If I am looking at a collection $\mathcal{C}$ of circles $\{C\_1,...,C\_n\}$ all of which have some radii $\{r\_1,...,r\_n\}$ where $r\_i\in\mathbb{R}^{+}$ for each $i \in[n]$. In $\mathcal{C}$, all the circles are pairwise intersecting i.e. each circle intersects another circle. We do not allow touchings (i.e. the circles only touch at one single point). Note that we also do not allow faces in these arrangements that are bounded by two arcs and two intersection points.


We define a $3$-cell in this collection of circles as a cell bounded by three arcs and three intersection points of the circles. Moreover, an empty $3$-cell is a $3$-cell such that no circle passes through it.


I am trying to prove or disprove the following: each circle $C\_i\in\mathcal{C}$ bounds at least two empty $3$-cells so that one of them is inside $C\_i$ and one of them on the outside of $C\_i$.


I have the some of these following ideas: fix $C\_i$ and give it a certain direction and split the intersection points of $C\_i$ with other circles into two sets: $\{p\_1,...,p\_{n-1}\}$ being the first intersections of $C\_i$ with $C\_j$ for $i\neq j$ and $\{q\_1,...,q\_{n-1}\}$ being the set of the second such intersections.


Now, it seems like any set of three circles creates exactly $7$ $3$-cells, however, in the entire $\mathcal{C}$ these are not necessarily empty, and this is the problem. Therefore as a whole, there are the total of $7\cdot \binom{n}{3}$ $3$-cells.


Another thing worth noting looking at is how the circles behave on the inner and the outer side of the fixed $C\_i$ based on the order of their intersection with $C\_i$, but I am not entirely sure how to go from here.


Perhaps



Pairwise intersecting circles in the plane
First, it is sufficient to show that there is always an empty 3-cell inside each circle. This is because we can always do a Möbius transformation to move the outside of a circle inside of it, which will preserve intersections between circles. The same argument can then be used.


Now pick some circle $C$. First, there must exist $C\_i$ and $C\_j$ that form an empty 3-cell in $C$, call it $T$. If this was not the case, no pair of the other circles could intersect each other inside of $C$. But then there would be a face bounded by two arcs in $C$.


Next consider what happens as you draw another circle, say $C\_k$. If it intersects $T$ at 0 or 1 points, $T$ remains a 3-cell. If it intersects two different edges of $T$, you get a smaller 3-cell inside of $T$. If it intersects one edge of $T$ twice, then by the assumption that no faces are bounded by two arcs, there must be another circle $C\_\ell$ so that $\{C\_i,C\_j,C\_k,C\_\ell\}$ contain a 3-cell inside $T$. By repeating this argument until we are out of circles, we find that there there is some empty 3-cell.


