IMO 1995 Problem B1 Find the maximum value of x 0 for which there exists a sequence x 0 , x 1 , ... , x 1995 of positive reals with x 0 = x 1995 such that for i = 1, ... , 1995: x i-1 + 2/x i-1 = 2x i + 1/x i .
2 997 . Solution The relation given is a quadratic in x i , so it has two solutions, and by inspection these are x i = 1/x i-1 and x i-1 /2. For an even number of moves we can start with an arbitrary x 0 and get back to it. Use n-1 halvings, then take the inverse, that gets to 2 n-1 /x 0 after n moves. Repeating brings you back to x 0 after 2n moves. However, 1995 is odd! The sequence given above brings us back to x 0 after n moves, provided that x 0 = 2 (n-1)/2 . We show that this is the largest possible x 0 . Suppose we have a halvings followed by an inverse followed by b halvings followed by an inverse. Then if the number of inverses is odd we end up with 2 a-b+c- ... /x 0 , and if it is even we end up with x 0 /2 a-b+c- ... . In the first case, since the final number is x 0 we must have x 0 = 2 (a-b+-...)/2 . All the a, b, ... are non-negative and sum to the number of moves less the number of inverses, so we clearly maximise x 0 by taking a single inverse and a = n-1. In the second case, we must have 2 a-b+c- ... = 1 and hence a - b + c - ... = 0. But that implies that a + b + c + ... is even and hence the total number of moves is even, which it is not. So we must have an odd number of inverses and the maximum value of x 0 is 2 (n-1)/2 . 36th IMO 1995 © John Scholes jscholes@kalva.demon.co.uk 24 Oct 1998 Last corrected/updated 25 Nov 03