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Difference between revisions of "1965 IMO Problems/Problem 1"

(Created page with '== Problem == Determine all values <math>x</math> in the interval <math>0\leq x\leq 2\pi </math> which satisfy the inequality <cmath>2\cos x \leq \left| \sqrt{1+\sin 2x} - \sqrt{…')
 
(solution)
 
(9 intermediate revisions by 2 users not shown)
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<cmath>2\cos x \leq \left| \sqrt{1+\sin 2x} - \sqrt{1-\sin 2x } \right| \leq \sqrt{2}.</cmath>
 
<cmath>2\cos x \leq \left| \sqrt{1+\sin 2x} - \sqrt{1-\sin 2x } \right| \leq \sqrt{2}.</cmath>
  
 +
== Solution 1 ==
 +
We shall deal with the left side of the inequality first (<math>2\cos x \leq \left| \sqrt{1+\sin 2x} - \sqrt{1-\sin 2x } \right| </math>) and the right side after that.
  
−
== Solution ==
+
It is clear that the left inequality is true when <math>\cos x</math> is non-positive, and that is when <math>x</math> is in the interval <math>[\pi/2, 3\pi/2]</math>. We shall now consider when <math>\cos x</math> is positive. We can square the given inequality, and the resulting inequality will be true whenever the original left inequality is true. <math>4\cos^2{x}\leq 1+\sin 2x+1-\sin 2x-2\sqrt{1-\sin^2 2x}=2-2\sqrt{\cos^2{2x}}</math>. This inequality is equivalent to <math>2\cos^2 x\leq 1-\left| \cos 2x\right|</math>. I shall now divide this problem into cases.
−
{{solution}}
+
 
 +
Case 1: <math>\cos 2x</math> is non-negative. This means that <math>x</math> is in one of the intervals <math>[0,\pi/4]</math> or <math>[7\pi/4, 2\pi]</math>. We must find all <math>x</math> in these two intervals such that <math>2\cos^2 x\leq 1-\cos 2x</math>. This inequality is equivalent to <math>2\cos^2 x\leq 2\sin^2 x</math>, which is only true when <math>x=\pi/4</math> or <math>7\pi/4</math>.
 +
 
 +
Case 2: <math>\cos 2x</math> is negative. This means that <math>x</math> is in one of the interavals <math>(\pi/4, \pi/2)</math> or <math>(3\pi/2, 7\pi/4)</math>. We must find all <math>x</math> in these two intervals such that <math>2\cos^2 x\leq 1+\cos 2x</math>, which is equivalent to <math>2\cos^2 x\leq 2\cos^2 x</math>, which is true for all <math>x</math> in these intervals.
 +
 
 +
Therefore the left inequality is true when <math>x</math> is in the union of the intervals <math>[\pi/4, \pi/2)</math>, <math>(3\pi/2, 7\pi/4]</math>, and <math>[\pi/2, 3\pi/2]</math>, which is the interval <math>[\pi/4, 7\pi/4]</math>. We shall now deal with the right inequality.
 +
 
 +
As above, we can square it and have it be true whenever the original right inequality is true, so we do that. <math>2-2\sqrt{\cos^2{2x}}\leq 2</math>, which is always true. Therefore the original right inequality is always satisfied, and all <math>x</math> such that the original inequality is satisfied are in the interval <math>[\pi/4, 7\pi/4]</math>.
 +
 
 +
==Solution 2==
 +
Manipulate the inequality so that it becomes:
 +
<cmath>\sqrt{2}\cos x \leq \left| \sqrt{\frac{1+\cos(\frac{\pi}{2}-2x)}{2}} - \sqrt{\frac{1-\cos(\frac{\pi}{2}- 2x)}{2}} \right| \leq 1</cmath>
 +
Inside the absolute value, we identify the half-angle formulas. However, since we do not know the sign of the resultant expression, we have to use absolute value signs since the principal square root is always positive; then the inequality becomes
 +
<cmath>\sqrt{2}\cos x \leq \left| \left|\cos\left(\frac{\pi}{4}-x\right)\right| - \left|\sin\left(\frac{\pi}{4}-x\right)\right|\right| \leq 1</cmath>
 +
which, considering absolute value, is the same as
 +
<cmath>\sqrt{2}\cos x \leq \left| \left|\cos\left(x-\frac{\pi}{4}\right)\right| - \left|\sin\left(x-\frac{\pi}{4}\right)\right|\right| \leq 1</cmath>
 +
Since both inner absolute values range between <math>0</math> and <math>1</math>, their positive difference also ranges from <math>0</math> to <math>1</math>, so the right inequality is always true. Thus we take a look at the left inequality. For <math>x\in\left[\frac{\pi}{2},\frac{3\pi}{2}\right]</math>, the left-hand side is never positive; however, the right-hand side is always nonnegative due to absolute value, so the inequality holds for these <math>x</math>-values. As a result, we consider the remaining two sections of the interval.
 +
 
