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I want to find out the 'average distance' between many points on a 2D plain. That is, if you were standing on a point, what would be the average distance to all other points? For example, if there are three points A, B and C. AB = 5, AC = 10 and BC = 15, adding these up and dividing by three, we get 10. This is what I mean by the average distance.

Sort of like towns on a map. I'd like to be able to figure out an average distance between points. However, I don't actuallyknowwhere any of the points are, just the size of the map and the number of points. My thinking is that there is a relationship between the size of a map and the distance between points on said map.

So far, I have made some progress. A 'strange' form of Pythagoras' Theorem can be used to find the 'average distance' between points. I have to note that I assume the points always spread out as far as possible. Apologies in advance, but I am new to LaTeX. It seems that:

[itex]d = 2\sqrt{\frac{a}{p}}[/itex]

Where:

d = 'Average Distance'

a = area of shape

p = number of points

As a worked example:

Imagine a square, with side length 10 and there are two points. Going off the assumption that the points spread as far apart as possible, the two points will be on two opposite corners.

So, by Pythagoras' Theorem, they are 10√2 (14.14) apart.

[itex]10√2 = \sqrt{10^2+10^2}[/itex]

In my equation so far:

[itex] d = 2\sqrt{\frac{(10^2+10^2)}{2}}[/itex]

This also equals 10√2 (14.14).

With three points, I'm not sure how they would arrange. Possibly an equilateral triangle? Or two points on the vertices of one line and the other point in the middle of the opposite line (an isosceles triangle)? Either way, in the equation, it becomes:

[itex] d = 2\sqrt{\frac{(10^2+10^2)}{3}}[/itex]

Which equals [itex] \frac{20\sqrt{3}}{3}[/itex] (11.54)

As I say, I'm not sure of the exact 'average distance', but this seems about right.

Now, if the shape is not a square. If we have a rectangle with lengths 5 and 10, and two points. Again, the two points will go in the opposite corners.

So, by Pythagoras' Theorem they will be:

[itex]5√5 = \sqrt{10^2+5^2}[/itex]

In my equation:

[itex] 5√5 = 2\sqrt{\frac{(10^2+5^2)}{2}}[/itex]

So, the equation seems to work out. If anyone could confirm this, or at least point me in the right direction, I would be very grateful.

Edit:

So, a little more playing around revealed that the equation I made does not work out the 'average distance' between many points. I used the idea of a square ABCD, with points on A, B, C, D and one in the centre, which we will call point E. Again, the square will have side length 10. So the distance between AB, BC, CD and AD will be 10. The distance between AE, BE, CE and DE will be 5√2. Finally, the distance between AC and BD will be 10√2. So, a simple averaging will give us 4+4√2:

[itex] 4+4\sqrt{2} = 4(10)+4(5\sqrt{2})+2(10\sqrt{2})[/itex]

However, in the equation I specified earlier on, it equals 4√5:

[itex] 4\sqrt{5} = 2\sqrt{\frac{(10^2+10^2)}{5}}[/itex]

Okay, now I'm really stumped.

I imagine that there is a relationship between the area of a shape and the distributed points in the shape. I think that this will require calculus to do. Maybe the sum of every possible point divided by the area squared?

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# 'Average Distance' Between Many Points

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