International Mathematical Olympiad Shortlisted Problems

For any set $S$ of $n=2000$ distinct real numbers, let $D_1⩽D_2⩽\cdots ⩽D_m$ be the distances between them, displayed with their multiplicities. Here $m=n(n-1)/2$. By rescaling the numbers, we may assume that the smallest distance $D_1$ between two elements of $S$ is $D_1=1$. Let $D_1=1=y-x$ for $x,y\in S$. Evidently $D_m=v-u$ is the difference between the largest element $v$ and the smallest element $u$ of $S$.

If $D_{i+1}/D_i<1+10^{-5}$ for some $i=1,2,\dots ,m-1$ then the required inequality holds, because $0⩽D_{i+1}/D_i-1<10^{-5}$. Otherwise, the reverse inequality $$\frac{D_{i+1}}{D_i}⩾1+\frac{1}{10^5}$$ holds for each $i=1,2,\dots ,m-1$, and therefore $$v-u=D_m=\frac{D_m}{D_1}=\frac{D_m}{D_{m-1}}\cdots \frac{D_3}{D_2}\cdot \frac{D_2}{D_1}⩾{\left(1+\frac{1}{10^5}\right)}^{m-1}$$ From $m-1=n(n-1)/2-1=1000\cdot 1999-1>19\cdot 10^5$, together with the fact that for all $n⩾1$, ${\left(1+\frac{1}{n}\right)}^n⩾1+\left(\genfrac{}{}{0ex}{}{n}{1}\right)\cdot \frac{1}{n}=2$, we get $${\left(1+\frac{1}{10^5}\right)}^{19\cdot 10^5}={\left({\left(1+\frac{1}{10^5}\right)}^{10^5}\right)}^{19}⩾2^{19}=2^9\cdot 2^{10}>500\cdot 1000>2\cdot 10^5$$ and so $v-u=D_m>2\cdot 10^5$.

Since the distance of $x$ to at least one of the numbers $u,v$ is at least $(u-v)/2>10^5$, we have $$|x-z|>10^5$$ for some $z\in \{u,v\}$. Since $y-x=1$, we have either $z>y>x$ (if $z=v$) or $y>x>z$ (if $z=u$). If $z>y>x$, selecting $a=z$, $b=y$, $c=z$ and $d=x$ (so that $b\ne d$), we obtain $$\left|\frac{a-b}{c-d}-1\right|=\left|\frac{z-y}{z-x}-1\right|=\left|\frac{x-y}{z-x}\right|=\frac{1}{z-x}<10^{-5}.$$ Otherwise, if $y>x>z$, we may choose $a=y$, $b=z$, $c=x$ and $d=z$ (so that $a\ne c$), and obtain $$\left|\frac{a-b}{c-d}-1\right|=\left|\frac{y-z}{x-z}-1\right|=\left|\frac{y-x}{x-z}\right|=\frac{1}{x-z}<10^{-5}.$$ The desired result follows.