To cover a solid

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.

 Nr. L H Square Cube Triangle Tetrahedron Octahedron Icosahedron 1 0 1 2 12 3 12 24 60 2 0 2 8 48 12 48 96 240 3 0 3 18 108 27 108 216 540 4 0 4 32 192 48 192 384 960 5 0 5 50 300 75 300 600 1500 6 0 6 72 432 108 432 864 2160 7 0 7 98 588 147 588 1176 2940 8 0 8 128 768 192 768 1536 3840 9 0 9 162 972 243 972 1944 4860 10 1 0 1 6 1 4 8 20 11 1 1 5 30 7 28 56 140 12 1 2 13 78 19 76 152 380 13 1 3 25 150 37 148 296 740 14 1 4 41 246 61 244 488 1220 15 1 5 61 366 91 364 728 1820 16 1 6 85 510 127 508 1016 2540 17 1 7 113 678 169 676 1352 3380 18 1 8 145 870 217 868 1736 4340 19 1 9 181 1086 271 1084 2168 5420 20 2 0 4 24 4 16 32 80 21 2 1 10 60 13 52 104 260 22 2 2 20 120 28 112 224 560 23 2 3 34 204 49 196 392 980 24 2 4 52 312 76 304 608 1520 25 2 5 74 444 109 436 872 2180 26 2 6 100 600 148 592 1184 2960 27 2 7 130 780 193 772 1544 3860 28 2 8 164 984 244 976 1952 4880 29 2 9 202 1212 301 1204 2408 6020 30 3 0 9 54 9 36 72 180 31 3 1 17 102 21 84 168 420 32 3 2 29 174 39 156 312 780 33 3 3 45 270 63 252 504 1260 34 3 4 65 390 93 372 744 1860 35 3 5 89 534 129 516 1032 2580 36 3 6 117 702 171 684 1368 3420 37 3 7 149 894 219 876 1752 4380 38 3 8 185 1110 273 1092 2184 5460 39 3 9 225 1350 333 1332 2664 6660 40 4 0 16 96 16 64 128 320 41 4 1 26 156 31 124 248 620 42 4 2 40 240 52 208 416 1040 43 4 3 58 348 79 316 632 1580 44 4 4 80 480 112 448 896 2240 45 4 5 106 636 151 604 1208 3020 46 4 6 136 816 196 784 1568 3920 47 4 7 170 1020 247 988 1976 4940 48 4 8 208 1248 304 1216 2432 6080 49 4 9 250 1500 367 1468 2936 7340 50 5 0 25 150 25 100 200 500 51 5 1 37 222 43 172 344 860 52 5 2 53 318 67 268 536 1340 53 5 3 73 438 97 388 776 1940 54 5 4 97 582 133 532 1064 2660 55 5 5 125 750 175 700 1400 3500 56 5 6 157 942 223 892 1784 4460 57 5 7 193 1158 277 1108 2216 5540 58 5 8 233 1398 337 1348 2696 6740 59 5 9 277 1662 403 1612 3224 8060 60 6 0 36 216 36 144 288 720 61 6 1 50 300 57 228 456 1140 62 6 2 68 408 84 336 672 1680 63 6 3 90 540 117 468 936 2340 64 6 4 116 696 156 624 1248 3120 65 6 5 146 876 201 804 1608 4020 66 6 6 180 1080 252 1008 2016 5040 67 6 7 218 1308 309 1236 2472 6180 68 6 8 260 1560 372 1488 2976 7440 69 6 9 306 1836 441 1764 3528 8820 70 7 0 49 294 49 196 392 980 71 7 1 65 390 73 292 584 1460 72 7 2 85 510 103 412 824 2060 73 7 3 109 654 139 556 1112 2780 74 7 4 137 822 181 724 1448 3620 75 7 5 169 1014 229 916 1832 4580 76 7 6 205 1230 283 1132 2264 5660 77 7 7 245 1470 343 1372 2744 6860 78 7 8 289 1734 409 1636 3272 8180 79 7 9 337 2022 481 1924 3848 9620 80 8 0 64 384 64 256 512 1280 81 8 1 82 492 91 364 728 1820 82 8 2 104 624 124 496 992 2480 83 8 3 130 780 163 652 1304 3260 84 8 4 160 960 208 832 1664 4160 85 8 5 194 1164 259 1036 2072 5180 86 8 6 232 1392 316 1264 2528 6320 87 8 7 274 1644 379 1516 3032 7580 88 8 8 320 1920 448 1792 3584 8960 89 8 9 370 2220 523 2092 4184 10460 90 9 0 81 486 81 324 648 1620 91 9 1 101 606 111 444 888 2220 92 9 2 125 750 147 588 1176 2940 93 9 3 153 918 189 756 1512 3780 94 9 4 185 1110 237 948 1896 4740 95 9 5 221 1326 291 1164 2328 5820 96 9 6 261 1566 351 1404 2808 7020 97 9 7 305 1830 417 1668 3336 8340 98 9 8 353 2118 489 1956 3912 9780 99 9 9 405 2430 567 2268 4536 11340

