perplexus.info :: Just Math : Dodecahedral Vertex Sums

Dodecahedral Vertex Sums (Posted on 2007-06-01)

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.

Submitted by brianjn

Rating: 4.0000 (1 votes)

Solution: (Hide)

Part 1:Twelve faces have a sum of 12*13/2 or 78
The average is 78/12 = 6.5
An average vertex sum is therefore 3*6.5 = 19.5
With 20 vertices to consider the total vertex sum is 390.

One could therefore assume optimum sums the values of 19 and 20.
That would mean that we could anticipate 10 vertices with a sum of 19 and 10 with a sum of 20.
Assumption destroyed.
Consider 1 to be the central pentagon of 6. If the vertex considered takes a 12 it must have 6 and 7 adjacent to it giving sums of 19 and 20. The 7 may only add 11 which makes 19. Now the 11 can only use an 8 [20] but that means that the remaining vertex has been automatically assigned the value of 15!
Consider the triplet of 1,9 and 10.
I can go nowhere with 1 + 9 because 8 only makes 18; 1 + 10 could only add 8, hence another case of limbo.
Some “solutions”:
With A to L being 1 to 12 there are 2 cases that are either 19 or 20 and the deviation is 64.
With A to L having values of 1,10,5,12,7,4,3,9,11,2,8,6 there are 6 19/20’s with a deviance of 24.5
With A to L having values of 5,4,12,1,11,7,2,9,10,8,3,6 there are 6 19/20’s with a deviance of 37, clearly not as good.
With A to L having values of 5,4,11,1,12,7,2,9,10,8,3,6 there are 7 19/20’s with a deviance of 37.5.
Note too that above I have recorded the sum of the deviations from the mean rather that providing an average as is is shown by respondents. I refer you to Charlie's response. I do note that Charlie did not indicate how many of his vertices actually landed on 19 or 20. However, a count of the 0.5 and -0.5's is 10 and there are 5 of each.

Comments: ( You must be logged in to post comments.)

Subject	Author	Date
Puzzle Thoughts	K Sengupta	2023-06-16 00:31:30
Vertices and deviations	brianjn	2007-06-02 23:43:16
computer solution	Charlie	2007-06-01 17:51:00
Trial and erorr	Leming	2007-06-01 15:57:26
first part	Charlie	2007-06-01 14:56:11

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