# 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 ${A_4}$.

The second model is my favorite Catalan solid, rhombic triacontahedron. It is formed by 30 golden-ratio rhombi. I folded it from a net created by Robert Webb. The net looks pretty cool itself:

Assembly took a lot more scotch tape (and patience) than the first model.