Kvant Math Problem 753

Verified: no
Verdicts: SKIP + SKIP
Solve time: 6m21s
Source on kvant.digital

Problem

The numbers $a$, $b$, $c$ lie in the interval $\left(0;\dfrac\pi2\right)$ and satisfy the equalities: $$\begin{align*} \cos a&=a,\ \sin\cos b&=b,\ \cos\sin c&=c. \end{align*}$$. Arrange these numbers in ascending order.

S. Hessen

All-Union School Mathematics Olympiad (1982, 10th grade)

Exploration

The problem gives three numbers $a$, $b$, $c$ in the interval $(0, \frac{\pi}{2})$ satisfying

$$\cos a = a, \quad \sin(\cos b) = b, \quad \cos(\sin c) = c,$$

and asks for their ascending order. Each equation defines a unique number because the functions involved are strictly monotone on $(0, \frac{\pi}{2})$. For small $x$, $\cos x \approx 1 - x^2/2$, $\sin x \approx x$, and $\sin(\cos x)$ and $\cos(\sin x)$ are also near $1$ for small $x$. This suggests $a$ is smaller than $b$ and $c$ because $\cos a = a$ forces $a < 1$, while $\sin(\cos b) = b$ and $\cos(\sin c) = c$ involve compositions of monotone functions close to $1$.

Numerical estimates can help. The function $f(x) = \cos x - x$ decreases from $\cos 0 - 0 = 1$ to $\cos(\pi/2) - \pi/2 = -\pi/2$, so $a \in (0,1)$. Similarly, $\sin(\cos x) - x$ starts at $\sin 1 \approx 0.8415$ at $x=0$ and decreases, suggesting $b$ is close to $0.7$, and $\cos(\sin x) - x$ starts at $1$ and decreases, suggesting $c$ is close to $0.8$. This hints at the ordering $a < b < c$.

The delicate point is confirming the monotonicity of each function to guarantee uniqueness and to rigorously compare $a$, $b$, $c$ without relying solely on approximations.

Problem Understanding

The problem asks to compare three numbers determined by nonlinear equations and arrange them in ascending order. This is a Type A problem: "Find all X" where $X$ is the relative order of $a$, $b$, $c$. The core difficulty lies in comparing numbers defined implicitly by strictly monotone functions. The preliminary intuition, supported by rough numerical estimates, suggests $a < b < c$, because $\cos x$ decreases faster than the other composed functions, and $x$ appears as a fixed point of increasingly larger functions.

Proof Architecture

Lemma 1: The equation $\cos x = x$ has a unique solution in $(0, \frac{\pi}{2})$, denoted $a$. Sketch: $f(x) = \cos x - x$ is strictly decreasing on $(0, \pi/2)$, changing sign from positive to negative.

Lemma 2: The equation $\sin(\cos x) = x$ has a unique solution in $(0, \frac{\pi}{2})$, denoted $b$. Sketch: $f(x) = \sin(\cos x) - x$ is strictly decreasing from $f(0) = \sin 1 > 0$ to $f(\pi/2) = \sin 0 - \pi/2 < 0$.

Lemma 3: The equation $\cos(\sin x) = x$ has a unique solution in $(0, \frac{\pi}{2})$, denoted $c$. Sketch: $f(x) = \cos(\sin x) - x$ is strictly decreasing from $f(0) = \cos 0 = 1$ to $f(\pi/2) = \cos 1 - \pi/2 < 0$.

Lemma 4: Comparing $a$ and $b$: show $a < b$ using $\cos x - x$ and $\sin(\cos x) - x$. Sketch: compare function values at $a$ and $b$.

Lemma 5: Comparing $b$ and $c$: show $b < c$ using $\sin(\cos x)$ versus $\cos(\sin x)$. Sketch: estimate or analyze derivatives to ensure monotone crossing.

The hardest direction is confirming $b < c$ rigorously without numerical approximations.

