What is the expected area of a cyclic quadrilateral inscribed in a unit circle?

Question

Choose four points randomly on the circumference of a circle with radius $1$. Connect them to form a quadrilateral. What is the expected area of this quadrilateral?

I have attempted to simulate to find an answer but not sure how to approach finding an exact value. The simulation fixes one of the points at $0$ and generates 3 other points uniformly around the circle between $0$ and $2\pi$. Then it orders the points and takes the differences between them to get the 4 central angles of the quadrilateral. From these 4 central angles it finds the length of each side $s_i$ using the formula $s_i=2sin\frac{\theta_i}{2}$. Once I have the four sides I can use Brahmagupta's Formula to find the area $K$ of the quadrilateral. I repeat this 100k times and take the average of $K$ and get $K\approx.96$.

Calculate the area via the central angles and then calculate $E[\sin \theta_i]$ — Calvin Lin, Dec 11 '19 at 17:34
Shouldn't the central angle be uniform on $[0, \Pi]$? You can consider one point as fixed and choose another one uniform at random on the circumference. — NonalcoholicBeer, Dec 11 '19 at 22:04
Can you elaborate a little on what your simulation with Brahmagupta's formula consisted of? — Zacky, Dec 12 '19 at 01:05
Noticed this is a special case of a question asked here about expected area of cyclic n-gon. — Goldbug, Dec 13 '19 at 20:13

score 5 · Accepted Answer · edited Dec 11 '19 at 23:18

The four central angles have the same distribution, and the expected area is $4$ times the expected area of one of the four triangles spanned by the central angles.

The probability density of the central angle is proportional to the volume it leaves to the remaining two points: $f_\alpha(\alpha)\propto(2\pi-\alpha)^2$. Normalization yields $f_\alpha(\alpha)=\frac3{(2\pi)^3}(2\pi-\alpha)^2$.

The area of the triangle spanned by the central angle $\alpha$ is $\frac12\sin\alpha$. Thus the expected area of the quadrilateral is

$$ 4\int\limits_0^{2\pi}\frac3{(2\pi)^3}(2\pi-\alpha)^2\frac12\sin\alpha\,\mathrm d\alpha=\frac3\pi\approx0.955\;. $$

Yes, this distribution matches my simulation perfectly. Thank you for your solution. — Goldbug, Dec 12 '19 at 00:16

What is the expected area of a cyclic quadrilateral inscribed in a unit circle?

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