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I am trying to create a circular fractal in which each circle is composed by a given number $n$ of smaller circles.

It would look something like this for $n = 8$: (The smaller circles should touch each other, however)

However, I don't know how to calculate the radius of the smaller circles.

Of course I know that the distance from the centre of the bigger circle to the centres of the smaller circles is $r_1 - r_2$, where $r_1$ is the radius of the bigger circle and $r_2$ is the radius of the smaller ones, and that their radius is $\frac{d}{2}$ where $d$ is the distance between two adjacent circles' s centres.

$d$ is for sure related to $n$ but I don't know how to calculate it.

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  • $\begingroup$ Is $d$ the distance between small circle centers? Are inner circles touching? (In your diagram they are not). $\endgroup$
    – Narasimham
    Commented Jul 12, 2016 at 8:41
  • $\begingroup$ @Narasimham yes, it is, and yes, they should touch, thanks for asking. $\endgroup$
    – user6245072
    Commented Jul 12, 2016 at 8:42
  • $\begingroup$ Are you trying to use an Iterated Function System to create something like this? $\endgroup$
    – Mark McClure
    Commented Jul 12, 2016 at 11:25
  • $\begingroup$ @Mark McClure yes, but I've removed the outline and added a circle of radius d - r2 in the centre, so it looks more like this. $\endgroup$
    – user6245072
    Commented Jul 12, 2016 at 11:50
  • $\begingroup$ @user6245072 So, the resulting fractal loses the "just touching" effect that the initiator had. Not surprising, as the limit set for the IFS is independent of the seed. Perhaps, it would make sense to think in terms of a so-called polygasket, as in this answer? $\endgroup$
    – Mark McClure
    Commented Jul 12, 2016 at 13:21

1 Answer 1

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Consider the triangle formed by the centers of two adjacent small circles and by the center of the outer circle; if the inner circles touch, then one has: $$ (r_1-r_2)\sin{\pi\over n}=r_2. $$ From that you can compute $r_2$: $$ r_2=r_1{\sin(\pi/n)\over1+\sin(\pi/n)}. $$

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  • $\begingroup$ ....how? I never studied equations with monomials at the denominator, sorry. Could you solve it for me? $\endgroup$
    – user6245072
    Commented Jul 12, 2016 at 8:48
  • $\begingroup$ See my edited answer. $\endgroup$
    – Intelligenti pauca
    Commented Jul 12, 2016 at 8:53
  • $\begingroup$ Crazy. Thanks a lot. $\endgroup$
    – user6245072
    Commented Jul 12, 2016 at 9:05
  • $\begingroup$ It works! And it's quite beautiful, I must say. $\endgroup$
    – user6245072
    Commented Jul 12, 2016 at 10:41

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