Category Archives: Geometry
Cornucopious
The Choice of the Circle
Here’s an elementary mathematical problem: how many ways are there to choose three numbers from a set of six numbers? If the set is (1, 2, 3, 4, 5, 6), these are the possible choices (or combinations):
(1, 2, 3), (1, 2, 4), (1, 2, 5), (1, 2, 6), (1, 3, 4), (1, 3, 5), (1, 3, 6), (1, 4, 5), (1, 4, 6), (1, 5, 6), (2, 3, 4), (2, 3, 5), (2, 3, 6), (2, 4, 5), (2, 4, 6), (2, 5, 6), (3, 4, 5), (3, 4, 6), (3, 5, 6), (4, 5, 6) (c = 20)
So 6C3 = 20 (C stands for “combination”). The general formula is nCr = (n! / (n-r)!) / r!, where n is the number to choose from, r is the number of choices and n! is factorial n, or n multiplied by all numbers less than itself. For example, 6! = 6 * 5 * 4 * 3 * 2 * 1 = 720. When n = 6 and c = 3, 6C3 = (6! / (6-3)!) / 3! = (720 / 6) / 6 = 20.
There isn’t much visual appeal in the choices above, but there’s a simple way to change that. Take the ways of choosing two numbers from a set of ten. They start like this:
(1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9), (1, 10), (2, 3), (2, 4), (2, 5), (2, 6), (2, 7), (2, 8), (2, 9), (2, 10), (3, 4), (3, 5), (3, 6)…
Suppose each choice represents the midpoint of two points chosen from a set of ten points around a pentagon, so that (1, 2) is half-way between points 1 and 2, (3, 5) is half-way between points 3 and 5, and so on:
Now take the ways of choosing three numbers from a set of ten:
(1, 2, 3), (1, 2, 4), (1, 2, 5), (1, 2, 6), (1, 2, 7), (1, 2, 8), (1, 2, 9), (1, 2, 10), (1, 3, 4), (1, 3, 5), (1, 3, 6), (1, 3, 7), (1, 3, 8), (1, 3, 9), (1, 3, 10)…
Now the pentagon looks like this, with (1, 2, 3) representing the point midway between 1, 2 and 3, (1, 3, 9) representing the point midway between 1, 3 and 9, and so on:
Now here are 10C4, 10C5 and 10C6 for the pentagon:
You can also generate the points 5C4 = 5, then add them to the original five points and generate 10C4:
5C4
10C4
And here are 5C5, 6C5 and 12C5:
Here are 7C7 and 8C8, adding points as for 5C4:
And here is 12C6 using a dodecagon:
And various nCr for dodecagons and other polygons:
This method can also be used to represent the partitions of n, or the number of sets whose members sum to n. The partitions of 5 are these:
(5), (4, 1), (3, 2), (3, 1, 1), (2, 2, 1), (2, 1, 1, 1), (1, 1, 1, 1, 1)
There are seven partitions, so p(5) = 7. Partitions start small and get very large, starting with p(1), p(2), p(3) and so on:
1, 2, 3, 5, 7, 11, 15, 22, 30, 42, 56, 77, 101, 135, 176, 231, 297, 385, 490, 627, 792, 1002, 1255, 1575, 1958, 2436, 3010, 3718, 4565, 5604, 6842, 8349, 10143, 12310, 14883, 17977, 21637, 26015, 31185, 37338, 44583, 53174, 63261, 75175, 89134, 105558, 124754, 147273, 173525, 204226, 239943, 281589, 329931, 386155, 451276, 526823, 614154, 715220, 831820, 966467, 1121505, 1300156…
Suppose the partitions of n are treated as sets of points around a polygon with n vertices. Each set is then used to generate the point midway between its members. For example, (5, 4, 4, 2) is one partition of 15 and would represent the point midway between 5, 4, 4 and 2 of a pentadecagon. Here is a graphical representation of p(30):
Here are graphical representations for the partitions 5 to 15, then 15 to 60 in increments of 5 (15, 20, 25, etc):
And here are some close-ups for the partitions of 35 and 40:
Post-Performative Post-Scriptum…
The title of this incendiary intervention wrily references the Bible:
The flowers appear on the earth; the time of the singing of birds is come, and the voice of the turtle is heard in our land — The Song of Solomon, 2:12
Yale Straight Un!
Yale tablet, YBC 7289, c. 1700 BC.
Over Again
In Boldly Breaking the Boundaries, I looked at the use of squares in what I called over-fractals, or fractals whose sub-divisions reproduce the original shape but appear beyond its boundaries. Now I want to look at over-fractals using triangles. They’re less varied than those involving squares, but still include some interesting shapes. This is the space in which sub-triangles can appear, with the central seeding triangle coloured gray: 
Here are some over-fractals based on the pattern above: 
Boldly Breaking the Boundaries
In “M.I.P. Trip”, I looked at fractals like this, in which a square is divided repeatedly into a pattern of smaller squares:
As you can see, the sub-squares appear within the bounds of the original square. But what if some of the sub-squares appear beyond the bounds of the original square? Then a new family of fractals is born, the over-fractals:
Lette’s Roll
A roulette is a little wheel or little roller, but it’s much more than a game in a casino. It can also be one of a family of curves created by tracing the path of a point on a rotating circle. Suppose a circle rolls around another circle of the same size. This is the resultant roulette:
The shape is called a cardioid, because it looks like a heart (kardia in Greek). Now here’s a circle with radius r rolling around a circle with radius 2r:
That shape is a nephroid, because it looks like a kidney (nephros in Greek).
This is a circle with radius r rolling around a circle with radius 3r:
And this is r and 4r:
The shapes above might be called outer roulettes. But what if a circle rolls inside another circle? Here’s an inner roulette whose radius is three-fifths (0.6) x the radius of its rollee:
The same roulette appears inverted when the inner circle has a radius two-fifths (0.4) x the radius of the rollee:
But what happens when the circle rolling “inside” is larger than the rollee? That is, when the rolling circle is effectively swinging around the rollee, like a bunch of keys being twirled on an index finger? If the rolling radius is 1.5 times larger, the roulette looks like this:
If the rolling radius is 2 times larger, the roulette looks like this:
Here are more outer, inner and over-sized roulettes:
And you can have circles rolling inside circles inside circles:
And here’s another circle-in-a-circle in a circle:
M.i.P. Trip
The Latin phrase multum in parvo means “much in little”. It’s a good way of describing the construction of fractals, where the application of very simple rules can produce great complexity and beauty. For example, what could be simpler than dividing a square into smaller squares and discarding some of the smaller squares?
Yet repeated applications of divide-and-discard can produce complexity out of even a 2×2 square. Divide a square into four squares, discard one of the squares, then repeat with the smaller squares, like this:
Increase the sides of the square by a little and you increase the number of fractals by a lot. A 3×3 square yields these fractals:
And the 4×4 and 5×5 fractals yield more:
The Hex Fractor
A regular hexagon can be divided into six equilateral triangles. An equilateral triangle can be divided into three more equilateral triangles and a regular hexagon. If you discard the three triangles and repeat, you create a fractal, like this:
Adjusting the sides of the internal hexagon creates new fractals:
Discarding a hexagon after each subdivision creates new shapes:
And you can start with another regular polygon, divide it into triangles, then proceed with the hexagons:
Performativizing Papyrocentricity #29
Papyrocentric Performativity Presents:
• Sky Story – The Cloud Book: How to Understand the Skies, Richard Hamblyn (David & Charles 2008)
• Wine Words – The Oxford Companion to Wine, ed. Janice Robinson (Oxford University Press 2006)
• Nu Worlds – Numericon, Marianne Freiberger and Rachel Thomas (Quercus Editions 2014)
