Hi guys! Unable to get to the answer of this question from problem set 7, q4:

Problem 4. (Friday, 5 points)
Find the average area of an inscribed triangle in the unit circle. Assume that each vertex of the triangle is equally likely to be at any point of the unit circle and that the location of one vertex does not affect the likelihood the location of another in any way. (Note that, as seen in Problem set 4, the maximum area is achieved by the equilateral triangle, which has side length √3 and area 3√3/4. How does the maximum compare to the average?)

Question

Hi guys! Unable to get to the answer of this question from problem set 7, q4:

Problem 4. (Friday, 5 points)
Find the average area of an inscribed triangle in the unit circle. Assume that each vertex of the triangle is equally likely to be at any point of the unit circle and that the location of one vertex does not affect the likelihood the location of another in any way. (Note that, as seen in Problem set 4, the maximum area is achieved by the equilateral triangle, which has side length √3 and area 3√3/4. How does the maximum compare to the average?)

anonymous · Answer

$$\int\limits_{-\pi/2}^{\pi/2}\int\limits_{-\pi/2}^{\pi/2}(2a^2\cos \theta _{1}\cos \theta _{2}\sin( \theta _{1}+\theta _{2}))d \theta _{1}d \theta _{2}$$

anonymous · Answer

this is the integral i got for the summation of areas of all triangles when circle is centred at (1,0) and the fixed point is at the origin while the other 2 points are moving.
this would of course be divided by the number of triangles.