Prove the center of the unit circle has the highest average ray length?

If the average ray length is the average distance of the segments from a point inside the circle to points evenly distributed on the boundary:

Prove the center of the unit circle has the highest average ray length.

My Attempt

Convert the circle $x^2+y^2=1$ into polar coordinates so the distances are evenly distrbuted. Since distance $r$ is centered at the origin we must move point $(u,v)$ in the circle to the origin.

$$(x+u)^2+(y+v)^2=1$$

$$(r\cos(\theta)+u)^2+(r\sin(\theta)+v)^2=1$$

$$r^2+2r(u\cos(\theta)+v\sin(\theta))+u^2+v^2-1$$

Solving for $r$ and simplifying give us

$$r=-\left(u\cos(\theta)+v\sin(\theta)\right)\pm\sqrt{(u\cos(\theta)+v\sin(\theta))^2-(u^2+v^2-1)}$$

Since $r$ must be positive, we get the average radius is

$$\frac{1}{2\pi}\int_{0}^{2\pi}\left|-\left(u\cos(\theta)+v\sin(\theta)\right)\pm\sqrt{(u\cos(\theta)+v\sin(\theta))^2-(u^2+v^2-1)}\right| d\theta$$

Then solve the integral and find the maximum in terms of $(u,v)$

The problem is I'm not sure if the integral is solvable. Is there another way of approaching this problem?

• I think this is wrong. For the center of the unit circle the average ray length is $1$, for a point on the boundary it is ${4\over\pi}$. – Christian Blatter Apr 1 '18 at 20:09
• Indeed the center is the point with the minimum average distance from the boundary. Please replace highest with lowest. – Jack D'Aurizio Apr 1 '18 at 20:32

Here is an easy elementary argument. Let $p$ be the point we are interested in. Instead of considering the average value of $px$ as $x$ runs through points on the circle, consider the average value of the sum $px + p\overline x$, where $\overline x$ is the opposite point on the circle to $x$.

$px+p\overline x=2$ iff $p$ is on the diameter from $x$ to $\overline x$; otherwise it is greater. Thus, the average value of $px+p\overline x$ will be greater than 2 unless $p$ is always on the diameter, which only happens if $p$ is the centre of the circle.

By symmetry, the average distance from the boundary of $\|z\|\leq 1$ is a function of the distance from the origin. If $x\in(0,1)$ its average distance from the the boundary of the unit circle centered at the origin is given by $$\frac{1}{2\pi}\int_{0}^{2\pi}\sqrt{(x-\cos\theta)^2+\sin^2\theta}\,d\theta =\frac{\sqrt{1+x^2}}{\pi}\int_{0}^{\pi}\sqrt{1-\frac{2x}{1+x^2}\cos\theta}\,d\theta$$ and by letting $\lambda=\frac{2x}{1+x^2}\in(0,1)$ we have $$\int_{0}^{\pi}\sqrt{1-\lambda\cos\theta}\,d\theta=\int_{0}^{\pi/2}\sqrt{1-\lambda\cos\theta}+\sqrt{1+\lambda\cos\theta}\,d\theta=\int_{0}^{1}\frac{\sqrt{1-\lambda u}+\sqrt{1+\lambda u}}{\sqrt{1-u^2}}\,du$$ clearly leading to an elliptic integral of the second kind. It is not difficult to devise tight algebraic approximations for these objects (see, for instance, the dedicated section in my notes), and according to Mathematica's notation we have that the average distance of $x\in(0,1)$ from the boundary of the unit circle centered at the origin is $$\frac{2}{\pi}(x+1)\cdot E\left(\frac{4x}{(x+1)^2}\right)=1+\frac{x^2}{4}+\frac{x^4}{64}+\frac{x^6}{256}+\frac{25 x^8}{16384}+\ldots$$ where all the involved coefficients of the Maclaurin series are non-negative, implying that the LHS is increasing over $(0,1)$.

Now a very tricky elementary approach. The average distance of $x\in(0,1)$ from the boundary of the unit circle is given by $\frac{1}{2\pi}$ times the perimeter of an ellipse with semiaxis lenghts $1-x$ and $1+x$. If $A,B$ are two bounded, convex sets in $\mathbb{R}^2$ and $A\subsetneq B$, then the perimeter of $A$ is less than the perimeter of $B$. This implies that the previous average distance / ellipse perimeter is an increasing function of the $x$ variable over the interval $(0,1)$.

Yet another elementary approach by convexity. The function giving the distance from a fixed point is convex and the sum of convex functions is convex. Convex functions over convex domains attain their maximum at the boundary. So we have that over $x^2+y^2\leq 1$ the average distance from $x^2+y^2=1$ is a radial and convex function. It is pretty obviously differentiable over $x^2+y^2<1$ (we already wrote an explicit integral representation) and the origin is a stationary point: in order to prove that the origin is an absolute minimum it is enough to show that the average distance is not constant over $x^2+y^2\leq 1$, and that is trivial.

There is a much more obvious way of answering this question. Take any point in the interior of the circle, pass a chord through that point. The average of the 2 rays that make up this chord is 1/2 the length of the chord. All chords except the diameter are shorter than the diameter. For the center of the circle, all chords are diameters. For all other points inside the circle, the chords (other than the diameter) are shorter, therefore the average is less than the average for the center.

• Still you have to deal with a measure. If you draw a random chord through an off-centered point, its endpoints are not uniformly distributed over the boundary of the unit circle, hence you might be dealing with a different average in your approach. – Jack D'Aurizio Apr 1 '18 at 21:12
• As originally stated, the measure is not defined. My approach assumes a measure defined as uniform distribution of angles around each point. – herb steinberg Apr 3 '18 at 2:49
• All right, but that is not OP's framework, where the involved distribution is uniform over $x^2+y^2=1$. So you have proved that the origin has a minimum average distance according to a different notion of average. – Jack D'Aurizio Apr 3 '18 at 2:56
• My answer stated that the origin has the maximum average distance using the distribution of equal angle with the ray origin. As others have shown, when using points on the circle to be uniformly distributed, then the center has the minimum average. – herb steinberg Apr 3 '18 at 18:22