# Smoothed octagon

A smoothed octagon.
The family of maximally dense packings of the smoothed octagon.

The smoothed octagon is a region in the plane conjectured to have the lowest maximum packing density of the plane of all centrally symmetric convex shapes.[1] It is constructed by replacing the corners of a regular octagon with a section of a hyperbola that is tangent to the two sides adjacent to the corner and asymptotic to the sides adjacent to these.

The smoothed octagon has a maximum packing density given by

$\frac{ 8-4\sqrt{2}-\ln{2} }{2\sqrt{2}-1} \approx 0.902414 \, .$[2]

This is lower than the maximum packing density of circles, which is

$\frac{\pi}{\sqrt{12}} \approx 0.906899.$

The maximum packing density of the ordinary regular octagon is

$\frac{4 + 4 \sqrt{2}}{5 + 4 \sqrt{2}} \approx 0.906163,$

also slightly less than the maximum packing density of circles, but higher than that of the smoothed octagon.[3]

The smoothed octagon achieves its maximum packing density, not just for a single packing, but for a 1-parameter family. All of these are lattice packings.[4]

In three dimensions, Ulam's packing conjecture states that no convex shape has a lower maximum packing density than the ball.

## Construction

The corners of the smoothed octagon can be found by rotating three regular octagons whose centres form a triangle with constant area.

By considering the family of maximally dense packings of the smoothed octagon, the requirement that the packing density remain the same as the point of contact between neighbouring octagons changes can be used to determine the shape of the corners. In the figure, three octagons rotate while the area of the triangle formed by their centres remains constant, keeping them packed together as closely as possible. For regular octagons, the red and blue shapes would overlap, so to enable the rotation to proceed the corners are clipped by a point that lies halfway between their centres, generating the required curve, which turns out to be a hyperbola.

Construction of the smoothed octagon (black), the tangent hyperbola (red) and the asymptotes of this hyperbola (green), and the tangent sides to the hyperbola (blue).

The hyperbola is constructed tangent to two sides of the octagon, and asymptotic to the two adjacent to these. The following details apply to a regular octagon of circumradius $\sqrt{2}$ with its centre at the point $(2+\sqrt{2},0)$ and one vertex at the point $(2,0)$. We define two constants, and m:

$\ell = \sqrt{2} - 1$
$m = \sqrt[4]{\frac{1}{2}}$

The hyperbola is then given by the equation

$\ell^2x^2-y^2=m^2$

or the equivalent parameterization (for the right-hand branch only):

$\begin{cases}x=\frac{m}{\ell} \cosh{t}\\y = m \sinh{t} \end{cases}-\infty

The portion of the hyperbola that forms the corner is given by

$-\frac{\ln{2}}{4}

The lines of the octagon tangent to the hyperbola are

$y= \pm \left(\sqrt{2} + 1 \right) \left( x-2 \right)$

The lines asymptotic to the hyperbola are simply

$y = \pm \ell x.$