1987 IMO Problems/Problem 5

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Problem

Let $n$ be an integer greater than or equal to 3. Prove that there is a set of $n$ points in the plane such that the distance between any two points is irrational and each set of three points determines a non-degenerate triangle with rational area.

Solution

Consider the set of points $S = \{ (x,x^2) \mid 1 \le x \le n , x \in \mathbb{N} \}$ in the $xy$ -plane.

The distance between any two distinct points $(x_1,x^2_1)$ and $(x_2,x^2_2)$ in $S$ (with $x_1 \neq x_2$ ) is:

$d = \sqrt{(x_1-x_2)^2+\left(x^2_1-x^2_2\right)^2} =$ $\sqrt{(x_1-x_2)^2+(x_1-x_2)^2(x_1+x_2)^2} = |x_1-x_2|\sqrt{1+(x_1+x_2)^2}$ .

Since $1+(x_1+x_2)^2$ is an integer and not a perfect square, $\sqrt{1+(x_1+x_2)^2}$ is irrational. Since $|x_1-x_2|$ is a non-zero integer, $d$ is irrational as desired.

All the points in $S$ lie on the parabola $y = x^2$ . Thus, it is impossible of any set of three points to be collinear, since no line can intersect a parabola at more than two points. Therefore, any triangle with all vertices in $S$ must be non-degenerate as desired.

Since all the points in $S$ are lattice points, by Pick's Theorem, the area of any triangle with all vertices in $S$ must be in the form $A = I + \dfrac{B}{2} - 1$ where $I$ and $B$ are integers. Thus, the area of the triangle must be rational as desired.

This completes the proof.

1987 IMO (Problems)
Preceded by Problem 4	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 6

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1987 IMO Problems/Problem 5

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Problem

Solution