Difference between revisions of "1960 IMO Problems/Problem 2"

Revision as of 11:07, 28 December 2007

For what values of the variable $x$ does the following inequality hold:

$\dfrac{4x^2}{(1 - \sqrt {2x + 1})^2} < 2x + 9 \ ?$

Set $x = -\frac{1}{2} + \frac{a^2}{2}$ , where $a\ge0$ . $\frac{4(-\frac{1}{2}+\frac{a^2}{2})^2}{(1-\sqrt{1+2(-\frac{1}{2}+\frac{a^2}{2})})^2}<2x+9$

After simplifying, we get $(a+1)^2<a^2+8$

So $a^2+2a+1<a^2+8$

Which gives $a<\frac{7}{2}$ and hence $x<\frac{45}{8}$ .

But $x=0$ makes the RHS indeterminate.

So, answer: $x<\frac{45}{8}$ , except $x=0$ .

@@ Line 5: / Line 5: @@
 ==Solution==
-{{solution}}
+Set <math>x = -\frac{1}{2} + \frac{a^2}{2}</math>, where <math>a\ge0</math>.
+<math>\frac{4(-\frac{1}{2}+\frac{a^2}{2})^2}{(1-\sqrt{1+2(-\frac{1}{2}+\frac{a^2}{2})})^2}<2x+9</math>
+After simplifying, we get
+<math>(a+1)^2<a^2+8</math>
+So
+<math>a^2+2a+1<a^2+8</math>
+Which gives <math>a<\frac{7}{2}</math> and hence <math>x<\frac{45}{8}</math>.
+But <math>x=0</math> makes the RHS indeterminate.
+So, answer: <math>x<\frac{45}{8}</math>, except <math>x=0</math>.
 ==See Also==

1960 IMO (Problems) • Resources
Preceded by Problem 1	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 3
All IMO Problems and Solutions