Circle inscribed between two curves Consider the plane region $S_n$ bounded from above and below for the graphs of $f_n(x)=x^{1/n}$ and $g_n(x)=x^n$, $0\le x\le1$. 
How to find the radious and center of the circle inscribed in $S_n$?
Intuitively, the center is on the set $\{(A,A):0\le A\le 1\}$, so we can think the problem finding the value of $A$ and $r$ such that the equation $(x-A)^2+(x^n-A)^2=r^2$ has only one solution for $x$.
Another approach is parametrization, but equations are not easy to solve.
Is there any hint?
 A: I believe that there are  many circles inscribed in that region.  As everybody else noticed, the diagonal $y=x$ is  axis of symmetry of the region.   Fix $x_0\in(0,1)$.  Imagine that a (circular) planar wave is born at the point $(x_0,x_0)$ on the diagonal. It evolves as a growing circle centered at $(x_0,x_0)$. At the first moment the wave touches  the two curves it becomes a circle  inscribed in that region.Due to symmetry it touches both  graphs at the same time) 
The MAPLE animation below illustrates the process. In the animation I chose  $n=4$, $x_0=0,75$.

So, if you choose the center of the circle  to be $(x_0,x_0)$ you are looking to find  the distance  from this point to the curve $y=x^n$.  In other words, you want to minimize the function
$$
f(x)=(x-x_0)^2+(x^n-x_0)^2.
$$
The radius of the inscribed circle centered at $(x_0,x_0)$ will then be
$$
r(x_0)=\Big(\;\min_{x\in [0,1]}(x-x_0)^2+(x^n-x_0)^2\;\Big)^{\frac{1}{2}}.
$$
The  equation 
$$
f'(x)=0  \;\Longleftrightarrow  2(x-x_0)+2nx^{n-1}(x^n-x_0)=0
\tag{1}
$$
describes a curve in the $(x,x_0)$ plane.  The  case $n=4$ is depicted below. (The parameter $x_0$ is measured on the vertical axis.) 

More  concretely  we can solve (1) for $x_0$
$$ x_0(1+nx^{n-1})= x+nx^{2n-1} \implies x_0=\underbrace{\frac{x+nx^{2n-1}}{1+nx^{n-1}}}_{c(x)}. $$
This shows that we can predict the location of the  center of the circle given the location  of its  tangencies with the curve $y=x^n$.  For $n=10$ the graph of the function $c(x)$ is depicted below.   

It shows that for $x_0$ (roughly) between $0.4$ and $0.55$ the function
$$ x\mapsto (x-x_0)^2+(x^n-x_0)^2 $$
has three critical points.  The figure below illustrates what happens when $n=10$  and $x_0=0.45$. We see that it has two local minima and a local max.   We depicted below the graph of this function for $n=10$ and $c=0.4$.

Remark 1. The  radius of the inscribed circle  has the form
$$
r(x)=\sqrt{\big(x-c(x)\big)^2+\big( x^n-c(x)\big)^2}.
$$
For the circle to be actually inscribed it is necessary that  $r(x)\leq c(x)$. 
For $n=10$ the function $r(x)$ is depicted below, in red,  and the function $c(x)$,  in blue. 

Remark 2.   The maximal radius $r_{\max}$ is the solution of the  min-max problem
$$
r_{\max}=\max_{y\in (0,1)}\min_{x\in (0,1)}\sqrt{F_n(x,y)},
$$
$$
F_n(x,y)=(x-y)^2+(x^n-y)^2
$$
This suggests looking  at the critical points of $F_n$. They are determined  by the equations
$$  (x-y)+nx^{n-1}(x^n-y)=0, \tag{2}$$
$$ y-x+y-x^n=0. \tag{3} $$
From (3) we deduce
$$ y=\frac{x+x^n}{2}. $$
Using this in (2) we deduce
$$ \frac{x-^n}{2}-nx^{n-1}\frac{x-x^n}{2} =0,$$
so 
$$x=\left(\frac{1}{n}\right)^{\frac{1}{n-1}}=:t_n.  $$
The  center is located at $(c(t_n),c(t_n))$.  Recalling that $nt_n^{n-1}=1$ we  deduce
$$c(t_n)=\frac{t_n\big(1+nt_n^{2(n-1)}\big)}{1+nt_n^{n-1}}=\frac{t_n(1+t_n^{n-1})}{2}=\frac{n+1}{2n}t_n.$$
For $n=10$ we  deduce
$$t_n\approx 0.7742,\;\;c(t_n)\approx 0.4258 $$
and
$$
r(t_n)\approx 0.4927.
$$ 
 The circle of radius $r(t_n)$ centered at $(c(t_n),c(t_n))$ is not even contained in the first quadrant, yet it is tangent to $y=x^n$.  Thus the min-max approach does not yield $r_{\max}$ as indicated in one of  @ksoriano's  comments; see the figure below . depicting the circle of radius $r(t_n)$ centered at $(c(t_n), c(t_n))$, $n=10$.

