most recent 30 from http://mathoverflow.net 2013-05-25T10:04:45Z http://mathoverflow.net/feeds/question/64422 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/64422/area-of-intersection-of-a-family-of-circles-in-the-plane csguy 2011-05-09T19:42:56Z 2011-05-10T14:10:59Z

<p>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 $f(r)$ ?</p> <p>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 .</p>

http://mathoverflow.net/questions/64422/area-of-intersection-of-a-family-of-circles-in-the-plane/64431#64431 Igor Rivin 2011-05-09T21:07:55Z 2011-05-09T21:07:55Z

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

http://mathoverflow.net/questions/64422/area-of-intersection-of-a-family-of-circles-in-the-plane/64500#64500 Sergei Ivanov 2011-05-10T14:10:59Z 2011-05-10T14:10:59Z

<p>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).</p> <p>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.</p> <p>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.</p> <p>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).</p>