4
$\begingroup$

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

Attempt:

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

$\endgroup$
3
$\begingroup$

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

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

or

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

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

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

or

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

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

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

enter image description here

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.