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Thread: Circle and square problem

  1. #1
    Senior Member Educated's Avatar
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    Question Circle and square problem



    There is a circle with a radius of 4cm and a square that is 5 * 5.

    2 corners (diagonally opposite) of the square are touching the circumference of the circle. Calculate the area overlap of the square and circle.

    At first when I read the problem I thought it was pretty simple. But I then realised it wasn't - there's 2 segments that stick out that you have to subtract (segment AC and BD).

    I just can't seem to find the area of the 2 segments that stick out. I can solve the rest. Can anyone show me or hint to me what I'm not seeing?
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  2. #2
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    Hi educated.

    Are you describing a square inscribed in a circle ?
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  3. #3
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    Hello, Educated!

    There is a circle with a radius of 4cm and a square that is 5-by-5.

    2 diagonally opposite corners of the square are on the circumference of the circle.

    Calculate the area overlap of the square and circle.

    The diagram looks like this:


    Code:
                  * * *
              *          /*
            *           /   *
           *          4/     *
                      /
          *     S ♥--/--------♥ P
          *       | o         *
          *       |           *
                  |           |
           *     5|          *|5
            *     |         * |
              *   |       *   |
                  ♥-*-*-------♥ Q
                  R     5

    Vertices $\displaystyle P$ and $\displaystyle R$ are on the circle.


    . . And, no . . . I haven't solved it yet.
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  4. #4
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    I don't see any neat way to do this. I tried putting the vertices of the square ABCD at the points $\displaystyle (0,\pm5/\sqrt2),\ (\pm5/\sqrt2,0)$. Then the centre S of the circle has to be at the point $\displaystyle (0,\sqrt{7/2})$. The overlap of the square and circle is then the sector DSB of the circle, with the two little protruding sections from D to Q and from P to B subtracted, together with the triangles ASB and ASD.

    If the angle DSB is $\displaystyle \theta$ then $\displaystyle \tan(\theta/2) = 5/\sqrt7$, from which you can find the area of the sector DSB. To subtract the two little protruding sections, you need to find the angle $\displaystyle \phi$ in the attached picture. Using the inner product for the vectors $\displaystyle \vec{SB}$ and $\displaystyle \vec{SP}$, I found that $\displaystyle \cos\phi = 5\sqrt7/16$ (correctness not guaranteed! I didn't check the calculation). Once you know $\displaystyle \phi$, you can find the area of the triangle SBP and subtract it from the area of the sector SBP to get the area of the sliver between P and B.

    Edit. I meant to add that P is the point $\displaystyle \bigl(\frac{\sqrt7}{\sqrt2},\frac{-5+\sqrt7}{\sqrt2}\bigr)$.
    Attached Thumbnails Attached Thumbnails Circle and square problem-sectors.jpg  
    Last edited by Opalg; Sep 3rd 2010 at 11:29 PM. Reason: extra information
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