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.

It is known that:

Angle APB = Angle BPC = Angle APC = 120 degrees, and:

Area of sector APB = pi/3 square units, and:

Area of sector BPC = pi/4 square units.

Determine the length of the straight line OP.

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 gives        V = 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:

            A(p, U) = 0.5[U - arcsin(p sin(U)) - p sin(U - arcsin(p sin(U)))]

‘Sector’ BPC is made up of two such areas (BPQ + CPQ) while ‘sector’ APB is the difference of two such areas (APQ - BPQ) as shown respectively, below:

Area of BPC =                A(p, U)  +  A(p, 2*pi/3 - U)  =  pi/4

Area of APB =                A(p, 2*pi/3 + U)  -  A(p, U)  =  pi/3

Solving these two simultaneous equations numerically, using Maple, gives:

p = 0.1760697601    so distance OP = 0.176..

U = 0.5676307755    so angle between OP and PB is 0.568 rad = 32.5..degrees

Posted by Harry on 2010-07-10 15:50:39