Best Answer bonanova, 01 July 2014 - 07:43 AM

Spoiler for a few moreAdmittedly this is brute force and not elegance, but it shows there are at least 29.

Hexagons.jpg

1. ABCDFE

2. ABCEDF

3. ABCEFD

4. ABCFED

5. ABCFDE

6. ABDECF

7. ABDEFC

8. ABDFEC

9. ABECFD

10. ABEDFC

11. ABEFCD

12. ABEFDC

13. ABFCDE

14. ABFEDC

15. ACBDEF

16. ACBEDF

17. ACBEFD

18. ACBFDE

19. ACBFED

ACDEFB (is equivalent to ABFEDC)

20. ACDFBE

ACDFEB (is equivalent to ABEFDC)

21. ACEBDF

ACEFDB (is equivalent to ABDFEC)

22. ACFBED

23. ACFEBD

ACFEDB (is equivalent to ABDEFC)

24. ADBCEF

25. ADBECF

ADBEFC (is equivalent to ACFEBD)

26. ADCBFE

27. ADCFBE

ADCFEB (is equivalent to ABEFCD)

28. ADEBCF

ADEBFC (is equivalent to ACFBED)

ADEFBC (is equivalent to ACBFED)

ADEFCB (is equivalent to ABCFED)

29. ADFCBE

And there aren't any more novel solutions with that orientation of points.

I haven't come up with a way to tell if that's the maximum possible; so far I can only guarantee that it's no greater than 48 if I understand the rules correctly.

By not inverting the internal triangle as I did, for no apparent (or useful) reason,

the outer vertices can access all the inner ones, leading to more cases that you found.

Keeping your numbering and pairing the mirror-derived cases,

28 cases, paired by Mirroring (B<->C and E<->F)

1. ABCDFE 15. ACBDEF = 1.M

2. ABCEDF 18. ACBFDE = 2.M

3. ABCEFD 19. ACBFED = 3.M

4. ABCFED 17. ACBEFD = 4.M

5. ABCFDE 16. ACBEDF = 5.M

6. ABDECF 20. ACDFBE = 6.M

7. ABDEFC 12. ABEFDC ~ 7.M

8. ABDFEC 14. ABFEDC ~ 8.M

9. ABECFD 22. ACFBED = 9.M

11. ABEFCD 23. ACFEBD = 11.M

13. ABFCDE 21. ACEBDF = 13.M

24. ADBCEF 26. ADCBFE = 24.M

25. ADBECF 27. ADCFBE = 25.M

28. ADEBCF 29. ADFCBE = 28.M

1 Unique symmetric case

10. ABEDFC ~ 10.M

Total: 29 Unique cases

Duplicate (equivalent to existing) cases

ACDEFB ~ ABFEDC = 14.

ACDFEB ~ ABEFDC = 12.

ACEFDB ~ ABDFEC = 8.

ACFEDB ~ ABDEFC = 7.

ADBEFC ~ ACFEBD = 23.

ADCFEB ~ ABEFCD = 11.

ADEBFC ~ ACFBED = 22.

ADEFBC ~ ACBFED = 19.

ADEFCB ~ ABCFED = 4.