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.
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