## 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 :-) |

# The Aha!" problems 4. Six integral points

### #1

Posted 27 April 2014 - 07:48 AM

Determine the coordinates of six points on the plane with the following properties.

- No three points are collinear.
- Every pairwise distance is an integer.

You may use sketchpad, compass, ruler, straight edge, whatever you think may be useful.

The answer will be six pairs of coordinates: (*x _{i }, y_{i}*).

*The greatest challenge to any thinker is stating the problem in a way that will allow a solution.*

- Bertrand Russell

### #2

Posted 27 April 2014 - 06:25 PM

There is an example with a set of seven coordinates on the plane such that no three points are

collinear and every pairwise distance is an integer. So, if you want a set of six coordinates, just

omit one of the seven. The example is at the top of the second page:

### #3

Posted 03 May 2014 - 01:00 PM

Some of the considerations discussed in that paper bear on the problem at hand.

But in the spirit of the Aha! theme, let's rule out exhaustive computer constructions and go with compass and straightedge.

*The greatest challenge to any thinker is stating the problem in a way that will allow a solution.*

- Bertrand Russell

### #4

Posted 06 May 2014 - 11:40 AM

*The greatest challenge to any thinker is stating the problem in a way that will allow a solution.*

- Bertrand Russell

### #5

Posted 07 May 2014 - 04:14 PM

### #6

Posted 07 May 2014 - 08:44 PM

### #7

Posted 07 May 2014 - 09:27 PM

**Edited by DeGe, 07 May 2014 - 09:29 PM.**

### #8

Posted 09 July 2014 - 06:41 AM

Spoiler for could work

**DeGe**, sorry for the delay in answering.

I think your right triangle A_{1}B_{1}C_{2 }will have sides of 4, 6 and sqrt(52).

That is, A_{1}C_{2} will be irrational. But did I get your construction right?

*The greatest challenge to any thinker is stating the problem in a way that will allow a solution.*

- Bertrand Russell

### #9

Posted 09 July 2014 - 05:44 PM Best Answer

### #10

Posted 09 July 2014 - 05:58 PM

Here is the solution I had in mind, and** k-man** found it.

*The greatest challenge to any thinker is stating the problem in a way that will allow a solution.*

- Bertrand Russell

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

0 members, 0 guests, 0 anonymous users