Consider two unshaded circles $C_r$ and $C_s$ with radii $r>s$ that touch at the origin of the complex plane. The shaded circles $C_1,C_2...C_7$ (labeled in counterclockwise direction sequentially) all touch $C_r$ internally and $C_s$ externally. $C_1$ also touches the real axis and $C_i$ and $C_{ i+1}$ touch for $i=1...6.$

Let $r_i$ denote the radius of $C_i$. Then show that for $i=1,2...,$ $$r_i^{-1} + 3r_{i+2}^{-1} = 3r_{i+1}^{-1} + r_{i+3}^{-1}$$

A picture is attached for clarity.

enter image description here


Under inversion, $C_s$ and $C_r$ are mapped to lines and $C_i, i \in \left\{1,...,6\right\}$ are mapped to circles. The map $f(z) = 1/z$ is conformal, so the angles are preserved wherever $f'(z) \neq 0$. Each circle has $4$ tangent points, (except $C_7$) so under the transformation these $\pi/2$ angles are preserved.

The resulting image I have after the transformation is that the $C_i$ are mapped to circles that lie within the two lines and touch the sides. The only way I see just now to preserve the angles is to have all the circles of the same radii. If $s$ is the radius of $C_s$ and $r$ the radius of $C_r$ then the centre between these two lines is $\frac{1}{2}\left(1/s - 1/2r\right) $. (But I do not think this makes sense.)

Many thanks


The two circles $C_r$ and $C_s$ are mapped to the parrallel lines $\left\{ \operatorname{Re} w = \frac{1}{2r}\right\}$ and $\left\{ \operatorname{Re} w = \frac{1}{2s}\right\}$ by the inversion. Since the inversion maps circles to circles (where a straight line is counted as a circle of infinite radius), it maps each of the $C_i$ to a circle touching both of these parallel lines. Also, the image of $C_1$ touches the real axis (since $C_1$ does, and the inversion maps the real axis $\cup \{\infty\}$ to itself), and further, each circle $C_k$ for $k > 1$ touches the circles $C_{k-1}$ and $C_{k+1}$, hence so do their images.

The distance between the two parallel lines is $\frac{1}{2s} - \frac{1}{2r}$, so the images of the $C_k$ have the common radius

$$R = \frac{1}{2}\left(\frac{1}{2s} - \frac{1}{2r}\right),$$

and their centres all lie on the line $$\left\{ \operatorname{Re} w = M\right\};\qquad M := \frac{1}{2}\left(\frac{1}{2s} + \frac{1}{2r}\right).$$

It follows that the centre of the image $C_k'$ of $C_k$ is

$$b_k = M - (2k-1)\cdot iR,$$

and hence the defining equation of $C_k'$ is

$$\lvert w-b_k\rvert^2 = R^2,$$


$$w\overline{w} - b_k\overline{w} - \overline{b_k}w + (\lvert b_k\rvert^2 - R^2) = 0.$$

Applying the inversion, we see that $C_k$ has the defining equation

$$1 - b_k z - \overline{b_kz} + (\lvert b_k\rvert^2-R^2)z\overline{z} = 0.$$

Since $\lvert b_k\rvert^2 = M^2 + (2k-1)^2R^2 > R^2$, we can divide and obtain the equivalent equation

$$z\overline{z} - \frac{\overline{b_k}}{\lvert b_k\rvert^2-R^2}\overline{z} - \frac{b_k}{\lvert b_k\rvert^2 - R^2} z + \frac{1}{\lvert b_k\rvert^2-R^2} = 0,$$


$$\left\lvert z - \frac{\overline{b_k}}{\lvert b_k\rvert^2-R^2} \right\rvert^2 = \left(\frac{R}{\lvert b_k\rvert^2-R^2}\right)^2.$$

The desired relation between the radii should not be difficult to obtain from that.

  • $\begingroup$ Very nice. Could you explain how you obtained the equation for the centre of the image $C_k'$ and what $R$ represents? $\endgroup$ – CAF Feb 12 '14 at 21:40
  • $\begingroup$ $R$ is half the distance between the two parallel lines. Since a circle that touches two parallel lines must have its centre half-way between the two lines, that is the radius of such a circle. So that gives us the real part of the centres, and for the imaginary part, we have $(k-1)$ circles between $C_k'$ and the real line, and for the centre, we must go one radius further, so $(2k-1)\cdot R$. And a minus sign, since the inversion maps the upper half-plane to the lower. $\endgroup$ – Daniel Fischer Feb 12 '14 at 21:46
  • $\begingroup$ Many thanks, I think your formulae are equivalent to the ones I am given for a general transformation - For a circle of radius $r$, centered at $z_o$, the image under inversion is also a circle of radius $R$ centered at $w_o$, where $$w_o = \frac{z_o^*}{|z_o|^2 - r^2} and R = \frac{r}{|z_o|^2 - r^2}$$ In this case, how would I obtain $z_o$? For sure I can divide both equations, but then I am still left with a $z_o^*$. $\endgroup$ – CAF Feb 12 '14 at 22:14
  • $\begingroup$ For the circle $C_k'$, we have $z_0 = b_k$. I gave the $b_k$ explicitly, you can just insert that. $\endgroup$ – Daniel Fischer Feb 12 '14 at 22:21
  • $\begingroup$ Yes, but $z_o$ is the centre of the circle before the transformation. In my notation, $w_o = b_k, $not $z_o$. Or did I misunderstand something? Thanks. $\endgroup$ – CAF Feb 12 '14 at 22:25

enter image description here

I'm trying to get an idea of the resultant picture after inversion. Is it something like this?

  • $\begingroup$ Hi Alan, that is exactly what I was thinking except shouldn't the circles be tending downwards? $\endgroup$ – CAF Feb 12 '14 at 20:58
  • $\begingroup$ I am not really sure how to progress to obtain the required identity. Do you have any ideas? $\endgroup$ – CAF Feb 12 '14 at 21:04
  • $\begingroup$ en.wikipedia.org/wiki/Pappus_chain and here:mathworld.wolfram.com/PappusChain.html $\endgroup$ – Alan Feb 12 '14 at 21:11

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