Average distance from point to unit circle

Question

If I measure the average distance from the origin to the points of the unit circle, obviously the answer is 1.

But if I instead measure the average distance from the point (0.1, 0) to the unit circle, I get this integral: $$\frac1{2\pi}\int_0^{2\pi} \sqrt{(\cos(t)-0.1)^2 + (\sin(t))^2} dt \neq 1 $$ Is there a nice intuitive explanation why this "average distance" is not 1? I think of this as averaging all rays from (0.1,0) to the circle- some of these are less than 1, and some are more than 1. Intuitively, I had expected this average to also be 1. Why is it different?

Answer 1

There is one immediate solution that exploits the symmetry as follows : twice the average is the average sum of distances from the given point to a point on the circle and its diametrically opposed, and each such sum is $> 2$ (except perhaps one).

Another way: consider a bounded figure $F$ in a normed space. For a point $P$ in the space consider $f(P)$ the average of the distance from $P$ to a point on $F$ ( we assume a positive measure on $F$). It is easy to see that $f(\cdot)$ is a convex function, and strictly convex, if $\|\cdot\|$ is strictly convex. Moreover, at $\infty$ the function $f$ becomes $+\infty$. Therefore, $f$ achieves a minimum at a point in the space. This point is unique if $f$ is strictly convex.

Now, assume moreover (on top of $f$ stricly convex) that $F$ is symmetric with respect to a point $O$, and the measure on $F$ is invariant under the symmetry through $O$. The the unique point of minimum of $f$ is also invariant under $O$, so it is $O$. We also see that $f$ is stricly increasing on every ray starting at $O$.

Note: this is the continuous analogue of finding a point whose sum of (weighted) distances to a family of points is smallest. If that set of points is invariant under some isometries, so is the point of minimum.

$\bf{Added:}$ @copper hat pointed out that it is not entirely obvious that there exists a point of minimum for the function $f$, if the ambient space is infinite dimensional ( Banach), something worth pondering on! However, if the problem has a center of symmetry $O$, then clearly $f(P) + f(P') \ge 2 f(O)$ ( perhaps strict), and moreover $f(P) = f(P')$, so no other argument is needed.

Answer 2

Let $d(t) = {1 \over 2 \pi} \int_0^{2 \pi} |t+e^{i \theta}| d \theta$, the average distance from the point $(t,0)$ to the circle. (Note that $d$ is an even function.)

Suppose $|t|<1$ and let $\phi(t,\theta) = |t+e^{i \theta}| = \sqrt{t^2+2 t\cos \theta + 1}$.

From this we get ${\partial \phi(t,\theta) \over \partial t} = {t + \cos \theta \over \sqrt{t^2+2 t\cos \theta + 1} }$ and ${\partial^2 \phi(t,\theta) \over \partial t^2} = {\sin^2\theta \over (t^2+2 t\cos \theta + 1)^{3 \over 2} }$, and so ${\partial \phi(0,\theta) \over \partial t} = \cos \theta$ and ${\partial^2 \phi(t,\theta) \over \partial t^2} >0$ except for $t \in \{0, \pi, 2 \pi\}$.

Leibniz gives $d'(t) = {1 \over 2 \pi} \int_0^{2 \pi} {\partial \phi(t,\theta) \over \partial t} d \theta$, $d''(t) = {1 \over 2 \pi} \int_0^{2 \pi} {\partial^2 \phi(t,\theta) \over \partial t^2} d \theta$ and so $d'(0) = 0$, $d''(t) >0$.

Hence for $|t| < 1$ and $t \neq 0$, Taylor shows that $d(t) > d(0) = 1$.

For $t= 0.1$ we have $d(t) \approx d(0)+ d'(0)t +{1 \over 2} d''(0) t^2 = 1.0025$

Answer 3

Sorry for the self-answer! Here is my explanation why the average distance at (0.1, 0) (or any point other than the origin) is greater than 1.

Let $C$ be the unit circle centered at the origin. The picture has $C$ as the solid colored line, and the unit circle centered at (0.1,0) is the dashed line. The points on $C$ with distance less than 1 to (0.1,0) are colored red, the ones with distance more than 1 are blue. You can see there are more blue ones than red ones.

If we slide the point (0.1,0) further to the right, the red zone only shrinks (and eventually disappears entirely), so this average distance will always be greater than 1, increasing as our point moves further away from the origin.

Answer 4

A good way to see this is not to place the point at $(0.1,0)$ but to place it at $(0.9,0)$. With the point placed there, plot the distance to $\cos\theta,\sin\theta)$ as a function of $\theta$. You don't get a symmetrical, sine-wave-like curve around the "mid-line" $y=1$. Instead you get broad ranges above this line punctuated by sharp dips below $y=1$ wgen $\theta$ is close to a multiple of $2\pi$. The broader ranges above $y=1$ makes the average distance greater than $1$.

The plot looks like that because only a small part of the circle is near the closest approach point $(1,0)$, whereas a large part on the "back side" is nearly as far away as the point $(-1,0)$.