[math-fun] Problem about tiling regular 2n-gon with rhombi
A recent local math olympiad problem showed how to tile a regular hexagon and octagon using rhombi (with each polygon edge coinciding with a rhombus edge), and asked "What is the number of rhombi needed for such a tiling of a 2n-gon ?" This wasn't hard to answer when the given tiling method was generalized to the 2n-gon. But implicit in the problem was the fact that *any* such rhombus tiling of a 2n-gon uses the same number of rhombi. I've since found the answer proved in passing in a book, but y'all, or y'some, may find this an interesting puzzle. (Also, I suspect there must be a more straightforward proof than the one I happened upon.) PUZZLE: Given a rhombus tiling of a 2n-gon (so that each polygon edge is a rhombus edge), show that the number of rhombi is determined by n. --Dan
asimovd@aol.com wrote:
...PUZZLE: Given a rhombus tiling of a 2n-gon (so that each polygon edge is a rhombus edge), show that the number of rhombi is determined by n.
The question of counting how many solutions there are, especially in the 3D case of filling a zonohedron with parallelepipeds, is an interesting open question. For a large model and some references, see: G. Hart, "A Color-Matching Dissection of the Rhombic Enneacontahedron", Symmetry: Culture and Science, vol. 11, 2000, pp. 183-199. online at: http://www.georgehart.com/private/hart-v1.doc George http://www.georgehart.com/
On Wed, Jun 11, 2003 at 06:34:03PM -0400, George W. Hart wrote:
asimovd@aol.com wrote:
...PUZZLE: Given a rhombus tiling of a 2n-gon (so that each polygon edge is a rhombus edge), show that the number of rhombi is determined by n.
The question of counting how many solutions there are, especially in the 3D case of filling a zonohedron with parallelepipeds, is an interesting open question. For a large model and some references, see: G. Hart, "A Color-Matching Dissection of the Rhombic Enneacontahedron", Symmetry: Culture and Science, vol. 11, 2000, pp. 183-199. online at: http://www.georgehart.com/private/hart-v1.doc
Any chance of a version that is not in .doc format (e.g., PostScript)? I don't have Microsoft Word, and for some reason the pictures of the sculptures didn't come out in my version. On trying to count efficiently: I wonder if it's possible to apply your technique of picking an equatorial band repeatedly, picking, say, two different directions and looking at all the possibilities for the two cases. Of course the two will interact, but maybe it's possible to control that. On the other hand, isn't it possible to think of a decomposition of a 3-d zonotope into parallelipipeds as a sequence of decompositions of a 2-d zonotope into parallelograms? I guess that's what this paper: http://www.liafa.jussieu.fr/~latapy/Publis/Pav/abstract http://www.liafa.jussieu.fr/~latapy/Publis/Pav/Pav.ps.gz is about. I'm surprised that it's not possible to extend the sequence at least one more step. Peace, Dylan Thurston
Dylan Thurston wrote:
..."A Color-Matching Dissection of the Rhombic Enneacontahedron"...
Any chance of a version that is not in .doc format (e.g., PostScript)?
I just put an html version at: http://www.georgehart.com/dissect-re/dissect-re.htm
On trying to count efficiently: I wonder if it's possible to apply your technique of picking an equatorial band repeatedly, picking, say, two different directions and looking at all the possibilities for the two cases. Of course the two will interact, but maybe it's possible to control that.
I don't see how to work out the details of this.
On the other hand, isn't it possible to think of a decomposition of a 3-d zonotope into parallelipipeds as a sequence of decompositions of a 2-d zonotope into parallelograms? I guess that's what this paper:
http://www.liafa.jussieu.fr/~latapy/Publis/Pav/abstract http://www.liafa.jussieu.fr/~latapy/Publis/Pav/Pav.ps.gz
Latapy didn't realize when he wrote his papers that the combinatorial form of the zonotope affects the answer. For example there are four combinatorially distinct 6-zone zonohedra, including a polar zonohedron (which has two vertices of order 6) and Kepler's rhombic triacontahedron (which doesn't). The number of dissections is different for these but Latapy assumed it only depended on the number of zones. Thus his entries in the EIS are incorrect/incomplete. I corresponded with him a year ago about this. He said:
Actually, I do not work on these topics anymore.
You might want to go back and clarify the information in Sloan's database. I certainly have to ! I'll try to do this soon.
George http://www.georgehart.com/
On Thu, Jun 12, 2003 at 08:56:46AM -0400, George W. Hart wrote:
Dylan Thurston wrote:
..."A Color-Matching Dissection of the Rhombic Enneacontahedron"...
Any chance of a version that is not in .doc format (e.g., PostScript)?
I just put an html version at: http://www.georgehart.com/dissect-re/dissect-re.htm
Thanks! I like the sculpture.
... I guess that's what this paper:
http://www.liafa.jussieu.fr/~latapy/Publis/Pav/abstract http://www.liafa.jussieu.fr/~latapy/Publis/Pav/Pav.ps.gz
Latapy didn't realize when he wrote his papers that the combinatorial form of the zonotope affects the answer. For example there are four combinatorially distinct 6-zone zonohedra, including a polar zonohedron (which has two vertices of order 6) and Kepler's rhombic triacontahedron (which doesn't). The number of dissections is different for these but Latapy assumed it only depended on the number of zones. Thus his entries in the EIS are incorrect/incomplete. I corresponded with him a year ago about this. He said:
Actually, I do not work on these topics anymore.
You might want to go back and clarify the information in Sloan's database. I certainly have to ! I'll try to do this soon.
I think you should go ahead and fix the entries in the EIS yourself. And add a link to your paper. Peace, Dylan
PUZZLE: Given a rhombus tiling of a 2n-gon (so that each polygon edge is a rhombus edge), show that the number of rhombi is determined by n.
Hmm. I don't think you need rhombi here, right? This is just a statement about any tiling by parallelograms. But you need to be a little careful that you're discussing *convex* 2n-gons -- otherwise, when you place a single tile down in your 2n-gon, you have a new (concave) 2n-gon which you can cover with one fewer tile :-). Okay, spoiler below. Or maybe spoiler above, if the idea that it's true in parallelogram generality gives it away... Fundamentally, parallelograms propagate edges: if you have an edge of your original 2n-gon with a given length and slope, then the parallelogram that uses that edge will leave you with a translation of that edge, and so on -- the only way you can ever have a tiling is if you eventually reach a matching edge, all the way across the 2n-gon. So in particular, the edges of the original 2n-gon come in matched pairs. Now if the 2n-gon is convex, then the parallelo-path from one edge to its mate has to cross the path from any other edge to its mate, since the slopes of the edges determine their cyclic order. Each one of those crossings is exactly one parallelogram. So there are n-choose-2 parallelograms, and their shapes (including orientation!) are exactly given by taking pairs of the original n distinct types of edges. Cool stuff! --Michael Kleber kleber@brandeis.edu
participants (4)
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asimovd＠aol.com -
Dylan Thurston -
George W. Hart -
Michael Kleber