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

Revision as of 13:10, 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.

@@ Line 15: / Line 15: @@
 If <math>x \le 1</math>, then we must clearly have <math>A^2 =2</math>.  Otherwise, we have
-<center>
-<math>x = \frac{A^2 + 2}{4} > 1,</math>
-<math>A^2 > 2 </math>
+<cmath>x = \frac{A^2 + 2}{4} > 1,</cmath>
-</center>
+<cmath>A^2 > 2 </cmath>
 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.

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

Revision as of 13:10, 15 December 2019

Problem

Solution

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