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#21 Prime

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Posted 03 December 2008 - 04:33 AM

My reasoning for the average distance between two random points is similar to that already posted here. However, I am going to show it anyway, since I think it may help to solve the distance between two points in a circle problem. And, in time, the original problem in the OP.

Spoiler for Solution by integration


From this we could move to constructing a similar solution for the distance between two points in a circle. And from there to the original problem -- distance between a random segment in a circle and anohter random point. From which we could find triangles' area and thus the probability of the 4th random point falling inside it.
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Past prime, actually.


#22 PVPA_Math

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Posted 09 December 2008 - 07:48 PM

Awesome problem!!!

Spoiler for My solution to the OP


:wub: :wub: :wub: This problem :wub: :wub: :wub:
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#23 Prime

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Posted 12 December 2008 - 05:29 AM

Awesome problem!!!

Spoiler for My solution to the OP


:wub: :wub: :wub: This problem :wub: :wub: :wub:

This is an interesting solution! Much simpler than I thought. But, I think it needs a bit more work.
The average segment length over an average chord seems to represent the average distance between two random points inside a circle. Although not all chords are of equal probability. When you consider a set of all parallel chords, each length occurs twice -- once on either side of the circle with a diameter as symmetry axis. The diameter, however, occurs only once. But since the diameter is just a line with a zero area, it may not affect the overall average length.

The second part -- distance between random point and an average length segment inside the circle may be a little off. I think that the actual average may be a bit bigger. The reason: not every possible distance belongs to a chord inside the circle. Here is an illustration:
circleprob.GIF
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Past prime, actually.


#24 PVPA_Math

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Posted 15 December 2008 - 01:29 AM

The second part -- distance between random point and an average length segment inside the circle may be a little off. I think that the actual average may be a bit bigger. The reason: not every possible distance belongs to a chord inside the circle.


Hrm...I've got to think on that a bit longer...something nags at me saying it's irrelevant...but....meh.
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#25 bonanova

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Posted 16 December 2008 - 07:32 AM

Determine the probability p(convex) that 4 random points placed on the unit disk form a convex quadrilateral.
Here is an algorithm for averaging the exact result for a million random triangles.

Spoiler for Algorithm


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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
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#26 bonanova

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Posted 18 December 2008 - 08:32 PM

Awesome problem!!!

I won't rehash the proof that the average distance between two points selected at random on a segment is 1/3 the length of the segment.
That's been shown.

So, knowing that, we can move on to the average distance between two points selected at random from the unit disc.
This must be 1/3 the average distance across the circle.
I decided to use all the vertical chords within the unit circle:

Avg chord length = (1/2)[Integral from -1 to 1 of 2(sqrt(1-x^2))] = Pi/2

Thus, the average distance between two points at random within the unit disc is (1/3)(Pi/2) = Pi/6.

The average length of the sides of the 3 million triangles of the previous post [random vertices] was .664....
That is just slightly less than 1/3 of the diameter, not of the average chord length.

If that helps our thinking.
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#27 Prime

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Posted 18 December 2008 - 11:36 PM

The average length of the sides of the 3 million triangles of the previous post [random vertices] was .664....
That is just slightly less than 1/3 of the diameter, not of the average chord length.

If that helps our thinking.

But the average length of a triangle side is not the same thing as average distance between 2 points inside a circle. Being a side of triangle makes additional constraints on the segment length. The solution path was: once the average segment length is found, find the distance to that segment from another random point, and not the other two sides of the triangle. So what would your simulator yield for the average distance between 2 points inside a unit circle? I think PVPA's solution to that may be correct.

Of course, in view of your correction on where the forth point may be, finding the average triangle area is not a solution to the OP, as we thought. The solution seems even more complex now.
But here is one observation: if we take any triangle on an infinite surface (not bounded by a circle), then the probability for the 4th pont to fall in the area where it would make a convex quadrangle is 1/2. (The sectors projected outside the triangle add up to 180o.) So for a very small triangle in the center of the circle, the probability tends to 1/2.
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Past prime, actually.


#28 bonanova

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Posted 19 December 2008 - 02:28 AM

But the average length of a triangle side is not the same thing as average distance between 2 points inside a circle.
Being a side of triangle makes additional constraints on the segment length.
The solution path was: once the average segment length is found, find the distance to that segment from another random point, and not the other two sides of the triangle.
So what would your simulator yield for the average distance between 2 points inside a unit circle?
I think PVPA's solution to that may be correct.

Of course, in view of your correction on where the forth point may be, finding the average triangle area is not a solution to the OP, as we thought. The solution seems even more complex now.
But here is one observation: if we take any triangle on an infinite surface (not bounded by a circle), then the probability for the 4th pont to fall in the area where it would make a convex quadrangle is 1/2. (The sectors projected outside the triangle add up to 180o.) So for a very small triangle in the center of the circle, the probability tends to 1/2.


Sure it is. All the points are randomly chosen.

I ran a program that just chose two points and verified this.


Agree with unbounded result of 1/2.
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#29 bonanova

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Posted 19 December 2008 - 02:41 AM

Spoiler for Here's something

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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

#30 unreality

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Posted 19 December 2008 - 02:56 AM

about the problem of a line with length x, and finding the average distance between two random points on the line, couldn't you do it

Spoiler for like this?

Edited by unreality, 19 December 2008 - 02:57 AM.

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