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

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== Solution ==
 
== Solution ==
  
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The square roots imply that <cmath>x\ge \frac{1}{2}</cmath> Square both sides to get <cmath> \Big( x + \sqrt{2x - 1}\Big)  + 2 \sqrt{x + \sqrt{2x - 1}}  \sqrt{x - \sqrt{2x - 1}} +  \Big( x - \sqrt{2x - 1}\Big) </cmath>and simplify to obtain <cmath>A^2 = 2(x+|x-1|)</cmath>  
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Firstly, the square roots imply that a valid domain for x  is <math>x\ge \frac{1}{2}</math>.
 +
 
 +
Square both sides of the given equation to get <cmath> \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</cmath>and simplify to obtain <cmath>A^2 = 2(x+|x-1|)</cmath>  
  
 
If <math>x \le 1</math>, then we must clearly have <math>A^2 =2</math>.  Otherwise, we have
 
If <math>x \le 1</math>, then we must clearly have <math>A^2 =2</math>.  Otherwise, we have

Revision as of 13:21, 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 to get \[\Big( x + \sqrt{2x - 1}\Big) + 2 \sqrt{x + \sqrt{2x - 1}} \sqrt{x - \sqrt{2x - 1}} + \Big( x - \sqrt{2x - 1}\Big) = A^2\]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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