On a unit disk, what is the expected distance between the midpoint of two uniformly random points, and the centre? On a unit disk, what is the expected distance between the midpoint of two uniformly random points, and the centre?
Context
I thought of this question after seeing a simpler question: on a unit disk, what is the expected distance between a random point, and the centre? (The answer is $2/3$.)
My attempt (at my question)
I started by considering the one-dimension case: on a line segment of length $2$ (analogous to a unit disk, which has diameter $2$), what is the mean distance between the midpoint of two uniformly random points and the centre? I first considered that the line segment is discrete, with $n$ points, worked out the probability distribution, found the expectation in terms of $n$, then took the limit as $n\to\infty$, and I got $\int_0^1 x(1-x)dx=1/6$. But when I jump to the two-dimensional case (unit disk), this approach doesn't seem to work, because there are directions involved, not just distances.
I tried to find the probability density function of $X$, the distance of the midpoint to the centre, but I have not been able to work out what it is.
This answer to the question, "What is the expected distance between two points inside unit circle?", suggests that polar coordinates may be useful, but I have not been able to obtain an integral expression using polar coordinates.
 A: (Self-answering)

The diagram shows points two uniformly random points $A$ and $B$ on the disk, and their midpoint $M$. Their distances to the centre $O$ are $a$, $b$ and $m$, respectively. $\theta=\angle{AOB}$.
Intuitively, we can guess that $E(m)$ is exactly half of the expected distance between two random points on a unit disk. That is, $E(m)=\dfrac{64}{45\pi}$. This is because the average value of $\theta$ is $\pi/2$, and the average value of $a/b$ is $1$, and with these values we have $m=\frac{1}{2}AB$.
In fact, using excel, I generated about $300,000$ pairs of random points on the unit disk, and found that $E(m)\approx 0.453$, which is $\dfrac{64}{45\pi}$ to 3 decimal places.
As for a proof that $E(m)=\dfrac{64}{45\pi}$, it can be shown that
$$m=\frac12\sqrt{5a^2+b^2-2ab\cos{\theta}-4a|b\cos{\theta}-a|}$$
The probability density functions for $a$, $b$ and $\theta$ are $2a$, $2b$ and $\frac{1}{2\pi}$, respectively. Therefore,
$$E(m)=\int_0^1\int_0^1\int_0^{2\pi} \frac{ab}{\pi}\sqrt{5a^2+b^2-2ab\cos{\theta}-4a|b\cos{\theta}-a|}\text{d}\theta\text{d}a\text{d}b$$
I have not evaluated this integral, and I do not have the technology to approximate it, but based on the above evidence, I am confident that it equals $\dfrac{64}{45\pi}$.
