4th APMO 1992 Problem 4 Find all possible pairs of positive integers (m, n) so that if you draw n lines which intersect in n(n-1)/2 distinct points and m parallel lines which meet the n lines in a further mn points (distinct from each other and from the first n(n-1)/2) points, then you form exactly 1992 regions. Solution
: (1, 995), (10, 176), (21, 80). n lines in general position divide the plane into n(n+1)/2 + 1 regions and each of the m parallel lines adds a further n+1 regions. So we require n(n+1)/2 + 1 + m(n+1) = 1992 or (n+1)(2m+n) = 3982 = 2·11·181. So n+1 must divide 3982, also (n+1)n < 3982, so n ≤ 62. We are also told that n is positive Thus n = 0 is disallowed. The remaining possibilities are n+1 = 2, 11, 2·11. These give the three solutions shown above. 4th APMO 1992 © John Scholes jscholes@kalva.demon.co.uk 11 Apr 2002