6. Place point $C$ at the origin of the Cartesian coordinate system, and point $B$ on the x-axis, as shown in the figure. Then the points $C, B, A$ will have coordinates $C(0,0), B(2,0), A(1, \sqrt{3})$. The geometric locus of points $K(x, y)$ such that $B K: C K=1: 2$ is a circle. Indeed,

$\frac{B K^{2}}{C K^{2}}=\frac{(x-2)^{2}+y^{2}}{x^{2}+y^{2}}=\frac{1}{4}$,

which is equivalent to

$$
\left(x-\frac{8}{3}\right)^{2}+y^{2}=\frac{16}{9}
$$

Similarly, the geometric locus of points $M$ such that $A M: M C=1: 2$ is a circle given by the equation

$$
\left(x-\frac{4}{3}\right)^{2}+\left(y-\frac{4 \sqrt{3}}{3}\right)^{2}=\frac{16}{9}
$$

To maximize the distance between points $K$ (lying on circle (1)) and $M$ (on circle (2)), these points should be placed on the line passing through the centers of the circles. (Note that the jury decided not to reduce the score for the problem if the student used this fact but did not provide a proof.)

Thus, the sought maximum distance is the distance between the centers of the circles (1), (: $M K_{\text {max }}=\frac{8}{3}+\frac{4}{3}+\frac{4}{3}=\frac{16}{3}$.
Answer: $16 / 3$.

![](https://cdn.mathpix.com/cropped/2024_05_06_616ccf7773361e641efcg-3.jpg?height=634&width=653&top_left_y=1428&top_left_x=2269)