Problem about distances of points in the Euclidean plane Consider the sets $\mathcal{A}$ and $\mathcal{B}$ of points in the Euclidean plane, such that for each pair $P\in \mathcal{A}, Q\in \mathcal{B}$, we have $\overline{PQ}\leq 1$. Prove that either all points in $\mathcal{A}$ or all points in $\mathcal{B}$ fit into a disc of diameter $\sqrt{2}$.
How do we solve this problem? I tried to use Jung's theorem, which yields that if the maximum distance in $\mathcal{A}$ is less than or equal to $\sqrt{3/2}$, we are done. Also, I found that if the maximum distance is greater than or equal to $\sqrt{2}$, we are done as well. How do we deal with the cases if the maximum distance is in between?
 A: Spose $Q,S \in B$ and $QS > \sqrt 2$.  Assuming $A$ is not empty (if $A$ or $B$ are empty the conjecture is vacuously true) then we know $\sqrt 2 < QS \le 2$ as for any $P\in A$ we have $QS \le QP + PS \le 1+1=2$.
Now construct the midpoint $M$ of $QS$ and a disc of radius $\frac {\sqrt{2}}2$ or diameter of $\sqrt 2$ centered at $M$.
Let $P\in A$  so  $PS \le 1$ and $PQ \le 1$.
If $PS=PQ$ then $P$ is on the perpendicular bisector of $QS$ and $MP^2 + MQ^2 = PQ^2$.  $MQ >\frac {\sqrt 2}2$ so $MQ^2 > \frac 12$ and $PQ \le 1$ so $PQ^2 \le 1$ and $MP^2 < 1-\frac 12$ so $MP < \frac {\sqrt 2}2$. And $P$ is in the disc.
If $PS > PQ$ construct a line through $P$ parallel to the perpendicular bisector $\overline{QS}$ and have it intersect $\overline{QS}$ at $K$.
Consider the right triangles $\triangle SKP$ and $\triangle MKP$ so $MP^2= MK^2 + KP^2$ and $PS^2 = SK^2 + MK^2=(SM + MK)^2 + MK^2$.  If $MP > \frac {\sqrt 2}2$ and $SM >\frac {\sqrt 2}2$ we'd get a contradiction that $1 \ge PS^2 = KS^2 + KP^2 > SM^2 +MS^2 > \frac 12 + \frac 12 = 1$.  $MP < \frac {\sqrt 2}2$ and $P$ is in the disk.
Similar argument would show that  if $PS < PQ$ that $P$ is in the disk.
So if so if the points of $B$ are not within a disk of diameter of  $\sqrt 2$ then $A$ is within a disk of diameter $\sqrt 2$.  Similar argument would show if the points of $A$ are not within a disk of radius $2$ then the points of $B$ will be.