Solution

Consider the function $f(x) = \cos x - x$ on $(0, \frac{\pi}{2})$. Its derivative $f'(x) = -\sin x - 1 < 0$, so $f$ is strictly decreasing. At $x=0$, $f(0) = 1 > 0$, and at $x = \pi/2$, $f(\pi/2) = 0 - \pi/2 < 0$. By the intermediate value theorem, there exists a unique $a \in (0, \pi/2)$ such that $\cos a = a$.

Next, define $g(x) = \sin(\cos x) - x$ on $(0, \pi/2)$. Its derivative $g'(x) = -\cos(\cos x)\sin x - 1 < 0$, so $g$ is strictly decreasing. At $x=0$, $g(0) = \sin 1 > 0$, and at $x = \pi/2$, $g(\pi/2) = \sin 0 - \pi/2 = -\pi/2 < 0$. By the intermediate value theorem, there exists a unique $b \in (0, \pi/2)$ such that $\sin(\cos b) = b$.

Similarly, define $h(x) = \cos(\sin x) - x$ on $(0, \pi/2)$. Its derivative $h'(x) = -\sin(\sin x)\cos x - 1 < 0$, so $h$ is strictly decreasing. At $x=0$, $h(0) = 1 > 0$, and at $x = \pi/2$, $h(\pi/2) = \cos 1 - \pi/2 < 0$. Therefore there exists a unique $c \in (0, \pi/2)$ such that $\cos(\sin c) = c$.

To compare $a$ and $b$, consider $f(x) - g(x) = \cos x - \sin(\cos x)$. Let $\phi(x) = \cos x - \sin(\cos x)$. Its derivative $\phi'(x) = -\sin x + \cos(\cos x)\sin x = \sin x(\cos(\cos x) - 1) \le 0$, so $\phi$ is decreasing. Evaluate $\phi(0) = 1 - \sin 1 > 0$ and $\phi(\pi/2) = 0 - \sin 0 = 0$. Since $\phi$ decreases from positive to zero, it follows that $\phi(a) > 0$, so $\cos a - \sin(\cos a) > 0$. But $\cos a = a$, so $a - \sin(\cos a) = a - b > 0$, implying $a < b$.

To compare $b$ and $c$, consider $\psi(x) = \cos(\sin x) - \sin(\cos x)$. Its derivative $\psi'(x) = -\sin(\sin x)\cos x - \cos(\cos x)\sin x < 0$, so $\psi$ is strictly decreasing. At $x=0$, $\psi(0) = 1 - \sin 1 > 0$, and at $x = \pi/2$, $\psi(\pi/2) = \cos 1 - 0 > 0$. However, $b$ and $c$ are solutions to $x = \sin(\cos x)$ and $x = \cos(\sin x)$. Evaluate $\psi$ at $b$: $\psi(b) = \cos(\sin b) - b$. Since $b < 1$, $\sin b < b$, so $\cos(\sin b) > \cos b > \sin(\cos b) = b$, implying $\psi(b) > 0$, hence $b < c$.

Consequently, the numbers satisfy $a < b < c$.

The ascending order of the numbers is

$$\boxed{a < b < c}.$$

Verification of Key Steps

The crucial step is comparing $b$ and $c$. To verify, note that $b$ satisfies $b = \sin(\cos b) < \cos(\sin b)$ because $\sin(\cos b) < \cos(\sin b)$ for $b \in (0, \pi/2)$. This follows from the inequality $\cos y > \sin y$ for $y \in (0, \pi/4)$ and $b \in (0,1)$, confirmed by direct evaluation: $\cos(\sin b) - b = \cos(\sin b) - \sin(\cos b) > 0$, so $b < c$.

The other delicate step is verifying uniqueness of solutions. Each function $f$, $g$, $h$ is strictly decreasing on $(0, \pi/2)$ and changes sign from positive to negative, guaranteeing exactly one zero in the interval. This rules out multiple intersections or equality failures.

Alternative Approaches

An alternative approach is to iteratively approximate $a$, $b$, $c$ using fixed-point iteration: $x_{n+1} = \cos x_n$, $x_{n+1} = \sin(\cos x_n)$, $x_{n+1} = \cos(\sin x_n)$. The limits converge rapidly due to derivatives less than one in magnitude