Geometric Geometry!

This is the Sangaku problem that determines the relationship between tangential circles of different sizes. You may also see want to see Daniel Liu's Ford circles .

The formula is $\dfrac{1}{\sqrt{r_{\text{left}}}} + \dfrac{1}{\sqrt{r_{\text{right}}}} = \dfrac{1}{\sqrt{r_{\text{middle}}}}$ Since the largest radius is $1$ , we can assume that the value of $r$ is strictly decreasing. So let $r$ denote the common ratio, which represents the radius of a circle. Consider $r_{\text{left}} = 1$ , $r_{\text{right}} = r$ and $r_{\text{middle}} = r^2$ . Then, $1 + \dfrac{1}{\sqrt{r}} = \dfrac{1}{r}$ Solving for $r$ , $\begin{array}{rl} r\cdot \left(1 + \dfrac{1}{\sqrt{r}} \right) &= \dfrac{1}{r} \cdot r\\ r + \sqrt{r} &= 1\\ \left(\sqrt{r}\right)^2 + \sqrt{r} + \left(\dfrac{1}{2}\right)^2 &= 1 + \left(\dfrac{1}{2}\right)^2\\ \left(\sqrt{r} + \dfrac{1}{2}\right)^2 &= \dfrac{5}{4}\\ \sqrt{r} + \dfrac{1}{2} &= \pm\dfrac{\sqrt{5}}{2}\\ \sqrt{r} &= \dfrac{-1 \pm \sqrt{5}}{2} \end{array}$ Either $r = \dfrac{3 + \sqrt{5}}{2}$ or $r = \dfrac{3 - \sqrt{5}}{2}$ . Because the value of $r$ is strictly decreasing, neglect $r = \dfrac{3 + \sqrt{5}}{2}$ . Therefore, the solution of the equation is $r = \dfrac{3 - \sqrt{5}}{2}$ .

Since the progression is geometric, we can see that for $1 + \dfrac{1}{\sqrt{r}} = \dfrac{1}{r}$ if we divide both sides by $r^{n/2}$ , where $n$ is positive integer, then the relationship still holds. Thus, the total area is $\begin{array}{rl} \pi \sum\limits_{n=0}^{\infty} \left(\dfrac{1}{2}\left(3 - \sqrt{5}\right)\right)^{2n} &= \pi \sum\limits_{n=0}^{\infty} \left(\dfrac{1}{2}\left(7 - 3\sqrt{5}\right)\right)^n\\ &= \dfrac{\left(5 + 3\sqrt{5}\right)\pi}{10} \approx \boxed{3.68} \end{array}$

For proof of the Sangaku identity, see here