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

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==Problem==
 
==Problem==
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For what values of the variable <math>x</math> does the following inequality hold:
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<cmath>\dfrac{4x^2}{(1 - \sqrt {2x + 1})^2} < 2x + 9 \ ?</cmath>
  
 
==Solution==
 
==Solution==
{{solution}}
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Set <math>x = -\frac{1}{2} + \frac{a^2}{2}</math>, where <math>a\ge0</math>.
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<math>\frac{4\left(-\frac{1}{2}+\frac{a^2}{2}\right)^2}{\left(1-\sqrt{1+2\left(-\frac{1}{2}+\frac{a^2}{2}\right)}\right)^2}<2\left(-\frac{1}{2}+\frac{a^2}{2}\right)+9</math>
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After simplifying, we get
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<math>(a+1)^2<a^2+8</math>
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So
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<math>a^2+2a+1<a^2+8</math>
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Which gives <math>a<\frac{7}{2}</math> and hence <math>-\frac{1}{2} \le x<\frac{45}{8}</math>.
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But <math>x=0</math> makes the LHS indeterminate.
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So, answer: <math>-\frac{1}{2} \le x<\frac{45}{8}</math>, except <math>x=0</math>.
  
 
==See Also==
 
==See Also==
  
 
{{IMO box|year=1960|num-b=1|num-a=3}}
 
{{IMO box|year=1960|num-b=1|num-a=3}}
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[[Category:Olympiad Algebra Problems]]
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[[Category:Olympiad Inequality Problems]]

Latest revision as of 11:41, 19 June 2017

Problem

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

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

Solution

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

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 $-\frac{1}{2} \le x<\frac{45}{8}$.

But $x=0$ makes the LHS indeterminate.

So, answer: $-\frac{1}{2} \le x<\frac{45}{8}$, except $x=0$.

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