Given $x^2 + y^2 = 1$, fix a point $X = (-1,0)$, then draw some $\theta \sim U(0,\pi)$ and create a second point $Y = (cos \theta,sin\theta)$. I then pick a third point (called $Z$) on the circumference such that the area of the inscribed triangle is maximised. What's the expected value of this area?
Can you please help me? The euclidian distance of the base is clearly $b=\sqrt{(cos\theta + 1)^2 + sin\theta^2}$ which will help in my calculation of $A = 0.5bh$.
But what's the height? One approach seems to be to find the distance from midpoint to origin, and add that to the radius (which will give the perpendicular line distance from line $XY$ to point $Z$). So far I have that the midpoint is $((cos\theta - 1)/2,sin\theta/2)$, which should mean that the height of the triangle is $h = 1 + \sqrt{((cos\theta - 1)/2)^2+(sin\theta/2)^2}$.
Have I made any mistakes so far? Is there a better approach? I don't want to plug this into $\frac{1}{2\pi}\int_{0}^{\pi}bh d\theta$ to find the expectation yet, seems really messy.
Given $X=(-1,0)$ and $Y=(\cos\theta,\sin\theta)$ the inscribed triangle of maximal area is isosceles with base $XY$. Its area $a(\theta)$ is given by $$a(\theta)={1\over 2}|XY| \bigl(|MO|+1\bigr)\ ,$$ where $M$ is the midpoint of $XY$. As $\angle(OYX)={\theta\over2}$ we therefore obtain$$a(\theta)=\cos{\theta\over2}\bigl(\sin{\theta\over2}+1\bigr)\ ,$$ hence $$E(a)={1\over\pi}\int_0^\pi\cos{\theta\over2}\bigl(\sin{\theta\over2}+1\bigr)\>d\theta={3\over\pi}\ .$$