IMO 1965 Problem 1

Proposed by: -
Verified: yes
Verdicts: PASS + PASS
Solve time: 8m00s

Problem

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

Exploration

The expression involves $\sin 2x$ inside two square roots of complementary forms, suggesting the standard algebraic identity

$$\sqrt{1+u}-\sqrt{1-u}=\frac{2u}{\sqrt{1+u}+\sqrt{1-u}}.$$

This converts the absolute value into a rational expression in $\sin 2x$, whose sign is controlled by $\sin 2x$.

A second simplification is to parametrize $u=\sin 2x$, reducing the problem to bounding a function of $u\in[-1,1]$ and then translating back to $x$ via $2x$ periodicity.

The inequality also involves $2\cos x$ on the left, which depends on $\cos x$ rather than $\sin 2x$, so the structure is not purely in $2x$. This suggests splitting into regions where $\sin 2x$ has fixed sign and then expressing everything in terms of $\sin x$ and $\cos x$.

A key observation is that $\sqrt{1+\sin 2x}$ and $\sqrt{1-\sin 2x}$ are well-defined for all $x$, since $|\sin 2x|\leq 1$.

The most delicate part is comparing $2\cos x$ with a function of $\sin 2x$, which naturally leads to rewriting everything in terms of $\sin x$ and $\cos x$ using $\sin 2x=2\sin x\cos x$.

The likely successful path is to simplify the middle expression completely, reduce the inequality to an algebraic condition in $\sin x$ and $\cos x$, and then solve on $[0,2\pi]$ by sign analysis.

Problem Understanding

The task is to determine all angles $x$ between $0$ and $2\pi$ for which a three-part inequality holds: a trigonometric expression involving $\cos x$ is bounded above and below by an expression involving square roots of $1\pm \sin 2x$.

This is a Type A problem, requiring a full characterization of all admissible $x$.

The core difficulty is that the middle expression mixes nonlinear square-root structure with $\sin 2x$, while the left term involves $\cos x$. A direct comparison is obstructed by incompatible trigonometric arguments. The problem is designed so that the square-root difference simplifies to an expression linear in $\sin 2x$ after rationalization.

The expected solution is that the inequality reduces to a simple trigonometric condition on $\cos x$, and the solution set will be an explicit union of intervals in $[0,2\pi]$.

Proof Architecture

Lemma 1. For all $x$,

$$\left|\sqrt{1+\sin 2x}-\sqrt{1-\sin 2x}\right| =\frac{2|\sin 2x|}{\sqrt{1+\sin 2x}+\sqrt{1-\sin 2x}}.$$

This follows from rationalizing the difference of square roots.

Lemma 2. For all $x$, the expression in Lemma 1 satisfies

$$\left|\sqrt{1+\sin 2x}-\sqrt{1-\sin 2x}\right|=\sqrt{2-2\sqrt{1-\sin^2 2x}}=\sqrt{2-2|\cos 2x|}.$$

This follows from squaring and simplifying using $(1+\sin 2x)(1-\sin 2x)=1-\sin^2 2x$.

Lemma 3. The inequality

$$2\cos x \leq \left|\sqrt{1+\sin 2x}-\sqrt{1-\sin 2x}\right|$$

is equivalent to a piecewise inequality depending on the sign of $\cos x$.

This is because the left-hand side changes sign while the right-hand side is nonnegative.

Lemma 4. The condition

$$\left|\sqrt{1+\sin 2x}-\sqrt{1-\sin 2x}\right|\leq \sqrt{2}$$

holds for all $x$.

This will follow from Lemma 2 and the bound $|\cos 2x|\geq -1$.

The hardest step is Lemma 2, since it requires careful algebraic elimination of radicals.

Solution

Lemma 1

Let $A=\sqrt{1+\sin 2x}$ and $B=\sqrt{1-\sin 2x}$. Then

$$|A-B|=\frac{|A^2-B^2|}{A+B}.$$

Since $A^2-B^2=(1+\sin 2x)-(1-\sin 2x)=2\sin 2x$, it follows that

$$|A-B|=\frac{2|\sin 2x|}{\sqrt{1+\sin 2x}+\sqrt{1-\sin 2x}}.$$

This establishes the required rationalized form, since the denominator is strictly positive for all $x$.

Certification: this step isolates the dependence on $\sin 2x$ and removes subtraction of radicals, which otherwise prevents algebraic comparison.

Lemma 2

Squaring the expression from Lemma 1 gives

$$|A-B|^2=A^2+B^2-2AB.$$

Compute

$$A^2+B^2=2,\qquad AB=\sqrt{(1+\sin 2x)(1-\sin 2x)}=\sqrt{1-\sin^2 2x}=|\cos 2x|.$$

Hence

$$|A-B|^2=2-2|\cos 2x|,$$

$$|A-B|=\sqrt{2-2|\cos 2x|}.$$

Certification: this step eliminates all nested radicals and reduces the problem to a single absolute value in $\cos 2x$.

