Circles in Geometric Progression? 2

Due to symmetry, the centers of all the blue circles except $C_0$ are on a straight line perpendicular to the base line and the line joining the centers of the red circle and $C_0$ . The length of this line from the base line to the contact point of the red circle and $C_0$ (or the line joining the two centers) is 1. Consider the $n$ th blue circle $C_n$ with radius $r_n$ , we have from the top of this line to the bottom.

$\begin{aligned} \sqrt{(1+r_n)^2 -1} + r_n + 2 \blue{\sum_{k=1}^{n-1} r_k} & = 1 & \small \blue{\text{Let }S_n = \sum_{k=1}^n r_k} \\ \sqrt{(1+r_n)^2 -1} & = 1 - r_n - 2 \blue{S_{n-1}} & \small \blue{\text{Squaring both sides}} \\ r_n^2 + 2r_n & = r_n^2 - 2r_n + 1 - 4S_{n-1}(1-r_n) + 4S_{n-1}^2 \\ 4(1-S_{n-1})r_n & = 1 - 4S_{n-1}(1-S_{n-1}) \\ r_n & = \frac 1{4(1-S_{n-1})} - S_{n-1} \\ \implies S_n & = \frac 1{4(1-S_{n-1})} \end{aligned}$

Calculations reveal that $S_1 = \frac 14, S_2 = \frac 26, S_3 = \frac 38, \cdots$ . It appears that $S_n = \frac n{2(n+1)}$ which can be readily proven by induction (see Note). Then $r_n$ is given by:

$\begin{aligned} r_n & = S_n - S_{n-1} = \frac n{2(n+1)} - \frac {n-1}{2n} = \frac 1{2n(n+1)} \end{aligned}$

And the total area of all blue circles is:

$\begin{aligned} A & = \pi (1^2) + \sum_{n=1}^\infty \pi r_n^2 = \pi + \sum_{n=1}^\infty \frac \pi{4n^2(n+1)^2} \\ & = \pi + \frac \pi 4 \sum_{n=1}^\infty \left(\frac 1{n^2} + \frac 1{(n+1)^2} - \frac 2n + \frac 2{n+1} \right) \\ & = \pi + \frac \pi 2 \blue{\sum_{n=1}^\infty \frac 1{n^2}} - \frac \pi 4 - \frac \pi 2 & \small \blue{\text{Riemann zeta function }\zeta(s) = \sum_{k=1}^\infty \frac 1{k^s}} \\ & = \frac \pi 4 + \frac {\pi^3}{12} & \small \blue{\text{and }\zeta(2) = \frac {\pi^2}6} \end{aligned}$

Therefore $a+b+c = 4+3+12 = \boxed{19}$ .

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Reference: Riemann zeta function

Note: To prove the claim $S_n = \frac n{2(n+1)}$ is true for all $n \ge 1$ by induction

For $n=1$ , $S_1 = \frac 1{4(1-0)} = \frac 14$ . Therefore the claim is true for $n=1$ . Assuming the claim is true for $n$ . then

$\begin{aligned} S_{n+1} & = S_n + r_{n+1} = S_n + \frac 1{4(1-S_n)} - S_n = \frac 1{4-\frac {4n}{2(n+1)}} = \frac {n+1}{2(n+2)} \end{aligned}$

The claim is true for $n+1$ and hence true for all $n\ge 1$ .