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Math Help - 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 P and 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 (0,\pm5/\sqrt2),\ (\pm5/\sqrt2,0). Then the centre S of the circle has to be at the point (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  \theta then \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  \phi in the attached picture. Using the inner product for the vectors \vec{SB} and \vec{SP}, I found that \cos\phi = 5\sqrt7/16 (correctness not guaranteed! I didn't check the calculation). Once you know  \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 \bigl(\frac{\sqrt7}{\sqrt2},\frac{-5+\sqrt7}{\sqrt2}\bigr).
    Attached Thumbnails Attached Thumbnails Circle and square problem-sectors.jpg  
    Last edited by Opalg; September 4th 2010 at 12:29 AM. Reason: extra information
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