Some circles are placed on the plane in such a way that there are exactly 12 points where at least two circles intersect. What is the minimum number of regions into which these circles divide the plane?
Let's call a set of circles connected if from any point on any circle, we can reach every other circle by traveling along the arcs. In the following, we will prove the following statement:

If the set of circles $M$ breaks down into exactly $m$ connected components, and exactly $k$ points lie on at least two circles, then these circles divide the plane into at least $k+m+1$ regions.

First, let $m=1$, that is, $M$ is connected. Start drawing the circles one by one so that the resulting set of circles is always connected. This is possible because of the connectedness of $M$; at each step, there will be an un-drawn circle that intersects or touches one of the already drawn circles.

![](https://cdn.mathpix.com/cropped/2024_05_02_ad8ed6bcd1c9967242d6g-1.jpg?height=659&width=694&top_left_y=451&top_left_x=694)

Figure 1

Let the $n$-th circle drawn have $l$ points in common with the previously drawn circles; then $l \geq 1$. This circle is divided into $l$ arcs by these $l$ points, and each arc divides a closed region of the plane into two, or encloses a closed region from the infinite region (Figure 1). Therefore, the number of regions in the plane increases by $l$ in the $n$-th step, while the number of intersection or tangency points of the circles increases by at most $l$, since among the $l$ points of intersection or tangency created by the $n$-th circle with the previously drawn ones, there may be points that already belonged to at least two circles before the $n$-th step.

In the first step, the single circle drawn divides the plane into two regions, and the number of points lying on at least two circles is 0. According to the above, the number of regions increases by at least as much as the number of intersection or tangency points of the circles in each step, so - after a total of $k$ points lie on at least two circles - the number of regions created is at least $k+2$.

Switching to the case $m>1$, if the $i$-th connected component $(i=1,2, \ldots, m)$ has $k_{i}$ intersection or tangency points of the circles, then at least $k_{i}+2$ regions are created in this component (including the infinite region). Considering that $\sum_{i=1}^{m} k_{i}=k$, and that the infinite region was counted in each of the $m$ connected components, the total number of regions created is at least

$$
=\sum_{i=1}^{m} k_{i}+2 m-(m-1)=k+m+1
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_ad8ed6bcd1c9967242d6g-1.jpg?height=466&width=480&top_left_y=2071&top_left_x=801)

In the problem, $k=12$, so the minimum number of regions is 14. This many regions are created, for example, when we have four circles, each pair of which intersects (Figure 2).