# First Whirled Warp

Imagine two points moving clockwise around the circumference of a circle. Find the midpoint between the two points when one point is moving twice as fast as the other. The midpoint will trace this shape: Midpoint of two points moving around circle at speeds s and s*2

(n.b. to make things easier to see, the red circle shown here and elsewhere is slightly larger than the virtual circle used to calculate the midpoints)

Now suppose that one point is moving anticlockwise. The midpoint will now trace this shape: Midpoint for s, -s*2

Now try three points, two moving at the same speed and one moving twice as fast: Midpoint for s, s, s*2

When the point moving twice as fast is moving anticlockwise, this shape appears: Midpoint for s, s, -s*2

Here are more of these midpoint-shapes: Midpoint for s, s*3 Midpoint for s, -s*3 Midpoint for s*2, s*3 Midpoint for s, -s, s*2 Midpoint for s, s*2, -s*2 Midpoint for s, s*2, s*2 Midpoint for s, -s*3, -s*5 Midpoint for s, s*2, s*3 Midpoint for s, s*2, -s*3 Midpoint for s, -s*3, s*5 Midpoint for s, s*3, s*5 Midpoint for s, s, s, s*3 Midpoint for s, s, s, -s*3 Midpoint for s, s, -s, s*3 Midpoint for s, s, -s, -s*3

But what about points moving around the perimeter of a polygon? Here are the midpoints of two points moving clockwise around the perimeter of a square, with one point moving twice as fast as the other: Midpoint for square with s, s*2

And when one point moves anticlockwise: Midpoint for square with s, -s*2

If you adjust the midpoints so that the square fills a circle, they look like this:  Midpoint for square with s, s*2, with square adjusted to fill circle

When the red circle is removed, the midpoint-shape is easier to see: Midpoint for square with s, s*2, circ-adjusted

Here are more midpoint-shapes from squares: Midpoint for s, s*3 Midpoint for s, -s*3 Midpoint for s, s*4

And some more circularly adjusted midpoint-shapes from squares:     Finally (for now), let’s look at triangles. If three points are moving clockwise around the perimeter of a triangle, one moving four times as fast as the other two, the midpoint traces this shape: Midpoint for triangle with s, s, s*4

Now try one of the points moving anticlockwise: Midpoint for s, s, -s*4 Midpoint for s, -s, s*4

If you adjust the midpoints so that the triangular space fills a circle, they look like this: Midpoint for s, s, s*4, with triangular space adjusted to fill circle Midpoint for s, -s, s*4, circ-adjusted Midpoint for s, s, -s*4, circ-adjusted

There are lots more (infinitely more!) midpoint-shapes to see, so watch this (circularly adjusted) space.

We Can Circ It Out — more on converting polygons into circles

# 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: 