10. The Right Stars (from 8th grade, 3 points). Let natural numbers $k$ and $n$ be coprime, with $n \geq 5$ and $k < n / 2$. A proper $(n ; k)$-star is defined as a closed broken line that results from replacing every $k$ consecutive sides in a regular $n$-gon with a diagonal having the same endpoints. For example, the $(5 ; 2)$-star shown in the figure has 5 points of self-intersection, which are the bold points in the figure. How many self-intersections does the star $(2018 ; 25)$ have?
Solution. We will solve the problem in general for the $(n ; k)$-star. Let the vertices of the $n$-gon be denoted by $A_{0}, A_{1}$, and so on up to $A_{n-1}$. Take two consecutive segments of the star: $A_{0} A_{k}$ and $A_{1} A_{k+1}$. They intersect at a circle of radius $r_{1}$ with the center at the center of the $n$-gon. By rotating by $360\%$, we obtain another $n-1$ points lying on the same

![](https://cdn.mathpix.com/cropped/2024_05_06_a933850cc044137d5d9dg-06.jpg?height=721&width=420&top_left_y=248&top_left_x=293)

circle. In total, there are $n$ points on this circle, located at the vertices of a smaller $n$-gon.

Now take the pair of segments $A_{0} A_{k}$ and $A_{2} A_{k+2}$. They have a common point on a circle of radius $r_{2} > r_{1}$. There are $n$ such points.

In the diagram, the case $k=3$ is shown: the intersection point of the segments $A_{0} A_{3}$ and $A_{1} A_{4}$ lies on a circle of radius $r_{1}$. On this same circle, the intersection point of the segments $A_{1} A_{4}$ and $A_{2} A_{5}$ is also located. The intersection point of the segments $A_{0} A_{4}$ and $A_{2} A_{5}$ lies on a larger circle of radius $r_{2}$.

Proceeding in this manner, we obtain a total of $k-1$ circles of different radii, each of which contains exactly $n$ intersection points. Therefore, no two points coincide. This way, all intersection points of the segments are enumerated, and there are a total of $n(k-1)$ of them.

For $n=2018, k=25$, we get that the number of self-intersections is $2018 \cdot 24 = 48432$.

Other solutions are possible, but in any case, it is necessary to explain why no two intersection points coincide.

Answer: 48432.