 +
For <math>x\in\left[0,\frac{\pi}{4}\right]</math> and <math>x\in\left[\frac{7\pi}{4},2\pi\right]</math>, when considering the <math>-\frac{\pi}{4}</math>, inside the absolute value, cosine is positive but sine is negative; thus our inequality becomes
 +
\begin{align*}
 +
\sqrt{2}\cos x &\le \left|\cos\left(x-\frac{\pi}{4}\right) +\sin\left(x-\frac{\pi}{4}\right)\right|\\
 +
\cos x &\le \left|\cos\left(x-\frac{\pi}{4}\right)\cdot\frac{\sqrt{2}}{2} +\sin\left(x-\frac{\pi}{4}\right)\cdot\frac{\sqrt{2}}{2}\right|\\
 +
\cos x &\le \left|\cos\left(x-\frac{\pi}{4}\right)\sin\left(\frac{\pi}{4}\right) +\sin\left(x-\frac{\pi}{4}\right)\cos\left(\frac{\pi}{4}\right)\right|\\
 +
\cos x &\le \left|\sin\left(\left(x-\frac{\pi}{4}\right)+\frac{\pi}{4}\right)\right|\\
 +
\cos x &\le |\sin x|\\
 +
\end{align*}
 +
If <math>x\in\left[0,\frac{\pi}{4}\right]</math>, then due to the interval, <math>\sin x</math> and <math>\cos x</math> are both never negative; thus,
 +
\begin{align*}
 +
\cos x &\le \sin x\\
 +
1&\le\tan x\\
 +
\end{align*}
 +
But this is never true in our interval (except at one endpoint, which is included in the interval below), so the inequality does not hold.
 +
 
 +
If <math>x\in\left[\frac{7\pi}{4},2\pi\right]</math>, then due to the interval, <math>\sin x</math> is negative and <math>\cos x</math> is positive; thus,
 +
\begin{align*}
 +
\cos x &\le -\sin x\\
 +
1&\le-\tan x\\
 +
-1&\ge\tan x\\
 +
\end{align*}
 +
But this is also never true in our interval (except at one endpoint, which is included in the interval below), so the inequality does not hold.
 +
 
 +
For <math>x\in\left[\frac{\pi}{4},\frac{\pi}{2}\right]</math>, inside the absolute value, when considering the <math>-\frac{\pi}{4}</math>, both cosine and sine are positive; thus our inequality becomes
 +
\begin{align*}
 +
\sqrt{2}\cos x &\le \left|\cos\left(x-\frac{\pi}{4}\right) -\sin\left(x-\frac{\pi}{4}\right)\right|\\
 +
\cos x &\le \left|\cos\left(x-\frac{\pi}{4}\right)\cdot\frac{\sqrt{2}}{2} -\sin\left(x-\frac{\pi}{4}\right)\cdot\frac{\sqrt{2}}{2}\right|\\
 +
\cos x &\le \left|\cos\left(x-\frac{\pi}{4}\right)\cos\left(\frac{\pi}{4}\right) -\sin\left(x-\frac{\pi}{4}\right)\sin\left(\frac{\pi}{4}\right)\right|\\
 +
\cos x &\le \left|\cos\left(\left(x-\frac{\pi}{4}\right)+\frac{\pi}{4}\right)\right|\\
 +
\cos x &\le |\cos x|\\
 +
\end{align*}
 +
This is always true, so the inequality holds.
 +
 
