What is the maximum number of triangles in a planar graph with n vertices? The answer is obvious for small numbers of nodes:
$$n&lt;3: 0\\
n=3: 1\\
n=4: 3\\
n=5: 5  (see below)$$
I will suppose you are only counting triangles with no nodes inside.  Let's also suppose both that the graph is completely triangulated so there are no polygons with more edges on the inside and that the nodes are not colinear previnting triangles(otherwise we could create more triangles).

Suppose there are $t$ triangles in total, $e$ edges, and $n$ nodes or vertices.  Suppose that of these, $p$ nodes and edges are on the outer boundary.  Then we have:

$$t+n=e+1$$ $$3t=2(e-p)+p$$

These give $t=2n-2-p$.  But $p \ge 3$ for $n\ge 3$ so $$\max(t) = 2n-5.$$  

For $n=5$,   $\max(t) = 5$ not $4$
I will suppose you are only counting triangles with no nodes inside.  Let's also suppose both that the graph is completely triangulated so there are no polygons with more edges on the inside and that the nodes are not colinear previnting triangles(otherwise we could create more triangles).

Suppose there are $t$ triangles in total, $e$ edges, and $n$ nodes or vertices.  Suppose that of these, $p$ nodes and edges are on the outer boundary.  Then we have:

$$t+n=e+1$$ $$3t=2(e-p)+p$$

These give $t=2n-2-p$.  But $p \ge 3$ for $n\ge 3$ so $$\max(t) = 2n-5.$$  

For $n=5$,   $\max(t) = 5$ not $4$