 (No. 436) (Jun 1982) 100-104. Consider a tetrahedron of spheres with 2n on an edge. The quadrisection described above gives four pyramids whose layers are the squares 1, 4, ..., n². Hence four times the sum of the first n squares is the tetrahedral number for 2n, i.e. 4 [1 + 4 + ... + n²]  =  BC(n+2, 3).