For any given set of 13 distinct real numbers, prove we can always find two numbers $x$ and $y$ that $0<\frac{x-y}{1+xy}\leq 2-\sqrt{3}$.

For any given set of 13 distinct real numbers, prove we can always find two numbers x and y that $0<\dfrac{x-y}{1+xy}\leq 2-\sqrt{3}$.

I knew we can always make $0<\dfrac{x-y}{1+xy}$ happen. Since x and y are distinct, we can just switch the order of x and y to change the sign of $\dfrac{x-y}{1+xy}$, but what should I do for the right hand side?

Solution 1:

Put $x = \tan(u)$ and $y = \tan(v)$ for some $u, v$ considered modulo $\pi$ (since $\tan$ has period $\pi$). Then $\frac{x-y}{1+xy}$ is $\tan(u-v)$.

So the key is to recognize that two of the angles must be within $\pi/12$ of each other. This is by a pigeonhole principle where we are trying to slot 13 numbers into 12 sectors given by $[0, \pi/12]$, $[\pi/12, 2\pi/12]$, etc.

Solution 2:

Hints: $\tan(\alpha-\beta) = \dfrac{\tan\alpha - \tan\beta}{1+\tan\alpha \tan\beta}$ and $\tan\dfrac{\pi}{12} = 2-\sqrt{3}$.