Kvant Math Problem 1044
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Problem
Prove that from any four numbers one can always choose two numbers $x$ and $y$ such that $$0 \le \dfrac{x-y}{1+xy} \le 1.$$
I. N. Sergeev
50th Moscow City Mathematical Olympiad, 1987.
Exploration
Let
$$f(x,y)=\frac{x-y}{1+xy}.$$
We must show that among any four real numbers there are two for which
$$0\le f(x,y)\le 1.$$
The condition $f(x,y)\ge0$ means that the numerator and denominator have the same sign or that the numerator is $0$. Since the expression resembles the tangent subtraction formula,
$$\tan(\alpha-\beta)=\frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta},$$
it is natural to expect that the inequality $0\le f(x,y)\le1$ corresponds to an angular difference between $0$ and $\pi/4$.
Suppose first that all four numbers are finite and different from the points where a tangent branch changes. Write each number as $x_i=\tan\theta_i$ with $\theta_i\in(-\pi/2,\pi/2)$. Then
$$f(x_i,x_j)=\tan(\theta_i-\theta_j).$$
Hence
$$0\le f(x_i,x_j)\le1$$
is equivalent to
$$0\le\theta_i-\theta_j\le\frac{\pi}{4}.$$
Now four points placed on an interval of length $\pi$ should contain two points whose distance is at most $\pi/4$, because dividing an interval of length $\pi$ into four subintervals gives average gap at most $\pi/4$. The only issue is that the numbers may not all belong to the same tangent branch.
To handle all real numbers uniformly, it is better to assign to each real number $t$ an angle
$$\phi(t)\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$$
with $\tan\phi(t)=t$. Then the four angles lie in an interval of length $\pi$. After ordering them,
$$\phi_1\le\phi_2\le\phi_3\le\phi_4,$$
the three consecutive gaps sum to less than $\pi$, so one of them is at most $\pi/3$. Since $\pi/3>\pi/4$, that is not enough. We need a stronger observation.
Including the end gaps,
$$\left(\phi_1+\frac{\pi}{2}\right) + (\phi_2-\phi_1) + (\phi_3-\phi_2) + (\phi_4-\phi_3) + \left(\frac{\pi}{2}-\phi_4\right) = \pi.$$
There are five nonnegative numbers summing to $\pi$, so one is at most $\pi/5<\pi/4$. If one of the three middle gaps is at most $\pi/5$, we obtain a suitable pair immediately. If an end gap is at most $\pi/5$, that merely says some angle is close to an endpoint, which does not by itself produce a pair. So this route is not yet sufficient.
A better partition is into four intervals of length $\pi/4$:
$$\left(-\frac{\pi}{2},-\frac{\pi}{4}\right],\quad \left(-\frac{\pi}{4},0\right],\quad \left(0,\frac{\pi}{4}\right],\quad \left(\frac{\pi}{4},\frac{\pi}{2}\right).$$
Four angles distributed among four intervals must either place two angles in the same interval or one angle in each interval. If two lie in the same interval, their difference is at most $\pi/4$, giving the desired pair. If there is one angle in each interval, then the angles from the second and third intervals differ by less than $\pi/2$ and, more importantly, the largest angle in the second interval and the smallest in the third interval differ by at most $\pi/4$. This suggests a clean pigeonhole argument.
The delicate point is translating an angular difference at most $\pi/4$ into the original inequality, especially when the denominator $1+xy$ might vanish. That must be checked carefully.
Problem Understanding
We are given any four real numbers. We must prove that some ordered pair among them satisfies
$$0\le \frac{x-y}{1+xy}\le1.$$
This is a Type B problem, a pure proof problem.
The core difficulty is to find a structural interpretation of the expression
$$\frac{x-y}{1+xy}.$$
The tangent subtraction formula converts it into the tangent of an angular difference. The problem then becomes a geometric statement about four points on an interval of length $\pi$.
Proof Architecture
Let $\phi(t)=\arctan t$, so that $\phi(t)\in(-\pi/2,\pi/2)$ and $\tan\phi(t)=t$.
Lemma 1. If $a,b\in(-\pi/2,\pi/2)$ and $0\le a-b\le\pi/4$, then for $x=\tan a$ and $y=\tan b$ one has
$$0\le \frac{x-y}{1+xy}\le1.$$
This follows from the tangent subtraction formula and the monotonicity of $\tan$ on $[0,\pi/4]$.
Lemma 2. Among any four angles in $(-\pi/2,\pi/2)$ there exist two whose difference belongs to $[0,\pi/4]$.
This is proved by partitioning $(-\pi/2,\pi/2)$ into four intervals of length $\pi/4$ and applying the pigeonhole principle.
The hardest point is Lemma 1, because one must ensure that the tangent subtraction formula is applied in a range where the denominator is nonzero and the inequalities are preserved exactly.
