Can we always draw $n/3$ disjoint triangles from $n$ points in the plane in general position? Suppose we are given $n$ points in the plane, where $n$ is a multiple of $3$ and no three of these points lie on a line. Is it possible to group all of these points into sets of three, so that if we draw the triangles formed by these sets, then no two of the triangles intersect?



Hi guys, I can visualise this question, but I have no idea how to approach it mathematically, any help?
Given $n = 3k$ points, there are $\frac{n(n-1)}{2}$ ways to join them into lines and form unit vectors (2 for each line).
Pick a coordinate system such that none of these unit vectors are parallel to $y$-axis. The $x$-coordinates of these $n$ points will be distinct. You can sort the points according to their $x$-coordinates and group them to groups of 3.
There are finitely many slopes between pairs of points, so we may choose a line whose slope is not any of these. For ease of visualization we may choose it arbitrarily close to vertical.

Now sweep the line across the set of points from left to right. Since we will cross the points one at a time, we can cut the plane along this line as we pass each set of three points. Since each piece of the plane is convex, the triangle formed between the three points it contains lies within it. And since the pieces are disjoint, so are the triangles.