 +
For <math>x\in\left[\frac{3\pi}{2},\frac{7\pi}{4}\right]</math>, inside the absolute value, when considering the <math>-\frac{\pi}{4}</math>, both cosine and sine are negative; thus our inequality becomes
 +
\begin{align*}
 +
\sqrt{2}\cos x &\le \left|-\cos\left(x-\frac{\pi}{4}\right) +\sin\left(x-\frac{\pi}{4}\right)\right|\\
 +
\sqrt{2}\cos x &\le \left|-\left(\cos\left(x-\frac{\pi}{4}\right) -\sin\left(x-\frac{\pi}{4}\right)\right)\right|\\
 +
\sqrt{2}\cos x &\le \left|\cos\left(x-\frac{\pi}{4}\right) -\sin\left(x-\frac{\pi}{4}\right)\right|\\
 +
\end{align*}
 +
This is the same as the case for <math>x\in\left[\frac{\pi}{4},\frac{\pi}{2}\right]</math>, so the inequality holds.
 +
 
 +
Combining our findings, we find that the solutions are <math>\boxed{x\in\left[\frac{\pi}{4},\frac{7\pi}{4}\right]}</math>.
 +
 
 +
~eevee9406
 +
 
 +
{{IMO box|year=1965|before=First Question|num-a=2}}

Latest revision as of 19:01, 15 March 2025

Problem

Determine all values $x$ in the interval $0\leq x\leq 2\pi$ which satisfy the inequality \[2\cos x \leq \left| \sqrt{1+\sin 2x} - \sqrt{1-\sin 2x } \right| \leq \sqrt{2}.\]

Solution 1

We shall deal with the left side of the inequality first ($2\cos x \leq \left| \sqrt{1+\sin 2x} - \sqrt{1-\sin 2x } \right|$) and the right side after that.

It is clear that the left inequality is true when $\cos x$ is non-positive, and that is when $x$ is in the interval $[\pi/2, 3\pi/2]$. We shall now consider when $\cos x$ is positive. We can square the given inequality, and the resulting inequality will be true whenever the original left inequality is true. $4\cos^2{x}\leq 1+\sin 2x+1-\sin 2x-2\sqrt{1-\sin^2 2x}=2-2\sqrt{\cos^2{2x}}$. This inequality is equivalent to $2\cos^2 x\leq 1-\left| \cos 2x\right|$. I shall now divide this problem into cases.

Case 1: $\cos 2x$ is non-negative. This means that $x$ is in one of the intervals $[0,\pi/4]$ or $[7\pi/4, 2\pi]$. We must find all $x$ in these two intervals such that $2\cos^2 x\leq 1-\cos 2x$. This inequality is equivalent to $2\cos^2 x\leq 2\sin^2 x$, which is only true when $x=\pi/4$ or $7\pi/4$.

Case 2: $\cos 2x$ is negative. This means that $x$ is in one of the interavals $(\pi/4, \pi/2)$ or $(3\pi/2, 7\pi/4)$. We must find all $x$ in these two intervals such that $2\cos^2 x\leq 1+\cos 2x$, which is equivalent to $2\cos^2 x\leq 2\cos^2 x$, which is true for all $x$ in these intervals.

Therefore the left inequality is true when $x$ is in the union of the intervals $[\pi/4, \pi/2)$, $(3\pi/2, 7\pi/4]$, and $[\pi/2, 3\pi/2]$, which is the interval $[\pi/4, 7\pi/4]$. We shall now deal with the right inequality.

As above, we can square it and have it be true whenever the original right inequality is true, so we do that. $2-2\sqrt{\cos^2{2x}}\leq 2$, which is always true. Therefore the original right inequality is always satisfied, and all $x$ such that the original inequality is satisfied are in the interval $[\pi/4, 7\pi/4]$.

Solution 2

Manipulate the inequality so that it becomes: \[\sqrt{2}\cos x \leq \left| \sqrt{\frac{1+\cos(\frac{\pi}{2}-2x)}{2}} - \sqrt{\frac{1-\cos(\frac{\pi}{2}- 2x)}{2}} \right| \leq 1\] Inside the absolute value, we identify the half-angle formulas. However, since we do not know the sign of the resultant expression, we have to use absolute value signs since the principal square root is always positive; then the inequality becomes \[\sqrt{2}\cos x \leq \left| \left|\cos\left(\frac{\pi}{4}-x\right)\right| - \left|\sin\left(\frac{\pi}{4}-x\right)\right|\right| \leq 1\] which, considering absolute value, is the same as \[\sqrt{2}\cos x \leq \left| \left|\cos\left(x-\frac{\pi}{4}\right)\right| - \left|\sin\left(x-\frac{\pi}{4}\right)\right|\right| \leq 1\] Since both inner absolute values range between $0$ and $1$, their positive difference also ranges from $0$ to $1$, so the right inequality is always true. Thus we take a look at the left inequality. For $x\in\left[\frac{\pi}{2},\frac{3\pi}{2}\right]$, the left-hand side is never positive; however, the right-hand side is always nonnegative due to absolute value, so the inequality holds for these $x$-values. As a result, we consider the remaining two sections of the interval.

