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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.

  Submitted by brianjn    
Rating: 4.0000 (2 votes)
Solution: (Hide)
My original solution statement:
The sum from 1 to 12 is 156, divided by six gives a mean of 26.

For a difference of 2
Since there are 6 values to consider none can be 26 and so must be spread either side of it, and are therefore 21,23,25,27,29 and 31.

           12                         9
          / \                        / \
     11--1---2---7              8---1---2---10
      \ /     \ /                \ /     \ /
       6       3                  6       3
      / \     / \                / \     / \
     10--5---4---8              7---5---4---11
          \ /                        \ /
           9                          12
Interestingly, the outer vertices are in the mirrored position in the second instance.

For a difference of 4
Similarly the 6 values here are:
16,20,24,28,32 and 36.
           10                         1
          / \                        / \
     8---11--1--12              11--10--12--3
      \ /     \ /                \ /     \ /
       9       3                  8       2
      / \     / \                / \     / \
     6---7---5---2              9---6---4---5
          \ /                        \ /
           4                          7
Consider the perimeter of these two stars. The numeral 1 resides in the first position while 2 is in the fourth. Backtracking 3 is in the third and then 4 is in the sixth and so on. This vertex perimeter and is rotated one position so that the inner vertices values are now at the star points to form the second solution.
The placement could have been described as: place a digit, 'leap-frog' the next vertex and place the next digit. Return to the 'leap-frogged' vertex and repeat the same action.

6 vertex sums of 26.

                                           1
A C F H       1 9 11 5  26                / \
A D G K       1 12 6 7  26           3---9---12--2
B C D E       3 9 12 2  26            \ /     \ /
B F I L       3 11 4 8  26             11      1
E G J L       2 6 10 8  26            / \     / \
H I J K       5 4 10 7  26           5---4---10--7
                                          \ /
                                           8

Might I now refer my reader to the comments list.

Pcbouhid has some valuable observations which are akin to my thoughts above.

As always, where a problem lends itself to a computer program analysis one is likely to uncover more solutions than one might have realistically considered.
Take note of Charlie's commentaries on this problem!

Comments: ( You must be logged in to post comments.)
  Subject Author Date
re: Can we prove? - I canpcbouhid2008-06-03 22:00:08
QuestionCan we prove?pcbouhid2008-06-03 14:51:29
SolutionMight as well diagram the 94 difference-4, adjacent-vertex solutions.Charlie2008-06-02 22:29:47
So Manybrianjn2008-06-02 22:14:19
re: Some thoughts (without reading Charlie´s comments)brianjn2008-06-02 20:46:16
re(2): A method to find solutions for the tease - need improvement - one step aheadpcbouhid2008-06-02 19:10:00
re: A method to find solutions for the tease - need improvementpcbouhid2008-06-02 14:10:30
A method to find solutions for the teasepcbouhid2008-06-02 12:03:21
SolutionThe final "tease" solutions; worked out correctly this time, I hope.Charlie2008-06-02 09:49:16
Some ThoughtsSome thoughts (without reading Charlie´s comments)pcbouhid2008-06-02 09:30:49
re(2): computer solution; spoilerCharlie2008-06-02 00:24:25
Solutionre: computer solution; spoilerCharlie2008-06-02 00:06:57
Solutioncomputer solution; spoilerCharlie2008-06-01 23:56:16
Stupid! Stupid! Stupid!!brianjn2008-06-01 22:32:06
re: Possible mistake?brianjn2008-06-01 21:59:36
re: Possible mistake?Charlie2008-06-01 13:22:39
QuestionPossible mistake?pcbouhid2008-06-01 12:39:22
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