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?


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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 $n=4$, $x_0=0,75$.

enter image description here

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.)

enter image description here

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.

enter image description here

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 $c=0.4$.

enter image description here

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.

enter image description here

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

$$ \frac{x-^n}{2}-nx^{n-1}\frac{x-x^n}{2} =0,$$


$$x=\left(\frac{1}{n}\right)^{\frac{1}{n-1}}=:t_n. $$

The center is located at $(c(t_n),c(t_n))$. Recalling that $nt_n^{n-1}=1$ we deduce


For $n=10$ we deduce

$$t_n\approx 0.7742,\;\;c(t_n)\approx 0.4258 $$


$$ r(t_n)\approx 0.4927. $$ 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.

enter image description here

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 $n=10$.

enter image description here

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.

  • $\begingroup$ Thanks. But what if I'm searching for $r_{\max}$? That is, the maximum circle incribed. $\endgroup$ – ksoriano Apr 20 at 3:41
  • $\begingroup$ I've added a remark about $r_{\max}$. $\endgroup$ – Liviu Nicolaescu Apr 20 at 10:24

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 $q$

  • $\begingroup$ for n=10 it fails. $\endgroup$ – ksoriano Apr 20 at 4:23
  • 3
    $\begingroup$ Yes, I added a similar answer before noticing that while I was writing it this one was already posted. But it is possible that the largest circle has four contact points with the two curves rather than contacting it only at the two points of a diameter, and when this happens the solution is more complicated. $\endgroup$ – David Eppstein Apr 20 at 4:35
  • $\begingroup$ @DavidEppstein yes, I completely missed the check for the curvature in the contact points; I'll leave my answer as an instructive example of how important careful analysis is... $\endgroup$ – Manfred Weis Apr 20 at 5:15

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