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Math Help - proportions of circle

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    proportions of circle

    Prove that there is a magnitude p such that the area of a circle of radius r is to the area of a square of side r as p is to 1.
    Last edited by rmpatel5; September 9th 2008 at 08:53 AM.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by rmpatel5 View Post
    Prove that there is a magnitude p such that the area of a circle of radius r is to the area of a square of side r as p is to 1.
    Is "p" supposed to be \pi? If so then we can solve this. The area of a circle of radius r is \pi r^2. The area of a square of side r is r^2, so the ratio of these two areas is
    \frac{\pi r^2}{r^2} = \frac{\pi}{1}

    -Dan
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    Quote Originally Posted by topsquark View Post
    Is "p" supposed to be \pi? If so then we can solve this. The area of a circle of radius r is \pi r^2. The area of a square of side r is r^2, so the ratio of these two areas is
    \frac{\pi r^2}{r^2} = \frac{\pi}{1}

    -Dan
    no..
    The proof depends on three earlier results in the Elements:
    1. Given two magnitudes AB > CD, if from AB a quantity more than half is removed, and from the remainder a a quantity more than half is removed, and so on, eventually what is left will be less than CD.
      Proof One of the axiomatic properties of magnitudes is that there is a positive integer n such that n CD > AB, and a property of intgers is that there is a positive integer m such that 2m > n. Thus there is an m such that CD : AB > 1:2m. Then the remainder after the m -th cut off is < 1/(2m)AB < CD.

    2. Let P and Q be regular similar polygons with correspnding side lengths p and q . Then Area P : Area Q = p2 : q2.
      Since this is true for similar triangles, by the construction of Thales, it is true for all similar polygons by decomposition.

    3. The Law of Trichotomy (not explicitly stated, but used throughout): If x and y are magnitudes, then x < y or x = y or x > y . To us, this is an axiom of the real numbers; to the Greeks it was an unstated axiom of magnitudes.
      Circles C and D are to one another as the squares on their diameters c and d .
      Either Area C : Area D = c2 : d2 or < c2 : d2 or > c2 : d2.
      Suppose Area C : Area D < c2 : d2, so Area D > Area C.(d2/c2) . Let S = Area C.(d2/c2) , so S < Area D .
      By Proposition 1 of Book X, we can continually remove from Area D larger and larger polygons inscribed in D until we get a polygon P such that S < Area P < Area D, so Area P > Area C. d2/c2.
      Now inscribe in C a polygon Q similar to P so Area Q < Area C and Area Q : Area P = c2:d2. Thus Area P = Area Q . (d2/c2) < Area C. (d2/c2), a contradiction.
      A similar contradiction arises if Area C : Area D > c2 : d2 and we inscribe a polygon in C .
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