# Shrinking some polygons to make the containing polygon connected

Inside a public area $C$ (a polygon), there are several private land-plots $C_1,\dots,C_n$ (pairwise-disjoint simple polygons):

Currentlly, the public area that is outside the private land-plots (the set $C\setminus (C_1\cup\dots\cup C_n)$) is not connected - one cannot always walk from one part of the park to another without crossing a private plot.

The government wants to make the park path-connected by confiscating a tiny fraction of each of the private land-plots. Is this possible? I.e, is it possible, for every $\epsilon>0$, to remove a fraction $\epsilon$ of the area of each $C_i$, such that the set $C\setminus (C_1\cup\dots\cup C_n)$ becomes connected?

Intuitively, the government can "shave" a tiny fraction of the perimeter of each $C_i$, with width $\epsilon / Perimeter_i$. It seems "obvious" that the remainder will be path-connected. But how to prove this formally?

• Is a line connecting two points an accepted path? The width is 0 - so you really do not "cut" anything.
– Moti
Mar 27, 2016 at 16:04
• @Moti this is interesting. I would really like that the path has some width, but I am not sure how to define this accurately. Mar 27, 2016 at 17:37

Just walk, for example, in clockwise orientation, along the boundary of public area, and when you come at some point that is common to public area and private area you then remove some subset of private area that has that point in itself, for example a triangle of area $< \epsilon$ that has that point as vertex. And when you travel along boundary of public area and arrive at some line-segment that is common to public and private area you then remove some subset of private area that has that side in itself, for example a triangle of area $<\epsilon$ that has that line-segment as its side.
So you can do that for all $\epsilon$ with removal of only triangles from the whole land. Since it can be done for every $\epsilon$ it can be as small as you want.