A circle has a radius of 1 unit with its center located at O and P is an arbitrary point inside the circle. Three straight lines are drawn from P to meet the circle’s circumference respectively at the points A, B and C.

I agree with previous postings that the nature of these eccentric (my word) sectors makes it very difficult to find a closed solution. So here’s a numerical approach.

Let OPQ be a radius of the circle, with angle BPQ = U and angle BOQ = V.

Denoting the distance OP by p and using the Sine Rule in triangle BOP:

p = sin(U - V)/sin(U)which givesV = U - arcsin(p sin(U))(1)

Area of BPQ = Area of BOQ-Area of BOP= 0.5(V - p sin(V))

Now, using (1), this area can be written as a function of p and U as follows: