Dilating the Delta

A circle with a radius of one unit has an area of exactly π units = 3.141592… units. An equilateral triangle inscribed in the unit circle has an area of 1.2990381… units, or 41.34% of the area of the unit circle.

In other words, triangles are cramped! And so it’s often difficult to see what’s going on in a triangle. Here’s one example, a fractal that starts by finding the centre of the equilateral triangle:

Triangular fractal stage #1


Next, use that central point to create three more triangles:

Triangular fractal stage #2


And then use the centres of each new triangle to create three more triangles (for a total of nine triangles):

Triangular fractal stage #3


And so on, trebling the number of triangles at each stage:

Triangular fractal stage #4


Triangular fractal stage #5


As you can see, the triangles quickly become very crowded. So do the central points when you stop drawing the triangles:

Triangular fractal stage #6


Triangular fractal stage #7


Triangular fractal stage #8


Triangular fractal stage #9


Triangular fractal stage #10


Triangular fractal stage #11


Triangular fractal stage #12


Triangular fractal stage #13


Triangular fractal (animated)


The cramping inside a triangle is why I decided to dilate the delta like this:

Triangular fractal

Circular fractal from triangular fractal


Triangular fractal to circular fractal (animated)


Formation of the circular fractal (animated)


And how do you dilate the delta, or convert an equilateral triangle into a circle? You use elementary trigonometry to expand the perimeter of the triangle so that it lies on the perimeter of the unit circle. The vertices of the triangle don’t move, because they already lie on the perimeter of the circle, but every other point, p, on the perimeter of the triangles moves outward by a fixed amount, m, depending on the angle it makes with the center of the triangle.

Once you have m, you can move outward every point, p(1..i), that lies between p on the perimeter and the centre of the triangle. At least, that’s the theory between the dilation of the delta. In practice, all you need is a point, (x,y), inside the triangle. From that, you can find the angle, θ, and distance, d, from the centre, calculate m, and move (x,y) to d * m from the centre.

You can apply this technique to any fractal created in an equilateral triangle. For example, here’s the famous Sierpiński triangle in its standard form as a delta, then as a dilated delta or circle:

Sierpiński triangle

Sierpiński triangle to circular Sierpiński fractal


Sierpiński triangle to circle (animated)


But why stop at triangles? You can use the same elementary trigonometry to convert any regular polygon into a circle. A square inscribed in a unit circle has an area of 2 units, or 63.66% of the area of the unit circle, so it too is cramped by comparison with the circle. Here’s a square fractal that I’ve often posted before:

Square fractal, jump = 1/2, ban on jumping towards any vertex twice in a row


It’s created by banning a randomly jumping point from moving twice in a row 1/2 of the distance towards the same vertex of the square. When you dilate the fractal, it looks like this:

Circular fractal from square fractal, j = 1/2, ban on jumping towards vertex v(i) twice in a row


Circular fractal from square (animated)


And here’s a related fractal where the randomly jumping point can’t jump towards the vertex directly clockwise from the vertex it’s previously jumped towards (so it can jump towards the same vertex twice or more):

Square fractal, j = 1/2, ban on vertex v(i+1)


When the fractal is dilated, it looks like this:

Circular fractal from square, i = 1


Circular fractal from square (animated)


In this square fractal, the randomly jumping point can’t jump towards the vertex directly opposite the vertex it’s previously jumped towards:

Square fractal, ban on vertex v(i+2)


And here is the dilated version:

Circular fractal from square, i = 2

Circular fractal from square (animated)


And there are a lot more fractals where those came from. Infinitely many, in fact.

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:

pent_10_2

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:

pent_10_3

Now here are 10C4, 10C5 and 10C6 for the pentagon:

pent_10_4

pent_10_5

pent_10_6

You can also generate the points 5C4 = 5, then add them to the original five points and generate 10C4:

pent4_1

5C4


pent4_2

10C4


And here are 5C5, 6C5 and 12C5:

pent5

Here are 7C7 and 8C8, adding points as for 5C4:

hept7

octo8

And here is 12C6 using a dodecagon:

dodeca_6

And various nCr for dodecagons and other polygons:

various

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):

partition30

Here are graphical representations for the partitions 5 to 15, then 15 to 60 in increments of 5 (15, 20, 25, etc):

partitions5_60

And here are some close-ups for the partitions of 35 and 40:

partitions40

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:
roulette1

roulette1static
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:
roulette2

roulette2static

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:
roulette3

roulette3static
And this is r and 4r:
roulette4

roulette4static
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:
roulette5

roulette5static
The same roulette appears inverted when the inner circle has a radius two-fifths (0.4) x the radius of the rollee:
roulette5a
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:
roulette6
If the rolling radius is 2 times larger, the roulette looks like this:
roulette2over

Here are more outer, inner and over-sized roulettes:

roulette_outer

roulette_inner

roulette_over

And you can have circles rolling inside circles inside circles:

roulette7

roulette0616

roulette0616all

And here’s another circle-in-a-circle in a circle:

roulette07c015c