Pascal's Triangle and The Enneacontahedron

[kirby urner]

Best to archive this somewhere.  Low bus numbers 'n all.

Kirby



---------- Forwarded message ----------
From: David Koski <dbk...@bitstream.net>
Date: Tue, Nov 18, 2014 at 3:52 AM
Subject: Re: Pascal's Triangle and The Enneacontahedron
To: coyote_starship urner <kirby...@gmail.com>


No just a quick dash off

On Nov 17, 2014, at 11:07 PM, kirby urner <kirby...@gmail.com> wrote:

Did you post this as a followup to your Poly list polyhedron of two days ago?  I think it might be useful to them too.

Kirby

On Mon, Nov 17, 2014 at 9:06 PM, David Koski <dbk...@bitstream.net> wrote:

I know it is confusing for me to use 10 unique directions instead of the 20 that correspond to an icosahedron.  But opposite faces or strut projecting out in the opposite direction do not count.

We will be referencing the webpage:

http://www.orchidpalms.com/polyhedra/rhombic/rh90/rh90.

The 10 unique directions that are based on the orientation of the faces of an icosahedron generate an Enneacontahedron.  Using the parallel projection process we can construct the enneaconathedron in a number of unique ways.

This distribution is fully expressed quantitatively on the 10 the row of Pascal’s triangle:

1,10,45,120,210,252,210,120,45,10,1

On the Enneacontahedron and its Relations webpage we see various unique combinations starting at 3 axes all the way to 10 axes, that latter being the enneacontahedron, itself.

We cans correspond all eleven numbers on the 10th of Pascal’s Triangle (PT) to the webpage by adding axes 0,1, & 2

Axes  Unique forms PT#  

0 1 1

1 1 10

2 2 45

3 5 120

4 7 210

5 8 252

6 7 210

7 5 120

8 2 45

9 1 10

10 1 1 

For example 2 axes has two unique forms the fat and thin rhombus, 3 axes 5 unique forms

What is the quantitative quality or how many of each unique forms for a given PT # 

We find that there are twice as many fat as thin rhombi on the Enneacontahedron 30 thin + 60 fat = 90 rhombic faces

So, there are 15 thin and 30 fat making the PT# 45

The PT# is 45, and  we find that we only need  a sample of 9, since 9/45 is 21/5 or 2/10

AB = thin, AC = thin, AD = thin, AE = fat, AF = fat, AG =fat, AH = fat, AI = fat, AJ = fat

3 thin and 6 fat, multiply both by 2/10 equal 15 thin and 30 fat 

We also know that there are Baer (ABCD&E) cells named after the 5 rhombic hexahedron shown with 3 axes, and that the distribution is:

10 A, 20 B, 30 C,D & E which add up to 120 the 4th PT#

The distribution can be predicted by a sample of only 3/10 or 36/120

ABC  = D ACD = D  ADE = E AEF = B AFG = B AGH = E AHI = B  AIJ = B

ABD  = D ACE = C ADF = C AEG = B AFH  = A AGI = A AHJ= B

ABE = D ACF = D  ADG  = D AEH = C AFI =C AGJ = C

ABF = E ACG = E ADH = E AEI = E AFJ = E

ABG = C ACH  = D ADI = D AEJ = A

ABH = C ACI = C ADJ = C

ABI = E ACJ = E

ABJ = D

Simple accounting nets : 3 A, 6 B, 9 C, 9 D, 9 E

Since this is 3/10 we get the full amount as listed about

For 4 unique directions we need a sample 84/210 or 2/5.  And for 5 directions 126/252 or 1/2

Note that row 9 of PT is :  1,9,36,84,126,126,84,36,9,1

row 9/row 10:  1/10, 9/45 or 2/10, 36/120 or 3/10, 84/210 or 4/10, 126/252 or 5/10, 126/210 or 6/10, 84/120 or 7/10, 36/45 or 8/10, 9/10, 1/1 or 10/10