Monday, July 20, 2015

regular polygons, in rings

Looking at the Kepler pentagonal tiling, you may notice the nice looking rings of pentagons around the decagons.


You can also make up other tilings with these rings of pentagons - to get the one below to work you have to sneak in some dented or overlapping pentagons.


But which regular n-gons can form rings like this? You obviously can't do it with a square.


And some regular n-gons, like heptagons, nonagons, decagons, and hendecagons (11-gons) don't work either.


All the angles of the regular n-gon are (n-2)pi/n - so the angles of the polygon in the center would have to be 4pi/n, but for that interior polygon to be a regular polygon itself, there must be some k for which the angle is also (k-2)pi/k. Equating these two values and solving for k gives k = 2n/(n-4). If we look for n that give integer values for k, then we have the n-gons that can form this sort of ring.


Which tells us that only the pentagon, hexagon, octagon, and dodecagon can form a ring around another regular n-gon (the regular decagon, hexagon, quadrilateral, and triangle, respectively). Coincidentally, these are the same polygons that can form a dented pinwheel, as described here.

But what if we skip over another edge (so 2 are skipped over) while forming the ring? We end up getting a star instead of a polygon in the center, and the smallest regular polygon this works for is the heptagon:


With a little bit of work, you may believe that this will work for n that give integer values for k = 2n/(n-6), and this turns out that those n values correspond to the regular heptagon, octagon, nonagon, decagon, dodecagon and octadecagon (18-gon).



As with the first kind of ring, the hendecagon fails:


But we can go further, and skip over another edge (3 now) when forming the ring of polygons. The center is no longer a star, but  a bumpy gear-like polygon, and the smallest regular polygon that can do this is the nonagon:


What other polygons can form this third kind of ring where 3 edges are skipped? Our function is now k = 2n/(n-8), and we get integer values for n = 9, 10, 12, 16, and 24.


Our hendecagons still won't form a ring when skipping 3 edges, but will once we start skipping 4.


If we skip edges when putting the ring together, we can find the number of regularn-gons that will form the ring using the formula k = 2n/(n-2(m+1)), and will only get closed rings when k takes on integer values.

From this relationship we can find out a few things about these rings. For example, for any odd n, where n is 5 or more, we can form a ring by skipping (n-3)/2 edges and have a ring of 2n: for regular pentagons, we skip 1 edge and get a ring of 10, for heptagons we skip 2 edges and get a ring of 14, and for regular hendecagons, we skip 4 edges to get a ring of 22. Another observation: the eminently factorable 12 allows the dodecagon to form rings of 3, 4, 6, or 12.

I was lead to this while playing with regular heptagon, having fun making rings (and rings of rings) like the ones below.



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