42nd IMO 2001 shortlist Problem N2 Find the largest real k such that if a, b, c, d are positive integers such that a + b = c + d, 2ab = cd and a ≥ b, then a/b ≥ k. Solution
: k = 3 + 2√2. A few examples of such a, b, c, d are: 2 + 12 = 6 + 8, 2.2.12 = 6.8, so a/b = 6 2 + 15 = 5 + 12, 2.2.15 = 5.12, so a/b = 7.5 3 + 20 = 8 + 15, 2.3.20 = 8.15, so a/b = 6.67 We have (a+b) 2 - 8ab = (c+d) 2 - 4cd, so (a/b) 2 - 6(a/b) + 1 = ((c-d)/b) 2 ≥ 0. The plot of x 2 - 6x + 1 is shown below. It has zeros at x = 3 - 2√2  3 + 2√2. The harder part is showing that we can get as close as we like to 3 + 2√2 and hence that k = 3 + 2√2. The idea is to show that we can make (c-d)/b as small as we like. Note first that (a+b) 2 - 4ab = (c+d) 2 - 2cd, so (a-b) 2 = c 2 + d 2 . In other words, a-b, c, d is a Pythagorean triple. It is well-known that r 2 + s 2 , r 2 - s 2 , 2rs are such triples for any integers r, s. Then we have a+b = c+d = r 2 + 2rs - s 2 . Hence a = r 2 + rs, b = rs - s 2 . It is now easy to check that a = r 2 + rs, b = rs - s 2 , c = r 2 - s 2 , d = 2rs satisfy a + b = c + d, 2ab = cd. Suppose we specialise to c - d = 1 (then if we could get arbitrarily large b we would be home). That requires r 2 - s 2 - 2rs = 1, or (r - s) 2 - 2s 2 = 1. We recognise that as a Pell equation. It has a solution r - s = 3, s = 2. But if (R, S) is a solution to x 2 - 2y 2 = 1, so is (3R+4S, 2R+3S). So we can get arbitrarily large r - s and s such that c - d = 1. Then b = rs - s 2 = (r - s)s, which can therefore be made arbitrarily large as required. For example, r - s = 3, s = 2, gives r = 5, so a = 35, b = 6, c = 21, d = 20. Then a/b = 5.8333. The next solution is r - s = 17, s = 12, so r = 29, so a = 1189, b = 204, c = 697, d = 696 and a/b = 5.828431 (cf 3 + 2√2 = 5.828427). 42nd IMO shortlist 2001 © John Scholes jscholes@kalva.demon.co.uk 14 Oct 2002