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

 
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== Problem ==
 
== Problem ==
  
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For what real values of <math>\displaystyle x</math> is
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For what real values of <math>x</math> is
  
 
<center>
 
<center>
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</center>
 
</center>
  
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given (a) <math>A=\sqrt{2}</math>, (b) <math>\displaystyle A=1</math>, (c) <math>\displaystyle A=2</math>, we only non-negative real numbers are admitted for square roots?
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given (a) <math>A=\sqrt{2}</math>, (b) <math>A=1</math>, (c) <math>A=2</math>, we only non-negative real numbers are admitted for square roots?
  
 
== Solution ==
 
== Solution ==
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<center>
 
<center>
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<math>\displaystyle A^2 = 2(x+|x-1|)</math>
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<math>A^2 = 2(x+|x-1|)</math>
 
</center>
 
</center>
  
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If <math>\displaystyle x \le 1</math>, then we must clearly have <math>\displaystyle A^2 =2</math>.  Otherwise, we have
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If <math>x \le 1</math>, then we must clearly have <math>A^2 =2</math>.  Otherwise, we have
  
 
<center>
 
<center>
 
<math>x = \frac{A^2 + 2}{4} > 1,</math>
 
<math>x = \frac{A^2 + 2}{4} > 1,</math>
  
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<math>\displaystyle A^2 > 2 </math>
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<math>A^2 > 2 </math>
 
</center>
 
</center>
  
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Hence for (a) the solution is <math> x \in \left[ \frac{1}{2}, 1 \right]</math>, for (b) there is no solution, since we must have <math>\displaystyle A^2 \ge 2</math>, and for (c), the only solution is <math> x=\frac{3}{2}</math>.  Q.E.D.
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Hence for (a) the solution is <math> x \in \left[ \frac{1}{2}, 1 \right]</math>, for (b) there is no solution, since we must have <math>A^2 \ge 2</math>, and for (c), the only solution is <math> x=\frac{3}{2}</math>.  Q.E.D.
  
 
{{Alternate solutions}}
 
{{Alternate solutions}}
  
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== Resources ==
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{{IMO box|year=1959|num-b=1|num-a=3}}
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* [[1959 IMO Problems]]
 
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* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=341492#p341492 Discussion on AoPS/MathLinks]
 
  
 
[[Category:Olympiad Algebra Problems]]
 
[[Category:Olympiad Algebra Problems]]

Revision as of 20:21, 25 October 2007

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) $A=2$, 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

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

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.

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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