10th Balkan 1993 Problem 1 Given reals a 1 ≤ a 2 ≤ a 3 ≤ a 4 ≤ a 5 ≤ a 6 satisfying a 1 + a 2 + a 3 + a 4 + a 5 + a 6 = 10 and (a 1 - 1) 2 + (a 2 - 1) 2 + (a 3 - 1) 2 + (a 4 - 1) 2 + (a 5 - 1) 2 + (a 6 - 1) 2 = 6, what is the largest possible a 6 ? Solution
: a 6 = 3 1/3. Adding twice the first equation to the second gives a 1 2 + a 2 2 + ... + a 6 2 = 20. Hence (a 1 - 4/3) 2 + (a 2 - 4/3) 2 + (a 3 - 4/3) 2 + (a 4 - 4/3) 2 + (a 5 - 4/3) 2 + (a 6 - 4/3) 2 = 20 - 10·8/3 + 6·16/9 = 4. So a 6 - 4/3 ≤ 2, or a 6 ≤ 3 1/3. It is easy to check that a 1 = a 2 = a 3 = a 4 = a 5 = 4/3, a 6 = 3 1/3 is indeed a solution. Comment. Note that this is similar to 85/2 10th Balkan 1993 © John Scholes jscholes@kalva.demon.co.uk 11 Apr 2002