Infinite Platonic Shapes??

webdnd · 2021-12-12 12:27:55

While working on some random stuff I happened on an interesting property of the platonic solids—that ~~( edges * 2 = sides_of_edges * edges )~~ ( edges * 2 = sides_of_faces * faces ). So I extended it using the following formulas:

number_in_series:

~~sides_of_edges~~ sides_of_faces:

???_units:

z = 2xy

vertices:

v = x(y - 2) + 2 = xy - 2x + 2

edges:

e = z / 2

faces:

f = z / y

I found that with these formulas, shape ( x ) is platonic if ( z mod v = 0 ). Also the number of polygons used for each 'super-vertices' is equal to ( z / v ). In addition to the traditional platonic shapes, there are also an infinite series of dual-polygons that satisfy the requirements to be platonic (top and bottom, but without height), and an infinite series of overlapping lines that would create 3-dimensional shapes without any width or height.

The table for the first 10 triangular shapes are:

Number | Units | Vertices | Edges| Faces | Super-Vertices
-----------------------------------------------------------------
   0   |   0   |     2    |  0   |   0   |
   1   |   6   |     3    |  3   |   2   |      2 (dual-triangle)
   2   |   12  |     4    |  6   |   4   |      3 (tetrahedron)
   3   |   18  |     5    |  9   |   6   |      
   4   |   24  |     6    |  12  |   8   |      4 (octahedron)
   5   |   30  |     7    |  15  |   10  |      
   6   |   36  |     8    |  18  |   12  |      
   7   |   42  |     9    |  21  |   14  |      
   8   |   48  |     10   |  24  |   16  |      
   9   |   54  |     11   |  27  |   18  |      
   10  |   60  |     12   |  30  |   20  |      5 (icosahedron)

The only problem that I have with these tables is when ( y = 1 ). When ( y = 1 ) there are negative values for the vertices and negative values for the super-vertices.

Since I am not a mathematician, and only practice it as a hobby...I was hoping that there might be someone here who could help define a better formula for calculating the vertices, or more generally know a theorem that proves this is geometrically true.

I would also be interested if anyone has any idea how to link this process to the table of simplexes made by pascal's triangle (columns: 'space', vertices, edges, faces, cells, etc...) since a tetrahedron is part of it (at row 4). The parts of a cube ( x = 3, y = 4 ) can also be found by multiplying the values of pascal's triangle by:

value * dimension^(row - column)

with this formula, the square is

dimension = 2, row = 4

and a hyper-cube is

dimension = 2, row = 5

I hope I got all of that right, and thanks to anyone who knows what I meant

Last edited by webdnd (2021-12-15 05:10:56)

Bob · 2021-12-13 01:06:23

hi webdnd

Welcome to the forum.

Since I am not a mathematician, and only practice it as a hobby

I do not agree that <only practice it as a hobby> implies <not a mathematician>

There are plenty of us 'hobbyists' who still think we're mathematicians.

Besides, you have just proposed some new math. That certainly makes you one!

I had a spot of difficulty understanding what 'sides of edges' and 'units' mean. So I started with what I know about tetra, octa and icosahedrons (ie. v, e and f) and worked backwards.

For any solid made from edges and vertices the following is true v + f = e + 2. (Euler I think)

Did you use this to get v = xy + 2 - 2x ? I did!

I'm also unclear what the other entries in that table are. Please explain.

Bob

webdnd · 2021-12-13 13:59:35

Thank you for the welcome Bob, and for the reply.

The original post had the mistake ( sides_of_edges ) instead of ( sides_of_faces ), I also corrected the headers of the table to match typical jargon.

After I made the post I did do some research into formulas used by the platonic solids, and realized that Euler's Law probably had something to do with it because of the ( +2 ) everywhere, but I initially just stumbled on the fact that all platonic shapes had the property ( e * 2 = f * sides_of_faces = units ). I had got ( xy + 2 - 2x ) by simplying ( x(y - 2) + 2 ).

As far as the table goes...

( number ) is an arbitrary incremental value used to find the number of faces and edges, ( units ) are a value that for some unknown reason can be used to determine platonicness, ( super-vertices ) are the number of polygon-faces that unite at one vertices. I have changed ( vertices, lines, and areas ) to ( vertices, edges, and faces ), which have the typical meaning.

I hope that better explains it and corrects some of my mistakes.

Michael

Bob · 2021-12-13 22:34:00

hi Michael,

Thanks, that has helped. I'm still puzzled about what the non-platonic lines in the table represent.

Let's take number 3 for example; V = 5 E = 9 F = 4.

It doesn't 'obey' Euler's relation so it's not a usual solid. There's a thing called a Schlegel diagram that allows a 2D representation of a 3D object so I had a go at drawing it. I put 5 vertices and joined them with 9 edges. I ended up with F = 6 which is what Euler would predict.

There are two more platonic solids: the cube and the dodecahedron. Will these fit into your pattern in some way?

Bob

Bob · 2021-12-13 23:07:11

I'm just making my own version of your table with columns x,y,z,f,e,v.

I've spotted an error in your faces column. You have 4 twice and all the numbers thereafter need to be adjusted.

Applying that to the above post, we get f=6, e=9 and v=5 and 6+5 = 9+2.

I'll see if I can 'draw' the actual shape.

B

webdnd · 2021-12-14 10:40:39

Yes. Obviously my forum posting needs some work , I will try to get it right this time.

