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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{…')
 
(ok THIS time I got it.)
 
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Determine all values <math>x</math> in the interval <math>0\leq x\leq 2\pi </math> which satisfy the inequality
 
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{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>
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== Solution ==
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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.
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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.
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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>.
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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.
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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.
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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>.
  
  
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== Solution ==
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{{IMO box|year=1965|before=First Question|num-a=2}}
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{{solution}}
 

Latest revision as of 10:40, 9 July 2010

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

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 $[\pi/4, 7\pi/4]$.


1965 IMO (Problems) • Resources
Preceded by
First Question 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 2
All IMO Problems and Solutions
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