1. H. T. & E. hte
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    12 Jan '05 11:44
    This is for the mathematically oriented ones. You have a circle of a certain radius say r . Two points are taken in the interior of the circle completely randomly. The question is to find out the average distance (expectation of the distance) between the two randomly chosen points, in terms of r .
  2. Standard memberTheMaster37
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    12 Jan '05 19:10
    This is going to be interesting...I will not answer above question but i will post another one, related to this.

    Say you have a circle, and in it a triangle with equal sides, so that the point of the triangle are on the circle. Make a line between two random points on the circle. What is the chance that this line is longer then a side of the triangle?

    The reason i posted this is that i've seen the answer 1/2, 1/3 and 1/4 for it. I have reasonings for 1/2 and 1/3.
  3. Joined
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    13 Jan '05 17:09
    Originally posted by ranjan sinha
    This is for the mathematically oriented ones. You have a circle of a certain radius say r . Two points are taken in the interior of the circle completely randomly. The question is to find out the average distance (expectation of the distance) between the two randomly chosen points, in terms of r .
    How are the random points generated? The answer is dependent on this.
  4. Standard memberScheel
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    13 Jan '05 20:46
    Originally posted by TheMaster37
    This is going to be interesting...I will not answer above question but i will post another one, related to this.

    Say you have a circle, and in it a triangle with equal sides, so that the point of the triangle are on the circle. Make a line between two random points on the circle. What is the chance that this line is longer then a side of the triangle? ...[text shortened]... d this is that i've seen the answer 1/2, 1/3 and 1/4 for it. I have reasonings for 1/2 and 1/3.
    The answer to your problem is 1/3

    Place your two random points: a and b.
    Turn your circle with the inscribed triangle so that one of the triangles corners coincide with your random point a.
    Note that the two random points will only have inter-distance greater than the sides of the triangle if point b is located on the part of the circle directly opposite of a between the two remaining corners of the triangle.
    As this section of the circle has length 1/3 of the total circle, the answer follows.

    The remaining question is then: what kind of reasoning do you have for the answer 1/2 ?
  5. DonationAcolyte
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    13 Jan '05 21:05
    For ranjan's original question: I'll assume that the points are distributed uniformly by area. We can express a point as (r,a) where r is the distance from the centre and a is an angular parameter, and the corresponding random variables as (Ri,Ai).

    Ai is clearly uniform by symmetry. But Ri is not; in fact, Si = Ri^2 is uniform, because the proportion of points within a distance of r from the origin is r^2.

    The distance between (R1,A1) and (R2, A2) is given by the cos rule (B = A1 - A2):
    D^2 = (R1cosB -R2)^2 + (R1sinB)^2 = R1^2 + R2^2 -2R1*R2*cosB

    I can't see how to integrate this easily right now, but it can probably be done with a substitution.
  6. Standard memberScheel
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    13 Jan '05 22:35
    Originally posted by iamatiger
    How are the random points generated? The answer is dependent on this.
    Does not matter how the points are generated, what matters is what we mean by random. And I believe that was what you meant.

    The natural definition of random distribution of points within a circle would be that any two areas within the circle (of same area) has the same probability of containing a point.
    The interesting question is then how do we obtain a parametrization of the points that preserves this randomness.

    My initial idea was to go for polar coordinates : (x,y) = (r cos(t), r sin(t)) and let r be uniform on [0,R] and t uniform on [-pi,pi] (R=radius)

    but that would not preserve the above mentioned randomness. So we have to find an other distribution of r.
    It should be fairly simple.
  7. H. T. &amp; E. hte
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    15 Jan '05 11:22
    Originally posted by iamatiger
    How are the random points generated? The answer is dependent on this.
    Random distribution means that the probability of the point lying inside any two regions of equal area, is equal .

    For a point , its probability of lying in a certain region is equal to , the area of the region in question, divided by the total area of the circle.

    In other words, the probability of the point lying in a certain area element dA is equal to dA/(pi* r^2).
  8. Standard memberTheMaster37
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    15 Jan '05 12:24
    Originally posted by Scheel
    The answer to your problem is 1/3

    Place your two random points: a and b.
    Turn your circle with the inscribed triangle so that one of the triangles corners coincide with your random point a.
    Note that the two random points will only have inter-distance greater than the sides of the triangle if point b is located on the part of the circle directly opposit ...[text shortened]... lows.

    The remaining question is then: what kind of reasoning do you have for the answer 1/2 ?
    Take two random points the circle and rotate until your line AB is horizontal. Make another circle: then biggest circle completely inside the triangle (it'll touch the triangle halfway each side).

    The line AB is longer then a side of the triangle exactly then when then middle of the line AB (call it M) is in the second circle. This point M is on the vertical diameter of the big circle. Above condition is nothing more then the point M being on the vertical diameter of the smaller circle. This diameter is half of the diameter of the big circle.
  9. Standard memberScheel
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    15 Jan '05 15:04
    Originally posted by TheMaster37
    Take two random points the circle and rotate until your line AB is horizontal. Make another circle: then biggest circle completely inside the triangle (it'll touch the triangle halfway each side).

    The line AB is longer then a side of the triangle exactly then when then middle of the line AB (call it M) is in the second circle. This point M is on the ...[text shortened]... ertical diameter of the smaller circle. This diameter is half of the diameter of the big circle.
    OK I see.
    The problem with this argument is that M is not uniform distributed over the vertical diameter.
    Since M is not uniform on the vertical diameter, we can not simply compare the lengths to get the probability.

    I also think I can see the argument for 1/4. If you again draw your smaller circle and draw the line between your points A and B, but this time don't turn it just mark the midle M, then AB is longer than the sides of the triangle if M lies within the smaller circle.
    The small circle has an area of 1/4 of the larger circle. But again M is not uniform over the area of the large circle so we can not just compare the areas to get the probability.
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