Given 8 fixed points
such that no 3 points are collinear (meaning on the same line); none of the
lines joining the given points are parallel; and other than at the 8 given
points, no 3 lines joining the given points are concurrent (meaning passing
through the same point). Find the number of points of intersections (not
including the 8 given points) for all the lines.

[Problem
submitted by Steve Lee, LACC Professor of Mathematics.]

**Solution for Problem 9:**

One particular line
joining 2 of the given points intersects with all the lines joining the other 6
points (there are _{} = 15 of them). _{} There are 15 points of
intersections on each line. Since there are _{}= 28 lines, the total number of intersections is 28*15. But
every point of intersection is counted twice. For example, when we count the
intersections on the line_{}, the point common to _{} and _{} is
counted once. Then when we count the intersections on the line
_{}, the same point is counted again. _{} The answer is ½*28*15
= 210.