Q23

Q23 - Let AB and CD be two parallel chords in a circle with radius 5 such that the centre O lies between these chords. Suppose AB = 6, CD = 8. Suppose further that the area of the part of the circle lying between the chords AB and CD is (m*pi + n)/k, where m, n, k are positive integers with gcd(m, n, k) = 1. What is the value of m + n + k ?

Answer: 75.

Solution:
Let's calculate the area below both the chords and subtract from the total circle area to get the area between these 2 chords.
Area below AB:
If AB subtends angle x at the center then
sin(x/2) = 3/5
=> x = 2.arcsin(3/5)
=> area captured by the arc AB = (x/2Pi)*Pi*r^2 = (x/2)*r^2 = 25*arcsin(3/5)

But this is the area captured by the arc AB.
We just want the area below the chord AB.
So we need to subtract the area captured by the triangle OAB.
[OAB] = (1/2)*perpendicular from O on AB*AB = (1/2)*4*6 = 12.
So finally area below the chord AB = 25*arcsin(3/5) - 12
Similarly area below the chord CD = 25*arcsin(4/5) - 12

So the area which we want is:
25Pi - (25.arcsin(3/5) - 12 + 25.arcsin(4/5) - 12)
= 25Pi + 24 - 25*Pi/2
= 25Pi/2 + 24
= (25Pi + 48)/2
So m + n + k = 75.

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