Improved space-filling iteration scheme for Pentagrams.

I wanted to eliminate the gaps appearing in the earlier post “Iteration of pentagrams“. The pentagon that surrounds the pentagram should ultimately be filled up with pentagrams of different sizes.

The six pentagrams of the earlier iteration scheme are shown in dark red. Between the outer five of them there are five gaps with the shape of pentagons. There I am adding five smaller pentagrams, shown in blue.

This can only be done because of an interesting relation between their sizes. The larger (red) pentagrams are smaller than the original pentagram by a factor of

cos(2*pi/5)/[cos(pi/5)]=0.38197.

For the smaller (blue) pentagrams the factor is

1-cos(2*pi/5)/[cos(pi/5)]=0.23606.

The third and second powers of these factors have the same value

0.38197*0.38197*0.38197=0.23606*0.23606=0.05572

Result of two iterations.

and thus there are essentially only two different pentagram sizes at each iteration. Without, the number of different sizes would double at each iteration step, resulting in chaos. Well, this has been a hard bit of maths and combinatorics. It is time for some results.

Four iterations.

From a small part of this image we get a finite part of a quasiperiodic tiling with five-fold symmetry. It is essentially a special decoration of the Penrose tiling.

Penrose tiling resulting from this iteration of pentagrams.

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