The next figure  depicts in the case $n=10$ an inscribed  circle of radius $\approx 0.45$.  It  is tangent to $y=x^n$ at $x=0.9$  and $c(0.9)\approx 0.46$. Note that the radius of an inscribed circle is at most $0.5$ so the  situation below is nearly optimal.

It seems to me that, for large $n$,  there is a unique  largest inscribed  circle,  which  has four points of tangencies  with the graphs and  as $n\to \infty$ its radius converges  to $0.5$, the radius of the circle inscribed in the square $[0,1]\times [0,1]$.
Remark 3. . Recall that the map $x\mapsto c(x)$  determines the location of the center tangent to $y=x^n$ at the point $(x,x^n)$. Its radius is $r(x)$. The plot of the $(c, r)$ curve $\newcommand{\bR}{\mathbb{R}}$
$$
[0,1]\ni x\mapsto \big(\; c(x), r(x)\;\big)\in\bR^2,
$$
is   very revealing. Below we depict the case $n=10$.

The location of the center  is tracked on the horizontal axis.  As the center moves  along the  diagonal from $(0,0)$  to $(1,1)$ we notice various "phase transitions" or bifurcations. Imagine a vertical line moving from left to right in the above figure and keep track of the number of  points of intersection with  the $(c,r)$ curve.  
Depending on the location of the center,  there are typically, either one, or three circles tangent to $y=x^n$.  To  detect $r_{\max}$ we need to keep track of the smallest of these circles. In the above plot  this means looking only at the bottom triangle of the $(c,r)$-curve.  The altitude of the top  vertex of this triangle is $r_\max$.    
In any case,  the above figure suggests that the function 
$$
y\mapsto \min_{x\in[0,1]} F_n(x,y)
$$ 
is rather regular. 
There exist  numbers $y_{\min}<y_{\max}$ in $(0,1)$ so that for $y\in[0,1]\setminus [y_{\min},y_{\max}]$ there exists only one circle centered at $(y,y)$ and tangent to the boundary of the region. $\newcommand{\pa}{\partial}$
As $y$ crosses  $y_{\min}$ increasingly the equation
$$
\pa_x F_n(x,y)=0
$$ 
undergoes a qualitative change: it goes from having one solution to having  three solutions.   It is a "birth" process.
As $y$ crosses  $y_{\max}$ increasingly, the above equation    goes from having three solutions to having one solution. It is a "death" process.
The points $y_{\min}$ and $y_{\max}$  are determined by solving the nonlinear system
$$
\pa_x F_n(x,y)=\pa^2_{xx}F_n(x,y)=0,
$$
i.e.,
$$
nx^{2n-1}-nx^{n-1}y+x-y=0,\;\;n(2n-1)x^{2n-2}+n(n-1)x^{n-2}y+1=0.
$$
I believe that for  $n$  large this system has three solutions
$$
(x_{\min},y_{\min}),\;\;(x_*,y_*),\;\;(x_{\max},y_{\max}),
$$
$$
y_{\min}<y_*<y_{\max}.
$$
The center of the larges inscribed  circle is located  $(y_*,y_*)$ and  its radius  is
$$
r_{\max}=r(x_*).
$$
The number of solutions of the system is  the number of roots in $(0,1)$ of the equation
$$
(2n-1)x^{2n-2}+(n-1)x^{n-2}c(x)+\frac{1}{n}=0.
$$
This reduces to a polynomial equation. Theoretically, the number of solutions of a real polynomial equation in a given interval can be determined using the classical Sturm theorem.
A: If the curves constitute to a function and its inverse, as is the case in your example, then you have to check pairs of points $p:=(x_a,f(x_a)),\ q:=(x_b,f^{-1}(x_b))=(f(x_a),x_a)$, so that:
$$\frac{d}{dx}f(x_a)=1=\frac{d}{dx}f^{-1}(x_b)\quad \wedge\quad \frac{f^{-1}(x_b)-f(x_a)}{x_b-x_a}=-1$$
if, as is the case in your example, the region defined by the two curves is also convex, that pair of points is unique and thus $(p,q)$ marks a diameter of the largest contained circle and thus $p=((\frac{1}{n})^\frac{1}{n-1},(\frac{1}{n})^{\frac{n}{n-1}}),\ q=((\frac{1}{n})^{\frac{n}{n-1}},(\frac{1}{n})^\frac{1}{n-1})$ define the pair of points in which the largest contained circle touches both curves; the diameter equals the distance of $p$ and $q$
