Projection method for 5-fold rotational symmetry

I am looking again at projection methods. In the post “breaking the 10-fold rotational symmetry” I have destroyed the 10-fold rotational symmetry of the five sets of stripes by shifting their positions back or forth. This gave a five-fold rotational symmetry. Searching for ten-pointed stars as before, I then got quasiperiodic designs with five-fold rotational symmetry for some special values of the shift. However, it is strange to look for ten-pointed stars if one wants a five-fold rotational symmetry.

Searching for Pentagrams, which are five-pointed stars, seems to be more natural. We use sets of lines similarly as in “projection method for 10-fold rotational symmetry“, but now we want a simple n-fold rotational symmetry for n sets of parallel lines. If n is an odd number, such as n=5, we use an angle of 2*Pi/n (that is 360/n degrees) between two sets of lines. As before, the distance between parallel lines is equal to one. If some line of each set passes through the origin we still get a line pattern of 2n-fold rotational symmetry. Similarly as before, we have to break this symmetry by shifting the lines by a distance s away from the origin. This gives us a star with n points (a pentagram for n=5) around the origin and a n-fold rotational symmetry. Actually this gives for odd n the same sets of lines as in “breaking the 10-fold rotational symmetry“. But searching for n-pointed stars we get different tiling points.

You will find more details in the code of the next post. It produces the result shown here for n=5 and s=0.42. You can get very different result with only small changes in s !

oddFiveHere too, the center of the perfect 5-fold symmetry lies at the upper right corner. It is the pentagon surrounded by five rhombs and pentagons.



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