1965 IMO Problems/Problem 1

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 2cosx|cos(xπ4)+sin(xπ4)|cosx|cos(xπ4)22+sin(xπ4)22|cosx|cos(xπ4)sin(π4)+sin(xπ4)cos(π4)|cosx|sin((xπ4)+π4)|cosx|sinx| If $x\in\left[0,\frac{\pi}{4}\right]$, then due to the interval, $\sin x$ and $\cos x$ are both never negative; thus, cosxsinx1tanx 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, cosxsinx1tanx1tanx 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 2cosx|cos(xπ4)sin(xπ4)|cosx|cos(xπ4)22sin(xπ4)22|cosx|cos(xπ4)cos(π4)sin(xπ4)sin(π4)|cosx|cos((xπ4)+π4)|cosx|cosx| 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 2cosx|cos(xπ4)+sin(xπ4)|2cosx|(cos(xπ4)sin(xπ4))|2cosx|cos(xπ4)sin(xπ4)| 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
Preceded by
First Question
1 2 3 4 5 6 Followed by
Problem 2
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