## Welcome to BrainDen.com - Brain Teasers Forum

 Welcome to BrainDen.com - Brain Teasers Forum. Like most online communities you must register to post in our community, but don't worry this is a simple free process. To be a part of BrainDen Forums you may create a new account or sign in if you already have an account. As a member you could start new topics, reply to others, subscribe to topics/forums to get automatic updates, get your own profile and make new friends. Of course, you can also enjoy our collection of amazing optical illusions and cool math games. If you like our site, you may support us by simply clicking Google "+1" or Facebook "Like" buttons at the top. If you have a website, we would appreciate a little link to BrainDen. Thanks and enjoy the Den :-)
Guest Message by DevFuse

# Max. # of convex/concave hexagons possible using 6 points

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

Spoiler for a few more

Spoiler for Agree with your result
Go to the full post

7 replies to this topic

### #1 Perhaps check it again

Perhaps check it again

Junior Member

• Members
• 95 posts
• Gender:Male

Posted 25 June 2014 - 07:57 PM

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

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 by Perhaps check it again, 25 June 2014 - 07:59 PM.

• 0

### #2 bonanova

bonanova

bonanova

• Moderator
• 6153 posts
• Gender:Male
• Location:New York

Posted 26 June 2014 - 07:48 PM

Spoiler for first thoughts

Edited by bonanova, 26 June 2014 - 10:50 PM.

• 0

Vidi vici veni.

### #3 bonanova

bonanova

bonanova

• Moderator
• 6153 posts
• Gender:Male
• Location:New York

Posted 27 June 2014 - 04:03 AM

• 0

Vidi vici veni.

### #4 plasmid

plasmid

Senior Lolcat

• VIP
• 1484 posts
• Gender:Male

Posted 27 June 2014 - 05:17 AM

Spoiler for a few more

• -1

### #5 Perhaps check it again

Perhaps check it again

Junior Member

• Members
• 95 posts
• Gender:Male

Posted 27 June 2014 - 06:03 AM

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 by Perhaps check it again, 27 June 2014 - 06:07 AM.

• 0

### #6 bonanova

bonanova

bonanova

• Moderator
• 6153 posts
• Gender:Male
• Location:New York

Posted 01 July 2014 - 07:43 AM   Best Answer

Spoiler for a few more

Spoiler for Agree with your result

• 1

Vidi vici veni.

### #7 bonanova

bonanova

bonanova

• Moderator
• 6153 posts
• Gender:Male
• Location:New York

Posted 08 July 2014 - 06:18 PM

plasmid should get the solve.

I only arranged his examples in symmetric pairs.

• 0

Vidi vici veni.

### #8 Perhaps check it again

Perhaps check it again

Junior Member

• Members
• 95 posts
• Gender:Male

Posted 09 July 2014 - 05:27 AM

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 by Perhaps check it again, 09 July 2014 - 05:34 AM.

• 0

#### 0 user(s) are reading this topic

0 members, 0 guests, 0 anonymous users