Consider three concentric circles with center O and radii 1,2 and 3 respectively.
Consider point A on the circle of radius 1, B on the circle with radius 2 and C on the circle with radius 3.
- What is the expected area of △ABC?
- What is the maximum area of △ABC?
I have tried assigning complex numbers to coordinates, proceeding by fixing angles, but nothing seems to work.
Assume uniform distribution for the point on each circle. We can fix one of the points because area is rotationally invariant. Fix the inner most point to be $(1,0)$.
Denote $x$ and $y$ the phase angles of the points on the circles with radii $2$ and $3$ respectively. Using a formula for the area of a triangle given the cartesian coordinates of its vertices we get that the expected area is
$$\frac{1}{8\pi^2}\int_0^{2\pi}\int_0^{2\pi}\left|2\sin\left(x\right)-3\sin\left(y\right)+6\cos\left(x\right)\sin\left(y\right)-6\cos\left(y\right)\sin\left(x\right)\right|dxdy$$
$$\approx 2.0829$$
UPDATE:
The formula for the area of triangle $(A, B, C)$ is
$$\frac{|A_x(B_y-C_y) + B_x(C_y-A_y) + C_x(A_y-B_y)|}{2}$$
We have $A=(1,0)$, $B=2(\cos(x), \sin(x))$ and $C=3(\cos(y), \sin(y))$. A factor of $\frac{1}{2\pi}$ comes from the uniform distribution for both $x$ and $y$. (We also assume independency.) Just place the values for the point coordinates in and the formula follows. (Notice, how the values $A_x=1$ and $A_y=0$ simplify the formula.)
For the maximum area, Wolfram Alpha gives $4.90482$ at $x=2.3882$ and $y=4.2046$. I don't know if exact forms can be found.