Argue, using symmetry, that the average distance between two points on a unit circle must exceed 1 Textbook problem:

Two points are chosen at random on the circumference of a circle of radius $1$. Let $D$ be the average distance between them. Argue, using principles of symmetry, that $D$ must exceed $1$.

Textbook solution:

Suppose the first point chosen is at the north pole of the circle. I assert that the average vertical distance to the second point will be exactly $1$. Indeed, the possible second points can be paired up: a point at distance $h$ below the north pole paired up with a corresponding point at distance $h$ above the south pole, and the average vertical distance of these two points from the north pole is $1$. Since the average distance between the points must exceed the vertical distance, it must exceed $1$.

Question: Why is it valid to average the vertical distance of these two points to get the average vertical distance of any two points (the step in bold above)?  I understand the rest of the argument, but this step I do not see exactly why it holds.
To get the average vertical distance of these two points for this step I did the following:

*

*North pole point has coordinates $(0,1)$

*Second chosen point (in the first or second quadrant) on the circle has coordinates $(x,1-h)$ for some $x$

*The paired-up corresponding point above the south pole has coordinates $(x,h-1)$

*From the north pole, averaging the vertical distance of the two points is then $\frac{1-(1-h)+1-(h-1)}{2} = 1$ since $1-(1-h)$ is the vertical distance from the north pole to one point and $1-(h-1)$ is the vertical distance from the north pole to the other point


Note that I'm able to use the integral definition of average value of a function to find that the exact answer is $4/\pi$.
 A: This is a valid proof because of the nature of right triangles.
Whichever point is chosen first, we can rotate the circle to make that the north pole. Since the distance to the south pole is two, the average vertical distance, as pointed out in the proof, is one. But the distance to any of the other points on the circle is the hypotenuse of a triangle with one leg equal to the vertical distance. Since the hypotenuse of the triangle is larger than any of its two legs, the average distance must be greater than the average vertical distance. Since the average vertical distance is one, the average distance must be greater than one.
EDIT: Through the discussion in the comments, Ungar's real difficulty was understanding why the average vertical distance is $1$. At his request, I paste the following comment from below into this answer.
We are trying to show that the average distance from the north pole to the south pole is $1$. Forget the circle and concentrate on the interval between $0$ and $2$. There is one point that is a distance of one away. For every other point, e.g. $\frac{1}{4}$, there is a point $\frac{7}{4}$. Their average is 1. For $\frac{1}{10}$, there is $\frac{19}{10}$ and their average is $1$. That's why the average is $1$. Every point that is not a distance of $1$ away is paired with a point whose distance is such that the average is $1$.
