• *Apud me omnia fiunt Mathematicè in Natura* — René Descartes (1596-1650).

• For me, all things in nature occur mathematically. — Correspondence with Martin Mersenne (1640).

# Monthly Archives: January 2016

# The Art Grows Onda

Anyone interested in recreational mathematics should seek out three compendiums by Ian Stewart: *Professor Stewart’s Cabinet of Mathematical Curiosities* (2008), *Professor Stewart’s Hoard of Mathematical Treasures* (2009) and *Professor Stewart’s Casebook of Mathematical Mysteries* (2014). They’re full of ideas and puzzles and are excellent introductions to the scope and subtlety of maths. I first came across Alexander’s Horned Sphere in one of them. I also came across this simpler shape that packs infinity into a finite area:

I call it a horned triangle or unicorn triangle and it reminds me of a wave curling over, like Katsushika Hokusai’s *The Great Wave off Kanagawa* (c. 1830) (“wave” is *unda* in Latin and *onda* in Spanish).

To construct the unicorn triangle, you take an equilateral triangle with sides of length 1 and erect a triangle with sides of length 0.5 on one of its corners. Then on the corresponding corner of the new triangle you erect a triangle with sides of length 0.25. And so on, for ever.

When you double the sides of a polygon, you quadruple the area: a 1×1 square has an area of 1, a 2×2 square has an area of 4. Accordingly, when you halve the sides of a polygon, you quarter the area: a 1×1 square has an area of 1, a 0.5 x 0.5 square has an area of 0.25 or 1/4. So if the original triangle of the unicorn triangle above has an area of 1 rather than sides of 1, the first triangle added has an area of 0.25 = 1/4, the next an area of 0.0625 = 1/16, and so on. The infinite sum is this:

1/4 + 1/16 + 1/256 + 1/1024 + 1/4096 + 1/16384…

Which equals 1/3. This becomes important when you see the use made of the shape in Stewart’s book. The unicorn triangle is a rep-tile, or a shape that can be divided into smaller copies of the same shape:

An equilateral triangle can be divided into four copies of itself, each 1/4 of the original area. If an equilateral triangle with an area of 4 is divided into three unicorn triangles, each unicorn has an area of 1 + 1/3 and 3 * (1 + 1/3) = 4.

Because it’s a rep-tile, a unicorn triangle is also a fractal, a shape that is self-similar at smaller and smaller scales. When one of the sub-unicorns is dropped, the fractals become more obvious:

Elsewhere other-posted:

# Performativizing Papyrocentricity #45

Papyrocentric Performativity Presents:

• Plants on Paper – *Drawing and Painting Plants*, Christina Brodie (A & C Black 2006)

• Lewminiferous – *Guide to Garden Wildlife*, Richard Lewington (British Wildlife Publishing 2008)

• Old Gold – *Puskás: Madrid, the Magyars and the Amazing Adventures of the World’s Greatest Goalscorer*, György Szöllős (Freight Books 2015)

• Rosetta Rok – *Rok 1984*, George Orwell (MUZA SA, Warszawa 2001)

Or Read a Review at Random: RaRaR

# Shareway to Seven

An adaptation of an interesting distribution puzzle from Joseph Degrazia’s *Math is Fun* (1954):

After a successful year of plunder on the high seas, a pirate ship returns to its island base. The pirate chief, who enjoys practical jokes and has a mathematical bent, hands out heavy bags of gold coins to his seven lieutenants. But when the seven lieutenants open the bags, they discover that each of them has received a different number of coins.

They ask the captain why they don’t have equal shares. The pirate chief laughs and tells them to re-distribute the coins according to the following rule: “At each stage, the lieutenant with most coins must give each of his comrades as many coins as that comrade already possesses.”

The lieutenants follow the rule and each one in turn becomes the lieutenant with most coins. When the seventh distribution is over, all seven of them have 128 coins, the coins are fairly distributed, and the rule no longer applies.

The puzzle is this: How did the pirate captain originally allocate the coins to his lieutenants?

If you start at the beginning and work forward, you’ll have to solve a fiendishly complicated set of simultaneous equations. If you start at the end and work backwards, the puzzle will resolve itself almost like magic.

The puzzle is actually about powers of 2, because 128 = 2^7 and when each of six lieutenants receives as many coins as he already has, he doubles his number of coins. Accordingly, before the seventh and final distribution, six of the lieutenants must have had 64 coins and the seventh must have had 128 + 6 * 64 coins = 512 coins.

At the stage before that, five of the lieutenants must have had 32 coins (so that they will have 64 coins after the sixth distribution), one must have had 256 coins (so that he will have 512 coins after the sixth distribution), and one must have had 64 + 5 * 32 + 256 coins = 480 coins. And so on. This is what the solution looks like:

128, 128, 128, 128, 128, 128, 128

512, 64, 64, 64, 64, 64, 64

256, 480, 32, 32, 32, 32, 32

128, 240, 464, 16, 16, 16, 16

64, 120, 232, 456, 8, 8, 8

32, 60, 116, 228, 452, 4, 4

16, 30, 58, 114, 226, 450, 2

8, 15, 29, 57, 113, 225, 449

So the pirate captain must have originally allocated the coins like this: 8, 15, 29, 57, 113, 225, 449 (note how 8 * 2 – 1 = 15, 15 * 2 – 1 = 29, 29 * 2 – 1 = 57…).

The puzzle can be adapted to other powers. Suppose the rule runs like this: “At each stage, the lieutenant with most coins must give each of his comrades __twice__ as many coins as that comrade already possesses.” If the pirate captain has six lieutenants, after each distribution each of five will have *n* + 2*n* = three times the number of coins that he previously possessed. The six lieutenants each end up with 729 coins = 3^6 coins and the solution looks like this:

13, 37, 109, 325, 973, 2917

39, 111, 327, 975, 2919, 3

117, 333, 981, 2925, 9, 9

351, 999, 2943, 27, 27, 27

1053, 2997, 81, 81, 81, 81

3159, 243, 243, 243, 243, 243

729, 729, 729, 729, 729, 729

For powers of 4, the rule runs like this: “At each stage, the lieutenant with most coins must give each of his comrades __three times__ as many coins as that comrade already possesses.” With five lieutenants, each of them ends up with 1024 coins = 4^5 coins and the solution looks like this:

16, 61, 241, 961, 3841

64, 244, 964, 3844, 4

256, 976, 3856, 16, 16

1024, 3904, 64, 64, 64

4096, 256, 256, 256, 256

1024, 1024, 1024, 1024, 1024

For powers of 5, the rule runs like this: “At each stage, the lieutenant with most coins must give each of his comrades __four times__ as many coins as that comrade already possesses.” With four lieutenants, each of them ends up with 625 coins = 5^4 coins and the solution looks like this:

17, 81, 401, 2001

85, 405, 2005, 5

425, 2025, 25, 25

2125, 125, 125, 125

625, 625, 625, 625