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Math Help - Dissected Circle Formula Help

  1. #1
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    Dissected Circle Formula Help

    Problem:
    Find the distance, from the circle center to a line, where the line dissects the circle into areas of 3/4 and 1/4.

    Any help with a formula, such that I can calculate this distance for various circle diameters would be greatly appreciated. I could add a pictures, if I have not described my problem clearly.

    Regards,
    Ken
    Last edited by Ken010101; January 14th 2008 at 11:37 AM. Reason: clearer description
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    Quote Originally Posted by Ken010101 View Post
    Problem:
    Find the distance, from the circle center to a line, where the line dissects the circle into areas of 3/4 and 1/4.

    ...
    Hello,

    1. I've attached a drawing.

    2. Using Pythagorean theorem you can derive a set of useful formula:

    h = r - \frac12 \cdot \sqrt{4r^2-s^2}

    s = 2\sqrt{2hr-h^2}

    A_{segment} = A_{sector} - A_{triangle}

    If you are allowed to use trigonometric functions the area of the segment can be calculated by:

    A_{segment}=\frac{r^2}{2}\cdot \left(\frac{\pi \alpha}{180^\circ}-\sin(\alpha)\right)
    Attached Thumbnails Attached Thumbnails Dissected Circle Formula Help-krs_abschnitt.gif  
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    Thank you very much earboth, for the quick reply.
    Quote Originally Posted by earboth View Post
    Using Pythagorean theorem you can derive a set of useful formula:

    h = r - \frac12 \cdot \sqrt{4r^2-s^2}

    s = 2\sqrt{2hr-h^2}
    So, for me to solve for h, am I understanding this correctly?
    h = r - \frac12 \cdot \sqrt{4r^2-(2\sqrt{2hr-h^2})^2}

    Is it possible for you to further assist me in simplifying this, so I can solve for h? Sorry, but I just don't understand where to begin...

    I cannot tell you how appreciative I am.
    Ken
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Ken010101 View Post
    Thank you very much earboth, for the quick reply.

    So, for me to solve for h, am I understanding this correctly?
    h = r - \frac12 \cdot \sqrt{4r^2-(2\sqrt{2hr-h^2})^2}

    Is it possible for you to further assist me in simplifying this, so I can solve for h? Sorry, but I just don't understand where to begin...

    I cannot tell you how appreciative I am.
    Ken
    h = r - \frac{1}{2} \cdot \sqrt{4r^2 - 4(2hr - h^2)}

    h - r + \frac{1}{2} = \sqrt{4r^2 - 4(2hr - h^2)}

    \left ( h - r + \frac{1}{2} \right ) ^2 = 4r^2 - 4(2hr - h^2)

    (I made minor mistake after this point, which would be too long to correct. Please refer to earboth's post below for some of the next steps.)

    -Dan
    Last edited by topsquark; January 15th 2008 at 07:00 AM.
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  5. #5
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    Quote Originally Posted by topsquark View Post
    h = r - \frac{1}{2} \cdot \sqrt{4r^2 - 4(2hr - h^2)}

    h - r + \frac{1}{2} = \sqrt{4r^2 - 4(2hr - h^2)}

    \left ( h - r + \frac{1}{2} \right ) ^2 = 4r^2 - 4(2hr - h^2)

    h^2 + r^2 + \frac{1}{4} - 2hr + \frac{h}{2} - r = 4r^2 - 8hr - 4h^2

    ......

    -Dan
    Good morning,

    I don't want to pick at you but you have made a minor mistake ....

    \left ( h - r + \frac{1}{2} \right ) ^2 = 4r^2 - 4(2hr - h^2)

    h^2+r^2+\frac14+h-r-2hr=4r^2-8hr+4h^2

    4h^2-8hr+4h+4r^2-4r=16h^2-32hr+16r^2-1

    After a few steps you'll get:

    h=\frac{6r-1}{6}~\vee~h=\frac{2r+1}{2}

    EDIT: In my opinion this result is much too simple so there must be a mistake in the original equation
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  6. #6
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    Quote Originally Posted by Ken010101 View Post
    Problem:
    Find the distance, from the circle center to a line, where the line dissects the circle into areas of 3/4 and 1/4.

    Any help with a formula, such that I can calculate this distance for various circle diameters would be greatly appreciated. I could add a pictures, if I have not described my problem clearly.

    Regards,
    Ken
    Hello,

    all those nice formulae I gave you lead to monstrous equation which I cann't handle anymore.

    So I decided to go back to the roots: I considered a half-circle to be the graph of a function

    c(x) = \sqrt{r^2-x^2}

    Then I used integration

    \int_u^r \left(\sqrt{r^2-x^2} \right)dx = \frac18 \cdot \pi \cdot r^2

    Then the height of the segment must be h = r-u

    I've attached a screenshot of the calculations (which were made by my computer)
    Attached Thumbnails Attached Thumbnails Dissected Circle Formula Help-krsabschn_int_num.gif  
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by earboth View Post
    Good morning,

    I don't want to pick at you but you have made a minor mistake ....

    \left ( h - r + \frac{1}{2} \right ) ^2 = 4r^2 - 4(2hr - h^2)

    h^2+r^2+\frac14+h-r-2hr=4r^2-8hr+4h^2

    4h^2-8hr+4h+4r^2-4r=16h^2-32hr+16r^2-1

    After a few steps you'll get:

    h=\frac{6r-1}{6}~\vee~h=\frac{2r+1}{2}

    EDIT: In my opinion this result is much too simple so there must be a mistake in the original equation
    Please, go ahead and "pick" at me! I'm always glad to have someone check my work. And good spot.

    -Dan
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  8. #8
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    Quote Originally Posted by topsquark View Post
    h = r - \frac{1}{2} \cdot \sqrt{4r^2 - 4(2hr - h^2)}

    h - r \boxed{+ \frac{1}{2}} = \sqrt{4r^2 - 4(2hr - h^2)}

    \left ( h - r + \frac{1}{2} \right ) ^2 = 4r^2 - 4(2hr - h^2)

    (I made minor mistake after this point, which would be too long to correct. Please refer to earboth's post below for some of the next steps.)

    -Dan
    Hello,

    as you have certainly spotted yourself I was blinded a little bit. The mm occured some steps earlier. \boxed{\text{I have boxed the algebraic accident}}
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