Hexiamonds on an Octahedron

January 26th, 2010 by munizao Leave a reply »

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?

2 comments

  1. Bryce Herdt says:

    For #7, the answer is unfortunately in the negative. Here’s the address of the picture I made for this proof (since I don’t know your policy on embedding images): http://i107.photobucket.com/albums/m298/Cy-Reb_Jr/Puzzles/7impossible.gif?t=1284909845

    The orange one, listed as the “rhomboid” in your link above to Torbijn’s 24 enlarged octiamonds, has essentially one place where it folds over one edge and blocks no other edges. This forces the two yellow hexiamonds over the two edges as shown, which means that the blue strings of five are each part of single hexiamonds: the club and the crown, in some order (which, despite being illustrated, is irrelevant).

    Now, to avoid duplication, the yellow hexiamonds must be a pistol and a snake (as shown, but this time it’s relevant). The snake isolates the upper purple shape, and this time there is no way to add a distinct hexiamond that only folds over one edge.

  2. munizao says:

    Nice proof. A weaker version of the problem could be to require each of the hexiamonds to fold over exactly one edge, but allow some of the edges to have multiple hexiamonds on them, while others have none. (In fact, the problem as I wrote it doesn’t require the relation of hexiamonds to edges to be one to one, even though the way you took the problem was certainly what I meant and implied.)

    As for embedding images, I don’t have a problem with it, although I should institute a policy of needing permission to host my own my own copy of an image and swap in the copy for the original, so that the record of the conversation can’t deteriorate when somebody else’s site does.

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