Problem:
Circle C0 has radius 1, and the point A0 is a point on the circle. Circle C1 has radius r<1 and is internally tangent to C0 at point A0. Point A1 lies on circle C1 so that A1 is located 90∘ counterclockwise from A0 on C1. Circle C2 has radius r2 and is internally tangent to C1 at point A1. In this way a sequence of circles C1,C2,C3,… and a sequence of points on the circles A1,A2,A3,… are constructed, where circle Cn has radius rn and is internally tangent to circle Cn−1 at point An−1, and point An lies on Cn90∘ counterclockwise from point An−1, as shown in the figure below. There is one point B inside all of these circles. When r=6011, the distance from the center of C0 to B is nm, where m and n are relatively prime positive integers. Find m+n.
Solution:
Let C0 be centered at 0 in the complex plane, and let A0=1. Circle C1 has radius r and center 1−r. It follows that A1=(1−r)+ir, and because circle C2 has radius r2,C2 has center A1−ir2=(1−r)+ir(1−r). In general, if Cn has center (1−r)+ir(1−r)+(ir)2(1−r)+⋯+(ir)n−1(1−r) and radius rn, then An will be in the direction rotated 90∘ counterclockwise from the direction of in−1, which is in the direction of in. Thus An=(1−r)+ir(1−r)+(ir)2(1−r)+⋯+(ir)n−1(1−r)+inrn, and the center of Cn+1 will be An−inrn+1=(1−r)+ir(1−r)+(ir)2(1−r)+⋯+(ir)n−1(1−r)+(ir)n(1−r). Hence, by induction, it follows that the center of circle Cn is (1−r)+ir(1−r)+(ir)2(1−r)+⋯+(ir)n−1(1−r). The point common to all the circles is the limit of this sequence of center points, which is an infinite geometric series with sum 1−ir1−r. This number is a distance 1+r21−r from 0. So if r=6011, the required distance is
