Richard Guy writes:

> The case n = 7 mod 8 is left to the reader.

I think odd-sized pyramids are impossible, by a 4-coloring:

layer 1: A

layer 2: B
C D

layer 3: A
D C
B A B

layer 4: B
C D
A B A
D C D C

layer 5: A
D C
B A B
C D C D
A B A B A

layer 6: B
C D
A B A
D C D C
B A B A B
C D C D C D

layer 7: A
D C
B A B
C D C D
A B A B A
D C D C D C
B A B A B A B

layer 8: B
C D
A B A
D C D C
B A B A B
C D C D C D
A B A B A B A
D C D C D C D C

and so on. A 4-sphere "tetrad" will have one of each color
regardless of where it is in this grid. To be solvable,
any size of pyramid must have an equal number of each color.

From Richard's packing, we know that all even-order pyramids
are solvable, thus must contain an equal number of all colors.

Consider an odd-numbered layer. Parse it into rhombi, as in
this example of layer 7:

A

-------
/D C/
/B A/ B
-------
------- -------
/C D/ /C D/
/A B/ /A B/ A
------- -------
------- ------- -------
/D C/ /D C/ /D C/
/B A/ /B A/ /B A/ B
------- ------- -------

All the rhombi are color-balanced, but the right edge contains
only A's and B's, so the layer as a whole cannot be color-balanced.
So, adding an odd-numbered layer to an even-sided (solvable)
pyramid makes it un-solvable.

-- Mike

Richard Guy <rkg
@cpsc.ucalgary.ca>
Sent by: math-fun-bounces+mbeeler=csc.com

02/09/2004 05:38 PM
Please respond to math-fun

To: Daniel Asimov <dasimov@earthlink.net>, math-fun <math-fun@mailman.xmission.com>
cc:
Subject: Re: [math-fun] Two questions in geometric combinatorics

A sufficient condition is n even.
For n = 2 it's trivial. For larger
even n, partition the tetrahedron
into pairs of layers, of edge 2k
and 2k-1, 0 < k < (n+2)/2. E.g., k = 3
with the b-layer sitting on the a-layer.

/a\
/ b \
/a a\
-------
\b b/
/a\ \ a / /a\
/ b \ \b/ / b \
/a a\ /a a\
------- -------
\b b/ \b b/
/a\ \ a / /a\ \ a / /a\
/ b \ \b/ / b \ \b/ / b \
/a a\ /a a\ /a a\
------- ------- -------

The case n = 7 mod 8 is left to the
reader. R.

On Sun, 1 Feb 2004, Daniel Asimov wrote:

> ----- Original Message -----
> From
> To: <math-fun@mailman.xmission.com>
> Sent: Sunday, February 01, 2004 2:24 PM
> Subject: Re: [math-fun] Two questions in geometric combinatorics
>
> -------------------------------------------------------------------------
> Dylan asked:
> <<
> On Fri, Jan 30, 2004 at 05:17:01PM -0500, asimovd@aol.com
> wrote:
> <<
> Let A_n denote a regular tetrahedral arrangement of points in
> 3-space, n to an edge. (The layers successively contain
> 1, 3, 6,... points.)
> >>
>
> Can you be more explicit what this arrangement is? I am
> completely baffled.
> >>
> ---------------------------------------------------------------------------
>
> The arrangement A_n of points in 3-space is just a regular tetrahedron's
> worth of the face-centered cubic lattice (from the triangular viewpoint).
> I.e., starting from the top, the (k)th layer is an k-sided equilateral
> triangle
> of k(k+1)/2 points arranged as in the triangular lattice. The layers are
> spaced so each point's nearest neighbors are (let's say) exactly one unit
> apart from it.
>
> The problem (to repeat) is, For which n can A_n be partitioned into disjoint
> "tetrads", where a tetrad here means 4 mutually nearest points) ? An
> obvious
> necessary condition is that 4 divides #(A_n), which happens exactly when
> n = 0,2,4,6, or 7 (mod 8).
>
> --Dan
> _______________________________________________
> math-fun mailing list
> math-fun@mailman.xmission.com
> http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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