Assume you've settled on some particular tile set (e.g. "Kites and Darts"). Then of the continuum of globally distinct but locally congruent Penrose tilings, just two (modulo rotation) have pentagonal (D_10) symmetry. Specifically, these are the infinite inflations of Conway's "Sun" and "Star" configurations, where 5 kites or 5 darts resp. meet at a central vertex. More significantly though, the (infinitely inflated) Cartwheel has almost decagonal (D_20) symmetry: the only non-conforming subset comprises 10 semi-infinite worms radiating like spokes from its centre. So the fundamental rotational symmetry of any Penrose tiling is 10-fold, rather than 5-fold. Fred Lunnon
* Fred lunnon <fred.lunnon@gmail.com> [Apr 30. 2010 20:23]:
Back around 1980, a group of us at Cardiff (John Rigby, Peter Pleasants, and myself) embarked on an attempt to generalise the Penrose tiling to 7-fold symmetry, in the course of which a number of apparently promising designs were produced. All proved to be what I christened at the time "doilies", with dihedral symmetry D_{2k}, but only about the origin. If I've interpreted your construction correctly, what you have here is (yet another) doily.
What is remarkable about the Penrose tiling is not its pentagonal (indeed --- though less obviously --- decagonal) dihedral symmetry alone, but the combination with "quasi-crystallographic" symmetry. This last is not trivial to pin down (I can dig up a reference if required), but it broadly involves maximally dense repetition of any finite region: so the tiling has as nearly as possible translation symmetry, without actually attaining it.
John Rigby in particular was not amused by our persistant shifting of the goalposts whenever he produced a beautifully-executed drawing of his latest candidate; but the trouble was that we were still trying to establish where the goalposts should be. I think that's clearer these days; but I don't think we ever managed definitively to exclude the possibility of 7-fold quasi-crystallographic symmetry in the plane.
Fred Lunnon
On 4/30/10, Joerg Arndt <arndt@jjj.de> wrote:
I always thought that 5-fold symmetry can only be obtained by sophisticated constructions (as with Penrose tiles, see the image on the top right of http://en.wikipedia.org/wiki/Penrose_tiling )
Now a childishly simple contruction gives 5-fold symmetry with just one tile:
Our tile is a triangle:
/\ / \ /====\ The two angles at the base shall equal 2*Pi/k for k>=5. (Drawing obviously not to scale).
Assemble tiles as follows:
/\====/\====/\====/\== / \ / \ / \ / \ etc. ad inf. /====\/====\/====\/====\/
This is "plank A"
Now arrange planks into an infinite wedge:
/ /\====/\ etc ad inf. to right and top / \ / \ / \/ /\====/\====/\====/\== / \ / \ / \ / \ / \/ \/ \/ /\====/\====/\====/\== / \ / \ / \ / \ / \/ \/ \/ /\====/\====/\====/\=== / \ / \ / \ / \ /====\/====\/====\/====\/==
This is "wedge A"
Arrange k such wedges around the origin to fill the plane. This arrangement has k-fold rotational symmetry but (except for k=6) no translational symmetry.
Rotational symmetry can be avoided by taking for one wegde the "wedge B" obtained by reflecting wedge A over its horizontal base (we still use one tile only, reflection of tiles is obtained by rotation them by 180 degrees). Wedge B cannot be transformed to wedge A by a rotation.
This is surely known since about Silurian years, but why doesn't it appear anywhere, e.g. in http://tilings.math.uni-bielefeld.de/ ?
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun