1959 IMO Problems/Problem 2

Problem

For what real values of $x$ is

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

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

Solution

Firstly, the square roots imply that a valid domain for x is $x\ge \frac{1}{2}$ .

Square both sides of the given equation: $\Big( x + \sqrt{2x - 1}\Big) + 2 \sqrt{x + \sqrt{2x - 1}} \sqrt{x - \sqrt{2x - 1}} + \Big( x - \sqrt{2x - 1}\Big) = A^2$

Add the first and the last terms to get $2x + 2 \sqrt{x + \sqrt{2x - 1}} \sqrt{x - \sqrt{2x - 1}} = A^2$

Multiply the middle terms, and use $(a + b)(a - b) = a^2 - b^2$ to get: $2x + 2 \sqrt{x^2 - 2x + 1} = A^2<cmath>

Since the term inside the square root is a perfect square, and by factoring 2 out, we get </cmath>2(x + \sqrt{(x-1)^2}) = A^2<cmath> Use the property that </cmath>\sqrt{x^2}=x<cmath> to get </cmath>A^2 = 2(x+|x-1|)$ (Error compiling LaTeX. Unknown error_msg) $If$ x \le 1 $, then we must clearly have$ A^2 =2$. Otherwise, we have

Hence for (a) the solution is$ (Error compiling LaTeX. Unknown error_msg) 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.

~flamewavelight (Expanded)

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

1959 IMO (Problems) • Resources
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1959 IMO Problems/Problem 2

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