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1960 IMO Problems/Problem 2: Difference between revisions

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So, answer: <math>-\frac{1}{2} \le x<\frac{45}{8}</math>, except <math>x=0</math>.
So, answer: <math>-\frac{1}{2} \le x<\frac{45}{8}</math>, except <math>x=0</math>.
==Solution 2==
If <math>x \neq 0</math>, 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*}
Plugging this into the inequality, we find
<cmath>2x + 2\sqrt{2x + 1} + 2 < 2x + 9.</cmath>
or
<cmath>\sqrt{2x + 1} < \frac{7}{2}.</cmath>
The original inequality therefore holds whenever <math>2x + 1 < 49/4</math>, i.e. <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 if <math>x</math> is in <math>[-1/2, 0) \cup (0, 45/8).</math>


==See Also==
==See Also==

Revision as of 19:24, 6 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*}

Plugging this into the inequality, we find \[2x + 2\sqrt{2x + 1} + 2 < 2x + 9.\] or \[\sqrt{2x + 1} < \frac{7}{2}.\]

The original inequality therefore holds whenever $2x + 1 < 49/4$, i.e. $x < 45/8$. But If $x < -1/2$ then the inequality makes no sense, since $\sqrt{2x + 1}$ is imaginary. So the original inequality holds if $x$ 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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