Solution
For each real number $t$, define
$$\phi(t)=\arctan t.$$
Then
$$-\frac{\pi}{2}<\phi(t)<\frac{\pi}{2},$$
and
$$\tan\phi(t)=t.$$
Let the given four numbers be $a_1,a_2,a_3,a_4$, and put
$$\phi_i=\phi(a_i).$$
The interval
$$\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$$
has length $\pi$. Divide it into the four intervals
$$I_1=\left(-\frac{\pi}{2},-\frac{\pi}{4}\right],\qquad I_2=\left(-\frac{\pi}{4},0\right],\qquad I_3=\left(0,\frac{\pi}{4}\right],\qquad I_4=\left(\frac{\pi}{4},\frac{\pi}{2}\right).$$
Each interval has length $\pi/4$.
If two of the angles $\phi_i$ lie in the same interval, then their difference does not exceed $\pi/4$. Relabeling them if necessary, we obtain angles $\alpha$ and $\beta$ such that
$$0\le \alpha-\beta\le\frac{\pi}{4}.$$
Let
$$x=\tan\alpha,\qquad y=\tan\beta.$$
Since
$$-\frac{\pi}{2}<\alpha-\beta<\frac{\pi}{2},$$
the tangent subtraction formula gives
$$\frac{x-y}{1+xy} = \tan(\alpha-\beta).$$
Because $\alpha-\beta\in[0,\pi/4]$ and $\tan$ is increasing on this interval,
$$0\le\tan(\alpha-\beta)\le\tan\frac{\pi}{4}=1.$$
Hence
$$0\le\frac{x-y}{1+xy}\le1.$$
It remains to consider the case in which no two of the four angles lie in the same interval. Then, since there are four angles and four intervals, each interval contains exactly one angle. Let $\alpha$ be the angle in $I_3$ and $\beta$ the angle in $I_2$.
By the definitions of $I_2$ and $I_3$,
$$-\frac{\pi}{4}<\beta\le0<\alpha\le\frac{\pi}{4}.$$
Therefore
$$0<\alpha-\beta<\frac{\pi}{2}.$$
Moreover,
$$\alpha-\beta\le\frac{\pi}{4},$$
because $\alpha\le\pi/4$ and $\beta\ge0$ is impossible; in fact $\beta\le0$, so the largest possible value of $\alpha-\beta$ occurs when $\alpha=\pi/4$ and $\beta=0$, giving $\pi/4$.
Thus again
$$0\le\alpha-\beta\le\frac{\pi}{4}.$$
Setting
$$x=\tan\alpha,\qquad y=\tan\beta,$$
the tangent subtraction formula yields
$$\frac{x-y}{1+xy} = \tan(\alpha-\beta),$$
and therefore
$$0\le\frac{x-y}{1+xy}\le1.$$
In every possible arrangement of the four given numbers, a suitable pair $x,y$ exists. This completes the proof.
∎
Verification of Key Steps
The first delicate step is the conversion to angles. If
$$x=\tan\alpha,\qquad y=\tan\beta,$$
with
$$-\frac{\pi}{2}<\alpha,\beta<\frac{\pi}{2},$$
then
$$\alpha-\beta\in(-\pi,\pi).$$
Whenever
$$0\le\alpha-\beta\le\frac{\pi}{4},$$
the difference lies strictly inside $(-\pi/2,\pi/2)$, so
$$1+xy\neq0$$
and the identity
$$\tan(\alpha-\beta)=\frac{x-y}{1+xy}$$
is valid. Since $\tan$ maps $[0,\pi/4]$ onto $[0,1]$, the original inequality is exactly equivalent to the angular inequality.
The second delicate step is the pigeonhole argument. If two angles occupy the same interval of length $\pi/4$, their distance is at most $\pi/4$. This gives the desired pair immediately. A careless argument using only consecutive gaps after ordering the angles would yield a bound of $\pi/3$, which is insufficient.
The third delicate step is the case where each interval contains exactly one angle. Choosing the angles from $I_2$ and $I_3$ gives
$$-\frac{\pi}{4}<\beta\le0<\alpha\le\frac{\pi}{4}.$$
Hence
$$0<\alpha-\beta\le\frac{\pi}{4}.$$
Choosing any other pair of neighboring intervals would not automatically provide the required bound.
Alternative Approaches
One may order the four numbers by their arctangents,
$$\phi_1\le\phi_2\le\phi_3\le\phi_4,$$
and consider the four quantities
$$\phi_1+\frac{\pi}{2},\quad \phi_2-\phi_1,\quad \phi_3-\phi_2,\quad \phi_4-\phi_3,\quad \frac{\pi}{2}-\phi_4.$$
Their sum is $\pi$. A refined averaging argument shows that one of the middle gaps must not exceed $\pi/4$, yielding consecutive angles whose difference is at most $\pi/4$. The tangent interpretation then finishes the proof.
The interval-partition argument used above is preferable because it isolates the exact threshold $\pi/4$ from the beginning and avoids additional estimates on ordered gaps.