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Thread: Area of Square in Circle

  1. #16
    MHF Contributor MarkFL's Avatar
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    Re: Area of Square in Circle

    Quote Originally Posted by harpazo View Post
    Improve our bounds???
    Yes, at the moment we have:

    $\displaystyle 2<\pi<4$

    Surely, we can narrow the gap...
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  2. #17
    Senior Member harpazo's Avatar
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    Re: Area of Square in Circle

    Quote Originally Posted by MarkFL View Post
    Yes, at the moment we have:

    $\displaystyle 2<\pi<4$

    Surely, we can narrow the gap...
    Mark,

    I would like to keep in touch with you via email. I can PM my personal email to you. I would like to keep all your replies to my questions on file as reference notes. Is this ok with you?
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  3. #18
    MHF Contributor MarkFL's Avatar
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    Re: Area of Square in Circle

    Quote Originally Posted by harpazo View Post
    Mark,

    I would like to keep in touch with you via email. I can PM my personal email to you. I would like to keep all your replies to my questions on file as reference notes. Is this ok with you?
    I prefer using a forum for mathematical discourse, primarily because LaTeX is available. I can barely keep up with emails as it is.

    Can you think of a way to improve the bounds on pi we have so far?
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  4. #19
    Senior Member harpazo's Avatar
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    Re: Area of Square in Circle

    Quote Originally Posted by MarkFL View Post
    I prefer using a forum for mathematical discourse, primarily because LaTeX is available. I can barely keep up with emails as it is.

    Can you think of a way to improve the bounds on pi we have so far?
    How can we improve the bounds on pi here?
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  5. #20
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    Re: Area of Square in Circle

    An easy way to determine the area is to see the square as a rhombus with the circle's diameter as the rhombus' diagonal. That way you can get its area without using those pesky square roots.
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  6. #21
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    Re: Area of Square in Circle

    Quote Originally Posted by harpazo View Post
    How can we improve the bounds on pi here?
    is there a simple formula for the area of a regular polygon?
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  7. #22
    MHF Contributor MarkFL's Avatar
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    Re: Area of Square in Circle

    Quote Originally Posted by harpazo View Post
    How can we improve the bounds on pi here?
    Suppose we use hexagons rather than squares:



    Let the radius of the circle be 1, so that its area is $\displaystyle \pi$ units squared. The area $\displaystyle A_S$ of the smaller hexagon is:

    $\displaystyle A_S=6\left(\frac{1}{2} \sin\left(\frac{2\pi}{6}\right)\right)= \frac{3\sqrt{3}}{2}\approx2.6$

    And the area $\displaystyle A_L$ of the larger hexagon is:

    $\displaystyle A_L=6\left(\frac{1}{2} \left(\frac{2}{\sqrt{3}}\right)^2\sin\left(\frac{2 \pi}{6}\right)\right)= 2\sqrt{3}\approx3.5$

    What do you think would happen if we used $\displaystyle n$-gons having more and more sides? Let's let $\displaystyle A_n$ be the area of an $\displaystyle n$-gon circumscribed by the circle...we have:

    $\displaystyle A_n=\frac{n}{2}\sin\left(\frac{2\pi}{n}\right)$

    Using a computer, we find:

    $\displaystyle n$ $\displaystyle A_n$
    10 2.938926261462366
    100 3.1395259764656687
    1000 3.1415719827794755
    10000 3.141592446881286
    1000000 3.141592653569122

    It appears that as $\displaystyle n\to\infty$ we have $\displaystyle A_n$ approaching some fixed finite value.
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  8. #23
    Senior Member harpazo's Avatar
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    Re: Area of Square in Circle

    Quote Originally Posted by MarkFL View Post
    Suppose we use hexagons rather than squares:



    Let the radius of the circle be 1, so that its area is $\displaystyle \pi$ units squared. The area $\displaystyle A_S$ of the smaller hexagon is:

    $\displaystyle A_S=6\left(\frac{1}{2} \sin\left(\frac{2\pi}{6}\right)\right)= \frac{3\sqrt{3}}{2}\approx2.6$

    And the area $\displaystyle A_L$ of the larger hexagon is:

    $\displaystyle A_L=6\left(\frac{1}{2} \left(\frac{2}{\sqrt{3}}\right)^2\sin\left(\frac{2 \pi}{6}\right)\right)= 2\sqrt{3}\approx3.5$

    What do you think would happen if we used $\displaystyle n$-gons having more and more sides? Let's let $\displaystyle A_n$ be the area of an $\displaystyle n$-gon circumscribed by the circle...we have:

    $\displaystyle A_n=\frac{n}{2}\sin\left(\frac{2\pi}{n}\right)$

    Using a computer, we find:

    $\displaystyle n$ $\displaystyle A_n$
    10 2.938926261462366
    100 3.1395259764656687
    1000 3.1415719827794755
    10000 3.141592446881286
    1000000 3.141592653569122

    It appears that as $\displaystyle n\to\infty$ we have $\displaystyle A_n$ approaching some fixed finite value.
    What a great explanation! Nicely done!
    Thanks from MarkFL
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