Problem:
Nine pairwise noncongruent circles are drawn in the plane such that any two circles intersect twice. For each pair of circles, we draw the line through these two points, for a total of lines. Assume that all 36 lines drawn are distinct. What is the maximum possible number of points which lie on at least two of the drawn lines?
Solution:
The lines in question are the radical axes of the 9 circles. Three circles with noncollinear centers have a radical center where their three pairwise radical axes concur, but all other intersections between two of the lines can be made to be distinct. So the answer is
by just counting pairs of lines, and then subtracting off double counts due to radical centers (each counted three times).