IMO 1959/A2 solution

IMO 1959

Problem A2

For what real values of x is √(x + √(2x-1)) + √(x - √(2x-1)) = A, given (a) A = √2, (b) A = 1, (c) A = 2, where only non-negative real numbers are allowed in square roots and the root always denotes the non-negative root?

Answer

(a) any x in the interval [1/2,1]; (b) no solutions; (c) x=3/2.

Solution

Note that we require x ≥ 1/2 to avoid a negative sign under the inner square roots. Since (x-1)² ≥ 0, we have x ≥ √(2x-1), so there is no difficulty with √(x - √(2x-1)), provided that x ≥ 1/2.

Squaring gives 2x + 2√(x²-2x+1) = A². Note that the square root is |x-1|, not simply (x-1). So we get finally 2x + 2|x-1| = A². It is now easy to see that we get the solutions above.

Solutions are also available in: Samuel L Greitzer, International Mathematical Olympiads 1959-1977, MAA 1978, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.