1959 IMO Problems/Problem 2

Problem

For what real values of $\displaystyle x$ is

$\sqrt{x+\sqrt{2x-1}} + \sqrt{x-\sqrt{2x-1}} = A,$

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

Solution

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

$\displaystyle A^2 = 2(x+|x-1|)$

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

$x = \frac{A^2 + 2}{4} > 1,$

$\displaystyle A^2 > 2$

Hence for (a) the solution is $x \in \left[ \frac{1}{2}, 1 \right]$ , for (b) there is no solution, since we must have $\displaystyle A^2 \ge 2$ , 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.

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1959 IMO Problems/Problem 2

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