Fibonacci Circles

Let the red circle be $C_1$ , the blue circle be $C_2$ , radius of $C_n$ be $r_n$ , and the length $T_jT_k = \ell_{jk}$ .

Then $\ell_{12} = \sqrt{(r_1+r_2)^2-(r_1-r_2)^2} = 2\sqrt{r_1r_2}$ and $\ell_{13} + \ell_{23} = \ell_{12} \implies 2\sqrt{r_1r_3} + 2\sqrt{r_2r_3} = 2\sqrt{r_1r_2} \implies \sqrt {r_3} = \dfrac {\sqrt{r_1r_2}}{\sqrt{r_1}+\sqrt{r_2}}$ . Putting $\sqrt{r_1}=1$ and $\sqrt{r_2} = x$ , we have:

$\begin{aligned} \sqrt{r_3} & = \frac {\sqrt{r_1r_2}}{\sqrt{r_1}+\sqrt{r_2}} = \frac x{1+x} \\ \sqrt{r_4} & = \frac {\sqrt{r_2r_3}}{\sqrt{r_2}+\sqrt{r_3}} = \frac x{2+x} \\ \sqrt{r_5} & = \frac {\sqrt{r_3r_4}}{\sqrt{r_3}+\sqrt{r_4}} = \frac x{3+2x} & \small \blue{\text{By proof by induction}} \\ \implies \sqrt{r_n} & = \frac {\sqrt{r_{n-2}r_{n-1}}}{\sqrt{r_{n-2}}-\sqrt{r_{n-1}}} \\ & = \frac {\frac x{F_{n-3} + F_{n-4}x} \times \frac x{F_{n-2} + F_{n-3}x}}{\frac x{F_{n-3} + F_{n-4}x} + \frac x{F_{n-2} + F_{n-3}x}} & \small \blue{\text{where }F_k \text{ is the }k\text{th Fibonacci number.}} \\ & = \frac x{F_{n-1} + F_{n-2}x} \end{aligned}$

Let the distance of $T_n$ from $T_1$ be $d_n$ and to simplify, $\ell_n = \ell_{n, n+1}$ , then

$\begin{aligned} d_1 & = \ell_1 = 2\sqrt{r_1r_2} = 2x \\ d_2 & = \ell_1 - \ell_2 = 2x \left(1- \frac x{1+x}\right) = \frac {2x}{1+x} \\ d_3 & = \ell_1 - \ell_2 + \ell_3 = \frac {2x}{1+x} \left(1 + \frac x{2+x} \right) = \frac {4x}{2+x} \\ d_4 & = \frac {2x}{2+x} \left(2-\frac x{3+2x} \right) = \frac {6x}{3+2x} \\ d_5 & = \frac {2x}{3+2x} \left(3+\frac x{5+3x} \right) = \frac {10x}{5+3x} & \small \blue{\text{By induction}} \\ \implies d_n & = \frac {2F_{n-1}x}{F_{n-1}+F_{n-2}x} \end{aligned}$

Since $d_\infty = \dfrac {\ell_1}2 = x$ ,

$\begin{aligned} x & = \lim_{n \to \infty} \frac {2F_{n-1}x}{F_{n-1}+F_{n-2}x} & \small \blue{\text{Divide up and down by }F_{n-2}} \\ & = \lim_{n \to \infty} \frac {2x\times \frac {F_{n-1}}{F_{n-2}}}{\frac {F_{n-1}}{F_{n-2}} + x} & \small \blue{\text{Note that }\lim_{n \to \infty} \frac {F_{n+1}}{F_n} = \varphi \text{, the golden ratio}} \\ & = \frac {2\varphi x}{\varphi + x} \\ \varphi x + x^2 & = 2\varphi x \\ \implies x & = \varphi \end{aligned}$

Note that $x = \sqrt{\dfrac {r_2}{r_1}}$ , therefore $\dfrac {r_1}{r_2} = \dfrac 1{x^2} = \dfrac 1{\varphi^2} = \boxed{\frac 1{\varphi + 1}}$ .

Let radii of $C_1, C_2$ be $r_1, r_2$ . Then

$\varphi ^2 = \dfrac{r_2}{r_1} = \varphi + 1$