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.