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. 

Tiling the plane

 

While at GMU, Tyler White worked with me on his senior thesis. He then went on to get a PhD at George Washington University on mathematical questions for certain tilings in the plane. He asked if I could print a set of these three different tiles that appear in his PhD thesis. It looks really cool because though it is two-dimensional, it is a projection of a three-dimensional object, so gives the impression of depth. In fact, this picture above is so misleadingly three-dimensional looking that I had to print another one with my hand in it to make it clear that the shapes really are flat. 

The way I arranged it is boring from a mathematical standpoint, since it is completely uniform. In his thesis, he studied aperiodic tilings using this set of tiles. I've asked him to explain the details. Here's a portion of an aperiodic arrangement: 

Here's the full figure from the PhD thesis: 

Here is Tyler White's full explanation of what this represents:

This tiling is part of an infinite tiling of the plane.  This tiling was originally constructed by Richard Kenyon in his 1996 paper called “The Construction of Self-Similar Tilings” published in the journal of Geometric and Functional Analysis.  This tiling is aperiodic in the sense that if a fixed reference point is chosen, and then the tiling is shifted in any direction, it is impossible to have the exact same tiling you started with.  Kenyon’s goal in the paper was to demonstrate the construction of a self-similar tiling (this tiling is only pseudo self-similar) by using a generalized substitution.  Though the tiling is aperiodic, it is generated in an algorithmic manner by a tiling substitution.  As a result, the tiling as a certain amount of regularity to it.  In fact, for any pattern found in the tiling, there is always another copy of the pattern within a fixed finite distinct (the distance you would have to look depends on the size of the pattern you are looking for), this property is known as bounded gaps.  Former George Mason University undergraduate Mathematics major Tyler White in his 2012 Ph.D. dissertation in Mathematics at George Washington University under the direction of E. Arthur Robinson, Jr. showed that under certain conditions (the shown tiling satisfies these conditions) the tiling is also topologically mixing, which is a stronger type of regularity conditions than the bounded gap property mentioned before.