Descartes Circle Theorem

Chloroform · #1 $Old$ October 29th, 2009, 03:35 PM

Hey. I have a question on Descartes Circle Theorem:

$O_1(r)$ is a circle $O_1$ with radius r.

$O_1(r)$ and $O_2(r)$ are two equal circles that touch at a point T, and with a common tangent L. The circle $O_3(r_3)$ touches both $O_1(r)$ and $O_2(r)$ externally and also shares the common tangent L. There exists a chain of circles $O_n(r_n)$ for $n = 4...infinity$ such that $O_n(r_n)$ touches $O_1(r), O_2(r)$ and $O_{n-1}(r_n-1)$ externally. Find $r_n$ in terms of r.

I shall try and supply a picture.

Using inversion, I've worked out that $r_n = r/(2(n-2)(n-1))$ . However, I would like to derive this using Descartes Circle Theorem. Any ideas?

Opalg · #2 $Old$ November 4th, 2009, 08:45 AM

Quote:

Originally Posted by Chloroform $View Post$

Hey. I have a question on Descartes Circle Theorem:

$O_1(r)$ is a circle $O_1$ with radius r.

$O_1(r)$ and $O_2(r)$ are two equal circles that touch at a point T, and with a common tangent L. The circle $O_3(r_3)$ touches both $O_1(r)$ and $O_2(r)$ externally and also shares the common tangent L. There exists a chain of circles $O_n(r_n)$ for $n = 4...infinity$ such that $O_n(r_n)$ touches $O_1(r), O_2(r)$ and $O_{n-1}(r_n-1)$ externally. Find $r_n$ in terms of r.

I shall try and supply a picture.

Using inversion, I've worked out that $r_n = r/(2(n-2)(n-1))$ . However, I would like to derive this using Descartes Circle Theorem. Any ideas?

Using Descartes' circle theorem, it's easier to work with the curvature $k_n = 1/r_n$ rather than the radius. Let $k=1/r$ . Then the Descartes theorem tells you that $k_{n+1} = 2k+k_n\pm2\sqrt{k(k+2k_n)}$ (see equation (2) in the Wikipedia article). You want the circle with the smaller radius, therefore the larger curvature, so take the positive square root: $k_{n+1} = 2k+k_n + 2\sqrt{k(k+2k_n)}$ .

It's now easy to prove by induction that $k_n = 2(n-2)(n-1)/r$ . You get the base case k_3 = 4/r by taking $k_2=0$ in the formula $k_3 = 2k + k_2 + 2\sqrt{k(k+2k_2)}$ . The inductive step goes like this.

$\begin{aligned}k_{n+1} &= \frac2r + \frac{2(n-2)(n-1)}r + 2\sqrt{\frac{1+4(n-2)(n-1)}{r^2}} \\&= 2\bigl(n^2 - 3n + 3 + \sqrt{4n^2-12n+9}\bigr)/r \\&= 2\bigl(n^2 - 3n + 3 + (2n-3)\bigr)/r = 2(n-1)n/r\end{aligned}$