[seqfan] Re: Slicing a tetrahedron

[Veikko Pohjola]

I give an example to show what we are actually hunting here.

Take a tetrahedron sliced twice, that is, by planes dividing the edges into 
n=3 equal parts. Let it stand on its face such that it can be visualized to 
have three horizontal layers. We have now 1 big tetra (the original one), 4 
partially overlapping midsize tetras, 3 of which stand on the base level and 
1 one level up, and 10 small tetras having the same orientation as the 
original one. The total number of all these tetras is thus 15. This is the 
number of our interest at n=3. The whole sequence is equal to A000332. So, 
not too much new so far.

Within the same tetra at n=3 we find also 4 big octas, and 1 big tetra 
having the reverse orientation. The smaller and smaller size octas appear at 
n=4,6,8,. and reverse tetras at n=6,9,12,..

I have constructed the sequences and their closed forms for the number of 
all sizes of tetras having the reverse orientation, for the number of all 
sizes of tetras having any orientation, for the number of all sizes of 
octas, etc. I have not found these sequences in OEIS.
Veikko

----- Original Message ----- 
From: "William Rex Marshall" <w.r.marshall at actrix.co.nz>
To: <seqfan at list.seqfan.eu>
Sent: Sunday, August 05, 2012 9:19 AM
Subject: [seqfan] Re: Slicing a tetrahedron


> On 5/08/2012 9:32 a.m., Ed Jeffery wrote:
>> Veikko,
>>
>> Let T denote the regular tetrahedron with unit edges. Let your slicing of 
>> T
>> corresponding to n be called an "n-slicing." An n-slicing produces n 
>> layers
>> (easiest when viewed parallel to the base of T) in which each layer
>> contains a number of congruent (again) regular tetrahedra t_n whose edges
>> are of length 1/n, and a number of congruent octahedra o_n each with 
>> eight
>> faces matching those of t_n. These pieces are easily counted as follows.
>>
>> Let V_n(T) denote the total number of pieces (counting both types 
>> together)
>> after an n-slicing. For k in {1, ..., n}, the k-th layer contains
>> A000217(k) of the t_n and A000217(k-1) of the o_n, where A000217 =
>> {0,1,3,6,10,15,...} = Triangular numbers. Summing the layers for each 
>> piece
>> type then gives a total of A000292(n) of the t_n and A000292(n-1) of the
>> o_n, where A000292 = {0,1,4,10,20,35,...} = Tetrahedral numbers. Hence
>>
>> V_n(T) = A000292(n) + A000292(n-1) = A000330(n),
>>
>> where A000330 = {1,5,14,30,55,...} = Square pyramidal numbers.
>
> Incorrect, as 10 tetrahedra and 4 octahedra only fill 26/27 of the order-3 
> slicing. (The regular octahedron has a volume exactly four times that of 
> the regular tetrahedron with the same edge length.) You also need to take 
> account of A000292(k-2) inverted tetrahedra in the k-th layer (for k >= 
> 3), so the total number of pieces is actually 1, 5, 15, 34, 65, 111, 175, 
> ... (A006003).
>
> _______________________________________________
>
> Seqfan Mailing list - http://list.seqfan.eu/
>