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Math Help - Descartes Circle Theorem

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    Descartes Circle Theorem

    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?
    Last edited by Chloroform; October 29th 2009 at 01:57 PM.
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  2. #2
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    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}} \\<br />
&= 2\bigl(n^2 - 3n + 3 + \sqrt{4n^2-12n+9}\bigr)/r \\<br />
&= 2\bigl(n^2 - 3n + 3 + (2n-3)\bigr)/r = 2(n-1)n/r\end{aligned}
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