Problem:
Given a circle of radius 13, let A be a point at a distance 4+13 from the center O of the circle. Let B be the point on the circle nearest to point A. A line passing through the point A intersects the circle at points K and L. The maximum possible area for △BKL can be written in the form da−bc, where a,b,c, and d are positive integers, a and d are relatively prime, and c is not divisible by the square of any prime. Find a+b+c+d.
Solution:
Assume, without loss of generality, that point K lies between points L and A. Observe that
[BAK][OAK]=[BAL][OAL]=44+13
so
[BKL][OKL]=[BAL]−[BAK][OAL]−[OAK]=44+13
and
[BKL]=[OKL]4+134
Then [OKL]=213sin∠KOL, and therefore the maximum possible value for [OKL] occurs when ∠KOL=90∘ and [OKL]=213. Thus the maximum value for [BKL] is 213⋅4+134=3104−2613. The requested sum is 104+26+13+3=146.