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.

I wonder how closely Leming's spreadsheet resembled mine? He has not said, but his data has 6 vertices being either 19 or 20.

Charlie also did not claim that he had 10 vertices landing on 19 or 20.

I note Leming's max/min vertex range was one inside Charlie's; nice work guys.

By the way Charlie, I had considered that a program would solve it but realising the significance of 12! and trying to map a result to 20 strings seemed like a task that would lock-up a computer for hours using some form of Basic.

How long did it take?  Firstly to devise and write the program and then the runtime?

My solution page is on-line.

I copied and pasted the text of Charlie's program to a Notepad text file and had QBasic open it.  It took about 3/4 hr on my machine.

Edited on June 4, 2007, 4:35 am

Posted by brianjn on 2007-06-02 23:43:16