At 05:17 PM 1/30/04, 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.) Define a "tetrad" to be any four mutually adjacent points of A_n.
QUESTION: For which n is it possible to partition A_n into disjoint tetrads ???
Call such an n "tetrizable". Clearly, any such n = 0 (mod 4).
You mean #(A_n) = 0 (mod 4).
It's not hard to show that the number of points #(A_n) is divisible by 4 exactly when n = 0,2,4,6, or 7 (mod 8).
QUESTION: More concretely, n = 2 is the first tetrizable number. What is the next one (if any) ???
Every even n is tetrizable, every odd n is not. Given a partition for some even n - 2, you can obtain one for n by partitioning the two added layers in a fairly obvious way, illustrated here for n = 6: 1 / \ 1 1---1 7---7 2 7 3 \ / / \ / \ 2 7 3 2---2 3---3 8---8 9---9 4 8 5 9 6 \ / \ / / \ / \ / \ 4 8 5 9 6 4---4 5---5 6---6 On the other hand, examining a drawing that I can't reproduce here makes it clear that any point on an edge of the big tetrahedron can only be in a tetrad that contains another point on the same edge. (In fact, the four adjacent points not on the edge form a square.) This pairing means that the number of points on the edge must be even. -- Fred W. Helenius <fredh@ix.netcom.com>