1220. A sangaku

    Pepe Chapuzas has become fond of geometric sangakus. I think he even invents them... If he finds out some difficult sangaku, he proposes it to his classmates as a challenge. This is the last one:

    A circular sector has a radius of 1 foot and it can be opened and closed as a fan. When drawing the chord of the arc of the sector, this is cut into two pieces: a circular segment (blue) and an isosceles triangle (green). If we inscribe a circle (red) in each part as shown in the drawing..., what should be the angle of the sector so that the circles are equal?

    Nina Guindilla began...

    Dear Teacher:
    If the angle to be calculated measures  2x , then the diameter of the upper circle is the sagitta of the circular segment and measures (in feet):

Diameter  =  1 − cos x

    The lower circle is inscribed in the isosceles triangle, so its diameter can be calculated from the area and the perimeter of the triangle as follows: 

Diameter  =  4 · Area / Perimeter  =  2 · sin x · cos x / (1 + sin x)

    If both diameters are equal we have a trigonometric equation:

1 − cos x  =  2 · sin x · cos x / (1 + sin x)

    Or better yet...

(1 + sin x) · (1 − cos x)  =  2 · sin x · cos x

    Here Nina got stuck... She tried it in several ways... unsuccessfully... This was one of those magical moments that only we teachers can taste: I told her about the change of variables: 

tan (x / 2)  =  t

    Hence:

sin x  =  2t / (1 + t²)

cos x  =  (1 − t²) / (1 + t²)

    Now she could go on ...

(1 + 2t / (1 + t²)) · (1 − (1 − t²) / (1 + t²))  =  2 · 2t / (1 + t²) · (1 − t²) / (1 + t²)

(1 + t² + 2t) / (1 + t²) · (1 + t² − 1 + t²) / (1 + t²)  =  4t / (1 + t²) · (1 − t²) / (1 + t²)

 

    Multiplying twice by  1 + t²  (≠ 0)

(1 + t² + 2t) · 2t²  =  4t · (1 − t²)

    Dividing by  2t  (≠ 0)

(1 + t² + 2t) · t  =  2 · (1 − t²)

t + t³ + 2t²  =  2 − 2t²

t³ + 4t² + t − 2  =  0

    Dividing by  t + 1  (≠ 0)

t² + 3t − 2  =  0

t  =  (− 3 + √17) / 2

t  ≠  (− 3 − √17) / 2

    And substituting back...

x / 2  =  arctan t  =  arctan ((− 3 + √17) / 2)

    The angle sought is...

2x  =  4 · arctan ((− 3 + √17) / 2)  =  117° 15' 58"

 

    Okay!

    Nina Guindilla won good grades...

1045. The periodic table of polyhedra

    I'd ordered my students to write an essay on regular polyhedra, also called Platonic solids, and most of them, as usual, copy&pasted from Wikipedia. Nevertheless, Pepe Chapuzas did an original work (too original) titled “La tabla periódica de los poliedros” that I translate next:

    The periodic table of polyhedra.

Dear Teacher:

    Plato had five regular polyhedra (Tetrahedron, Octahedron, Cube, Icosahedron and Dodecahedron) for the four elements of nature (Fire, Air, Earth and Water), so he had to resort to the mysterious Ether to be able to elaborate his precious theory of polyhedral atoms.

    If Plato had known that none of his elements was a true element, and that there were more than five, and more than a hundred, which polyhedra would he have chosen to elaborate such a theory...? But, in fact, how many elements are there? The last chemical element to be baptized, Oganesson (Og), occupies since 2016 the box no. 118 of the periodic table. (The proof of the existence of an element is that it has gotten a name, a symbol and a place in this famous table.) A comment: I think the adjective periodic is used wrongly... In Maths, a period is a constant amount, however, in the periodic table the period grows: 2, 8, 18 and 32. It is rather a stepped table as we can see in Janet’s table of elements... Then..., periodic or stepped? That’s the question... Anyway, I won’t change now the title of my work...

    I found Janet's table on the Internet... With these stairs, sometimes, the rule for filling orbitals (s, p, d and f) and electron configurations of atoms are explained... Janet, in his table, unlike the standard periodic table, places the f-block to the left and the s-block to the right, and banishes Helium (He) from the group of noble gases. Since there are 4 steps and each step has 2 periods, in this table there is room for 120 elements. Would we need at least 120 polyhedra?

