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Thread: how is a circle "greater than" a square of equal area

  1. #1
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    how is a circle "greater than" a square of equal area

    I trying to formalize the measure by which a circle covers "more" space than a square of equal area, in following sense.

    Let G be a 1000 unit area in orthogonal dimensions x and y, with a corner point p, and whose interior points are all points {n,m} denoting the point n positive units from p in x, and m positive units from p in y. Let C be a circle, and S be square both with A=10, and {500,500} as their centers.

    Let CD be a set of distinct randomly chosen points from the perimeter of C; SD be a set of randomly chosen points from the perimeter of S, and N be the cardinality of both. Let A(CD) and be a function that returns the average of the distance of each point in CD from {500,500}, the center of C, and likewise for A(SD).

    Intuitively (to me, and please disprove formally if wrong):

    as N approaches infinity A(CD)>A(SD)

    If this is correct, can someone please formalize general numerical relationship between circles and squares from which it can be derived?
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  2. #2
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    Re: how is a circle "greater than" a square of equal area

    It's a well known fact that the circle maximizes area for a given perimeter.

    Is that what you're trying to say?
    Thanks from topsquark
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    Re: how is a circle "greater than" a square of equal area

    No, that is a measure, say P, the perimeter function, by which circle C and square S of equal area are such that:

    P(C)<P(S)

    I was asking for a measure function by which C > S. I did my best to describe the basis of such a measure, roughly as the limit of average of the lengths of the line segments connecting the center of C to a perimeter point of C, and likewise for S. I just don't know formally check or represent that.
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    Re: how is a circle "greater than" a square of equal area

    Quote Originally Posted by jackvio829 View Post
    No, that is a measure, say P, the perimeter function, by which circle C and square S of equal area are such that:

    P(C)<P(S)

    I was asking for a measure function by which C > S. I did my best to describe the basis of such a measure, roughly as the limit of average of the lengths of the line segments connecting the center of C to a perimeter point of C, and likewise for S. I just don't know formally check or represent that.
    The average distance from the center to a point on the perimeter of a square with area $s^2$ is

    $\bar{\ell} = \dfrac{4}{\pi}\displaystyle{\int_0^{\pi/4}}~\dfrac{s}{2 \cos(\theta)}~d\theta = \dfrac{s \ln\left(3+2\sqrt{2}\right)}{\pi} \approx 0.5611 s$

    Clearly the average distance from the center of a circle to a point on the perimeter is $r$, so for a circle of area $s^2$ we have

    $A = s^2 = \pi r^2$

    $r = \dfrac{s}{\sqrt{\pi}} \approx 0.5642 s$

    So yes, the average distance from the center to a point on the perimeter is a bit larger for a circle than for a square of the same area.
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    Re: how is a circle "greater than" a square of equal area

    Just what I wanted and perfectly clear - thanks so much!
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