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A Tri Star Issue (Posted on 2008-06-01) Difficulty: 3 of 5
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.

See The Solution Submitted by brianjn    
Rating: 4.0000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Some Thoughts Some thoughts (without reading Charlie´s comments) | Comment 8 of 17 |

The sum of the numbers from 1 to 12 is 78. Since each one appears twice in the diagram, we have:

a) for the tease: no need to say that the sum must be equal 26. It has to be. The total sum is equal to 2 x 78 = 156, so each line MUST sum 156/6 = 26.

b) for the arithmetic progression with difference 2: call the sum of the lines (x-5), (x-3), (x-1), (x+1), (x+3), (x+5). Total = 6x = 156.... x = 26. So the lines MUST sum: 21, 23, 25, 27, 29, 31.

c) for difference 4: lines sum equal to (x-10), (x-6), (x-2), (x+2), (x+6), (x+10)....... Total = 6x...... x= 26. The lines MUST  sum 16, 20, 24, 28, 32, 36. 

 


  Posted by pcbouhid on 2008-06-02 09:30:49
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