Difference between revisions of "1987 IMO Problems/Problem 5"

Latest revision as of 12:37, 26 July 2009

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) • Resources
Preceded by Problem 4	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 6
All IMO Problems and Solutions

@@ Line 3: / Line 3: @@
 ==Solution==
-{{solution}}
+Consider the set of points <math>S = \{ (x,x^2) \mid 1 \le x \le n , x \in \mathbb{N} \}</math> in the <math>xy</math>-plane.
+The distance between any two distinct points <math>(x_1,x^2_1)</math> and <math>(x_2,x^2_2)</math> in <math>S</math> (with <math>x_1 \neq x_2</math>) is:
+<math>d = \sqrt{(x_1-x_2)^2+\left(x^2_1-x^2_2\right)^2} =</math> <math>\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}</math>.
+Since <math>1+(x_1+x_2)^2</math> is an integer and not a perfect square, <math>\sqrt{1+(x_1+x_2)^2}</math> is irrational.
+Since <math>|x_1-x_2|</math> is a non-zero integer, <math>d</math> is irrational as desired.
+All the points in <math>S</math> lie on the parabola <math>y = x^2</math>. 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 <math>S</math> must be non-degenerate as desired.
+Since all the points in <math>S</math> are [[lattice points]], by [[Pick's Theorem]], the area of any triangle with all vertices in <math>S</math> must be in the form <math>A = I + \dfrac{B}{2} - 1</math> where <math>I</math> and <math>B</math> are integers. Thus, the area of the triangle must be rational as desired.
+This completes the proof.
 {{IMO box|num-b=4|num-a=6|year=1987}}
 [[Category:Olympiad Geometry Problems]]

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Difference between revisions of "1987 IMO Problems/Problem 5"

Latest revision as of 12:37, 26 July 2009

Problem

Solution