Problem 1694

To prove that there exists a circle with diameter $\sqrt{2}$ such that no two points outside of this circle have the same color, we will proceed with the following steps:

1. Assumptions and Definitions:
- Assume there are at least two points of each color (red and blue), as the other cases are trivial.
- Let $S$ be the set of all colored points.
- Define an $S$-tight circle as the smallest circle that contains all points in $S$. This circle is unique and has at least two points from $S$ on its boundary.

2. **Properties of $S$-tight Circles**:
- If the $S$-tight circle has exactly two points $A, B \in S$ on it, then $\overline{AB}$ is the diameter of the circle.
- If the $S$-tight circle has at least three points in $S$ on it, then we can find three of those points that form a right or acute triangle.

3. Case Analysis:
- Let $\omega$ be the $S$-tight circle with diameter $d$ and midpoint $M$.
- If $d \leq \sqrt{2}$, the proof is complete as $\omega$ itself satisfies the condition.
- Assume $d > \sqrt{2}$ from now on.

4. **Case 1: Exactly Two Points on $\omega$**:
- Let the two points on $\omega$ be $A$ and $B$. Since $d > \sqrt{2} > 1$, $A$ and $B$ must be of the same color (say red).
- Draw two circles with radius 1 centered at $A$ and $B$. All blue points must lie in the intersection of these circles.
- Let $M$ and $N$ be the intersection points of these circles. The distance $MN$ is given by:
$$MN = 2MH = 2\sqrt{AM^2 - AH^2} \leq 2\sqrt{1 - \left(\frac{\sqrt{2}}{2}\right)^2} = 2 \cdot \sqrt{1 - \frac{1}{2}} = 2 \cdot \frac{1}{\sqrt{2}} = \sqrt{2}.$$
- Thus, the circle with diameter $\overline{MN}$, where $MN \leq \sqrt{2}$, contains all blue points.

5. **Case 2: At Least Three Points on $\omega$:
- Claim 1**: All points on $\omega$ are of the same color.
- Assume for contradiction that this is not the case. By the pigeonhole principle, at least two points must be of the same color. Let $A$ and $B$ be red, and $C$ be blue.
- If there are exactly three points on the circle, they form a right or acute triangle. Let $\triangle ABC$ be such a triangle.
- Since $\angle AMB \leq 180^\circ$, one of the angles $\angle CMA$ or $\angle BMC$ is $\geq 90^\circ$. Assume $\angle CMA \geq 90^\circ$.
- By the Pythagorean inequality:
$$AC^2 \geq MA^2 + MC^2 > 2 \cdot \left(\frac{\sqrt{2}}{2}\right)^2 = 1 \implies AC > 1,$$
which violates the distance condition.
- If there are at least four points on the circle, each of the $\binom{4}{3}$ triangles contains two points of the same color. One of these triangles must be right or acute, leading to a contradiction as in the $k=3$ case.
- Thus, all points on $\omega$ must be of the same color.

6. Conclusion from Claim 1:
- We can find three points on $\omega$ that have the same color (say red) and form a right or acute triangle $\triangle ABC$.

7. Claim 2: All blue points have a distance less than $\frac{\sqrt{2}}{2}$ from $M$.
- Let $P$ be any blue point in $\omega$. Since $\triangle ABC$ is right or acute, $M$ lies inside $\triangle ABC$ or on one of its sides.
- Project $A$, $B$, and $C$ through $M$ onto $\omega$ to $A'$, $B'$, and $C'$.
- If $P$ lies in sector $C'B'$, one of the angles $\angle PMB$ or $\angle PMC$ is right or obtuse. By the Pythagorean inequality:
$$|PM|^2 + d^2 \leq \ell^2 \implies |PM|^2 \leq \ell^2 - d^2 < 1 - \frac{1}{2} = \frac{1}{2} \implies |PM| < \frac{\sqrt{2}}{2}.$$
- The same argument holds for sectors $B'A'$ and $A'C'$. Thus, all blue points lie in a circle with diameter $\sqrt{2}$ around $M$.

The last claim finishes the proof.