Some points in the plane are either colored red or blue. The distance between two points of the opposite color is at most 1. Prove that there exists a circle with diameter $\sqrt{2}$ such that no two points outside of this circle have same color. It is enough to prove this claim for a finite number of colored points.
Problem
Source: Bundeswettbewerb Mathematik 2022, Round 2, Problem 4
Tags: enclosing circles, colored points, geometry, combinatorics
02.09.2022 11:37
(This is a summary of the proof that I submitted for the contest, 9 DinA4 pages iin total): Assume that there are at least two points of each color, as the other cases are trivial. Then the problem becomes equivalent to showing that there is a circle with diameter $\sqrt{2}$ that contains all the colored points (so either in or on the circle). We will call a circle $S$-tight if it is the smallest circle that contains all points of a set $S$. We will take the following for granted: (1) Every set $S$ of countably many points in the plane has a $S$-tight circle, that is moreover unique. (2) $S$-tight circles has at least two points $A,B\in S$ on it. (3) If the $S$-tight circle has exactly two points $A,B\in S$ on it, then $\overline{AB}$ is the diameter of the circle. (4) If the $S$-tight circle at least three points in $S$ on it, then we can find three of those points that form a right or acute triangle. We will denote the tight circle with respect to the colored points by $\omega$ with diameter $d$ and Midpoint $M$. If $d\le \sqrt{2}$ we are easily done. Hence we may assume that $d>\sqrt{2}$ from now on. Case 1: There are exactly two colored points on $\omega$ (see second attachement}. Using (3) those two points must form the diameter of the circle and have to be of the same color, as otherwise the distance condition would be dissatisfied, because $d>\sqrt{2}>1$. Wlog let them be red and denoted by $A$ and $B$. We now toss two circles with radius $1$ around those two points, such that now all blue points must lie in the intersection of these circles in order to satisfy the distance condition. If $M$ and $N$ are the intersection points of the two triangles, $H\coloneq \overline{AB}\cap\overline{MN}$, then we see that $$MN=2MH=2\sqrt{AM^2-AH^2}\le 2\sqrt{1-\left(\frac{\sqrt{2}}{2}\right)^2}=2\cdot\sqrt{1-\frac{1}{2}}=2\cdot\frac{1}{\sqrt{2}}=\sqrt{2}.$$It is now easy to see that the circle with $\overline{MN}, MN\le\sqrt{2}$ contains all blue points. Case 2: There are at least three colored points on $\omega$.
With this claim we now know that we may find three points on $\omega$ that have the same color, wlog red, and form a right or acute triangle. Let this triangle be called $\triangle ABC$. We complete the proof with the following claim.
The last claim finishes the proof.
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