For $x\in\left[0,\frac{\pi}{4}\right]$ and $x\in\left[\frac{7\pi}{4},2\pi\right]$, when considering the $-\frac{\pi}{4}$, inside the absolute value, cosine is positive but sine is negative; thus our inequality becomes \begin{align*} \sqrt{2}\cos x &\le \left|\cos\left(x-\frac{\pi}{4}\right) +\sin\left(x-\frac{\pi}{4}\right)\right|\\ \cos x &\le \left|\cos\left(x-\frac{\pi}{4}\right)\cdot\frac{\sqrt{2}}{2} +\sin\left(x-\frac{\pi}{4}\right)\cdot\frac{\sqrt{2}}{2}\right|\\ \cos x &\le \left|\cos\left(x-\frac{\pi}{4}\right)\sin\left(\frac{\pi}{4}\right) +\sin\left(x-\frac{\pi}{4}\right)\cos\left(\frac{\pi}{4}\right)\right|\\ \cos x &\le \left|\sin\left(\left(x-\frac{\pi}{4}\right)+\frac{\pi}{4}\right)\right|\\ \cos x &\le |\sin x|\\ \end{align*} If $x\in\left[0,\frac{\pi}{4}\right]$, then due to the interval, $\sin x$ and $\cos x$ are both never negative; thus, \begin{align*} \cos x &\le \sin x\\ 1&\le\tan x\\ \end{align*} But this is never true in our interval (except at one endpoint, which is included in the interval below), so the inequality does not hold.

If $x\in\left[\frac{7\pi}{4},2\pi\right]$, then due to the interval, $\sin x$ is negative and $\cos x$ is positive; thus, \begin{align*} \cos x &\le -\sin x\\ 1&\le-\tan x\\ -1&\ge\tan x\\ \end{align*} But this is also never true in our interval (except at one endpoint, which is included in the interval below), so the inequality does not hold.

For $x\in\left[\frac{\pi}{4},\frac{\pi}{2}\right]$, inside the absolute value, when considering the $-\frac{\pi}{4}$, both cosine and sine are positive; thus our inequality becomes \begin{align*} \sqrt{2}\cos x &\le \left|\cos\left(x-\frac{\pi}{4}\right) -\sin\left(x-\frac{\pi}{4}\right)\right|\\ \cos x &\le \left|\cos\left(x-\frac{\pi}{4}\right)\cdot\frac{\sqrt{2}}{2} -\sin\left(x-\frac{\pi}{4}\right)\cdot\frac{\sqrt{2}}{2}\right|\\ \cos x &\le \left|\cos\left(x-\frac{\pi}{4}\right)\cos\left(\frac{\pi}{4}\right) -\sin\left(x-\frac{\pi}{4}\right)\sin\left(\frac{\pi}{4}\right)\right|\\ \cos x &\le \left|\cos\left(\left(x-\frac{\pi}{4}\right)+\frac{\pi}{4}\right)\right|\\ \cos x &\le |\cos x|\\ \end{align*} This is always true, so the inequality holds.

For $x\in\left[\frac{3\pi}{2},\frac{7\pi}{4}\right]$, inside the absolute value, when considering the $-\frac{\pi}{4}$, both cosine and sine are negative; thus our inequality becomes \begin{align*} \sqrt{2}\cos x &\le \left|-\cos\left(x-\frac{\pi}{4}\right) +\sin\left(x-\frac{\pi}{4}\right)\right|\\ \sqrt{2}\cos x &\le \left|-\left(\cos\left(x-\frac{\pi}{4}\right) -\sin\left(x-\frac{\pi}{4}\right)\right)\right|\\ \sqrt{2}\cos x &\le \left|\cos\left(x-\frac{\pi}{4}\right) -\sin\left(x-\frac{\pi}{4}\right)\right|\\ \end{align*} This is the same as the case for $x\in\left[\frac{\pi}{4},\frac{\pi}{2}\right]$, so the inequality holds.

Combining our findings, we find that the solutions are $\boxed{x\in\left[\frac{\pi}{4},\frac{7\pi}{4}\right]}$.

~eevee9406

1965 IMO (Problems) • Resources
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