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

Revision as of 18:13, 8 June 2008

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

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

@@ Line 7: / Line 7: @@
 </center>
-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?
+given (a) <math>A=\sqrt{2}</math>, (b) <math>A=1</math>, (c) <math>A=2</math>, where only non-negative real numbers are admitted for square roots?
 == Solution ==

Page

Toolbox

Search

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

Revision as of 18:13, 8 June 2008

Problem

Solution