# Packing disks on a cone, or: Garlands on a tree

'Twas the night before Christmas, and throughout the net
Not a question was stirring, at least---not yet.
The tree was all draped with garlands and lights,
With presents beneath for morning delights.

As I enjoyed this warm-hearted scene,
A question arose as if in a dream:
The tree is a cone, the garlands are O's.
I wonder how many can fit of those?

Has this been studied? Can you shed some light?
Merry Christmas to all! And to all a good night!


Q1. Have packings of congruent disks on (finite) cones been explored?

Notation. Let $C$ be a (finite) right circular cone. A generator of $C$ is a segment connnecting the apex $a$ to a point on the rim of the base. Let $\alpha$ be the angle of the cone at $a$ when cut open along a generator; so $\alpha$ is the total surface angle incident to $a$. Let $L$ be the length of a generator. The surface area of $C$ is $A=A(L,\alpha)=\pi L^2 (\alpha/(2 \pi))$.

Q2. For a fixed surface area $A$, which cone angle $\alpha$ permits the most unit-radius disks to be packed on $C$?

In the example above, $L= 3 \sqrt{2} \approx 4.14$ and $\alpha = \frac{2}{3} \pi = 120^\circ$; so $A=6 \pi$, which is the area of $6$ unit-radius disks, although I only packed $4$ disks (and so achieved a density of $\delta=\frac{2}{3}$).

As $\alpha \to 0$, no unit-radius disk can fit, and for $\alpha = 2 \pi$, we are packing disks in a circle of radius $L$, a heavily studied problem. In the absence of an answer to Q2, let me suggest:

Q3. Can the behavior of the packing density along the $A(L,\alpha) = \mathrm{const}$ hyperbola-like curves be described qualitatively? E.g., is the density $\delta$ unimodal along those curves, perhaps for sufficiently large $A$?

• MathOverflow — the new Ladies' Diary. [‟All mathematical Questions in The Diary were stated in verse until 1729.”] Dec 24, 2014 at 17:20
• On Q1. I heard about many different packing problems, but not about that. May be Erich Friedman knows more. Merry Christmas and Happy New Year! :-) Dec 24, 2014 at 17:25
• There's also covering n disks by the smallest wedge. That point of view may lead to quick optimal results. Dec 24, 2014 at 18:35
• @AlexRavsky Link to Erich Friedman's packing center is now erich-friedman.github.io/packing Dec 7, 2020 at 0:52
• @ZsbánAmbrus Thanks. Also I thank you for fixing links to Erich’s pages in my posts, which I lost. Merry Christmas and Happy New Year! :-) Dec 31, 2020 at 13:01

Circle packings have been considered for $\alpha=2\pi$, which is the planar case; in that case the densest packing are the incircles of a hexagonal mesh.
While the planar case is trivial, two non-planar cases with an exact answer can be derived from it: place the apex at a point, that is the common corner of three of the meshes' hexagons and cut out a sector of either $\frac{2\pi}{3}$ or $\frac{4\pi}{3}$ and glue together the remains along the cuts.
For arbitrary $\alpha$ one could take inspiration from nature's way of packing seeds in circular disks, the most prominent example being the arrangement of the sunflower's seeds.
In the planar case, the $n$-th circle center would be placed at $\left(x(n),y(n)\right) := \left(c*n*\phi\cos(n*\phi),c*n*\phi*\sin(n*\phi)\right)$,
where $\phi:=2\pi\left(1-\frac{\sqrt{5}-1}{2}\right), n\in\mathbb{N}, c\in\mathbb{R}^+$.
• "in that case the densest packing are the incircles of a hexagonal mesh": For the infinite plane, Yes. But disks packing in a circle are more complicated. E.g., here is 20 disks. So I think your remarks concerning $\frac{2}{3}\pi$ and $\frac{4}{3}\pi$, only hold for infinite cones, or in the limit $L \to \infty$. Dec 25, 2014 at 12:53