Maximum Number Of Points (Combinatorics)

The problem is like the following.

Let $n$ blue lines, no two of which are parallel and no three concurrent, be drawn on a plane.
An intersection of two blue lines is called a blue point.

Through any two blue points that have not already been joined by a blue line, a red line is drawn.

An intersection of two red lines is called a red point,
and an intersection of red line and a blue line is called a purple point.

What is the maximum possible number of purple points?

P.S.
This is from Turkey National Olympiad Second Round 1994.
I have no solution and can't come up with anything.
Thank you in advance.

Each pair of blue lines intersect in a blue point, so there are $\frac{n(n-1)}{2}$ blue points. Each blue point is connected to $\frac{n(n-1)}{2} - (2n-3)$ other blue points with red lines, so there are $\frac{n(n-1)}{2} \left(\frac{\frac{n(n-1)}{2} - (2n-3)}{2}\right)$ red lines. Each red line intersects each blue line, but is concurrent with two blue lines at its two blue points, so each red line contributes $n-4$ purple points, plus the original $\frac{n(n-1)}{2}$ blue points, assuming $n > 3$.
The total number of purple points then becomes $\displaystyle \frac{n(n-1)}{2} \frac{1}{2}\left(\frac{n(n-1)}{2} - (2n-3)\right)(n-4) + \frac{n(n-1)}{2}$