Brodskii H.

Figure F represents the intersection of $n$ circles ($n \geq 2$, the radii are not necessarily the same). What is the maximum number of curved "sides" that figure F can have? (A curved side is a segment of the boundary of F that belongs to one of the circles and is bounded by intersection points with other

circles.)
Number the circles with numbers from 1 to $n$. The numbering of the sides of F is a circular arrangement of these numbers (with repetitions), satisfying the following conditions:

a) two identical numbers do not stand next to each other;

b) there are no fragments of the form ................. .

Indeed, consider two arcs $A B$ bounding the intersection of circles $a$ and $b$. If we take two points $A_{1}$ and $A_{2}$ on one of them (belonging to the boundary of circle $a$) and two points $B_{1}$ and $B_{2}$ on the other, then the chords $A_{1} A_{2}$ and $B_{1} B_{2}$ obviously do not intersect.

We will prove by induction on $n$ that the length $l$ of such an arrangement does not exceed $2 n-2$.

Base case ($n=2$) is obvious.

Inductive step. If all numbers occur no more than once, then everything is fine: $n \leq 2 n-2$.

Suppose one of the numbers (a) appears $k>1$ times. Break the arrangement at these points and form a separate arrangement from each part, "gluing" the ends. All of them satisfy the given conditions, so the length $l_{i}$ of the $i$-th arrangement is no more than $2 n_{i}-2$. Moreover, these arrangements have no common numbers other than $a$. Therefore, adding them up, we get $l \leq 2(n+k-1)-2 k=2 n-2$.

Arrangements of length $2 n-2$ exist, for example, $1,2, \ldots, n-1, n, n-1, \ldots, 2$ or $1, n, 2, n, \ldots, n-1, n$. It is not difficult to construct examples of figures F corresponding to these arrangements.

## Answer

$2 n-2$ sides.