# 2012 points

• Jul 11th 2012, 10:51 AM
Mhmh96
2012 points
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 point is rational number
• Jul 11th 2012, 11:39 AM
Mhmh96
Re: 2012 points
Perhaps we could try taking a circle with centre $\displaystyle c$ and 2011 equally spaced points $\displaystyle (a_1,a_2\; ... \; a_{2011})$ on the circle. we will have no three points on the same line since 2011 is prime. Suppose the circle has an irrational radius $\displaystyle r$, find the conditions such that any distance is irrational, then see if the area of any arbitrary triangle is necessarily rational?
• Jul 11th 2012, 02:55 PM
richard1234
Re: 2012 points
I would suggest trying small cases. Are there 3 points in the plane such that the above conditions are met?
• Jul 12th 2012, 03:03 AM
Mhmh96
Re: 2012 points
Quote:

Originally Posted by richard1234
I would suggest trying small cases. Are there 3 points in the plane such that the above conditions are met?

How about the points (x,x^2) the conditions are met in these points .
• Jul 12th 2012, 04:22 AM
emakarov
Re: 2012 points
Quote:

Originally Posted by Mhmh96
How about the points (x,x^2) the conditions are met in these points .

You must mean the points (n, n^2) where n is a natural number. I think this actually works. The area is rational by this formula, and the distances are irrational (after some calculations) because the sum of 1 and a positive perfect square is not a perfect square.
• Jul 12th 2012, 04:31 AM
Mhmh96
Re: 2012 points
Quote:

Originally Posted by emakarov
You must mean the points (n, n^2) where n is a natural number. I think this actually works. The area is rational by this formula, and the distances are irrational (after some calculations) because the sum of 1 and a positive perfect square is not a perfect square.

But how about that any three points are not in the same line ?
• Jul 12th 2012, 04:34 AM
emakarov
Re: 2012 points
Have you ever seen a straight line intersecting a parabola in three points?
• Jul 12th 2012, 05:12 AM
Mhmh96
Re: 2012 points
Quote:

Originally Posted by emakarov
Have you ever seen a straight line intersecting a parabola in three points?

You are right ,but why the set of points should be natural numbers ?
• Jul 12th 2012, 05:22 AM
emakarov
Re: 2012 points
Quote:

Originally Posted by Mhmh96
why the set of points should be natural number ?

Allow me to be a math Nazi for a moment. In the example above, the set of points is not a natural number; it's a set. It is not even a set of natural numbers; it a set of pairs of natural numbers.

The problem asks whether the claim "There exist 2012 points such that ..." holds. To prove an existential statement (i.e., a statement that starts with "there exists") it is sufficient to give a single example satisfying the requires conditions. If the claim were false, its negation "For any set of 2012 points, it is not the case that ..." would be true. To prove a universal statement (the one that starts with "for all"), one has to fix an arbitrary object (in this case, a set of 2012 points) and prove the rest of the statement (in this case, that it does not satisfy conditions 1 - 3 simultaneously).
• Jul 12th 2012, 09:29 AM
richard1234
Re: 2012 points
Quote:

Originally Posted by Mhmh96
You are right ,but why the set of points should be natural numbers ?

For this type of construction proof, it suffices to find one case where this works. You can assume that the 2012 points are lattice points (x-, y-coordinates are integers or natural numbers) and prove that this works. Then you are done.
• Jul 13th 2012, 04:26 AM
Mhmh96
Re: 2012 points
The set of points are parabola ?
• Jul 13th 2012, 04:31 AM
emakarov
Re: 2012 points
The suggested set of points is $\displaystyle \{(n,n^2)\mid n\in\mathbb{N}\}$, which is a proper subset of a parabola.