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?
 A: 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.
A: Hint: There are finitely many ways that we can 'triple' up the points to form $n/3$ triangles.
Hint: Consider the configuration which minimizes the total areas of all $n/3$ triangles. Claim: This works.
A: Hint: Can you find $3$ points whose triangle is guaranteed not to intersect any triangle formed from the other $n-3$ points?

 Find a line which separates $3$ of the points from the rest. Triangulate them. Proceed via induction.

A: List the points in lexical order, such that $(x_i < x_{i+1})$ or $(x_i=x_{i+1} \wedge y_{i} < y_{i+1})$ for $i\in\{1,2,\ldots n-1\}$.  Then the grouping $\{\{1,2,3\},\{4,5,6\},\ldots,\{n-2,n-1,n\}\}$ into sets of three contains no intersecting triangles.
For $n=3$, the theorem is trivially true.  Now assume the theorem is true for $n=3k$, and consider $3k+3$ points in general position.  The lexically smallest $3k$ points contain no intersecting triangles when grouped in this way, by the inductive hypothesis.  Moreover, the vertical line at $x=\frac{1}{2}(x_{3k} + x_{3k+1})$ separates the first $k$ triangles from the last triangle: all points but (at most) one of the first $k$ triangles lie to its left, and all points but (at most) one of the $(k+1)$-st triangle lie to its right.  We conclude that the final triangle cannot intersect any of the first $k$ triangles, and hence the theorem is true for $n=3k+3$.  The proof is complete by induction on $k$. 
A: Hall's marriage theorem trivialises the whole problem.
But that's my suggestion.
A: Sorry, the account I created to ask this was a guest account so I haven't been able to answer back to anything and ask other questions, and I don't think I have enough 'rep' to respond to answers directly.
achille hui, firstly, how did you derive the number of ways to join $3k$ points?
Secondly, can my co-ordinates consist of real numbers, or just natural ones, which would obviously make things much simpler.
And finally, why can't any two points create a vector parallel to x=0? I don't quite yet see how it would effect intersection, but I'm guessing that's what it's for.
Thanks, and again sorry for not being able to put my response where I need it to go.
A: 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.