Lemma 3

Since $|A-B|\ge 0$, the inequality

$$2\cos x \le |A-B|$$

splits according to the sign of $\cos x$.

If $\cos x\le 0$, the inequality holds automatically because the right-hand side is nonnegative.

If $\cos x>0$, squaring is valid and gives

$$4\cos^2 x \le 2-2|\cos 2x|.$$

Certification: this step reduces the mixed inequality to a purely algebraic condition in cosine functions with controlled squaring.

Lemma 4

From Lemma 2,

$$|A-B|^2=2-2|\cos 2x|\le 2,$$

since $|\cos 2x|\ge 0$. Hence $|A-B|\le \sqrt{2}$ for all $x$.

Certification: this shows the upper bound is automatically satisfied and removes it from further consideration.

Main argument

Lemma 4 implies the rightmost inequality holds for all $x\in[0,2\pi]$, so only the inequality

$$2\cos x \le |A-B|$$

remains.

If $\cos x\le 0$, the inequality holds for all such $x$. This gives the interval

$$x\in\left[\frac{\pi}{2},\frac{3\pi}{2}\right].$$

Assume now $\cos x>0$, so $x\in[0,\frac{\pi}{2})\cup(\frac{3\pi}{2},2\pi]$. Using Lemma 2, the condition becomes

$$2\cos x \le \sqrt{2-2|\cos 2x|}.$$

Squaring yields

$$4\cos^2 x \le 2-2|\cos 2x|.$$

Rearranging,

$$2\cos^2 x + |\cos 2x| \le 1.$$

Using $\cos 2x=2\cos^2 x-1$, we analyze cases.

If $\cos 2x\ge 0$, then $|\cos 2x|=2\cos^2 x-1$, so

$$2\cos^2 x + (2\cos^2 x-1)\le 1$$

gives

$$4\cos^2 x \le 2,$$

hence $\cos^2 x \le \frac{1}{2}$.

Together with $\cos x>0$ and $\cos 2x\ge 0$ (equivalently $\cos^2 x\ge \frac{1}{2}$), we obtain $\cos^2 x=\frac{1}{2}$, so $\cos x=\frac{1}{\sqrt{2}}$, yielding

$$x=\frac{\pi}{4},\quad x=\frac{7\pi}{4}.$$

If $\cos 2x<0$, then $|\cos 2x|=1-2\cos^2 x$, so

$$2\cos^2 x + (1-2\cos^2 x)\le 1,$$

which holds identically. Hence all $x$ with $\cos x>0$ and $\cos 2x<0$ are valid.

The condition $\cos x>0$ and $\cos 2x<0$ holds exactly for

$$x\in\left(\frac{\pi}{4},\frac{\pi}{2}\right)\cup\left(\frac{3\pi}{2},\frac{7\pi}{4}\right).$$

Combining with the earlier case $\cos x\le 0$ gives

$$x\in\left[\frac{\pi}{2},\frac{3\pi}{2}\right]\cup\left(\frac{\pi}{4},\frac{\pi}{2}\right)\cup\left(\frac{3\pi}{2},\frac{7\pi}{4}\right)\cup\left{\frac{\pi}{4},\frac{7\pi}{4}\right}.$$

This simplifies to

$$x\in\left[\frac{\pi}{4},\frac{7\pi}{4}\right].$$

Thus the solution set is

$$\boxed{\left[\frac{\pi}{4},\frac{7\pi}{4}\right]}.$$

Verification of Key Steps

The first delicate point is the claim $|A-B|^2=2-2|\cos 2x|$. This depends on correctly identifying $AB=\sqrt{1-\sin^2 2x}=|\cos 2x|$, which requires handling both signs of $\cos 2x$ since squaring removes sign information.

The second delicate point is the case split on $\cos 2x\ge 0$ versus $\cos 2x<0$. A failure to separate these cases would incorrectly drop the absolute value and produce an incorrect inequality region.

The third delicate point is the final interval simplification, where overlapping cases must be carefully unified; missing endpoint checks at $x=\frac{\pi}{4},\frac{3\pi}{2},\frac{7\pi}{4}$ would omit valid equality cases.

Alternative Approaches

A trigonometric substitution $t=\tan x$ transforms the inequality into a rational condition in $t$, since $\sin 2x$ and $\cos x$ both admit algebraic expressions in $t$. This yields a purely algebraic inequality in $t$ after clearing denominators, followed by a sign analysis on intervals corresponding to the monotonicity of $\tan x$.

A second approach uses the identity

$$\sqrt{1+\sin 2x}-\sqrt{1-\sin 2x}= \sqrt{2},\sin x$$

after choosing appropriate branches, which converts the problem directly into a linear inequality in $\sin x$, though this requires careful justification of sign consistency across intervals.