Show that the incircle is the largest circle which fits inside regular polygon Consider a regular polygon with $n \ge 3$ sides. The incircle of the regular polygon passes through the midpoint of each sides and is also tangent to the sides at the midpoints. Intuitively, the incircle should be the largest circle, in terms of the radius, that fits completely inside the regular polygon (formally the closure of the set of points contained inside the polygon). Is there a rigorous proof for this?
This should be true and it turns out to be useful for a very nice problem:

Problem: Show that there exists a set S of points in the plane that cannot be covered by a regular n polygon with side length 1 but any n points of S can be covered.


 Take the incircle and enlarge it ever so slightly about its center. The enlarged circle intersects each side at two points. We can make the total sum of the arcs that protrudes from the sides to be $< 0.001n^{-2}$ fraction of the total circumference. If the incircle is the largest circle that fits inside the regular polygon, then the set S of all the points on the circumference of the enlarged circle cannot fit in the polygon. However, if we take any $n$ points, there exists a rotation of the regular polygon such that all the points are inside it. To show this, just take a random rotation about the center of the circle and the expected number of points lying outside the polygon is $<1$.

 A: Let $O$ be the center of symmetry of your polygon $P$. Take any circle not centered at $O$. You can rotate the center by $2\pi/n$ about $O$ and the image of the circle remains inside $P$. In fact, you can rotate it $n-1$ times, forming a congruent polygon $Q\cong P$ containing $O$. Any point in the convex hull of the centers will house a circle of equal size, so you might as well place the circle at $O$. Symmetry tells you that the largest circle will intersect $P$ at the midpoints of the sides.
A: Here is a tedious solution:
By translating & rotating as necessary, we can assume that the polygon $P$ is centred at the origin with one point at $\rho e_1$, with $\rho >0$.
Let $\theta = {2 \pi \over n}$, $U_\alpha$ be a clockwise rotation by $\alpha$ and $v_k = \rho U_{k\theta} e_1$ (rotation of unit vector in $x$ direction by $k \theta$). Let $m_k = {1 \over 2} (v_k+v_{k+1})$, the midpoint of two adjacent vertices (with $v_n=v_0$ for notational convenience.)
Some work shows that $P= \operatorname{co} \{ v_k\}_k = \{x \mid \langle m_k, x \rangle \le \|m_k\|^2, \text{ for all } k \} $.
Consider the convex problem
$\sup \{ r \mid r \ge 0, c \in \mathbb{R}^2,\langle m_k, c+r U_\theta e_1 \rangle \le \|m_k\|^2, \text{ for all } k, \theta \} $.
Some more work shows that
$\langle m_k, c+r U_\theta e_1  \rangle \le \|m_k\|^2$ for all $ k,\theta $ iff
$ \langle m_k, c \rangle +r \|m_k\| \le \|m_k\|^2$ for all  $k$.
Hence we can write the problem as
$\min\{ -r \mid \langle m_k, c\rangle + r \|m_1\|\le \|m_1\|^2, \text{ for all } k \} $, and so a solution exists since the feasible set is compact.
If $(c,r)$ is a solution, so is $(U_{k\theta} c, r)$, hence the average (over $k$) is also a solution, and $(0,r)$ solves the problem, which gives $r=\|m_1\|$. It follows from this that $c=0$ for any solution $(c,r)$ and so the solution is $(0,\|m_1\|)$.
Hence the largest circle that fits inside $P$ is tangent to the midpoint of each side ($m_k$).
