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 mπ−nd, 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 AP≤BP and CP≤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,
OF=OB2−FB2=422−392=(42+39)(42−39)=93
Because OP=18, it follows that
FP=9 and ∠OPB=60∘
hence that
BP=BF+FP=39+9=48
and AP=30. A similar argument shows that ∠OPD=60∘,DP=48, and CP=30. Because △DPB is isosceles and ∠DPB=120∘, it follows that inscribed angle ABD is 30∘, and that central angle AOD is 60∘. Thus △AOD is equilateral, with AD=42. The desired area is the sum of the area of △APD and the area of the segment of the circle bounded by AD and minor arc AD. This is
==Area(△APD)+[ Area ( Sector AOD)−Area(△AOD)]21(30)(48)sin60∘+[61π(42)2−21(42)2sin60∘]294π−813.
