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Lets have a square with sides equal to a=13. From this square we can construct a dodecagon and an octagon which cover the plane. Please see the diagram below.enter image description here

If you prefer all the polygons to be convex then we have a dodecagon, a square and an isosceles triangle. My question is, is it possible to construct another periodic tiling of the plane with another regular polygon applying the method of Pythagorean triplets?

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Your picture is not very clear, so I don't know what you're doing here. You can inscribe a regular dodecagon in any square. Why 13? –  Robert Israel Jul 23 '12 at 7:15
The square is inside the dodecagon. As for 13, it is just a mathematical curiosity because we have two hypotenuses at the same time: 13 and 5. The numbers on the drawing are 13, 5, 4 and 3. I hope this clarifies. –  Vassilis Parassidis Jul 25 '12 at 3:22
What is unclear for me is what you mean by "the method of Pythagorean triplets." I don't know much about periodic tilings in general, so if there is a good link, that would help also. –  Eric Stucky Jul 29 '12 at 1:23
@ Eric Stucky.Take any regular polygon eg a regular pentagon or an equilateral triangle. Each side of the regular polygon has to have length twice the size of one of the vertical sides plus the hypotenuse of the Pythagorean triple eg 15+15+17=47 or 8+8+15=33. In general the new polygon, which is made from a regular polygon, has to have a triple number of sides all of which have to be equal; the new polygon has to be convex. –  Vassilis Parassidis Jul 29 '12 at 2:08
I hope you can see that there are a lot of lines missing from your diagram. You say "the square is inside the dodecagon", but I can't see any squares! Can't you get a better scan? Otherwise, you could try drawing the diagram on your computer directly, for example using GeoGebra. –  Rahul Jul 29 '12 at 2:31

1 Answer 1

The choice of the $13\times 13$ square and $(3,4,5)$ triangle seems arbitrary to me. It looks like you've slightly modified the regular tiling with squares and octagons by pushing in the sides of the cube. It seems that you would be able to change that $(3,4,5)$ into any other integer-sided right triangle by changing how much the sides of the square are "bent" and then scaling.

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