Performativizing Polyhedra

Τα Στοιχεία του Ευκλείδου, ια΄

κεʹ. Κύβος ἐστὶ σχῆμα στερεὸν ὑπὸ ἓξ τετραγώνων ἴσων περιεχόμενον.
κϛʹ. ᾿Οκτάεδρόν ἐστὶ σχῆμα στερεὸν ὑπὸ ὀκτὼ τριγώνων ἴσων καὶ ἰσοπλεύρων περιεχόμενον.
κζʹ. Εἰκοσάεδρόν ἐστι σχῆμα στερεὸν ὑπὸ εἴκοσι τριγώνων ἴσων καὶ ἰσοπλεύρων περιεχόμενον.
κηʹ. Δωδεκάεδρόν ἐστι σχῆμα στερεὸν ὑπὸ δώδεκα πενταγώνων ἴσων καὶ ἰσοπλεύρων καὶ ἰσογωνίων περιεχόμενον.

Euclid’s Elements, Book 11

25. A cube is a solid figure contained by six equal squares.
26. An octahedron is a solid figure contained by eight equal and equilateral triangles.
27. An icosahedron is a solid figure contained by twenty equal and equilateral triangles.
28. A dodecahedron is a solid figure contained by twelve equal, equilateral, and equiangular pentagons.

Sumbertime Views

Like 666 (see Revelation 13:18), 153 (see John 21:11) appears in the Bible. And perhaps for the same reason: because it is the sum of successive integers. 153 = 1+2+3+…+17 = Σ(17), just as 666 = Σ(36). So both numbers are sum-numbers or sumbers. But 153 has other interesting properties, including one that can’t have been known in Biblical times, because numbers weren’t represented in the right way. It’s also the sum of the cubes of its digits: 153 = 1^3 + 5^3 + 3^3 = 1 + 125 + 27. So 153 is a cube-sumber or 3-sumber. The other 3-sumbers are 370, 371 and 407. There are 4-sumbers too, like 1,634 = 1^4 + 6^4 + 3^4 + 4^4, and 5-sumbers, like 194,979 = 1^5 + 9^5 + 4^5 + 9^5 + 7^5 + 9^5, and 6-sumbers, like 548,834 = 5^6 + 4^6 + 8^6 + 8^6 + 3^6 + 4^6.

But there are no 2-sumbers, or numbers that are the sum of the squares of their digits. It doesn’t take long to confirm this, because numbers above a certain size can’t be 2-sumbers. 9^2 + 9^2 = 162, but 9^2 + 9^2 + 9^2 = 243. So 2-sumbers can’t exist above 99 and if you search that high you’ll find that they don’t exist at all. At least not in this house, but they do exist in the houses next door. Base 10 yields nothing, so what about base 9?

4^2 + 5^2 = 45[9] = 41[10]
5^2 + 5^2 = 55[9] = 50

And base 11?

5^2 + 6^2 = 56[11] = 61[10]
6^2 + 6^2 = 66[11] = 72

This happens because odd bases always yield a pair of 2-sumbers whose second digit is one more than half the base and whose first digit is the same or one less. See above (and the appendix). Such a pair is found among the 14 sumbers of base 47, which is the best total till base 157 and its 22 sumbers. Here are the 2-sumbers for base 47:

2^2 + 10^2 = 104
3^2 + 12^2 = 153
5^2 + 15^2 = 250
9^2 + 19^2 = 442
12^2 + 21^2 = 585
14^2 + 22^2 = 680
23^2 + 24^2 = 1,105
24^2 + 24^2 = 1,152
33^2 + 22^2 = 1,573
35^2 + 21^2 = 1,666
38^2 + 19^2 = 1,805
42^2 + 15^2 = 1,989
44^2 + 12^2 = 2,080
45^2 + 10^2 = 2,125

As the progressive records for 2-sumber-totals are set, subsequent bases seem to either match or surpass them, except in three cases below base 450:

2 in base 5
4 in base 7
6 in base 13
10 in base 43
14 in base 47
22 in base 157
8 in base 182*
16 in base 268*
30 in base 307
18 in base 443*

Totals for sums of squares in bases 4 to 450

Totals for sums-of–squares in bases 4 to 450 (click for larger image)

Appendix: Odd Bases and 2-sumbers

Take an even number and half of that even number: say 12 and 6. 12 x 6 = 11 x 6 + 6. Further, 12 x 6 = 2 x 6 x 6 = 2 x 6^2 = 6^2 + 6^2. Accordingly, 66[11] = 6 x 11 + 6 = 12 x 6 = 6^2 + 6^2. So 66 in base 11 is a 2-sumber. Similar reasoning applies to every other odd base except base-3 [update: wrong!]. Now, take 12 x 5 = 2 x 6 x 5 = 2 x (5×5 + 5) = 5^2+5 + 5^5+5 = 5^5 + 5^5+2×5. Further, 5^5+2×5 = (5+1)(5+1) – 1 = 6^2 – 1. Accordingly, 56[11] = 11×5 + 6 = 12×5 + 1 = 5^2 + 6^2. Again, similar reasoning applies to every other odd base except base-3 [update: no — 1^2 + 2^2 = 12[3] = 5; 2^2 + 2^2 = 22[3] = 8]. This means that every odd base b, except base-3, will supply a pair of 2-sumbers with digits [d-1][d] and [d][d], where d = (b + 1) / 2.

