Show that there are no seven lines in the plane such that there are at least six points which lie on the intersection of just three lines and at least four points which lie on the intersection of just two lines.
From 7 lines we can choose just 21 pairs. An intersection of 3 lines accounts for 3 distinct pairs of lines, an intersection of 2 lines for 1 pair. Hence the configuration given would have at least 6 x 3 + 4 x 2 = 22 distinct pairs of lines.
Comment. Note that the solution above is not only short, but the obvious approach. In other words, the problem is trivial. Curious, the A6 questions are usually harder. In fact the whole paper was easy. Every question yields to the obvious approach.
34th Putnam 1973
© John Scholes
22 Aug 2001