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Difference between revisions of "1960 IMO Problems/Problem 2"
 
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==Solution==
 
==Solution==
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{{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>
 +
 
 +
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>.
 +
 
 +
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>.
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 +
==Solution 2==
 +
 
 +
If <math>x \neq 0</math>, then the LHS is defined and rewrites as follows:
 +
 
 +
\begin{align*}
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\frac{4x^2}{(1 - \sqrt{2x + 1})^2} &= \biggl(\frac{2x}{1 - \sqrt{2x + 1}}\biggl)^2 \\
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&= \biggl( \frac{2x}{1 - \sqrt{2x + 1}} \cdot \frac{1 + \sqrt{2x + 1}}{1 + \sqrt{2x + 1}} \biggl)^2 \\
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&= (1 + \sqrt{2x + 1})^2 \\
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&= 2x + 2\sqrt{2x + 1} + 2.
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\end{align*}
 +
 
 +
The inequality therefore holds if and only if
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<cmath>2x + 2\sqrt{2x + 1} + 2 < 2x + 9.</cmath>
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or
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<cmath>\sqrt{2x + 1} < \frac{7}{2}.</cmath>
 +
 
 +
So <math>2x + 1 < 49/4</math> and therefore <math>x < 45/8</math>. But if <math>x < -1/2</math> then the inequality makes no sense, since <math>\sqrt{2x + 1}</math> is imaginary. So the original inequality holds iff <math>x</math> is in <math>[-1/2, 0) \cup (0, 45/8).</math>
  
 
==See Also==
 
==See Also==
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{{IMO box|year=1960|num-b=1|num-a=3}}
 
{{IMO box|year=1960|num-b=1|num-a=3}}
  
 +
[[Category:Olympiad Algebra Problems]]
 
[[Category:Olympiad Inequality Problems]]
 
[[Category:Olympiad Inequality Problems]]

Latest revision as of 04:51, 7 April 2024

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 $x=0$.

Solution 2

If $x \neq 0$, then the LHS is defined and rewrites as follows:

\begin{align*} \frac{4x^2}{(1 - \sqrt{2x + 1})^2} &= \biggl(\frac{2x}{1 - \sqrt{2x + 1}}\biggl)^2 \\ &= \biggl( \frac{2x}{1 - \sqrt{2x + 1}} \cdot \frac{1 + \sqrt{2x + 1}}{1 + \sqrt{2x + 1}} \biggl)^2 \\ &= (1 + \sqrt{2x + 1})^2 \\ &= 2x + 2\sqrt{2x + 1} + 2. \end{align*}

The inequality therefore holds if and only if \[2x + 2\sqrt{2x + 1} + 2 < 2x + 9.\] or \[\sqrt{2x + 1} < \frac{7}{2}.\]

So $2x + 1 < 49/4$ and therefore $x < 45/8$. But if $x < -1/2$ then the inequality makes no sense, since $\sqrt{2x + 1}$ is imaginary. So the original inequality holds iff $x$ is in $[-1/2, 0) \cup (0, 45/8).$

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
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