Two pictures of C20 Sangaku problem

I wrote a simple matlab script to generate the 3D structure of the C20 Sangaku problem. The first one is viewed from a 3-fold rotational axis and the second one is grom the 5-fold axis. Following the original problem as shown in the previous post, I used blue color to denote ten small spheres located on a great circle.

Ancient proof of R/r=sqrt{5}

In the Meiji era, mathematicians of Japan didn't use trigonometry to prove R/r=sqrt{5} for the problem about a big ball covered by 30 small balls. I doubt that they knew the trigonometry as we know today. Instead, a regular pentagon as shown in the following figure was recognized. Then the answer follows naturally. It is a smart proof, isn't it?

Proof of R/r=sqrt{5} for the geometry of bead C20

Even though the original drawing for this particular sangaku problem is done by the black-and-white style, it may not be a bad idea to make a colorful version for the cross section:

or may be like this one:

Sangaku problem and C20

In the book "算法助述" I got from Mr. Horibe, there is an interesting problem in pp. 51-52 as shown in the following picture. As I mentioned before, Mr. Horibe made two beautiful bead models with a large metal sphere inside. And more importantly, large and small balls have the correct ratio of radii, R/r=sqrt{5}. He told me that he gave one of these two models to Hidetoshi Fukagawa, the author of the book "Sacred Mathematica - Japanese Temple Geometry", because, through Dr. Fukagawa, he knew so much about these interesting Japanese sangaku problems.

This Sangaku problem was originally proposed by Ishikawa Nagamasa and written on a tablet hung in 1798 in Tokyo’s Gyuto Tennosha shrine. I was quite surprised by that I could understand this ancient problem without any translation. It was written in Chinese (or Kanji):

"今有以小球三十個如圖圍大球，小球者各切臨球四個與大球，小球徑三百零五寸，問大球徑幾何？

答曰：大球徑六百八十二寸。"

In English:

Thirty small balls cover one big ball where each small ball touches four other small balls and big balls.
The radius of small ball is 305 inches. Find the radius of large ball.

Answer: The radius of large ball is 682 inches.

In the page 52 of the book, there are some explanations on how to reach the answer by noting that thirty balls are located on the thirty vertices of an icosidodecahedron. It was written with some Japanese characters which I don't understand. But it is not hard to guess the rationale. One can see that if one starts from an arbitrarily chosen small ball, then moves to next ball along a fixed direction, and then the next one along the same direction. Eventually, one would trace over exactly 10 balls which are located on the large circle of a sphere with the radius equal to the sum of radii of small and large balls. Using some trigonometry, one can get the answer.

Sangaku and mathematical beading

Mr. Horibe, as a math teacher in Japan, is particularly interested in the connection between mathematical beading and the traditional Japanese temple mathematics, namely Sangaku （算額）. His teacher and colleague, Hidetoshi Fukagawa ( 深川英俊), is famous for his work on Sangaku and published a book, Sacred Mathematics: Japanese Temple Geometry, with Princeton's physicist Tony Rothman.

There are a few questions of Sangaku which are related to tangent spheres. For instance, the following problem appears in Fujita Kagen’s 1796 edition of Shinpeki Sanpo. (Collection of Fukagawa Hidetoshi.)



The translation of this problem given in the "Sacred mathematics" is
"Twenty small balls of radius r cover one big ball of radius R where each small ball touches three other small balls. Find R in terms of r."


Another problem related to the bead model of C₂₀ appeared in 1830 book Sanpo Kisho, or Enjoy Mathematics Tablets, by Baba Seitoku (1801–1860)



In this problem, small spheres correspond to beads that represent 30 edges of a dodecahedron.

To illustrate these two problems, Mr. Horibe made a beautiful, but non-standard bead model consisting of 20 ping-pong balls punctured with three holes at suitable places in order to connect them with an elastic rubber string. Of course, this model is a dodecahedron with ping-Pong balls located at the 20 vertices. Using the similar technique, he also made a bead model of an icosahedron consisting 12 Ping-Pong balls located at the vertices too.



To illustrate the second problem with 30 small spheres in touch with a large central sphere, Mr. Horibe did something even more amazing. He tried to make a composite bead model for the problem 2. This model consists of 30 small spheres outside (i.e. a dodecahedron) and a large sphere inside with small and large spheres satisfying the correct ratio of their radii, i.e. R=sqrt{5} r. This was a difficult task. Mr. Horibe quickly realized that it was very expensive to ask people to made two wooden beads exactly with this ratio. So he could only look around for different kind of balls in many stores in Japan, particularly he always brought a calculator and a ruler with him to measure if he was lucky enough to find out just the right beads that satisfied this condition. Quite fortunately, he found out a right size of metal ball for the central large sphere. So he made two nice models, one for Fukagawa and another for himself, for the famous Sangaku problem. In the following pictures I took at Nagoya's Children and Family Center, you can see how Mr. Horibe stretched the small spheres apart and pull the inner large sphere out. It was quite a show.

Another set of beaded models of platonic solids

Here is another set of beaded Platonic solids made with the same type of beads.

Platonic solids with cylindrical bonds

Cylindrical beads that have capsule shape are perfect pedagogical materials for creating physical models of fullerenes since they can look like the chemists' intuition of chemical bonds and, at the same time, effectively mimic the steric repulsion among different beads. To demonstrate these points, here, we use these beads to create the five Platonic solids. 

Platonic Solids

I bought these elongated beads last week, on sale, of course. The shape of these beads are closer to that of chemical bonds. I have used these beads to create five Platonic solids. They look great.

Tetrahedron


Cube



Octahedron




Dodecahedron



Icosahedron

Two more beaded C20s

I just made two more dodecahedral C20 in order to make my point clearer. The model in the left of the following figure is made from the standard spherical beads. Due to the hard-sphere repulsions among nearest beads, the resulting structure is quite stable even under moderate external pressure.

To some people, particularly some organic chemists, the most confusing part of our beaded representation of fullerenes is that the beads in the model stand for bonds instead of atoms as commonly used in stick-and-ball models. Thus we need thirty beads to make a C20. Since most people are used to the concept that any spherical object in a molecular model must correspond to an atom, thus our beaded models based on spherical beads may cause some confusion. The simplest way to solve this problem is that we should use beads with large aspect ratio to build our physical models. The model in the right of the following figure is constructed with this kind of beads. The beads as chemical bonds can be seen clearly. Unfortunately, this kind of beads cannot mimic repulsions among sp2 orbitals as good as spherical beads.

An effective strategy to improve this problem is based on the type of beading introduced in previous message.

Dodecahedral C20 with correct bond shape and force field

Previously, I have shown that beads with spherical shape can effectively mimic sp2 force field of fullerenes. However, these beads represent bonds instead of atoms. This may lead to confusion for students. We can avoid this problem by using beads with large aspect ratio, but the resulting structure usually has poor mechanical stability. Here Chuang has created a nice beaded fullerene of C20 (Ih) which can clearly exhibit the bond network of fullerene and at the same time possess great stability.