Pentagon tiling by Hope Roberts

The following post is by Hope Roberts as a part of George Mason University Math 493, Mathematics Through 3D Printing.

           

PENTAGONAL TILES

Hope Roberts

What Is Pentagonal Tiling?

Pentagonal tiling is a periodic tile method that covers the plane in such a way that there are no gaps.  Every individual piece of the tiling is part of a pentagon. 

In order to cover the plane with no gaps with a pentagon there are certain symmetries, angles and side length requirements that must be fulfilled.

Measurements of my constructed tile.

Brief History

Reinhardt discovered the first five of the convex shapes in 1918,

Kershner discovered the next 3 in 1968 who incorrectly stated he had completed the list. The tenth type is credited to James in 1975 and shortly after in 1977 a stay at home housewife with a fascination for math named Rice found four more. There was a lull with the fourteenth not being found till 1985. A fifteenth was found in October of 2015 by Mann/McLoud/Von Derau (2015) from computer algorithm.

The reflections the tile type 5.

The Tiles in General

The tiles are restricted by the internal angles and requirements of the edges having to be certain lengths.  The tiles also have certain wallpaper group symmetries and some of the symmetries may repeat in different tiles. Because the tile tessellates there are no spaces, which means that the internal angles must be designed in a way that all the pieces of the pentagon fit together. The pentagon has internal angles that add up to 540 degrees so when finding a new type of pentagon tiling the angels must be arranged so that they are not greater than 540 degrees which means that some of the edges are dependent upon this fact. Certain sides may also have different requirements of length.

My tiles in 6-tile primitive unit.

About My Tile

The tile I chose is type Monohedral convex pentagonal tiling #5. It was discovered by Reinhardt and was amongst the first of several tiles to be discovered. This tile differs from the other tiles such as p6(*632) in that it does not have glide reflections. This is one example of how the reflections for this tile are particular to this tile making it different from the others.  The tiles in general differ in the requirements of the angles and side lengths that they are restricted to. In the wallpaper group symmetry my tile is P6 (632) symmetry. This symmetry has one rotation center of order six differing every 60 degrees. There are two centers of rotations of order three differing by 120 degrees and three centers of rotations of order two differing by 180 degrees.

The angle and sides requirements of this tile are given by the equation;

 a=b d=e A=60 D=120 (B+C+E=360)

My angles and sides for my tile are;

Angles: 60, 136, 120, 72, 152

Sides:  a=b=6.14, d=e=5.26, c=6.53

(see picture)

I used a site Math is Fun to construct my tile. I insured first the required angles of 60 and 120 were and the required sides that needed to be equal.  After doing this the remaining sides and angles were given so that I did not need to further manipulate.

This pattern does not have reflections or glide reflections.

The tiling is periodic which means that there is translational symmetry and the pattern will repeat.

SOURCES;

https://en.wikipedia.org/wiki/Wallpaper_group#Group_p6_.28632.29

https://www.mathsisfun.com/geometry/pentagon.html

Pentagonal Tiling Type 3 - Nicole Van Oort

The following post is by Nicole Van Oort as a part of George Mason University Math 493, Mathematics Through 3D Printing.

    Nicole Van Oort-Pentagonal Tiling Type 3

The tile I created (shown above) I used to create a pentagonal tiling. There are 15 convex pentagons known, that tile the plane monohedrally (meaning using the one pentagon, the plane can be tiled infinitely). A regular pentagon on the other hand cannot be used as a tessellation. The pentagon tile I created is a type 3 convex pentagon. This type, along with the other first five, were found by German mathematician in 1918, Karl Reinhardt. Type 3 pentagons consist of 3, 120 degree angles and two reaming angles that summate to 180 degrees. The two supplementary angles are separated by one 120 degree angle on one side, and the remaining two 120 degree angles on the other side. Type 3 convex polygons also contain two adjacent sides that are congruent. The two congruent sides from the point of the pentagon, adjacent on both sides to the two supplementary angles. For a more vivid idea of what Type 3 pentagons look like, resource the image above obtained from Wikipedia’s article on pentagonal tiling.

      The Type 3 pentagonal tessellation I choose to create, illustrates the relationship between hexagons and pentagons, and the ability of a hexagon to be subdivided into 3, type 3 pentagons (all congruent). This therefore allows a pentagonal tessellation to be created in which an overlay of two regular hexagons of different sizes are tessellated repeatedly over the plane, thus tessellating the type 3 pentagon as well. As seen through the representative image of an example of this type of tiling below, using a hexagon as the lattice, the pentagon is rotated, reflected and translated throughout the plane and thus is a representation of a p6m, group 17 wallpaper group.

     To create this tessellation, the first step was to create a regular hexagon (all with the same side length-I chose 10). Then to dissect the hexagon into its 3 type 3 pentagons, I had to find the midpoint of the hexagon in which all 3 pentagons would converge to. Thus the 3 pentagons were formed by drawing a line from the midpoint of the top edge to the midpoint of the hexagon, and from the midpoint of the inferior two side edges to the midpoint of the hexagon. This will create a convex pentagon that is rotated twice around to form a hexagon. The difficulty was maintaining two congruent sides for each pentagon. The distance from the midpoint of the hexagon, straight up to the top edge of the hexagon, needed to be equal to the distance from the midpoint to any hexagon edge’s midpoint. In other words, the lines drawn inward to the hexagon’s midpoint were supposed to form the congruent sides of the pentagons. The website where I found my tiling, Wikipedia, does not explicitly state this so it took me a few tries to realize why my pentagons were not congruent and then took even more manipulating of numbers to get them to be. I had decided on side length 10 for my pentagon, with a center horizontally of zero. To find the proper midpoint, I used the idea that a hexagon can also be divided into equilateral triangles. Since an equilateral triangle would also use an edge of the hexagon as its side, it too would have side length 10. Therefore to find the midpoint (the height from the lower edge midpoint to the actual midpoint), I used Pythagorean’s theorem. Taking my equilateral triangle and dividing it in half gives me a right triangle whose vertical side in the height of the midpoint. Thus since the side length needs to be 10 and horizontal side is half of that (5), my formula resulted in 5^2+y^2=10^2, allowing me to figure out the height of the midpoint to be 8.6602…

    Allowing my three pentagons to converge on the proper midpoint guaranteed they were all congruent. Finally to finish the underlying hexagon, I used the congruent pentagon I had created and translated it, mirrored it, and reflected it to create the visible border of the larger hexagon surrounding the smaller one. These two pieces (or two hexagons together) I then translated as a whole to finish tessellating the plane.