Problem:
In a circle of radius 42, two chords of length 78 intersect at a point whose distance from the center is 18. The two chords divide the interior of the circle into four regions. Two of these regions are bordered by segments of unequal lengths, and the area of either of them can be expressed uniquely in the form mΟβndβ, where m,n, and d are positive integers and d is not divisible by the square of any prime number. Find m+n+d.
Solution:
Let the chords be denoted by AB and CD, and the intersection of the chords by P, where APβ€BP and CPβ€DP. Let O be the center of the circle, and F be the foot of the perpendicular from O to AB. By the Pythagorean Theorem,
and AP=30. A similar argument shows that β OPD=60β,DP=48, and CP=30. Because β³DPB is isosceles and β DPB=120β, it follows that inscribed angle ABD is 30β, and that central angle AOD is 60β. Thus β³AOD is equilateral, with AD=42. The desired area is the sum of the area of β³APD and the area of the segment of the circle bounded by AD and minor arc AD. This is
==βArea(β³APD)+[ Area ( Sector AOD)βArea(β³AOD)]21β(30)(48)sin60β+[61βΟ(42)2β21β(42)2sin60β]294Οβ813β.β
