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# 2012 points Tweet

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1. 2012 points
I think i posted many threads in short time but

are there 2012 points in the plane such that:

1-any three points are not in the same line

2-the distance between any two points from the 2012 points is irrational number

3-the area of any triangle that has vertices from these points is rational number
Last edited by MAA_96; 11-07-2012 at 20:20.
2. Re: 2012 points
what?
3. Re: 2012 points
could you not just chop up a circle into 2012 points?
if I get the gist
Last edited by sputum; 11-07-2012 at 20:23. Reason: no you can't. chop up exp?
4. Re: 2012 points
No, i mean in the plane
5. Re: 2012 points
(Original post by MAA_96)
I think i posted many threads in short time but

are there 2012 points such that:

1-any three points are not in the same line

2-the distance between any two points from the 2012 points is irrational number

3-the area of any triangle that has vertices from these points is rational number
Hm... Perhaps you could try taking a circle with centre and 2011 equally spaced points on the circle. You will have no three points on the same line since 2011 is prime. Suppose the circle has an irrational radius , find the conditions such that any distance is irrational, then see if the area of any arbitrary triangle is necessarily rational?

Edit: woops just realised the idea of the circle had already been posted!
Last edited by Lord of the Flies; 11-07-2012 at 20:44.
6. Re: 2012 points
(Original post by Lord of the Flies)
Hm... Perhaps you could try taking a circle with centre and 2011 equally spaced points on the circle. You will have no three points on the same line since 2011 is prime. Suppose the circle has an irrational radius , find the conditions such that any distance is irrational, then see if the area of any arbitrary triangle is necessarily rational?
But why 2011 and not 2012?
7. Re: 2012 points
There's clearly nothing special about 2012 here, so try the problem with "2012" replaced by "n". I haven't tried this, but it seems plausible, and to prove it, induction is probably a good idea here, the base case being n=3.
8. Re: 2012 points
(Original post by MAA_96)
No, i mean in the plane
hey circles are in the plane too

maybe a (pi,1/pi)(2pi,2/pi) kind of deely?
no maybe not
9. Re: 2012 points
(Original post by sputum)
hey circles are in the plane too

maybe a (pi,1/pi)(2pi,2/pi) kind of deely?
no maybe not
But can we use the idea of circle in 2012 points and not in 2011
10. Re: 2012 points
(Original post by nuodai)
There's clearly nothing special about 2012 here, so try the problem with "2012" replaced by "n". I haven't tried this, but it seems plausible, and to prove it, induction is probably a good idea here, the base case being n=3.
How to do that?What is the expression that i have to prove it by induction ?
11. Re: 2012 points
(Original post by MAA_96)
How to do that?What is the expression that i have to prove it by induction ?
"Prove by induction that for any there exist points in the plane such that ... (etc)".
12. Re: 2012 points
(Original post by nuodai)
"Prove by induction that for any there exist points in the plane such that ... (etc)".
But why the base case is 3,and how to do the induction ,by drawing or ..?
13. Re: 2012 points
For a quadrilateral of 4 points all sides and diagonals are irrational.
but the area of the quadrilateral and all interior triangles must be rational.

which makes the heights of the triangles irrational (such that bh is rational for each triangle)
so you have to be able to subdivide the quadrilateral by using the heights of the interior triangles and get 4 smaller quadrilaterals with irrational sides that sum to a rational area.

meh
Last edited by sputum; 11-07-2012 at 21:16. Reason: lo-fi maths :)
14. Re: 2012 points
(Original post by sputum)
For a quadrilateral of 4 points all sides and diagonals are irrational.
but the area of the quadrilateral and all interior triangles must be rational.

which makes the heights of the triangles irrational (such that bh is rational for each triangle)
so you have to be able to subdivide the quadrilateral by using the heights of the interior triangles and get 4 smaller quadrilaterals with irrational sides that sum to a rational area.

meh
How about " any three points not in the same line" ?
15. Re: 2012 points
(Original post by MAA_96)
But why the base case is 3,and how to do the induction ,by drawing or ..?
The base case is 3 because the question mentions triangles and you need at least three points to form triangles.
16. Re: 2012 points
(Original post by MAA_96)
But why the base case is 3,and how to do the induction ,by drawing or ..?
What's your level of maths so far? (Just for the sake of context.)
17. Re: 2012 points
(Original post by MAA_96)
But why 2011 and not 2012?
Mistake of mine - for some odd reason I disregarded the possibility of dividing the circle in 2012 since this would lead to 3 points in a line if you count the centre - then realised the centre does not need to be part of the problem. Stupid mistake really, sorry about the confusion.
18. Re: 2012 points
(Original post by Lord of the Flies)
The base case is 3 because the question mentions triangles and you need at least three points to form triangles.
But how to do it ?I know how to do induction but this is look different
19. Re: 2012 points
(Original post by nuodai)
What's your level of maths so far? (Just for the sake of context.)
In the last year of high school .
20. Re: 2012 points
(Original post by MAA_96)
But how to do it ?I know how to do induction but this is look different
Well, it's up to you to figure out an expression. I would do it but I have to go now. I'll post it tomorrow if someone hasn't already done so - I'm guessing you're 16? If so, where did you get this question from?
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