China Mathematical Olympiad

The answer is $0<t<\frac{1}{2}$.

Firstly, for $0<t<\frac{1}{2}$, choose $\lambda \in \left(0,\frac{1-2t}{2(1+t)}\right)$, let $x_i={\lambda }^i$, $X=\{x_1,x_2,\dots \}$. We claim that for all (not necessarily distinct) $x,y,z\in X$, all real numbers $a$ and all positive real numbers $d$, we have the following inequality: $$\max \{|x-(a-d)|,|y-a|,|z-(a+d)|\}>td.$$ Suppose on the contrary that there exists $a\in \mathbb{R}$, $d\in {\mathbb{R}}^{+}$ and $x_i,x_j,x_k$, such that $$\max \{|x_i-(a-d)|,|x_j-a|,|x_k-(a+d)|\}\le td.$$ Hence $$\{\begin{array}{l}-td\le x_i-(a-d)\le td,\\ -td\le x_j-a\le td,\\ -td\le x_k-(a+d)\le td,\end{array}$$ i.e. $$\{\begin{array}{l}{x}_i+(1-t)d\le a\le x_i+(1+t)d,\\ x_j-td\le a\le x_j+td,\\ x_k-(1+t)d\le a\le x_k-(1-t)d,\end{array}(\ast )$$ which implies that $$\{\begin{array}{l}{x}_k-(1+t)d\le a\le x_i+(1+t)d,\\ x_i+(1-t)d\le a\le x_j+td,\\ x_j-td\le a\le x_k-(1-t)d,\end{array}$$ note that $0<t<\frac{1}{2}$, it follows that

$$\{\begin{array}{l}d\ge \frac{x_k-x_i}{2(1+t)},\\ d\le \frac{x_j-x_i}{1-2t},\\ d\le \frac{x_k-x_j}{1-2t}.\end{array}$$ By the second and third inequalities and $d>0$, we get $x_i<x_j<x_k$, hence $i>j>k$, ${\lambda }^j+{\lambda }^{i+1}\le {\lambda }^{k+1}+{\lambda }^i$, we get

$$\frac{x_j-x_i}{x_k-x_i}=\frac{{\lambda }^j-{\lambda }^i}{{\lambda }^k-{\lambda }^i}\le \lambda .$$ By the first and second inequalities, we get $\frac{x_j-x_i}{1-2t}\ge \frac{x_k-x_i}{2(1+t)}$, hence $$\frac{x_j-x_i}{x_k-x_i}\ge \frac{1-2t}{2(1+t)}>\lambda ,$$ which contradicts the previous inequality! Thus proved our earlier claim about $X$.

Secondly, for $t\ge \frac{1}{2}$, we show that for any infinite set $X$, for any $x<y<z$ in $X$, we can choose $a\in \mathbb{R}$ and $d\in {\mathbb{R}}^{+}$ such that $$\max \{|x-(a-d)|,|y-a|,|z-(a+d)|\}\le td.$$ In fact, let $d=\frac{z-x}{2}$, hence $x+(1-t)d=z-(1+t)d$. Let $a=\max \{x+(1-t)d,y-td\}$. Since $t\ge \frac{1}{2}$, we obtain $$\{\begin{array}{l}y-x<2d\le (1+2t)d,\\ x-y<0\le (2t-1)d,\end{array}$$ i.e. $$\{\begin{array}{l}y-td\le x+(1+t)d,\\ x+(1-t)d\le y+td,\end{array}$$ hence $$\{\begin{array}{l}x+(1-t)d\le a\le x+(1+t)d,\\ y-td\le a\le y+td,\\ z-(1+t)d\le a\le z-(1-t)d,\end{array}$$ from which we conclude that $$\max \{|x-(a-d)|,|y-a|,|z-(a+d)|\}\le td.$$ So every $t\ge \frac{1}{2}$ does not satisfy the requirement of the problem.

In conclusion, the set of all required $t$ is $(0,\frac{1}{2})$.