Live and Let Dice

How many ways are there to die? The answer is actually five, if by “die” you mean “roll a die” and by “rolled die” you mean “Platonic polyhedron”. The Platonic polyhedra are the solid shapes in which each polygonal face and each vertex (meeting-point of the edges) are the same. There are surprisingly few. Search as long and as far as you like: you’ll find only five of them in this or any other universe. The standard cubic die is the most familiar: each of its six faces is square and each of its eight vertices is the meeting-point of three edges. The other four Platonic polyhedra are the tetrahedron, with four triangular faces and four vertices; the octahedron, with eight triangular faces and six vertices; the dodecahedron, with twelve pentagonal faces and twenty vertices; and the icosahedron, with twenty triangular faces and twelve vertices. Note the symmetries of face- and vertex-number: the dodecahedron can be created inside the icosahedron, and vice versa. Similarly, the cube, or hexahedron, can be created inside the octahedron, and vice versa. The tetrahedron is self-spawning and pairs itself. Plato wrote about these shapes in his Timaeus (c. 360 B.C.) and based a mathemystical cosmology on them, which is why they are called the Platonic polyhedra.

An animated gif of a tetrahedron

Tetrahedron


An animated gif of a hexahedron

Hexahedron

An animated gif of an octahedron

Octahedron


An animated gif of a dodecahedron

Dodecahedron

An animated gif of an icosahedron

Icosahedron

They make good dice because they have no preferred way to fall: each face has the same relationship with the other faces and the centre of gravity, so no face is likelier to land uppermost. Or downmost, in the case of the tetrahedron, which is why it is the basis of the caltrop. This is a spiked weapon, used for many centuries, that always lands with a sharp point pointing upwards, ready to wound the feet of men and horses or damage tyres and tracks. The other four Platonic polyhedra don’t have a particular role in warfare, as far as I know, but all five might have a role in jurisprudence and might raise an interesting question about probability. Suppose, in some strange Tycholatric, or fortune-worshipping, nation, that one face of each Platonic die represents death. A criminal convicted of a serious offence has to choose one of the five dice. The die is then rolled f times, or as many times as it has faces. If the death-face is rolled, the criminal is executed; if not, he is imprisoned for life.

The question is: Which die should he choose to minimize, or maximize, his chance of getting the death-face? Or doesn’t it matter? After all, for each die, the odds of rolling the death-face are 1/f and the die is rolled f times. Each face of the tetrahedron has a 1/4 chance of being chosen, but the tetrahedron is rolled only four times. For the icosahedron, it’s a much smaller 1/20 chance, but the die is rolled twenty times. Well, it does matter which die is chosen. To see which offers the best odds, you have to raise the odds of not getting the death-face to the power of f, like this:

3/4 x 3/4 x 3/4 x 3/4 = 3/4 ^4 = 27/256 = 0·316…

5/6 ^6 = 15,625 / 46,656 = 0·335…

7/8 ^8 = 5,764,801 / 16,777,216 = 0·344…

11/12 ^12 = 3,138,428,376,721 / 8,916,100,448,256 = 0·352…

19/20 ^20 = 37,589,973,457,545,958,193,355,601 / 104,857,600,000,000,000,000,000,000 = 0·358…

Those represent the odds of avoiding the death-face. Criminals who want to avoid execution should choose the icosahedron. For the odds of rolling the death-face, simply subtract the avoidance-odds from 1, like this:

1 – 3/4 ^4 = 0·684…

1 – 5/6 ^6 = 0·665…

1 – 7/8 ^8 = 0·656…

1 – 11/12 ^12 = 0·648…

1 – 19/20 ^20 = 0·642…

So criminals who prefer execution to life-imprisonment should choose the tetrahedron. If the Tycholatric nation offers freedom to every criminal who rolls the same face of the die f times, then the tetrahedron is also clearly best. The odds of rolling a single specified face f times are 1/f ^f:

1/4 x 1/4 x 1/4 x 1/4 = 1/4^4 = 1 / 256

1/6^6 = 1 / 46,656

1/8^8 = 1 / 16,777,216

1/12^12 = 1 / 8,916,100,448,256

1/20^20 = 1 / 104,857,600,000,000,000,000,000,000

But there are f faces on each polyhedron, so the odds of rolling any face f times are 1/f ^(f-1). On average, of every sixty-four (256/4) criminals who choose to roll the tetrahedron, one will roll the same face four times and be reprieved. If a hundred criminals face the death-penalty each year and all choose to roll the tetrahedron, one criminal will be reprieved roughly every eight months. But if all criminals choose to roll the icosahedron and they have been rolling since the Big Bang, just under fourteen billion years ago, it is very, very, very unlikely that any have yet been reprieved.