In an old problem the pentominoes are folded to cover a cube. Below there are two my solutions. Wen-Shan Kao has many others. The question now is: Are there other regular polyforms sets that cover a regular polyhedron?

In the pictures and in the table below you could read how many squares a square (or a cube area) contains exactly, and how many triangles are contained in a triangle or on an area of a tetrahedron, an octahedron or an icosahedron.

Now we could look for the number that coincides exactly with the number of squares or triangles in a polyominoes or polyiamonds set. I suggest the shapes below, without to guarantee the existence of a solution. However I'm optimist because no parity problems are present.

Sometime my problem is to name the new polyforms: in this case I inspired by "Chanel nr.5" and "Mambo nr.5".

They are the 120 Order 5 squared flippable Polysexes. They cover the cube nr.22 exactly. (We must work with this interesting tile, also on the plane! I had supposed his existence some months ago, but it passed unnoticed. They are 120, that is a wonderful number for a geometric puzzle! Click here if you want to see all the Sexominoes Number Five.)

The 30 two Flanges for edge Tubominoes cover exactly the cube nr.11. Do you want to minimize or to maximize the number of tubes? (I think it's a hard problem)

Here there are the 5 Tetrominoes with the 7 Biarc and the 22 Triarc by Henri Picciotto for a total of 60 squares! Do you want to cover the cube nr.21? I think it's possible.

.When I have seen the Picciotto's Page, I drew immediately the 24 Tri-Arc-iamonds.. I think they are nice but I don't know if they could cover the octahedron nr.30 (24x3=72 exactly).

Many Polyominoes can to tile a cube area. The Domino and each Triomino tile the elementary cube. The cube nr.1 (12 squares) is tiled by a Triomino, a Tetromino and a Hexomino. Many Pentominoes (not all!) tile the cube nr.11 (30 squares) that is tiled also by a 10-omino and a 15-omino. The cube nr.2 (48 squares) is tiled by 2-3-4-6-8-12-16-24-ominoes! Perhaps it's not easy to tile the cube nr.21 (60 squares) with a Hexomino. The Sexomino "FMFM" tiles all the cubes.

These are the Order 14 two-side Triangular Polysexes. They are obtained coloring with 7 colors the 4 forms below. Each edge could be colored with a different color. The complete set has 560 pieces [see Chris Hartman's email]. The rules of the apartheid (terrible!) impose that only the edges with the same color can to join. The set could cover exactly the Icosahedron nr.22.

Any racial tolerance (red with yellow, etc.) facilitates the solution. (The tolerance does easy the things!). If you use only 3 colors, you obtain a set of 56 pieces that cover exactly the Octahedron nr. 11. The Order 7 two-side Triangular Polysexes could cover exactly the Tetrahedron nr.31.

This could be an other chapter. The study below shows that regular hexagons can to cover a regular octahedron, but not a tetrahedron or an icosahedron. There are many combinations.

Do you want to solve the octahedron nr.2 with 12 of 13 full-sexed sexehexes? Or the octahedron nr.11 with 84/92 sexehexes?