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.

Integral Calculus 4

Here is another shape to find the volume for. The shape has a base which is a disk of radius 1, and each cross section is an equilateral triangle.

View of the base

Top view

Integral Calculus 3

Here is another shape for which calculus allows us to compute volume. It is a shape for which the base is the region bounded by y=4-x^2 and the x-axis

and every vertical cross section is a square:

Integral Calculus 2

This object illustrates the disk and shell methods for finding volumes of rotation using integration. Take the region bounded by the graph of the function f(x) = -x^2 - x + 8 and the x-axis from x=0 to x=2:

and rotate it about the y-axis. In the middle is the resulting solid of revolution 

On one side is the approximation of the volume using four disks, and on the other side is the approximation of the volume using four shells. To make it clear which is which, I have shown the same model with the pieces taken apart.

 This model is "Volumes of Hanoi" designed by mathgrrl.

Integral Calculus

I am preparing some integral calculus prints. The first is the solid obtained by revolving the region between the graph of y=4-x^2 and the x-axis on [0,2] around the y-axis, and the model on the right is an approximation of that solid using eight shells, designed by mathgrl.

Here is a view of all the calculus volumes collection. 

Four volumes

Pencil for scale

Tactile graphs for a blind math student

The last few weeks, I have been working with a colleague producing raised graphs for a blind student who is taking a math class, but who has never really been taught the details of plotting.  It has been a very interesting experience learning a few details of how to design such graphs.  So far the student has made excellent progress in understanding, since nobody has ever even tried to show this to him  before! The 3D printer makes for an excellent tool, since it allows relatively quick  of graphs tailored for the class, the student, and the situation. After the first few, it has taken 15-20 minutes to prepare a graph for printing, and around one hour to print.

Below the figures of the tactile plots, I have given the workflow and the details of what has worked and what has not. I'd be very interested to hear from anyone who tries this out!

x^2 compared to x^4

ln(x) compared to exp(x)

The top graph is a  polynomial and the bottom is 1/x

x^3 versus x^(1/3) and sin(x) on two different domains

Technical details: In all cases, my general workflow was as follows:

• Step 1. Produce  two JPG figures using Matlab: (1) A grid. I used linewidth 4, with a larger size dot at the origin for reference.  (2) A graph of a function. (Note that you could use any method at all of coming up with a graph and follow the next set of steps still.) 
• Step 2. Convert figures to vector graphics SVG format using the free program Inkscape. 

Step 1 Grid



Step 1 Graph of a function

• Step 3. Open the free program Tinkercad, and add a box approximately 80x80x1 mm.

Step 3 Box in Tinkercad around 80x80x1mm

• Step 4.  Import the SVG file of the grid into Tinkercad. It needs to fit on the box from the previous step. (In my case this involved shrinking to 8%.) It should be 2-3mm high so that it can be distinguished from the background.

Step 4 add the grid to Tinkercad

• Step 5. Import the SVG file of the function into Tinkercad. It needs to be exactly the same size as the grid so it matches up, so if you shrink one to 8%, make sure to shrink the other to 8%. However, for the graph, make it 1-1.5mm taller than the grid. At first I made it too tall, and the graph and the grid couldn't be felt simultaneously. 1mm sound small, but think about the thickness of Braille, and it won't seem so mind boggling.

Step 5 Add the graph to Tinkercad but make it 1-1.5mm taller than the grid. 

• Step 6.  Download for printing, using  file type STL.



Step 6 download for 3D printing

• Step 7. Print on printer, with a raft but without support. It takes around 1 hour to print each.