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Math Help - Sector of circle radius and areas

  1. #1
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    Sector of circle radius and areas

    Hi everybody,

    I could not solve the following problem.
    Look at the attachment.

    It is not homework.

    I need it to solve another problem.


    I did not succeed so I hope that you will.
    Attached Thumbnails Attached Thumbnails Sector of circle radius and areas-circlearea.gif  
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  2. #2
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    Re: Sector of circle radius and areas

    It seems that it is very hard problem.
    So for now I abandon.
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  3. #3
    Senior Member MacstersUndead's Avatar
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    Re: Sector of circle radius and areas

    (given) Area ABCD = Area CDEF - Area OEF = Area OCD - Area OEF - Area OEF = Area OCD - 2 Area OEF = (y^2)(a/2) - 2(x^2)(a/2)
    (derived) Area ABCD = Area OAB - Area OCD = (r^2)(a/2) - (y^2)(a/2)

    Hence
    (y^2)(a/2) - 2(x^2)(a/2) = (r^2)(a/2) - (y^2)(a/2)
    2(y^2 - x^2) = r^2
    y^2 - x^2 = (r^2)/2

    If x and y are integers, then the difference of squares is an integer, so r^2/2 is an integer, say n.

    EDIT:
    (r^2)/2 = (r/sqrt(2))^2
    Hence
    (r/sqrt(2))^2 + x^2 = y^2 and hence (r/sqrt(2),x,y) would be a Pythagorean triplet.

    that's all I have so far, but I hope that helps a little. my intuition says it's impossible, so I would continue down this line of reasoning to find a contradiction.
    Last edited by MacstersUndead; February 17th 2013 at 07:54 PM.
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  4. #4
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    Re: Sector of circle radius and areas

    Thank you very much.
    You gave me little hope.
    If we can not find x and y we could at least bound it.
    Here is my idea :
    The area CDEF is always equal to area ABCD + area OEF.
    We have to find DF such as DF = OF+DB
    The value of computed DF then will give us an upper bound for y-x
    Am I right?
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  5. #5
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    Re: Sector of circle radius and areas

    Another idea comes to my mind.
    We could find :
    - an upper bound (or lower?) to y-x by assuming that the area OEF goes to zero
    - and lower bound (or upper?) to y-x by assuming that the area ABCD goes to zero.

    So lower bound <(y-x)< upper bound.

    Is it easy to compute?
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  6. #6
    Senior Member MacstersUndead's Avatar
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    Re: Sector of circle radius and areas

    I do like that idea of boundedness and considering the difference to find either a contradiction or a bound.

    Suppose y-x = M for some positive integer M.
    then y^2 - x^2 = (M+x)^2 - x^2 = M^2 + 2Mx = \frac{r^2}{2}

    then x=\frac{r^2-2M^2}{4M} and y = \frac{r^2+2M^2}{4M}

    and since x and y are integers 4M must divide both r^2-2M^2 and r^2+2M^2

    Recall that if (ab) divides c, then a divides c, so 4 must divide both r^2-2M^2 and r^2+2M^2

    edit://

    so r^2-2M^2 = 4k for some integer k and r^2+2M^2=4m for some integer m.
    If k = m then M = 0. If m < k then r^2+2M^2 < r^2 - 2M^2, a contradiction.
    If k < m r^2 - 2M^2 < r^2 + 2M^2

    so now I would investigate the case of k<m, but I might just be complicating things or made a mistake. interesting problem though. I also considered looking at the problem when a = 360 degrees.
    Last edited by MacstersUndead; February 18th 2013 at 11:19 PM.
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