## Irrational sunflowers

A neat way to visualize a real number ${\alpha}$ is to make a sunflower out of it. This is an arrangement of points with polar angles ${2 \pi \alpha k}$ and polar radii ${\sqrt{k}}$ (so that the concentric disks around the origin get the number of points proportional to their area). The prototypical sunflower has ${\alpha=(\sqrt{5}+1)/2}$, the golden ratio. This is about the most uniform arrangement of points within a disk that one can get.

But nothing stops us from using other numbers. The square root of 5 is not nearly as uniform, forming distinct spirals.

The number ${e}$ begins with spirals, but quickly turns into something more uniform.

The number ${\pi}$ has stronger spirals: seven of them, due to ${\pi\approx 22/7}$ approximation.

Of course, if ${\pi}$ was actually ${22/7}$, the arrangement would have rays instead of spirals:

What if we used more points? The previous pictures have 500 points; here is ${\pi}$ with ${3000}$. The new pattern has 113 rays: ${\pi\approx 355/113}$.

Apéry’s constant, after beginning with five spirals, refuses to form rays or spirals again even with 3000 points.

The images were made with Scilab as follows, with an offset by 1/2 in the polar radius to prevent the center from sticking out too much.

n = 500
alpha = (sqrt(5)+1)/2
r = sqrt([1:n]-1/2)
theta = 2*%pi*alpha*[1:n]
plot(r.*cos(theta), r.*sin(theta), '*');
set(gca(), "isoview", "on")

## Golden ratio in stereometry

Is there really such a thing as icosahedron?

Euclid found this problem difficult enough to be placed near the end of the Elements, and few of his readers ever mastered his solution. A beautiful direct construction was given by Luca Pacioli, a friend of Leonardo da Vinci, in his book De divina proportione (1509).

from Mathematics and its History by John Stillwell

The model consists of three golden-ratio rectangles passing one through another cyclically. Besides the central slits, a topological obstruction requires a temporary cut in one rectangle, which is then taped over. The convex hull of the union is an icosahedron, its vertices being the ${3\cdot 4=12}$ vertices of the rectangles. Indeed, if the rectangles have dimensions ${2\varphi\times 2}$, then the vertices are ${(\pm \varphi,\pm 1,0)}$, ${(0,\pm \varphi,\pm 1)}$, ${(\pm 1,0,\pm \varphi)}$. To prove that the faces are regular triangles, it suffices to check that ${\varphi^2 + (1-\varphi)^2+1^2 =4}$, which quickly turns into ${\varphi^2=\varphi+1}$.

I used the Fibonacci approximation ${\varphi \approx F_{n+1}/F_n}$ to draw the rectangles (specifically, their dimensions are ${89\times 55}$). The central slit in rectangle of size ${F_{n+1}\times F_{n}}$ should begin at distance ${F_{n-1}/2}$ from the shorter side.
The group of rotational symmetries of the paper model is smaller than the icosahedral group ${A_5}$: it has order ${12}$ and acts freely on the vertices. Come to think of it, the group is ${A_4}$.