Difference between revisions of "1959 IMO Problems/Problem 2"

Revision as of 13:58, 15 December 2019

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: $A^2 = \Big( x + \sqrt{2x - 1}\Big) + 2 \sqrt{x + \sqrt{2x - 1}} \sqrt{x - \sqrt{2x - 1}} + \Big( x - \sqrt{2x - 1}\Big)$

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

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

Since the term inside the square root is a perfect square, and by factoring 2 out, we get $A^2 = 2(x + \sqrt{(x-1)^2})$ Use the property that $\sqrt{x^2}=x$ to get $A^2 = 2(x+|x-1|)$

Case I: If $x \le 1$ , then $|x-1| = 1 - x$ , and the equation reduces to $A^2 = 2$ . This is precisely part (a) of the question, for which the valid interval is now $x \in \left[ \frac{1}{2}, 1 \right]$

Case II: If $x > 1$ , we have $x = \frac{A^2 + 2}{4} > 1,$ $A^2 > 2$

Hence for , 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.

@@ Line 29: / Line 29: @@
 Case II: If <math>x > 1</math>, we have
 <cmath>x = \frac{A^2 + 2}{4} > 1,</cmath>
 <cmath>A^2 > 2 </cmath>

1959 IMO (Problems) • Resources
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Difference between revisions of "1959 IMO Problems/Problem 2"

Revision as of 13:58, 15 December 2019

Problem

Solution

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