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

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== Solution ==
 
== Solution ==
  
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We note that the square roots imply that <math>x\ge \frac{1}{2}</math>. We now square both sides and simplify to obtain
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== See Also ==
 
== See Also ==

Revision as of 12:58, 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

The square roots imply that \[x\ge \frac{1}{2}\] Square both sides and simplify to obtain

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

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

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

$A^2 > 2$

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

See Also

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