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

(New page: ==Problem== ==Solution== {{solution}} ==See Also== {{IMO box|year=1960|num-b=1|num-a=3}})
 
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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==
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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>.
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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>.
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==Solution 2==
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If <math>x \neq 0</math>, then the LHS is defined and rewrites as follows:
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\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*}
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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>
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 +
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>
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==Solution 3 (A bit of light bash)==
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 +
We consider when does the equality holds:
 +
 
 +
<cmath>\frac{4x^2}{(1 - \sqrt {2x + 1})^2} = 2x + 9 </cmath>
 +
 
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<cmath> 4x^2 = (2x+9)(1 - \sqrt{2x+1})^2 \Rightarrow 4x^2 = (2x+9)(2x+2-2\sqrt{2x+1})</cmath>
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 +
Noting that we can cancel out the <math>4x^2</math>, we expand RHS and obtain: <math>4x^2 = 4x^2 + 22x + 18 - (4x+18)\sqrt{2x+1}</math>, which simplifies to <math>(4x+18)\sqrt{2x+1} = 22x + 18</math>, squaring both sides again, we yield:
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<cmath>32x^3 + 304x^2 + 792x + 324 = 484x^2 + 792x + 324 \Rightarrow 32x^3 = 180x^2 \Rightarrow x^2(32x-180) = 0</cmath>
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 +
From here we get <math>x = \{0, \frac{45}{8}\}</math>, but when <math>x = 0</math>, the original expression is undefined as denominator is 0. Therefore, the only valid solution is <math>x = \frac{45}{8}</math>.
 +
 
 +
Then we consider the function <math>f(x) = \frac{4x^2}{(1 - \sqrt {2x + 1})^2} - (2x + 9) = (1+\sqrt{2x+1})^2-(2x+9)</math>, we aim to find the derivative of <math>f(x)</math>, which is:
 +
 
 +
<cmath>f'(x) = \frac{2(1+\sqrt{2x+1})}{\sqrt{2x+1}}-2 = \frac{2}{\sqrt{2x+1}}+2-2 = \frac{2}{\sqrt{2x+1}} > 0</cmath>
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Therefore we established that <math>f'(x)</math> is monotonous and increasing. We are asked to find the values of <math>x</math> such that <math>\frac{4x^2}{(1 - \sqrt{2x + 1})^2} < 2x + 9</math>, in other words, we are finding <math>f(x)<0</math>. By the monotonous and increasing property of the function <math>f</math>, <math>f(x)<0</math> when <math>x<\frac{45}{8}</math>
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However, there are two other special restrictions:
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1. <math>x</math> cannot be 0, as that will result in <math>(1 - \sqrt{2x + 1})^2 = 0</math>, making the fraction indeterminate.
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2. <math>x \geq -\frac{1}{2}</math>, as <math>2x+1 \geq 0</math> so that the square root is determinate.
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In conclusion, inequality hold for <math>\boxed{x \in [-\frac{1}{2}, 0) \cup (0, \frac{45}{8})}</math>.
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 +
~[[User:IDKHowtoaddsolution|IDKHowtoaddsolution]]
  
 
==See Also==
 
==See Also==
  
 
{{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]]

Latest revision as of 12:14, 4 September 2025

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 $[-1/2, 0) \cup (0, 45/8).$

Solution 3 (A bit of light bash)

We consider when does the equality holds:

\[\frac{4x^2}{(1 - \sqrt {2x + 1})^2} = 2x + 9\]

\[4x^2 = (2x+9)(1 - \sqrt{2x+1})^2 \Rightarrow 4x^2 = (2x+9)(2x+2-2\sqrt{2x+1})\]

Noting that we can cancel out the $4x^2$, we expand RHS and obtain: $4x^2 = 4x^2 + 22x + 18 - (4x+18)\sqrt{2x+1}$, which simplifies to $(4x+18)\sqrt{2x+1} = 22x + 18$, squaring both sides again, we yield:

\[32x^3 + 304x^2 + 792x + 324 = 484x^2 + 792x + 324 \Rightarrow 32x^3 = 180x^2 \Rightarrow x^2(32x-180) = 0\]

From here we get $x = \{0, \frac{45}{8}\}$, but when $x = 0$, the original expression is undefined as denominator is 0. Therefore, the only valid solution is $x = \frac{45}{8}$.

Then we consider the function $f(x) = \frac{4x^2}{(1 - \sqrt {2x + 1})^2} - (2x + 9) = (1+\sqrt{2x+1})^2-(2x+9)$, we aim to find the derivative of $f(x)$, which is:

\[f'(x) = \frac{2(1+\sqrt{2x+1})}{\sqrt{2x+1}}-2 = \frac{2}{\sqrt{2x+1}}+2-2 = \frac{2}{\sqrt{2x+1}} > 0\]

Therefore we established that $f'(x)$ is monotonous and increasing. We are asked to find the values of $x$ such that $\frac{4x^2}{(1 - \sqrt{2x + 1})^2} < 2x + 9$, in other words, we are finding $f(x)<0$. By the monotonous and increasing property of the function $f$, $f(x)<0$ when $x<\frac{45}{8}$

However, there are two other special restrictions:

1. $x$ cannot be 0, as that will result in $(1 - \sqrt{2x + 1})^2 = 0$, making the fraction indeterminate.

2. $x \geq -\frac{1}{2}$, as $2x+1 \geq 0$ so that the square root is determinate.

In conclusion, inequality hold for $\boxed{x \in [-\frac{1}{2}, 0) \cup (0, \frac{45}{8})}$.

~IDKHowtoaddsolution

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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