The following are the 8 tetracubes which can be made by combining 4 cubes. In all, 32 unit cubes are used. That would form two layers of a 4x4x4 cube.
If you are able to use each tetracube once to form this arrangement I'd be pleased to receive your solution. If it can't be done indicate why.
| | | | | | | | | | | | | | | | |
| A1 | A2 | A3 | A4 | | B1 | B2 | | | C1 | C2 | C3 | | D1 | D2 | |
| | | | | | | B3 | B4 | | C4 | | | | D3 | D4 | |
| | | | | | | | | | | | | | | | |
| E1 | E2 | E3 | | | F4 | F2 | | | G1 | G2 | | | H1 | H4 | |
| | E4 | | | | F3 | | | | G4 | | | | H3 | | |
| | | | | | | | | | | | | | | | |
Note: in the graphic F4, G4 and H4 are above the cubes that form the lower layer. Also, the G and H tetracubes 'screw' one to the left and the other to the right.
A solution might take the form of 2 4x4 grids using letters only; numerals really only assist in orienting oneself.
I can't decide which way looks better but here's a solution I found. I'm pretty sure there are others.
FFBB FBBA
FEEE CCEA
GDDH CDDA
GGHH CGHA
+---+---+ +-+---+-+
|F F|B B| |F|B B|A|
| +-+---+ +-+-+-+ |
|F|E E E| |C C|E|A|
+-+---+-+ | +-+-+ |
|G|D D|H| |C|D D|A|
| +-+-+ | | +-+-+ |
|G G|H H| |C|G|H|A|
+---+---+ +-+-+-+-+
|
|
Posted by Jer
on 2007-05-09 12:25:58 |
Found one