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.

Hexominoes
Cube of line 23 (if you don't use the cross)

My solution

Hexominoes + Pentominoes
Cube of line 33 (35x6+12x5=270 exactly)

Sept 2001 - Solution by Alessandro Fogliati

Fractal
Cube covered by 6 fractals on a grid of infinite number of squares.

Sexominoes
The 6 full-sexed Sexominoes cover the cube nr. 10 exactly (solution)

The 24 one-side Sexominoes cover the cube of line 20 exactly (solution)

Sexominoes Number Five

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.)

Sept 2001 - Solution by Alessandro Fogliati

Tubominoes

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)

Polyarc

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.

Sept 2001 - Solution by Alessandro Fogliati

.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).

Tiling

Sometime it isn't easy to tile a cube by a polyomino.

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.

Hexiamonds
Tetrahedron of line 12 and Octahedron of line 30

At left a solution by Alessandro Fogliati

Order 8 Isoperimetric Polyiamonds
Tetrahedrons of line 8 (exactly) and line 44 (with 4 holes)
You could find the Isoperimetric Polyforms on The Poly Pages by Andrew Clarke.

Sept 2001 - Solution by Alessandro Fogliati

Eptiamonds
Octahedron of line 31 exactly

Sept 2001 - Solution by Alessandro Fogliati

Eptiamonds + Hexiamonds
Icosahedron of line 2 (12x6+24x7=240 exactly)

Sept 2001 - Solution by Alessandro Fogliati

Octiamonds
Tetrahedron of line 54 (66x8+4=532)
You could find many information about octiamonds on Mathpuzzle.com by Ed Pegg Jr.

Sept 2001 - Solution by Alessandro Fogliati

Sexiamonds
The 4 Order 2 Triangular Polysexes cover the tetrahedron nr.10

Solution

Apartheid

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.

I'll die before see the solution of this puzzle.

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.

Polyhexes & SexeHexeS

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.

 Nr. L H Hex on Octahedron 1 1 0 4 2 1 1 12 3 2 0 16 4 2 1 28 5 2 2 48 6 3 0 36 7 3 1 52 8 3 2 76 9 3 3 108 10 4 0 64 11 4 1 84 12 4 2 112 13 4 3 148 14 4 4 192 15 5 0 100 16 5 1 124 17 5 2 156 18 5 3 196 19 5 4 244 20 5 5 300 21 6 0 144 22 6 1 172 23 6 2 208 24 6 3 252 25 6 4 304 26 6 5 364 27 6 6 432 28 7 0 196 29 7 1 228 30 7 2 268 31 7 3 316 32 7 4 372 33 7 5 436 34 7 6 508 35 7 7 588 36 8 0 256 37 8 1 292 38 8 2 336 39 8 3 388 40 8 4 448 41 8 5 516 42 8 6 592 43 8 7 676 44 8 8 768 45 9 0 324 46 9 1 364 47 9 2 412 48 9 3 468 49 9 4 532 50 9 5 604 51 9 6 684 52 9 7 772 53 9 8 868 54 9 9 972

Tetrahexes
Octahedron of line 4 (my solution)

Order 6 Hexastrips
Octahedron of line 21 (24x6=144 exactly)
The Polystrips are an invention of Miroslav Vicher.

SexeHexeS

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?

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First edition: Aug 15th, 2000 - Last revision: Sept 20th, 2001

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