You have a twenty-sided die (a regular icosahedron) with an arrow on each face. You play a little game with the die. You place the die flat on the table. You rotate the die in the direction of the arrow on the top face. This step is repeated each time by looking at the arrow on the top face of the die. The game is over when you see the same arrow pointing in the same direction twice.
If you can choose the directions of the arrows and the starting position of the die, what is the longest this game can last?
What is the farthest the die can go from the start to the end of the game?
The faces of a regular icosahedron correspond to the vertices of a regular dodecahedron. A Hamiltonian path on the dodecahedron will visit that shape's 20 vertices, and the corresponding path on the icosahedron will show all 20 faces.
When one such path is drawn on an icosahedron, the original face comes back in the same orientation as it started, so only 20 moves took place.
Might other paths exist where the orientation was off by 120 degrees? That would allow a 60-move solution. (off by 60 degrees would obviously be impossible, as each move alternates top-triangle-point-away with top-triangle-point-toward yourself.)
Alternatively, a path that did not include all 20 faces might wind up with a different orientation. In fact, it must, since moves alternate point-away and point toward. So that would give 19*6 moves, or 114 plus one more using the unused face to point into the repeating path.
Such a path should be found...
Posted by Charlie
on 2006-04-06 15:49:08