On the surface of a unit sphere, three points $A$, $B$ and $C$ are chosen in the following way:
- Points $A$ and $B$ are chosen randomly and independently on the whole surface
- After $A$ and $B$ are fixed, $C$ is a point on sphere that maximises the area of $\triangle ABC$.
Find the expected area of $\triangle ABC$.
(which I assume to be the triangle cutting through the sphere, not on the spherical surface)
I am looking for a solution for this question from a test, hopefully a "smart" one that uses a quick approach. This was my approach:
- I quickly chose $C$ to be the mid-point of the major arc of the great circle through $A$ and $B$.
- To find the area, first I attempted to find the distribution of the angle between two random points, $\angle AOB$ with $O$ being the centre of sphere.
- I tried to use the fact that bands of equal width on sphere have the same area, and hope that it would link area with the angle $\angle AOB$, but I failed to find the distribution.
- Lastly I planned to take the expectation of $$\frac12\sin\angle AOB+\sin\frac{2\pi-\angle AOB}2$$
Yet this approach appears to be too lengthy for my test.
Edit: I seem to find the distribution of $\Theta=\angle AOB$ as $$f_\Theta(\theta) = \begin{cases}\frac12\sin\theta&0\le\theta\le\pi\\ 0&\text{otherwise}\end{cases}$$
And the expected value to be $$\int_0^\pi\left(\frac12\sin\theta+\sin\frac{2\pi-\theta}2\right)\frac12\sin\theta\,d\theta = \frac\pi8+\frac23$$