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Consider two circles $C_1$ and $C_2$. $C_2$ lies within $C_1$. Let $S_1...S_n$ be circles lies between $C_1$ and $C_2$, and the following were satisfied:

(i):Each circle $S_i$ is tangent to C1 and C2

(ii):Each $S_i$ is tangent to $S_{i+1}$, $S_n$ is tangent to $S_1$.

The problem is to prove that the number n is independent of the choice of the circles {$S_i$}.

Attempt:

WIthout losing generality, we can assume that the big circle $C_1$ passes through the origin and orthogonal to real axis, then under conformal map 1\z, it becomes a vertical line. The small circle $C_2$ is mapped to another circle lie on one side of the vertical line. The small circles $S_i$ should be tangent to both the line and the image of $C_2$. But I don't know how to prov that the number of small circles are finite and independent of choice. Any help or hint is much appreciated.

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  • $\begingroup$ Do you mean that $C_1$ has its centre on the real axis? $\endgroup$ – user441558 Aug 27 '17 at 10:27
  • $\begingroup$ It's a valid assumption, isn't it? $\endgroup$ – scd Aug 27 '17 at 10:36
  • $\begingroup$ Yup, of course, because translation is a conformal mapping. $\endgroup$ – user441558 Aug 27 '17 at 10:38
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Use an inversia such that two given circles will be with common center.

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  • $\begingroup$ Sorry, I don't quite get what do you mean? $\endgroup$ – scd Aug 27 '17 at 10:40
  • $\begingroup$ @KIYORI If for given two circles there is a common center then your statement is obvious. About inversia see here: en.wikipedia.org/wiki/Inversive_geometry $\endgroup$ – Michael Rozenberg Aug 27 '17 at 10:44
  • $\begingroup$ Do you mean that if C1 and C2 are not co-centered, we can take inversion to make them co-centered? $\endgroup$ – scd Aug 27 '17 at 11:23
  • $\begingroup$ That's correct. $\endgroup$ – John Hughes Aug 27 '17 at 11:24
  • $\begingroup$ @KIYORI Y Yes of course. I am ready to explain how we can find the circle of the inversia. $\endgroup$ – Michael Rozenberg Aug 27 '17 at 11:40

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