Prove that given a finite set $A$ in the plane, there is an infinite family of points that have the same distance ($d_{\infty}$) to each point in A Take $n$ distinct points $a_1, \dots ,a_n \in \mathbb R^2$. Prove that there exists $b \in \mathbb R^2$ such that the set $ \{c \in \mathbb R^2 : d_{\infty}(c,a_i) = d_{\infty}(b,a_i) \text{ for all } i \} $ is infinite
I don't know what to do to be honest. The only obvious case is $n = 1$ where I can draw a ball (square) of any desired radius. The case $n=2$ is easy if the distances on the $x$ axis and on the $y$ axis are different - in this case I can draw a ball of radius $\frac {d_{\infty}(a_1,a_2)} 2$ around $a_1$ and around $a_2$ and notice that the intersection of their boundaries is infinite. However I have no idea what to do when these distances are the same, and I have even less of an idea on what to do when $n \ge 3$.
 A: Ok so without loss of generality we make the assumption that ${a_1}_y \le {a_2}_y \le \dots \le {a_n}_y$ and also that these numbers are not all equal (if they were we could make this argument on the x coordinate, since if they were all equal then the x coordinates can't be all equal because they are distinct points).
Let $b_y = {a_n}_y$. It is immediate that $\underset{1 \le i \le n}\max |b_y - {a_i}_y| = b_y - {a_1}_y$
Let $b_x$ be any number such that $b_x - {a_i}_x > b_y - {a_1}_y$ for all $i$. Such a number must exist.
Now the point $b$ satisfies $|b_y - {a_i}_y| \le |b_y - {a_1}_y| < b_x - {a_i}_x = |b_x - {a_i}_x|$
So $d_{\infty}(b, a_i) = b_x - {a_i}_x$ for all $i$
Now let's show that the family we wanted is infinite. Let $c_x = b_x$. Since we made the assumption that the y coordinates for our n points were not all equal, we get that $[{a_1}_y, {a_n}_y]$ is a proper interval. Take $c_y$ in said interval.
Then, we finally get $|c_y - {a_i}_y| \le {a_n}_y - {a_1}_y = b_y - {a_1}_y < b_x - {a_i}_x = c_x - {a_i}_x = |c_x - {a_i}_x|$
So $d_{\infty}(c, a_i) = |c_x - {a_i}_x| = |b_x - {a_i}_x| = d_{\infty}(b, a_i)$
And we had infinite choices for $c_y$
