## Area of intersection of a family of circles in the plane

Suppose you are given a family F of circles in the plane such that each circle has radius 1. Let G be the family of circles with same centers as in family F but now each circle has radius $r$. Let A be area of union of circles in family F and let B be area of union of circles in family G. Then can we upper bound the number $\frac{B}{A}$ by a function $f(r)$ ?

One trivial thing is that if the family F contains all disjoint circles then $\frac{B}{A}$ is at most $r^2$. But in general case the geometry is getting weird and complicated .

-
Is this about intersections of circles (title) or unions (text)? – Mark Bennet May 9 2011 at 21:08
@Mark: It just does not make sense for intersections. Hopefully the OP will edit the title eventually. – Sergei Ivanov May 10 2011 at 14:24

Do a google search on Kneser-Poulsen, and you will be sadder and wiser.

-
I thought that the Kneser-Poulsen conjecture (now theorem in the plane) involved moving the centers and fixing the radii, whereas this question is fixing the centers and altering the radii. But perhaps there is some relationship... – Joseph O'Rourke May 9 2011 at 21:44
A good personality test: if someone offered you the chance to become sadder and wiser, would you take it? – Tom Church May 10 2011 at 4:22
I think most people on this forum did, for better or worse... – Igor Rivin May 10 2011 at 14:15
@Joseph: just rescale the whole configuration to get the same radii. – Sergei Ivanov May 10 2011 at 14:28

The problem is much simpler than the general Kneser-Poulsen case. Here is an elementary proof of the fact that $B/A\le r^2$ for all $r\ge 1$. To simplify technicalities, I assume that the family of circles is finite (the general case follows as a limit).

Let $p_1,\dots,p_N$ be the centers of our circles. Let $V$ denote the union of the unit circles and $U$ the union of the circles of radius $r$. Divide $U$ into Voronoi regions $U_i$ of the points $p_i$. Namely, $U_i$ is the set of points $x\in U$ such that $p_i$ is nearest to $x$ among $p_1,\dots,p_N$. (For convenience, remove the points having more than one nearest center; this is a zero measure set.) Each $U_i$ is star-shaped w.r.t. the respective point $p_i$. Indeed, if $x\in U_i$, then $|p_ix|\le r$, hence the segment $[p_ix]$ is contained in $U$. And obviously $p_i$ is the nearest center for every point of this segment.

Now apply the $(1/r)$-homothety centered at $p_i$ to each set $U_i$. The resulting sets $U_i'$ are disjoint and contained in $V$, hence $$S(V) \ge\sum S(U_i') = \frac1{r^2} S(U_i) = \frac1{r^2} S(U)$$ where $S$ denotes the area. Thus $S(U) \le r^2 S(V)$, q.e.d.

The same argument works in $\mathbb R^n$ (with $r^n$ in place of $r^2$), in Alexandrov spaces of nonnegative curvature, and in Riemannian manifolds of nonnegative Ricci curvature (just combine it with the proof of Bishop-Gromov inequality).

-