Posts Tagged ‘octahedron’

This is how I roll: Magic dice, part 2

April 11th, 2015

One formulation of a magic cube simply numbers the vertices of a cube from one to eight, and requires that, for each face, the set of vertices that frame it have the same sum. There are three distinct magic cubes of this type:

magic-vertex-cubes-all
Now, first of all, in applying this to a design of dice, it seems that the ideal application would be octahedral dice, since the octahedron is the dual of the cube, and thus the numbers could simply be printed on different faces. The sets of faces with magic sums would then be the ones surrounding a vertex of the octahedron. But I don’t have an order for custom d8’s, I have one for custom d6’s. Given that, what is the best way to represent one of these figures on a cube? As before, we can use a bold font to highlight the number that is used as the result of a roll. This time, since we can’t print directly on the vertices, the numbers will be repeated on each face that borders a vertex.

I have two answers. The first places the numbers on their traditional faces (with opposite faces summing to seven.) Only the middle numbering above can be used in this manner.

magic-vertex-cube-1

The second is suggested by the figure on the right. Since the numbers we aren’t using (that is, 7 and 8) appear at opposite vertices, we can highlight a diagonal ring around the cube and catch all of the numbers from 1 to 6. This is nicely symmetrical, but it does not put the numbers in their traditional positions.

magic-vertex-cube-2

Hexiamonds on an Octahedron

January 26th, 2010

Here’s an interesting problem that seems not to have gotten as much attention as I think it deserves. The twelve hexiamonds contain a total of 72 triangular units. A regular octahedron with edges 3 units long can fit 9 triangles on each of its 8 faces, exactly enough to tile with the hexiamonds. Some individual solutions to this problem have been found. A solution at Livio Zucca’s site bears the label “Adrian Struyk 1963?” so we may assume the problem has been around at least since then. Another solution by Michael Dowle is here.

Notice that you can unfold an octahedron to produce a net in the form of an octiamond. This provides another source for solutions. The octahedron has 11 different octiamond nets. Pieter Torbijn found that 24 enlarged octiamonds could be tiled with the hexiamonds; of these, 5 are nets of an octahedron.

As far as I know, nobody has made an exhaustive computerized search for solutions to this problem. You can be the first!

#4: How many distinct ways can the hexiamonds tile an octahedron?

The octahedron has a large amount of symmetry compared to any planar figure that these pieces can tile. It has 48 automorphisms, or ways to map the solid onto itself. This would indicate a relatively small number of different solutions, since many solutions will be mappings of each other over the various ways of rotating and reflecting the octahedron. On the other hand, the shape lacks external borders, which ought to greatly increase the number of possibilities.

Some solutions have a piece that wraps around and caps a vertex. This could be considered an aesthetic flaw, because it would be impossible to tell which hexiamond the capping piece is just from knowing what triangles it occupies; you must also know its edges.

There are 7 different pieces that can cap a vertex, one of which, the pistol, can cap it in two distinct ways. Notice that due to the symmetry of the shape it makes when it caps a vertex, only the orientation of the sphinx is a riddle; its identity is never in question.


#5: How many solutions have a capped vertex?

#6: What is the largest number of vertices that can be capped in a solution? The ideal would be for all six vertices to be capped with all of the above hexiamonds except the sphinx.

The octahedron has twelve edges, the same as the number of hexiamonds. This suggests another problem:

#7: Is it possible to tile the octahedron so that each of the twelve hexiamonds is folded across exactly one of the edges of the octahedron?