Solution to Problem #9

If three points are chosen at random (and uniformly) on the circumference
of a circle of radius 1, what is the expected value of the area of the
resulting triangle?
Find the expectation value of the area of a triangle made by
3 random points on the circumference of the unit circle:

            Area= 3/(2*pi)

Proof: Take the origin at the centre and the x-axis through one point P.
Let the polar angles of points Q and R be a and b.
The coordinates of P, Q, R are now
                                    
             (1,0); (cos a, sin a) ; (cos b, sin b)

The determinantal formula for the area A gives: 
           
              2*A = sin a -sin b +cos a * sin b -sina * cos b    .....(1)
But this has to be counted positive when PQR are in anticlockwise order,
i.e. for b between a and 2*pi, and negative for clockwise order, i.e b
between 0 and a.

Now integrate (1) with respect to b at constant a, over the two ranges
separately and combine them after changing the sign  of the result for the
second part  of the range. After removing the factor 2 at the start of (1)_
the resulting integral I is
       
                 I = 2-a*sin a +pi*sin a -2*cos a   ......(2)

Now we integate (2) over all a from 0 to 2*pi.
The last 2 terms vanish and the first 2 yield 6*pi.

The average area 3/(2*pi)  is got by dividing twice by the integration
ranges 2*pi.

A computer simulation with 1 million random triangles gave 0.4772 = 1.499/pi.