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|Two Point Distance (Posted on 2010-06-25)
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.
Numerical (computer) solution
Comment 4 of 4 |
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
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