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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.)
Solution computer solution; spoiler | Comment 5 of 17 |

When the differences are 2, the totals are 21 through 31. The following program produces unique solutions as it requires that A+D+G+K be the one that adds to 21, and that A be less than K, removing the possibility of rotated or reflected variations of other solutions.

DIM taken(12)
tv(1) = 21: tv(2) = 23
tv(3) = 25: tv(4) = 27
tv(5) = 29: tv(6) = 31
FOR a = 1 TO tv(1) / 2
 taken(a) = 1
 FOR d = 1 TO tv(1) - 2 - a
  IF d < 13 THEN
  IF taken(d) = 0 THEN
   taken(d) = 1
   FOR g = 1 TO tv(1) - 1 - a - d
    IF g < 13 THEN
    IF taken(g) = 0 THEN
      taken(g) = 1
      k = tv(1) - a - d - g
      IF k > a AND k < 13 THEN
        IF taken(k) = 0 THEN
         taken(k) = 1
       
REDIM tTaken(6)
tTaken(1) = 1

FOR b = 1 TO 12
  IF taken(b) = 0 THEN
   taken(b) = 1
     FOR c = 1 TO 12
  IF taken(c) = 0 THEN
   taken(c) = 1
  FOR e = 1 TO 12
  IF taken(e) = 0 THEN
   taken(e) = 1

   total = b + c + d + e
   validTot = 0
   FOR i2 = 2 TO 6
    IF tv(i2) = total THEN validTot = 1: EXIT FOR
   NEXT
   IF validTot THEN
     IF tTaken(i2) THEN validTot = 0
   END IF
   IF validTot THEN
     tTaken(i2) = 1
   
   
FOR h = 1 TO 12
  IF taken(h) = 0 THEN
   taken(h) = 1
     FOR i = 1 TO 12
  IF taken(i) = 0 THEN
   taken(i) = 1
  FOR j = 1 TO 12
  IF taken(j) = 0 THEN
   taken(j) = 1

   total = h + i + j + k
   validTot = 0
   FOR i3 = 2 TO 6
    IF tv(i3) = total THEN validTot = 1: EXIT FOR
   NEXT
   IF validTot THEN
     IF tTaken(i3) THEN validTot = 0
   END IF
   IF validTot THEN
     tTaken(i3) = 1
  
FOR f = 1 TO 12
 IF taken(f) = 0 THEN
   taken(f) = 1

FOR l = 1 TO 12
 IF taken(l) = 0 THEN
   taken(l) = 1

   total = b + f + i + l
   validTot = 0
   FOR i4 = 2 TO 6
    IF tv(i4) = total THEN validTot = 1: EXIT FOR
   NEXT
   IF validTot THEN
     IF tTaken(i4) THEN validTot = 0
   END IF
  
   IF validTot THEN
   tTaken(i4) = 1

   total = h + f + c + a
   validTot = 0
   FOR i5 = 2 TO 6
    IF tv(i5) = total THEN validTot = 1: EXIT FOR
   NEXT
   IF validTot THEN
     IF tTaken(i5) THEN validTot = 0
   END IF
 
   IF validTot THEN
   tTaken(i5) = 1

   total = l + j + g + e
   validTot = 0
   FOR i6 = 2 TO 6
    IF tv(i6) = total THEN validTot = 1: EXIT FOR
   NEXT
   IF validTot THEN
     IF tTaken(i6) THEN validTot = 0
   END IF
 
   IF validTot THEN
   tTaken(i6) = 1

     PRINT a; b; c; d; e; f; g; h; i; j; k; l
     ct = ct + 1

   tTaken(i6) = 0
   END IF


   tTaken(i5) = 0
   END IF

   tTaken(i4) = 0
   END IF
   taken(l) = 0
 END IF
NEXT l

   taken(f) = 0
 END IF
NEXT f

  
     tTaken(i3) = 0
   END IF  'validTot

   taken(j) = 0
  END IF
  NEXT j
   taken(i) = 0
  END IF

     NEXT i
   taken(h) = 0
  END IF
NEXT h
   
   
     tTaken(i2) = 0
   END IF  'validTot

   taken(e) = 0
  END IF
  NEXT e
   taken(c) = 0
  END IF

     NEXT c
   taken(b) = 0
  END IF
NEXT b
      
         taken(k) = 0
        END IF
      END IF
      taken(g) = 0
    END IF
    END IF
   NEXT g
   taken(d) = 0
  END IF
  END IF
 NEXT d
 taken(a) = 0
NEXT a
PRINT ct

The above program finds 13,728 solutions. To narrow that down, I added code to make the arithmetic progression proceed either clockwise or counterclockwise around the figure, even though the puzzle explicitly states this is not necessary. The added code surrounds the printing of the answer:

     good = 0
     IF total6 > total5 AND total5 > total4 AND total4 > total3 AND total3 > total2 THEN
       good = 1
     END IF
     IF total6 < total5 AND total5 < total4 AND total4 < total3 AND total3 < total2 THEN
       good = 1
     END IF
     IF good THEN
       PRINT a; b; c; d; e; f; g; h; i; j; k; l
       ct = ct + 1
     END IF

(also added were the second of each of the pairs of lines:

   total = b + c + d + e
   total2 = total
  
   total = h + i + j + k
   total5 = total
  
   total = b + f + i + l
   total4 = total

   total = h + f + c + a
   total3 = total

   total = l + j + g + e
   total6 = total
)

This results in only 338 solutions, still barring rotations and reflections, so these are also unique.  

The last one found, listed from A to L, is

8  12  11  3  5  6  1  4  2  10  9  7

which looks like
         8
        
    12 11  3  5
   
      6      1
     
     4  2 10  9
    
         7
        


Again, 8+3+1+9 = 21, and the totals increase, in this case, clockwise.

The modification for a difference of 4 is merely:

tv(1) = 16: tv(2) = 20
tv(3) = 24: tv(4) = 28
tv(5) = 32: tv(6) = 36

as these comprise the necessary set of totals.

This results in 3216 unique solutions. Again asking for the successive arithmetic increments to be clockwise or counterclockwise in order, reduces the number of solutions to 94, the last of which is printed out as: 5  6  2  1  11  8  3  9  4  12  7  10, and in array is:

      5  
     
6   2  1  11
  8      3 
 
9   4  12  7
     10
    


In this case, the successive sums proceed counterclockwise from the 5+1+3+7=16 side.

Further modification finds those cases where each side adds to 26. There are 72 such cases, made unique by requiring that H, A and K be in ascending order and that the 1 be in position H, C, G or I.

The last one found was 11  9  5  2  10  6  1  4  3  7  12  8:

    11
   
9  5  2  10
  6     1 
 
4  3  7  12 
     8

  Posted by Charlie on 2008-06-01 23:56:16
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