Problem 1247

Prove (1) Take any six circles, and represent the six circles with six points.

If two circles are disjoint, connect them with a red line; if two circles are not disjoint, connect them with a blue line. It is easy to know that six points connected with two colors must contain a monochromatic triangle.

Given that there is no blue triangle, there must be a red triangle, i.e., among the six circles, there must be three circles that are pairwise disjoint.
(2) Let $\odot A$ be the circle with the smallest radius among 12 circles. It can be proven: The number of circles not disjoint from $\odot A$ is at most five.
If not, $\odot A$ is not disjoint from $\odot A_{1}, \odot A_{2}, \cdots, \odot A_{6}$.
Suppose one of $\odot A_{i}(i=1,2, \cdots, 6)$ is concentric with $\odot A$, without loss of generality, let $\odot A_{1}$ be concentric with $\odot A$. Then, since $\odot A_{2}$ is not disjoint from $\odot A$, and the radius of $\odot A_{1}$ is greater than the radius of $\odot A$, hence $\odot A, \odot A_{1}, \odot A_{2}$ are mutually not disjoint, which is a contradiction.
Thus, $\odot A_{i}$ are not concentric with $\odot A$.
Let the points $A_{1}$, $A_{2}, \cdots, A_{6}$ be arranged in a counterclockwise direction around point $A$. Since
$$\begin{array}{l} \angle A_{1} A A_{2}+\angle A_{2} A A_{3}+\angle A_{3} A A_{4}+ \\ \angle A_{4} A A_{5}+\angle A_{5} A A_{6}+\angle A_{6} A A_{1}=2 \pi, \end{array}$$

there must be one of the six angles that is no greater than $\frac{\pi}{3}$.
Without loss of generality, let $\angle A_{1} A A_{2} \leqslant \frac{\pi}{3}$. Then
$$A_{1} A_{2} \leqslant \max \left\{A_{1} A, A A_{2}\right\} \text {. }$$

Since $\odot A$ is not disjoint from $\odot A_{1}$ and $\odot A_{2}$, then $\odot A_{1}$ and $\odot A_{2}$ are disjoint. Therefore,
$$\begin{array}{l} r_{A_{1}}+r_{A_{2}}<A_{1} A_{2} \leqslant \max \left\{A A_{1}, A A_{2}\right\} \\ \leqslant \max \left\{r_{A}+r_{A_{1}}, r_{A}+r_{A_{2}}\right\}, \end{array}$$

where $r_{A_{1}}$ represents the radius of $\odot A_{i}$.
Without loss of generality, let $r_{A_{1}}+r_{A_{2}}<r_{A_{1}}+r_{A_{1}}$. Then $r_{A_{2}}<r_{A}$, which contradicts the fact that $\odot A$ is the circle with the smallest radius among the 12 circles.
Thus, the number of circles not disjoint from $\odot A$ is at most five.
Therefore, among the 12 circles, there are at least six circles that are disjoint from $\odot A$.

By (1), among the six circles disjoint from $\odot A$, there must be three circles that are pairwise disjoint. Adding $\odot A$ to these three circles, there are four circles that are pairwise disjoint.
Thus, the problem is proven.