Problem 5

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.

Step 4 of 5: No three chosen points are collinear

Previous step Next step

In plain words

A quadratic curve bends too much for one line to hit it three times.

\text{A line intersects }y=x^2\text{ in at most two points}

Detailed analysis

A line has equation $y=mx+b$ . Intersecting it with $y=x^2$ gives $x^2-mx-b=0$ , a quadratic equation with at most two real roots. Therefore no line contains three points of $S$ , so every three selected points form a non-degenerate triangle.