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Math Help - Inscribed Quadrilateral

  1. #1
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    Inscribed Quadrilateral

    If there are four different circles that are tangent both to the edges of an inscribed quadrilateral and to the circle around that inscribed quadrilateral, how should the inscribed quadrilateral be chosen so that the sum of the perimeters of those four different circles could be maximum or minimum?

    Case 1:
    If the inscribed quadrilateral is a "square" with the same side length a, then I find the sum of the perimeters of the "same" four circles P_1 = \frac{2\pi\cdot a}{1 + \sqrt{2}}

    Case 2:
    If the inscribed quadrilateral is a "rectangle" with the side lengths a and \frac{a}{2}, then I find the sum of the perimeters of the two bigger circles and the two smaller circles P_2 = \pi \cdot a \cdot \frac{2\sqrt{5} - 3}{2}

    Case 3:
    Let the side lengths of the rectangle which is the inscribed quadrilateral be a and b,
    the radius of the two bigger circles be r_1,
    the radius of the two smaller circles be r_2,
    the total sum of the perimeters of these circles be P_3,
    the radius of the largest circle that contains both the inscribed quadrilateral (rectangle) around which are those four circles be R.

    R = \sqrt{\frac{a^2}{4} + \frac{b^2}{4}}

    r_1 = \frac{R}{2} - \frac{a}{4}

    r_2 = \frac{R}{2} - \frac{b}{4}

     P = 2(2\pi \cdot r_1) + 2(2\pi \cdot r_2) = 4\pi \cdot R - \pi(a + b)

     P = 4\pi \cdot \sqrt{\frac{a^2}{4} + \frac{b^2}{4}} - \pi(a + b)

     P = 2\pi \cdot \sqrt{a^2 + b^2} - \pi(a + b)

    \frac{dP}{da} = \frac{2\pi \cdot a}{\sqrt{a^2 + b^2}} - \pi = 0

    \frac{2a}{\sqrt{a^2 + b^2}} - 1 = 0

     b = a\cdot \sqrt{3}

    and for this value, the sum of the perimeters is P_3 = \pi \cdot a(3 - \sqrt{3}) which is the minimum of all three cases.

    So, assuming that my approach and calculations to be correct, can I conclude as follows?

    I. When the inscribed quadrilateral is a "square", the sum of the perimeters of the circles is maximum.

    II. When the inscribed quadrilateral is a "rectangle" whose long side length is \sqrt{3} times its other (short) side length, the sum of the perimeters of the circles is minimum

    Thanks and regards.
    Last edited by Honore; October 17th 2011 at 04:11 AM.
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  2. #2
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    Re: Inscribed Quadrilateral

    Let me add that,

    \frac{d^2{P}}{da^2}=f''(a)=\frac{2\pi \cdot \sqrt{a^2+b^2}-2\pi{a}\cdot \frac{a}{\sqrt{a^2+b^2}}}{a^2+b^2}

    f''(\frac{b}{\sqrt{3}})=\frac{2\pi\cdot \sqrt{\frac{b^2}{3}+b^2}-\frac{2\pi\cdot b}{\sqrt{3}}\cdot \frac{\frac{b}{\sqrt{3}}}{\sqrt{\frac{b^2}{3}+b^2}  }}{\frac{b^2}{3}+b^2}

    f''(\frac{b}{\sqrt{3}})=\frac{\frac{4\pi{b}}{\sqrt  {3}}-\frac{2\pi b^2}{3}\cdot \frac{\sqrt{3}}{2b}}{\frac{4b^2}{3}}

    f''(\frac{b}{\sqrt{3}})=\frac{3\pi \sqrt{3}}{4b}>0 which is minimum for a=\frac{b}{\sqrt{3}}

    Thanks for any critiques, comments and suggestions. Regards.
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  3. #3
    MHF Contributor
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    Ottawa, Canada
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    Re: Inscribed Quadrilateral

    Just a couple of remarks:
    1: I think you need to specify that the point of tangency (circles to edges) is the midpoint
    of the edges; else you can "move" the circles thus making them smaller!
    2: I didn't see a calculation involving the 4 edges being different; like with edges a,b,c,d;
    your diagram clearly involves that also...
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  4. #4
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    Re: Inscribed Quadrilateral

    Thanks a lot for your remarks. I will work more to cover those two points. Regards.
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