    OK..., 120 is a number that I like because it’s the factorial of 5, that is, 5! = 5·4·3·2 = 120. (Don’t forget these numbers: 5, 4, 3 and 2). I believe that Plato would have also stopped here: 120 polyhedra worthy of being called elemental or atomic... provided they weren't too irregular. The least that could be demanded is that they all had only regular faces and that they all were convex. I searched on the Internet for these polyhedra and I discovered that, besides the 5 Platonic solids, there were the 13 Archimedean solids, the 92 Johnson solids, and the two infinite series of prisms and antiprisms. I also read that Plato's, Archimedes', and Johnson's polyhedra were composed exclusively of polygons with 3, 4, 5, 6, 8 and 10 sides (note that numbers 6, 8 and 10 are twice numbers 3, 4 and 5 respectively). Were other polygons forbidden? In that case, with the allowed polygons we must add 5 prisms and 5 antiprisms... Let’s sum: 5 + 13 + 92 + 5 + 5 = 120. We already have the 120 polyhedra we need! And what is better (or worse), we have an excuse to elaborate a new Platonic theory: the Chapuzonic theory... (May I invent that neologism?)

    I was afraid of this: I dream about electronic structures and Janet's stairs, vertically and horizontally, full of polyhedra...! I've become obsessed... (Mr. López. You should be more careful with the works you ask us...) In addition, the names of these polyhedra are horrific: orthobicupolae, hebesphenomegacoronnae, etc. Fortunately, Johnson indexed his solids from (J-1) to (J-92)... The other 28 polyhedra are uniform and are determined by the polygons that join in a vertex: the cube would be (4.4.4) because in each vertex there are 3 squares. I hope not to lose my mind...

    Matching 120 polyhedra with 120 chemical elements seemed an impossible task but some clue prompted me to begin... What peculiarity did the elements of each step have? The first step has only s-orbitals and only the fourth step has f-orbitals... Do you believe that I begin to find some similarity between orbitals and polyhedra?... I don’t know... What could be the relationship between polyhedra and orbitals? Surely none! But let's go on... Look, if we count the polyhedra containing pentagons and decagons (not forgetting the icosahedron whose regular pentagons are hidden), there are 64... This is just the right amount of boxes in the fourth step of Janet's table! On the other hand we have several families of 2, 3 and even 4 polyhedra: for example, the family of pyramids (triangular, square and pentagonal) would be a family of 3 polyhedra...

    You still remember numbers 5, 4, 3 and 2, don't you? I thought that if the fourth step was related in some way to the number 5 (pentagons and decagons), the third step would be related to the number 4 (squares and octagons), the second step to the number 3 (triangles and hexagons) and, of course, the first step to the number 2 (edges and squares). Squares again?... Let's see. Each family would have a polyhedron at each step... and interspersing two families we'd fill a column of Janet's table, that is, a group of elements, right?

    I was not very optimistic... but searching and researching I found that there were neither more nor less than 12 families of 3 polyhedra for the 36 p-block elements (in this block we find the Chinese wall separating metals and non-metals). I have already drawn 4 families of 3 polyhedra. Here are the other 8 families:

    And there were also 4 families of 4 polyhedra for the 16 s-block elements!

    For the f-block (the 28 rare earths), we needed polyhedra without relatives... And there they were!: the rotundae...

... and the rhombicosidodecahedra (we owe Kepler this name). Exactly 28 in whole! Some of these polyhedra are so similar that we need to observe carefully to distinguish them from each other...

    Finally, for the d-block (the 40 transition metals), I only found 11 couples (families of 2 polyhedra): I didn't have enough couples... because, by their shape, some polyhedra were..., how can I say it..., singular. I didn't know whether to laugh or cry...

    Then, I remembered that in the d-block (and in the f-block) there were exceptional chemical elements because they didn’t follow the rule for filling orbitals. Their electron configurations were anomalous in their groups... Eureka!... If singular polyhedra were placed in the boxes of exceptional elements (as you may have guessed), then I needed only 11 couples! This was crazy!...

    There was still a problem. An intruder had entered the third step: the polyhedon called snub disphenoid (or Siamese dodecahedron) (J-84), which didn't have either squares or octagons to be there... Well, I awarded it to Palladium (Pd), the most bizarre element of the periodic table, because it was the only (neutral) atom that had no electrons at its outermost shell or level (paradoxically): its (ground state) configuration was [Kr] 4d¹⁰ 5s⁰ instead of the expected [Kr] 4d⁸ 5s²... Thus, I gradually dropped the polyhedra on the table, distributing them by rows and columns, letting my intuition fly... and taking into account that this theory lacked a scientific basis, and that any resemblance to reality... would be pure coincidence...    

           Pepe Chapuzas.

    I was speechless. Moreover, on the last page, the exact location of the polyhedra in the periodic table was shown. (How strange table!) Even so, this paper seemed to me a kind of mathematical juggling... and a curious way of approaching the beautiful world of polyhedra...