Expected area of triangle inscribed in a circle

Question

On a unit circle , BC is a chord with length 4/π . Point A is picked randomly at its circumference . Find the expected area of △ ABC .

Answer 1

For easy reference let chord BC sits in the lower portion of the unit circle. It lies horizontally making angle $2\phi$ at the centre of the circle symmetric to the y axis.

Then the perpendicular distance of the chord from the centre is $D = \sqrt{(1 - (\frac{2}{\pi})^2}$ and $sin\phi = \frac{2}{\pi}$

Let the line joining A and the centre makes an angle $\theta$ with the x axis, then

For $\theta$ between $0$ and $\frac{\pi}{2}$,

$$Area = \frac{1}{2}\frac{4}{\pi}(sin\theta + D)$$

For $\theta$ between $0$ and $\frac{\pi}{2} - \phi$

$$Area = \frac{1}{2}\frac{4}{\pi}(D - sin\theta)$$

For $\theta$ between $\frac{\pi}{2} - \phi$ and $\frac{\pi}{2}$

$$Area = \frac{1}{2}\frac{4}{\pi}(sin\theta - D)$$

Hence the average area is

$$\frac{4}{2\pi^2}(\int_0^{\frac{\pi}{2}}(sin\theta + D)d\theta + \int_0^{\frac{\pi}{2} - \phi}(D - sin\theta)d\theta + \int_{\frac{\pi}{2} - \phi}^{\frac{\pi}{2}}(sin\theta - D)d\theta)$$

$$= \frac{2}{\pi^2}(\frac{2}{\pi} + D\pi - 2D\phi + sin\phi) = 0.533$$

I managed to borrow a PC and use 1000 random points and the result is around 5,1 which is very close to the calculation.

Also if the area is $\frac{3}{2\pi}$, I have used trial and error to find x to be about 1.54 which is your expected answer.