# some interesting puzzles

### elementary puzzles

Like all regular antidiamonds a cube can be cut into two identical polyhedra; the section goes through the midpoints of six edges, and we get __two heptahedra__ with one face which is a regular hexagon.

A regular tetrahedron can also be cut into __two identical pentahedra__; the section is a square whose vertices are the midpoints of four edges. Each pentahedron is the assemblage of a half octahedron and two regular tetrahedra.

We can cut a cube into __three identical pyramids__ with square base. As this pyramids have a plane of symmetry, we can too cut the cube into __six non regular tetrahedra__ with same base (three mirror image pairs).

At last, we can cut a cube into __six identical regular pyramids__ with square base (common vertex on the center of the cube). If we use the planes of symmetry of these pyramids, we can continue the cutting to create tetrahedra.
### assemblages of pyramids which build a cube

A pyramid is regular if its base is a regular polygon, and if all its lateral edges have same length (thus the lateral faces are identical isosceles triangles). The three pictures above show those which interest us now:

• the brave regular tetrahedron (Tr),

• the half octahedron (Po) which is a pyramid with a square base and all edges of the same length,

• the "cube corner" tetrahedron (Tc) which has an equilateral and three right-angled isosceles faces (regular right-angled tetrahedron).

We know that four vertices of a cube define a regular tetrahedron; by cutting four "corners" opposite by pairs of a cube, we can cut a cube into __five regular pyramids__ (one Tr and four Tc) to make a cubic jigsaw P[5].

The central picture above shows that a tetrahedron can be cut into 2x3=6 regular pyramids (two Po and four Tr). In this way we carry out a nice cubic jigsaw P[10] of __ten regular pyramids__ (4 Tc + 4 Tr + 2 Po).

A __cuboctahedron__ results of the cutting of the eight "corners" of a cube (truncation using the midpoints of the edges). The segments joining the twelve vertices to the center of symmetry define fourteen regular pyramids (six Po and eight Tr) whose bases are the faces of the cuboctahedron. This leads to a cubic jigsaw P[22] of __twenty-two regular pyramids__ (8 Tc + 8 Tr + 6 Po).

If we use the __anticube__ (stella octangula), we can achieve an other cutting of the cube into __twenty-two pyramids__ among which ten are regular (two Po and eight Tr which form the anticube); the twelve others are non regular tetrahedra but each of them is an assemblage of two Tc. Finally we get a cubic jigsaw P[34] of __thirty-four regular pyramids__ (24 Tc + 8 Tr + 2 Po).

In addition, Po can be cut into four Tc using two planes of symmetry containing the diagonals of the base square. Thus P[10] becomes P[16] (__sixteen regular pyramids__); P[22] and P[34] are then each made up of __forty regular pyramids__ (32 Tc + 8 Tr). This new cubic jigsaw P[40] can therefore be assembled in two different manners, depending on the arrangement of the Tc (an external face of the jigsaw appears differently according to the layout : P[22] or P[34], with 32 or 24 visible Tc).
The two arrangements are assemblages of eight cubic jigsaws P[5], and we switch from one arrangement to the other by exchanging the lower and upper "half cubes" (each consisting of four P[5]).

Let's notice that the Kepler star (8 Tc + 8 Tr) is formed by eight diamonds (1 Tc + 1 Tr) assembled around the center of the solid; the eight Tc form the central regular octahedron.

The volume of Tc, which is also a pyramid with base a half square, is easy to calculate; we can therefore deduce easily the volumes of the cuboctahedron, of the regular tetrahedron and of the anticube (respectively 5/6, 1/3 and 1/2 of the cube's volume).

To finish let's notice that it has been proven that it is not possible to cut a cube neither into regular tetrahedra nor into cubes all of different sizes.