1959 IMO Problems/Problem 2

Problem

For what real values of $x$ is

given (a) $A=\sqrt{2}$ , (b) $A=1$ , (c) $A=2$ , where only non-negative real numbers are admitted for square roots?

We note that the square roots imply that $x\ge \frac{1}{2}$ . We now square both sides and simplify to obtain

If $x \le 1$ , then we must clearly have $A^2 =2$ . Otherwise, we have

Hence for (a) the solution is $x \in \left[ \frac{1}{2}, 1 \right]$ , for (b) there is no solution, since we must have $A^2 \ge 2$ , and for (c), the only solution is $x=\frac{3}{2}$ . Q.E.D.

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

1959 IMO (Problems)
Preceded by Problem 1	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 3