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 $\begin{aligned} \sqrt{2} \cos x & \leq | \cos (x - \frac{π}{4}) + \sin (x - \frac{π}{4}) | \\ \cos x & \leq | \cos (x - \frac{π}{4}) \cdot \frac{\sqrt{2}}{2} + \sin (x - \frac{π}{4}) \cdot \frac{\sqrt{2}}{2} | \\ \cos x & \leq | \cos (x - \frac{π}{4}) \sin (\frac{π}{4}) + \sin (x - \frac{π}{4}) \cos (\frac{π}{4}) | \\ \cos x & \leq | \sin ((x - \frac{π}{4}) + \frac{π}{4}) | \\ \cos x & \leq | \sin x | \end{aligned}$ If $x\in\left[0,\frac{\pi}{4}\right]$ , then due to the interval, $\sin x$ and $\cos x$ are both never negative; thus, $\begin{aligned} \cos x & \leq \sin x \\ 1 & \leq \tan x \end{aligned}$ 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, $\begin{aligned} \cos x & \leq - \sin x \\ 1 & \leq - \tan x \\ - 1 & \geq \tan x \end{aligned}$ 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 $\begin{aligned} \sqrt{2} \cos x & \leq | \cos (x - \frac{π}{4}) - \sin (x - \frac{π}{4}) | \\ \cos x & \leq | \cos (x - \frac{π}{4}) \cdot \frac{\sqrt{2}}{2} - \sin (x - \frac{π}{4}) \cdot \frac{\sqrt{2}}{2} | \\ \cos x & \leq | \cos (x - \frac{π}{4}) \cos (\frac{π}{4}) - \sin (x - \frac{π}{4}) \sin (\frac{π}{4}) | \\ \cos x & \leq | \cos ((x - \frac{π}{4}) + \frac{π}{4}) | \\ \cos x & \leq | \cos x | \end{aligned}$ 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 $\begin{aligned} \sqrt{2} \cos x & \leq | - \cos (x - \frac{π}{4}) + \sin (x - \frac{π}{4}) | \\ \sqrt{2} \cos x & \leq | - (\cos (x - \frac{π}{4}) - \sin (x - \frac{π}{4})) | \\ \sqrt{2} \cos x & \leq | \cos (x - \frac{π}{4}) - \sin (x - \frac{π}{4}) | \end{aligned}$ 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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1965 IMO Problems/Problem 1

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