10.52 A circle contains 6000 points, no three of which are collinear.
（1）Is it possible to divide this circle into 2000 regions, each containing exactly three points? How can this be done?
(2) If the three points in each region satisfy: the distance between any two points is an integer and does not exceed 9, then using the three points in each region as vertices to form triangles, how many of these triangles are of the same size at minimum?
[Solution] (1) 6000 points inside a circle can determine $C_{6000}^{2}$ lines.
Since $C_{6000}^{2}$ is a finite number, there must exist a tangent line to the circle that is not parallel to any of the $C_{6000}^{2}$ lines. Let this tangent line be $l$.

Moving $l$ parallel to itself along the circle, it is clear that the 6000 points will be crossed one by one, because if $l$ crosses two points at the same time, then the line connecting these two points must be parallel to $l$, which contradicts the choice of $l$. Thus, when $l$ has crossed 3 points but not yet the 4th point, draw a chord $l_{1}$ of the circle. Similarly, when it crosses the 4th, 5th, 6th points, draw chords $\iota_{2}, \cdots$. In this way, 1999 chords can be drawn, dividing the circle into 2000 segments, each containing exactly three points.
(2) First, calculate the number of triangles with integer side lengths and the longest side not exceeding 9.
Let the side lengths be $a \leqslant b \leqslant c \leqslant 9, a, b, c$ be positive integers.
When $c=9$, noting that $a+b>c$, then
$c=9, b=5, a=5$, at this time there is 1 triangle.
$c=9, b=6, a=4,5,6$, at this time there are 3 triangles.
$c=9, b=7, a=3,4,5,6,7$, at this time there are 5 triangles.
$c=9, b=8, a=2,3, \cdots, 8$, at this time there are 7 triangles.
$c=9, b=9, a=1,2, \cdots, 9$, at this time there are 9 triangles.
Thus, when $c=9$, there are 25 triangles.
Similarly, when $c=8$, there are 20 triangles, and when $c=7,6,5,4,3,2,1$, the number of different triangles obtained are: $16,12,9,6,4,2,1$ respectively.
Therefore, the total number of different triangles with integer side lengths and the longest side not exceeding 9 is
$$
1+2+4+6+9+12+16+20+25=95 \text { triangles. }
$$

In 2000 triangles, the number of triangles that are exactly the same in size is at least
$$
\left[\frac{2000}{95}\right]+1=22
$$
triangles.