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 Cn​90∘ 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+r2​1−r​ from 0. So if r=6011​, the required distance is
