Geometric Property of intersection of Circle with Polygon Start with a simple polygon $P$. Choose a point $A$ in the polygon and draw a circle of radius $1$ around it. Now look at the intersection of the circle, the polygon and the points visible from $A$ in the polygon. Let's name this intersection $D$. I am interested in the following question:
What can we say about the largest circle that can fit in $D$? Can we estimate its size? For instance let's assume that $D$ has an area which is at least half of the circle, that is $\frac{\pi}{2}$. Can we argue that we can fit a circle of area $\frac{\pi}{4}$ in $D$?
 A: The answer is negative.
Start from the closed unit disk $\mathcal{D} = \bar{B}((0,0),1)$ centered at origin.
For any $r \in (0,\frac12)$, consider the line segment
$$\ell_0 \stackrel{def}{=} \bigg\{ (\rho, 0 ) : \rho \in ( r, 1 ] \bigg\}$$
If one look at those points in $\mathcal{D}$ at a distance less than $r$ from $\ell_0$, one will notice they fall inside the union of an open circle $B((r,0),r)$ and a rectangle $[r,1] \times (-r,r)$.
In particular, this include all points of the form
$(\rho\cos\theta,\rho\sin\theta)$ for $\rho \in (0,1]$ and $|\theta| < \sin^{-1} r$.
Pick an integer $N$ large enough so that $\frac{\pi}{N} < \sin^{-1}{r}$. Let
$\ell_1,\ldots,\ell_{N-1}$ be the line segments
$$\ell_k \stackrel{def}{=} \left\{ \left( \rho\cos\frac{2\pi k}{N}, \rho\sin\frac{2\pi k}{N} \right) : \rho \in (r,1] \right\}$$
Aside from the origin, all other points in $\mathcal{D}$ will be at a distance less than $r$ from at least one of the line segments $\ell_0,\ell_1,\ldots,\ell_{N-1}$. This implies the largest closed disk one can fit inside $\mathcal{D} \setminus \left(\bigcup_{k=0}^{N-1} \ell_k\right)$ is the disk $\bar{B}((0,0),r)$ centered at origin with radius $r$.
Replace the line segments with narrow triangles, we can find
polygon $P$ to make the area of corresponding $D$ as close to $\pi$ as possible and yet the largest circle that can fit inside $D$ remains to have radius $r$.
Since $r$ can be as small as we wish, it is impossible to lower bound the radius of largest circle using the area of $D$ alone.
