Very roughly this is the net of a dodecahedron; each letter represents a pentagon.

       A
       |
    B--C--D
      / \
     E   F
      \
  G   H
   \ /
 I--J--K
    |
    L

Consider each face to be numbered 1 through 12.

Each vertex is the intersection of 3 faces. The vertex sum is therefore the sum of the values of those three faces.

The faces making up the vertices in the diagram above are:

 1. ABC  2. ABI  3. ACD  4. ADL  5. AIL
 6. BCE  7. BEG  8. BGI  9. CDF 10. CEF
11. DKF 12. DKL 13. EFH 14. EGH 15. FHK
16. GHJ 17. GIJ 18. HJK 19. IJL 20. JKL.

What is the global vertex sum (20 vertices) and therefore the mean vertex sum?

How best can the faces be labeled so that the 20 vertices are as close as possible to the mean vertex sum?

“Close as possible” means that:
the sum of differences above (or below) the mean is at the optimum
or
the most vertex sums land on or have the best proximity to the mean as possible. The prior condition also applies; ie, any deviance is minimal.

Building on Charlies value for a mean of 19.5 . . .

An Excel spreadsheet approach.

I started with A through L = 1 through 12 respectively.  I then swapped values between letters with adjacent values.  The best solution I arrived at was:

A  B   C   D   E   F   G   H   I   J   K   L

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

This has a deviation of 1.7 and max/min of 23/16 for vertex sums.

 

Edited on June 1, 2007, 4:20 pm

Posted by Leming on 2007-06-01 15:57:26