• Thalassobiblion – Ocean: The Definitive Visual Guide, introduction by Fabien Cousteau (Dorling Kindersley 2014) (posted @ Overlord of the Über-Feral)
Or Read a Review at Random: RaRaR
N-route
In maths, one thing leads to another. I wondered whether, in a spiral of integers, any number was equal to the digit-sum of the numbers on the route traced by moving to the origin first horizontally, then vertically. To illustrate the procedure, here is a 9×9 integer spiral containing 81 numbers:
| 65 | 64 | 63 | 62 | 61 | 60 | 59 | 58 | 57 | | 66 | 37 | 36 | 35 | 34 | 33 | 32 | 31 | 56 | | 67 | 38 | 17 | 16 | 15 | 14 | 13 | 30 | 55 | | 68 | 39 | 18 | 05 | 04 | 03 | 12 | 29 | 54 | | 69 | 40 | 19 | 06 | 01 | 02 | 11 | 28 | 53 | | 70 | 41 | 20 | 07 | 08 | 09 | 10 | 27 | 52 | | 71 | 42 | 21 | 22 | 23 | 24 | 25 | 26 | 51 | | 72 | 43 | 44 | 45 | 46 | 47 | 48 | 49 | 50 | | 73 | 74 | 75 | 76 | 77 | 78 | 79 | 80 | 81 |
Take the number 21, which is three places across and up from the bottom left corner of the spiral. The route to the origin contains the numbers 21, 22, 23, 8 and 1, because first you move right two places, then up two places. And 21 is what I call a route number, because 21 = 3 + 4 + 5 + 8 + 1 = digitsum(21) + digitsum(22) + digitsum(23) + digitsum(8) + digitsum(1). Beside the trivial case of 1, there are two more route numbers in the spiral:
58 = 13 + 14 + 6 + 7 + 7 + 6 + 4 + 1 = digitsum(58) + digitsum(59) + digitsum(60) + digitsum(61) + digitsum(34) + digitsum(15) + digitsum(4) + digitsum(1).
74 = 11 + 12 + 13 + 14 + 10 + 5 + 8 + 1 = digitsum(74) + digitsum(75) + digitsum(76) + digitsum(77) + digitsum(46) + digitsum(23) + digitsum(8) + digitsum(1).
Then I wondered about other possible routes to the origin. Think of the origin as one corner of a rectangle and the number being tested as the diagonal corner. Suppose that you always move away from the starting corner, that is, you always move up or right (or up and left, and so on, depending on where the corners lie). In a x by y rectangle, how many routes are there between the diagonal corners under those conditions?
It’s an interesting question, but first I’ve looked at the simpler case of an n by n square. You can encode each route as a binary number, with 0 representing a vertical move and 1 representing a horizontal move. The problem then becomes equivalent to finding the number of distinct ways you can arrange equal numbers of 1s and 0s. If you use this method, you’ll discover that there are two routes across the 2×2 square, corresponding to the binary numbers 01 and 10:
Across the 3×3 square, there are six routes, corresponding to the binary numbers 0011, 0101, 0110, 1001, 1010 and 1100:
Across the 4×4 square, there are twenty routes:
(Please open in new window if it fails to animate)
Across the 5×5 square, there are 70 routes:
(Please open in new window etc)
Across the 6×6 and 7×7 squares, there are 252 and 924 routes:
After that, the routes quickly increase in number. This is the list for n = 1 to 14:
1, 2, 6, 20, 70, 252, 924, 3432, 12870, 48620, 184756, 705432, 2704156, 10400600… (see A000984 at the Online Encyclopedia of Integer Sequences)
After that you can vary the conditions. What if you can move not just vertically and horizontally, but diagonally, i.e. vertically and horizontally at the same time? Now you can encode the route with a ternary number, or number in base 3, with 0 representing a vertical move, 1 a horizontal move and 2 a diagonal move. As before, there is one route across a 1×1 square, but there are three across a 2×2, corresponding to the ternary numbers 01, 2 and 10:
There are 13 routes across a 3×3 square, corresponding to the ternary numbers 0011, 201, 021, 22, 0101, 210, 1001, 120, 012, 102, 0110, 1010, 1100:
And what about cubes, hypercubes and higher?
Overlord In Terms of Core Issues Around Maximal Engagement with Key Notions of the Über-Feral 