If $k$ circles have already been drawn, then the ( $k+1$ )-th circle is divided by them into no more than $2 k$ arcs.

## Solution

One circle divides the plane into two parts. Suppose $k$ circles have already been drawn. Consider the ( $k+1$ )-th circle. It intersects the previous $k$ circles in no more than $2 k$ points (each circle - in no more than two points). Therefore, the ( $k+1$ )-th circle is divided by the first $k$ circles into no more than $2 k$ arcs. Each arc divides one of the parts into which the plane was divided by the $k$ circles into two parts. Thus, each arc adds one part of the plane, and the ( $k+1$ )-th circle adds no more than $2 k$ parts of the plane. Moreover, the ( $k+1$ )-th circle adds exactly $2 k$ parts of the plane if and only if it intersects each of the previous circles in two points and all these points are distinct. Thus, $n$ circles divide the plane into no more than $2+(2+4+6+\ldots+2(n-1))=n(n-1)+2$ parts, and equality is achieved if each pair of circles intersects in two points and all these intersection points are distinct (i.e., no three circles pass through the same point).

An example of such a system of $n$ circles can be given: take one circle, and all the others are obtained from it by translations by vectors $\boldsymbol{a}, 2 \boldsymbol{a}, \ldots,(n-1) \boldsymbol{a}$, where $\boldsymbol{a}$ is a vector of sufficiently small length (so small that the first and last circles intersect in two points).

## Answer

Into $n(n-1)+2$ parts.
What is the maximum number of parts that $n$ circles can divide a plane into?

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