1
$\begingroup$

I'm pretty sure this is a graph theory problem :

I'm trying to figure out a formula for the number of triangles that can be created from n distinct points in the (Euclidean) Plane such that no 2 lines cross each other (if a formula is possible).

For my purposes, we can assume that no 3 points are on the same line, and of course, we can assume n > 2. Also, points, and lines, can be used for more than 1 triangle, but, as mentioned, lines cannot cross.

Is there a formula for this?

Also, a related question : Given those n points, and the triangles formed by them, is there a formula to find the number of distinct segments forming those triangles? (So, for 2 triangles that share 1 side, there are 5 segments, etc.)

Thanks in advance!

$\endgroup$

1 Answer 1

1
$\begingroup$

This follows from Euler's formula for planar graphs: $V-E+A=1$. Here $V$ is the number of vertices (points), $E$ the number of edges (lines) and $A$ the number of connected areas (not counting the exterior of the planar graph). Split $V$ as $V=V_1+V_2$ where $V_1$ is the number of points on the boundary of the convex hull of all points and $V_2$ is the number of points in the interior of the convex hull. If every connected area is a triangle then the key observation is $$E=\frac{3A+V_1}{2}$$ so from Euler's formula it follows that $A=V_1+2V_2-2$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .