Triangle Mania!

Let the number of points including the vertices on a slanting side of triangle be $n$ as shown in the diagram. We note that any two of the $n$ points together with the opposite vertex form $n-1$ triangles.

Therefore, the number of triangles by the $n$ points with the opposite vertex at the base is $\displaystyle {n \choose 2}(n-1)$ . Since there are two identical sets of points and lines, the total number of triangles is $\displaystyle 2(n-1){n \choose 2}$ .

But there are a set of triangles shared by the two base vertices. These triangles are those form by the two bases vertices with the points except the base vertex of one slanting sides. That is $\displaystyle (n-1){n-1 \choose 1}$ .

Therefore, the total number of triangles formed is

$\begin{aligned} N_n & = 2(n-1){n \choose 2} - (n-1){n-1 \choose 1} \\ & = 2(n-1)\times \frac {n(n-1)}2 - (n-1)^2 \\ & = (n-1)^3 \end{aligned}$

For $n = 5$ , we have $N_5 = (5-1)^3 = \boxed{64}$ .