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.

(In reply to Can we prove? by pcbouhid)

Say ONE vertex (say vertix A) is the only one which has an odd number in it, the FIVE others vertices, an even number. The two lines that cross in A, have in theirs extremes (H & K) 2 even numbers, so the two inners numbers in each line (C & F in one, D & G in the other) must be one even and one odd (so the sum in the two lines yields the even number, 26). This means that we would have 5 even numbers at the vertices, plus 2 even numbers, as inner numbers. Total: 7 even numbers. Impossible.

If we have THREE odd numbers at the vertices, if they are the vertices of a triangle, say A, H, and K, the 6 inner numbers (C, D, F, G, I, J) must be even numbers (so each line sum to an even number, 26), plus the other 3 even numbers at vertices B, E, L. Total: 9 even numbers. Impossible.

If we have FIVE odd numbers at the vertices, say A, B, E, H, K, (L even), the proof is similar...

Certainly, the proof is more simple than this.

Posted by pcbouhid on 2008-06-03 22:00:08