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Solving a Penrose puzzle

The Penrose tilings, P1, P2, and P3 have many different solutions. It could be different every time, although you are likely to see strong similarities in the patterns. Chet and Esther bought our P1 at the the Atlanta American Craft Council show, and wanted to know if there was a solution that used all the tiles. They sent us this image:

Because there are multiple solutions, we don’t select the number of tiles based on a particular solution. Instead, we choose the number of pieces based on the proportions of tiles that would appear if you tiled the entire, infinite plane. The idea is that on average, you can use most of the tiles. For Penrose P2 and P3, the larger tiles appear more often, and in the ratio of the golden mean, which is about 1.618 to 1. So, for example, if we put 100 smaller tiles in a box, we would put 162 larger tiles in the box.

For P1, it is a bit more complicated, since there are 6 different tile shapes: 3 are based on pentagons, one is a thin rhomb, one is based on a star shape (called a pentacle, and the other is part of a start (called a half-pentacle). Using the colors from Chet and Esther’s set, as a puzzle is expanded, the proportions will converge to 3.8% for the pentacle, 8.6% for the half-pentacle, 13.5% for the rhombus, 10.9% for the red pentagon, 27.8% for the green pentagon, and 35.4% for the blue pentagon.

So is there a solution that uses all the tiles? Maybe, but we don’t know!