# Number of disjoint circles in half plane minus a disk that touch both boundary components

Let $\Omega \subset \mathbb{C}$ be right half-plane, with the disc $D$ removed, where $D$ is the disk of radius $r=3$ centered at $z_0=5$. What is the maximum number of disjoint open disks in $\Omega$ whose boundary touch both boundary components of $\Omega$?

This is a question I had on my qualifying exam years ago; I couldn't solve it then, and today I stumbled upon it again and it nags me that I can't find a solution. Or at least, one without using any table or any map that I couldn't come up with on my own during an exam.

What I did (then and now) is to transform conformally the right half plane into the unit circle, and then see where the disk $D$ is mapped. Then my idea was to count the number of circles in this scenario. If the circles are not those coming from vertical lines in the right half plane, then they will come from circles in the original domain $\Omega$ (a simple test for a circle not to be a "bad" circle, is not to be centered on the real line, for example).

That is, $$f(z)=\frac{z-1}{z+1}$$ maps $\Omega$ to the unit disk with the disk $\tilde{D}$ removed, where $\tilde{D}$ is the disk centered at $w_0=\frac{11}{15}$ of radius $\frac{2}{15}$ (this follows from some elementary but tedious computations which I won't include in the question).

Now, I don't know how many disjoint circles touching both boundary components of $f(\Omega)$ I can fit in it, nor do I know how to guarantee that the answer is invariant under $f^{-1}$.

• Isn't invariance under $f^{-1}$ immediate from the fact that $f$ sends circles to circles? – Eric Wofsey Jun 30 '16 at 5:38
• Circles and lines to circle and lines, no? So some circles might come from vertical lines (which are mapped to circles). I might be very wrong, though. – Silvia Ghinassi Jun 30 '16 at 5:40
• Oh, right. But by a "general position" argument you should be able to assume none of your circles go to one of the two vertical lines that are tangent to $D$. (Since, for instance, automorphisms of $\Omega$ act transitively on $\partial D$.) – Eric Wofsey Jun 30 '16 at 5:42
• Anyways, my general strategy on this problem would be to choose $f$ so that it turns $\Omega$ into an annulus, so you have a lot more symmetry to exploit...although the answer is still not obvious for an annulus. – Eric Wofsey Jun 30 '16 at 5:44
• I agree on the general position, but better being safe. If I had an annulus, I'd have a fixed difference in radii and I could see how many circles of that diameter fits in it using the area, and then figure out if I'm not over counting. Still needs work, though. Also I'm not very good at coming up with such a map by hand? Maybe I was, at the beginning of grad school, surely not anymore. – Silvia Ghinassi Jun 30 '16 at 5:46

## 2 Answers

As already said in the comments, it is useful to map $\Omega$ with Möbius transformation $T$ to an annulus between two concentric circles, and we can assume that annulus is centered around $w=0$.

Some geometric observations help to find such a transformation $T$. Let $a$ and $b$ be the pre-images of $\infty$ and $0$, respectively.

• From the symmetry of $\Omega$ and the annulus with respect to the real axis we can conclude that both $a$ and $b$ are real.
• $a$ and $b$ must be symmetric with respect to the imaginary axis, i.e. $b = -\overline a$. For real $a, b$, this simplifies to $b = -a$.
• $a$ and $b$ must be symmetric with respect to the circle of radius $r=3$ centered at $z_0 = 5$, i.e. $(a - z_0)\overline{(b-z_0)} = r^2$. In our case, this simplifies to $(a-5)(b-5) = 3^2$.

It follows that $9 = (a-5)(-a-5) = -a^2 + 25$ and therefore $a = \pm 4$, so $$T(z) = \frac{z+4}{z-4}$$ is a candidate. Now verify that $T$ indeed maps $\Omega$ to an annulus with center $w=0$, inner radius $R_1= 1$ and outer radius $R_2 = 3$.

Disks touching both boundary components of $\Omega$ are mapped by $T$ to disks touching both inner and outer circle of the annulus. These image disks must have radius one and a center on $|w| = 2$. It is now easy to see that there can be at most six of them, compare Wikipedia: Steiner chain (link provided by @ccorn in a comment) or Wolfram World: Steiner chain.

• Thanks you so much for your answer and the construction of the map. About the number of circles (without mentioning Steiner chains): heuristically, $4\pi$ divided by $2$ is $6$, but I approximated twice in this computation and I don't know how to make it a full proof. – Silvia Ghinassi Jun 30 '16 at 14:37
• @SilviaGhinassi: Think of where the centers of the circles are. The inner circle of the annulus has radius 1, and the disk inside the annulus also have radius one. If you pack two disks inside the annulus, touching each other, then their centers together with $w=0$ form a equilateral triangle ... – Martin R Jun 30 '16 at 14:41
• Very nice and elementary. Thank you again! – Silvia Ghinassi Jun 30 '16 at 14:44

Just adding a graphical figure to Martin R's answer. The transformation used in the figure differs slightly from Martin's in that $$f(z)=4+\frac{16}{z-4}=\frac{4z}{z-4}=2\left(1+T(z)\right)$$ but both are equally usable. The important point is that the inversion circle has its center at $c=4$. Well, $c=-4$ would have worked as well, both possibilities follow from requiring constant annulus width: $$f(0)-f(z_0-r) = f(z_0+r)-f(\infty)$$ The solid black circles are $\partial D$ and an exemplary set of solution circles. The dash-dotted circle represents the inversion circle associated with $\bar{f}$. The gray circles are the boundaries of $f(\Omega)$. The dashed circles are the images of the solution circles under $f$, these form a Steiner chain with respect to the gray circles. Two of the dashed circles disappear under solid circles, those are just exchanged by $f$.

Note that the contents of the annulus could be rotated and still yield valid solution circles under the inverse transform, with one exception: No dashed circle shall touch the inversion center $c$, otherwise it maps back to a (vertical) line which does not touch the $y$-axis (in $\mathbb{C}$, that is).

• Nice graphics, may I ask how you produced it? – Martin R Jun 30 '16 at 9:49
• Geogebra Desktop version v3.2. – ccorn Jun 30 '16 at 9:52
• Thanks for the nice picture! Maybe it's too much, but would you able to get two separate pictures? – Silvia Ghinassi Jun 30 '16 at 14:38
• Transformed only and Transformed back. For further enhancements, it's probably best that you do the scene in Geogebra yourself, it's easy. If you are more inclined to describe the image with drawing commands and calculations (rather than constructing it the euclidean way), take at a look at Asymptote. – ccorn Jun 30 '16 at 15:13
• They look great! Current young grad students and future ones at my school will be grateful for these answers. Maybe I should learn how to produce pretty pictures one of these days, I'll keep in mind your advice. Thanks. – Silvia Ghinassi Jun 30 '16 at 16:45