Here is the table for the 4-shapes and 5-shapes:

Number | Units | Vertices | Edges| Faces | Super-Vertices
-----------------------------------------------------------------
   0   |    0  |     2    |   0  |   0  |
   1   |    8  |     4    |   4  |   2  |      2 (dual-square)
   2   |   16  |     6    |   8  |   4  |
   3   |   24  |     8    |  12  |   6  |      3 (cube)
   4   |   32  |    10    |  16  |   8  |
   5   |   40  |    12    |  20  |  10  |      
   6   |   48  |    14    |  24  |  12  |
   7   |   56  |    16    |  28  |  14  |      
   8   |   64  |    18    |  32  |  16  |      
   9   |   72  |    10    |  36  |  18  |      
  10   |   80  |    22    |  40  |  20  |


Number | Units | Vertices | Edges| Faces | Super-Vertices
-----------------------------------------------------------------
   0   |    0  |     2    |   0  |   0  |
   1   |   10  |     5    |   5  |   2  |      2 (dual-pentagon)
   2   |   20  |     8    |  10  |   4  |
   3   |   30  |    11    |  15  |   6  |      
   4   |   40  |    14    |  20  |   8  |
   5   |   50  |    17    |  25  |  10  |      
   6   |   60  |    20    |  30  |  12  |      3 (dodecahedron)
   7   |   70  |    23    |  35  |  14  |      
   8   |   80  |    26    |  40  |  16  |      
   9   |   90  |    29    |  45  |  18  |      
  10   |  100  |    32    |  50  |  20  |

As you can see, both the cube and dodecahedron follows the pattern that (units / vertices = whole_number ). Units is still just a means to find out if it is platonic or not. The number column is also directly related to faces, but there's no such thing as solids with only one side (or partial lines when using the formulas).

And the super-vertices are the number of polygons linked at any vertices. For a 3-shape, three form a thetrahedron, four form half of an octohexron, five form one pole of an icosahedron, six would form a flat hexigon (this is usually used to describe a platonic shape).

In this formula, for a 3-shape, 6 can never be achieved because the fraction between it and 5 only becomes more amd more infinitesimal when dividing (units / vertices ). This is also the same with 4-shapes and approaching 4, 5-shapes and approaching 4, and any shapes 6+ approaching 3.

Hope that makes it more clear.

webdnd · 2021-12-15 05:10:20

I finally sorted out why the formula worked and the mystery number I was getting from doubling the lines.

Here are the steps using a tetrahedron where x = 2, y = 3, z = 12, v = 4, e = 6, and f = 4:

1.) Take a hollow tetrahedron and separate each face from one another. This will give a total ( z = y * v = p) points, ( l = 2 * e ) lines, and ( f ) areas.

2.) Now for each vertices to have an equal number of lines at each vertex (defining a platonic solid), all points will have to divide evenly by the original number of vertices ( IF z MOD v = 0). In this case, ( (4 * 3) / 4 = 3 ). This also gives the number of faces that unite at each vertices when the tetrahedron is reconstructed.

The mystery unit was merely the value of expanding the shape out. I just took the liberty of creating a table that showed the the process being done on a number of even-faced shapes of different polyhedra. Even-faced probably because that gave an incremental value to the vertices of the table for triangular polyhedra.

Stumbling onto Euler's characteristic while I was designing the table was just something extra I found.

And so...if it was assumed that a two-dimensional triangle in three-dimensional space has two sides (a front and a back), that triangle would have ( v = 3, e = 3, f = 2 ) and could be deconstructed to give ( z = 6, l = 6 ). This could then be used to determine that each vertices would connect to exactly 2 different faces ( z / v = 2 ). The same would be true for all other dual-polygon shapes, as well as any number of lines overlapping in three-dimensional space.

Thanks for the help.

Last edited by webdnd (2021-12-15 06:57:26)

Bob · 2021-12-16 21:02:31

hi webdnd

This is great!. And I think it is what makes you a mathematician! Welcome to the club.

Bob

pamshaw · 2022-02-27 19:30:28

The discussion here on this link is about the author's calculations during his equations and table calculations. He found some interesting properties in the spiritual and extended them with the listed formula and calculations. But to his dismay, the procedure gave negative values when he used y = 1; the value for vertices and super vertices comes out to be negative. He further asked for linking the data with pascal's triangles.
The discussion on the forum starts with the comments on the statement of his occasional mathematician doing it for fun. The argument proceeds with the complexity of the question and the terms related to the octal and tetra vertices. Further, it was pointed out that this solid doesn't represent any regular solid and does not obey Euler's relation. To understand all this and your calculation, I had to draw 2D images, called Schlegel diagrams, where I made the drawing and placed vertices with nine edges, then I ended up with the F = 6, which Euler's relation would predict.

Math Is Fun Forum

#1 2021-12-12 12:27:55

Infinite Platonic Shapes??

#2 2021-12-13 01:06:23

Re: Infinite Platonic Shapes??

#3 2021-12-13 13:59:35

Re: Infinite Platonic Shapes??

#4 2021-12-13 22:34:00

Re: Infinite Platonic Shapes??

#5 2021-12-13 23:07:11

Re: Infinite Platonic Shapes??

#6 2021-12-14 10:40:39

Re: Infinite Platonic Shapes??

#7 2021-12-15 05:10:20

Re: Infinite Platonic Shapes??

#8 2021-12-16 21:02:31

Re: Infinite Platonic Shapes??

#9 2022-02-27 19:30:28

Re: Infinite Platonic Shapes??

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