Perhaps check it again Posted June 25, 2014 Report Share Posted June 25, 2014 (edited) In the plane you get to choose six points. By choosing these six points in one of an infinite number of appropriate configurations, what is the maximum number of combined convex and concave (but not self-intersecting) hexagons that can be formed, if each of the six points is to be one of the vertices for every hexagon so formed? For reference sake, the points could be labeled A, B, C, D, E, and, F. I am not asking for any coordinates. However, if you want to volunteer a set of them to illustrate your work/solution, then that would be fine. ---------------------------------------------------------------------------------------------------------------------- Here is an example with four points: A, B, C, and, D. Place them in the plane. You choose where. What is the maximum number of combined convex and concave (but not self-intersecting) quadrilaterals that can be formed, if each of the four points is to be one of the vertices for every quadrilateral so formed? If you do not choose all of the points to be distinct and/or you place at least three of them on the same lime, you won't get any quadrilaterals formed. If you place the four points so that they are the vertices of a convex quadrilateral, then you will get one total (convex) quadrilateral. If you place three of the points as vertices of a triangle and the fourth point as an interior point of that triangle, then three total (concave) quadrilaterals can be formed. "Three" is the answer, unless there is some other general configuration for four points that has been overlooked with a higher total number of quadrilaterals. . . . Edited June 25, 2014 by Perhaps check it again Quote Link to comment Share on other sites More sharing options...

0 bonanova Posted July 1, 2014 Report Share Posted July 1, 2014 Admittedly 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) casesACDEFB ~ 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. 1 Quote Link to comment Share on other sites More sharing options...

0 bonanova Posted June 26, 2014 Report Share Posted June 26, 2014 (edited) With a point inside a convex pentagon you get five so the answer has to be six or greater. With two points inside a square (hourglass), e.g. (+/-1, 0) inside (+/-3, +/-3) you get (at least) 9. I'll leave that as a target for now. Edit: (Inverted) triangle inside a triangle gets more, at least 22. I'm still counting. Edited June 26, 2014 by bonanova Add solution. Quote Link to comment Share on other sites More sharing options...

0 bonanova Posted June 27, 2014 Report Share Posted June 27, 2014 Triangle-in-triangle array of points. A D E F B C ADBFCEA - mirror and rotational symmetry (1) ABFDCEA - neither symmetry (6) ABFDECA - neither symmetry (6) ABCFDEA - neither symmetry (6) ABCFECA - mirror but not rotational symmetry (3)That gives 22 distinct hexagons, each having six nodes of degree 2. ABFDEFCA - mirror but not rotational symmetry (3). One can argue whether this is a hexagon. It uses all six points as nodes and has no intersection lines. But one of its nodes has degree 4.If allowed, the total grows to 25. Quote Link to comment Share on other sites More sharing options...

0 plasmid Posted June 27, 2014 Report Share Posted June 27, 2014 Admittedly this is brute force and not elegance, but it shows there are at least 29.1. ABCDFE2. ABCEDF3. ABCEFD4. ABCFED5. ABCFDE6. ABDECF7. ABDEFC8. ABDFEC9. ABECFD10. ABEDFC11. ABEFCD12. ABEFDC13. ABFCDE14. ABFEDC15. ACBDEF16. ACBEDF17. ACBEFD18. ACBFDE19. ACBFEDACDEFB (is equivalent to ABFEDC)20. ACDFBEACDFEB (is equivalent to ABEFDC)21. ACEBDFACEFDB (is equivalent to ABDFEC)22. ACFBED23. ACFEBDACFEDB (is equivalent to ABDEFC)24. ADBCEF25. ADBECFADBEFC (is equivalent to ACFEBD)26. ADCBFE27. ADCFBEADCFEB (is equivalent to ABEFCD)28. ADEBCFADEBFC (is equivalent to ACFBED)ADEFBC (is equivalent to ACBFED)ADEFCB (is equivalent to ABCFED)29. ADFCBEAnd 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. 1 Quote Link to comment Share on other sites More sharing options...

0 Perhaps check it again Posted June 27, 2014 Author Report Share Posted June 27, 2014 (edited) bonanova, for your #6 category that was up for debate for three more possible hexagons in your last spoiler, they won't fall under the convex/concave types. That node of degree four that you pointed out will have to disqualify them. Every convex/concave polygon that is not self-intersecting must have a degree of two for all of its nodes/vertices. Edited June 27, 2014 by Perhaps check it again Quote Link to comment Share on other sites More sharing options...

0 bonanova Posted July 8, 2014 Report Share Posted July 8, 2014 plasmid should get the solve. I only arranged his examples in symmetric pairs. Quote Link to comment Share on other sites More sharing options...

0 Perhaps check it again Posted July 9, 2014 Author Report Share Posted July 9, 2014 (edited) bonanova stated: "plasmid should get the solve. I only arranged his examples in symmetric pairs." ----------------------------------------------------------------------- No, plasmid's posts are not recognized on here by me for any credit. plasmid already knows he/she has to state to me a retraction of a post that plasmid made to me elsewhere that was/is out of line. And plasmid already knows that there are no conditions on the retraction, regardless that he/she is trying to place any on the retraction. And none of plasmid's other posts will be recognized by me for any credit on any other puzzle challenge threads that I start unless a retraction is made by plasmid to me in the forum. Edited July 9, 2014 by Perhaps check it again Quote Link to comment Share on other sites More sharing options...

## Question

## Perhaps check it again

In the plane you get to choose six points.

By choosing these six points in one of an infinite number of appropriate configurations,

what is the maximum number of combined convex and concave (but not self-intersecting)

hexagons that can be formed, if each of the six points is to be one of the vertices for every

hexagon so formed?

For reference sake, the points could be labeled A, B, C, D, E, and, F.

I am not asking for any coordinates. However, if you want to volunteer a set of them to

illustrate your work/solution, then that would be fine.

----------------------------------------------------------------------------------------------------------------------

Here is an example with four points: A, B, C, and, D.

Place them in the plane. You choose where.

What is the maximum number of combined convex and concave (but not self-intersecting)

quadrilaterals that can be formed, if each of the four points is to be one of the vertices for

every quadrilateral so formed?

If you do not choose all of the points to be distinct and/or you place at least three of them

on the same lime, you won't get any quadrilaterals formed.

If you place the four points so that they are the vertices of a convex quadrilateral, then

you will get one total (convex) quadrilateral.

If you place three of the points as vertices of a triangle and the fourth point as an interior

point of that triangle, then three total (concave) quadrilaterals can be formed.

"Three" is the answer, unless there is some other general configuration for four points that

has been overlooked with a higher total number of quadrilaterals.

.

.

.

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