15th APMO 2003 Problem 5 Find the smallest positive integer k such that among any k people, either there are 2m who can be divided into m pairs of people who know each other, or there are 2n who can be divided into n pairs of people who do not know each other.
k = m + n + max(m, n) - 1 Solution We have to find the smallest k so that any graph with k points has either m disjoint edges or n disjoint pairs of points with each pair not joined by an edge. Let the smallest k be f(m, n). For any graph G, there is another graph G' with the same points and the complementary set of edges (so that A, B are joined by an edge in G' iff they are not joined in G). This shows that f(m, n) = f(n, m), so there is no loss of generality in assuming that m ≤ n. Consider the graph G with m+2n-2 points defined as follows. Let A be a subset of m-1 points and B the remaining 2n-1 points. Every point of A is joined to every other point (in A or B) and there are no other edges. Now since all points in A are joined to every other point in G, if two points are not joined then they must both be in B. So there can be at most n-1 pairs of unjoined points. If two points are joined then at least one of them must be in A (since if both are in B they are unjoined). But A has less than m points, so if there are m pairs of joined points, then they must overlap. Thus G shows that f(m, n) > m+2n-2. We now prove by induction on m that any graph with m-1+2n points (and m  m and we have already established the result for n-1. Since m-1+2(n-1) = m we can continue in this way until we obtain m pairs of joined points (one of each pair in A and the other in B). Contradiction. So the result is true for n also. It remains to consider the case n = m. To get the n-1 pairs we use the induction on m. G has m-1+2m = 3m-1 points and does not have either m pairs of joined points or m pairs of unjoined points. We know by induction that f(m-1, m) = m-1+2m-1 = 3m-2, so f(m, m-1) = 3m-2 also. Since m-1+2m > 3m-2, G has either m pairs of joined points or m-1 points of unjoined points. It does not have the former (by assumption) so it must have the latter. We now proceed as before. So B is a set of m-1 pairs of unjoined points and A the set of the remaining m+1 points. We get m-1 pairs of joined points (one of each pair in A and the other in B). But now wehave run out of unmarked pairs in B. However, there are still two points in A. If they are not joined, then we would have a contradiction, because with the pairs in B they would give n pairs of unjoined points. So they are joined and hence give us an mth joined pair. Contradiction. Comment. This all seems complicated, and certainly much harder work than questions 1-3. So maybe I am missing something. Does anyone have a simpler solution? 15th APMO 2003 (C) John Scholes jscholes@kalva.demon.co.uk 14 Aug 2003 Last corrected/updated 14 Aug 03