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Pentagons in tiling

A regular pentagon cannot appear in any tiling of regular polygons. First, to prove a pentagon cannot form a regular tiling (one in which all faces are congruent, thus requiring that all the polygons be pentagons), observe that 360° / 108° = 3 1 ⁄ 3 (where 108° Is the interior angle), which is not a whole number; hence there exists no integer number of pentagons sharing a single vertex and leaving no gaps between them. More difficult is proving a pentagon cannot be in any edge-to-edge tiling made by regular polygons:

The maximum known packing density of a regular pentagon is (5 − 5) / 3 ≈ 0.921 {\displaystyle (5-{\sqrt {5}})/3\approx 0.921}, achieved by the double lattice packing shown. In a preprint released in 2016, Thomas Hales and Wöden Kusner announced a proof that this double lattice packing of the regular pentagon (known as the "pentagonal ice-ray" Chinese lattice design, dating from around 1900) has the optimal density among all packings of regular pentagons in the plane.

There are no combinations of regular polygons with 4 or more meeting at a vertex that contain a pentagon. For combinations with 3, if 3 polygons meet at a vertex and one has an odd number of sides, the other 2 must be congruent. The reason for this is that the polygons that touch the edges of the pentagon must alternate around the pentagon, which is impossible because of the pentagon's odd number of sides. For the pentagon, this results in a polygon whose angles are all (360 − 108) / 2 = 126°. To find the number of sides this polygon has, the result is 360 / (180 − 126) = 6 2 ⁄ 3, which is not a whole number. Therefore, a pentagon cannot appear in any tiling made by regular polygons.

There are 15 classes of pentagons that can monohedrally tile the plane. None of the pentagons have any symmetry in general, although some have special cases with mirror symmetry.