Problem:
A collection of circles in the upper half-plane, all tangent to the x-axis, is constructed in layers as follows. Layer L0β consists of two circles of radii 702 and 732 that are externally tangent. For kβ₯1, the circles in βj=0kβ1βLjβ are ordered according to their points of tangency with the x-axis. For every pair of consecutive circles in this order, a new circle is constructed externally tangent to each of the two circles in the pair. Laver Lkβ consists of the 2kβ1 circles constructed in this way. Let S=βj=06βLjβ, and for every circle C denote by r(C) its radius. What is
CβSββr(C)β1β?
Answer Choices:
A. 35286β
B. 70583β
C. 73715β
D. 14143β
E. 1461573β Solution:
Suppose that circles C1β and C2β in the upper half-plane have
centers O1β and O2β and radii r1β and r2β, respectively. Assume that C1β and C2β are externally tangent and tangent to the x-axis at X1β and X2β, respectively. Let C with center O and radius r be the circle externally tangent to C1β and C2β and tangent to the x-axis. Let X be the point of tangency of C with the x-axis, and let T1β and T2β be the points of tangency of C with C1β and C2β, respectively. Let M1β and M2β be the points on the x-axis such that M1βT1βββ₯O1βT1ββ and M2βT2βββ₯O2βT2ββ.
Because M1βX1ββ and M1βT1ββ are both tangent to C1β, it follows that X1βM1β=M1βT1β. Similarly, M1βT1ββ and M1βXβ are both tangent to C, and thus M1βT1β=M1βX. Because β OT1βM1β,β M1βX1βO1β,β M1βT1βO, and β OXM1β are all right angles and β T1βM1βX=Οββ X1βM1βT1β, it follows that quadrilaterals O1βX1βM1βT1β and M1βXOT1β are similar. Thus
where C1β and C2β are the consecutive circles in βj=0kβ1βLjβ that are tangent to C. Note that every circle in βj=0kβ1βLjβ appears twice in the sum on the right-hand side, except for the two circles in L0β, which appear only once. Thus
