The letters, A to L, within this star represent intersections of unique pairings of its 6 lines, and α, β, γ, δ, ε and ζ are sums of intersections defined as:

α = A + D + G + K     β = E + G + J + L    γ = K + J + I + H
δ = L + I + F + B     ε = H + F + C + A    ζ = B + C + D + E

           A α
          / \
  ζ  B---C---D---E  β
      \ /     \ /
       F       G
      / \     / \
  ε  H---I---J---K  γ
          \ /
           L  δ

Assign values from 1 to 12 to each of the locations A to L such that each sum is an element of an arithmetic progression with an arithmetic difference of two (2) but not necessarily as adjacent vertex values.

Secondly, attempt the same task but with a difference of four (4) as the outcome.

And for a tease... can you offer a solution if all such vertex sums are equal, ie, 26?

Note:
Discounting rotations and reflections, more than one possibility exists for each of the first two tasks.

26, that is the sum of the numbers in each line, can be partitioned as follows:

with 12 as greatest number: (12, 11, 2, 1) / (12, 10, 3, 1) / (12, 9, 4, 1) / (12, 9, 3, 2) / (12, 8, 5, 1) / (12, 8, 4, 2) / (12, 7, 6, 1) / (12, 7, 5, 2) / (12, 7, 4, 3) / (12, 6, 5, 3)

with 11 as greatest number: (11, 10, 4, 1) / (11, 10, 3, 2) / (11, 9, 5, 1) / (11, 9, 4, 2) / (11, 8, 6, 1) / (11, 8, 5, 2) / (11, 8, 4, 3) / (11, 7, 6, 2) / (11, 7, 5, 3) / (11, 6, 5, 4)

with 10 as greatest number: (10, 9, 6, 1) / (10, 9, 5, 2) / (10, 9, 4, 3) / (10, 8, 7, 1) / (10, 8, 6, 2) / (10, 8, 5, 3) / (10, 7, 6, 3) / (10, 7, 5, 4)

with 9 as greatest number: (9, 8, 7, 2) / (9, 8, 6, 3) / (9, 8, 5, 4) / (9, 7, 6, 4)

with 8 as greatest number: (8, 7, 6, 5)

Now, all we have to do is find six 4-plets (!!) that use 2 times each number.

I started with (12, 11, 2, 1) and (12, 10, 3, 1). Then, in the 11-greatest number list, I took (11, 8, 4, 3). Next in the 10-list, I took (10, 9, 5, 2). In the 9-list I took (9, 7, 6, 4) and finally the (8, 7, 6, 5).

In these six 4-plets, each number appears twice. Now, just place the 4-plets properly in the diagram, and you have one solution, of many.

Edited on June 2, 2008, 12:05 pm

Edited on June 2, 2008, 6:58 pm

Posted by pcbouhid on 2008-06-02 12:03:21