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There's something I am missing comparing Descartes' theorem for three isometric circles here and this wiki post on circle packing of 3 circles here.

From my calculation:

$$ r_{ext} = \frac{r_{int}}{{(3\pm2\sqrt3)}}, \tag{1} $$

where rext is the external radius and rint the internal radius. In the second article it seems:

$$ r_{ext} = r_{int}(1+2\frac{\sqrt3}{3}). \tag{2} $$

(Probably is crappy mathematics of my own.)

For the sake of completness. The generic formula I derived from Descartes' theorem is (for different radii):

$$ r_{ext} = \frac{r_{1}r_{2}r_{3}}{(r_{1}r_{2}+r_{2}r_{3}+r_{1}r_{3} \pm 2{\sqrt{r_{1}r_{2}r_{3}(r_{1}+r_{2}+r_{3})}{}})}. \tag{3} $$

what am I missing?

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    $\begingroup$ It is not clear what you denote by $r_{ext}$ and $r_{int}$. Please define/describe the problem precisely, including notation. $\endgroup$
    – GH from MO
    May 16, 2016 at 15:51

2 Answers 2

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The $k$s in the Wiki article are the curvatures (the inverses of the radii). If you correct for that, the two numbers will agree.

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  • $\begingroup$ I've already done that. I could show you the complete formula for radii instead curvatures. My calculations seems right I don't know where my mistake is. $\endgroup$ May 16, 2016 at 16:08
  • $\begingroup$ Mathematica disagrees... $\endgroup$
    – Igor Rivin
    May 16, 2016 at 16:33
  • $\begingroup$ I imagine that, but I uninstalled mathematica a long time ago, i'm more interested in what my mistake was. $\endgroup$ May 16, 2016 at 16:39
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My bad. I misunderstood the second article:

$$ 1+2\frac{\sqrt{(3)}}{3} $$

is the coefficient in a linear expression. Both the solutions are the same, you can derive (2) from (1) using square difference formula.

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