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Math Help - AMC problem, finding the area inside a circle but outside a square

  1. #1
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    AMC problem, finding the area inside a circle but outside a square

    Hi everyone. I'm a little stuck on how to solve this problem.

    2010 AMC 10B Problems - AoPSWiki

    Basically, there's a circle and a square with the same center. I can't just take the area of the circle minus the area of the square, since some of the square lies outside of the circle. How do I take that into account?

    The answer, by the way, should be "B."
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Lord Darkin View Post
    Hi everyone. I'm a little stuck on how to solve this problem.

    2010 AMC 10B Problems - AoPSWiki

    Basically, there's a circle and a square with the same center. I can't just take the area of the circle minus the area of the square, since some of the square lies outside of the circle. How do I take that into account?

    The answer, by the way, should be "B."
    I presume you are talking about problem 16? This will help you out.

    -Dan
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    Yes, that's the problem (16) and thanks for the link.

    So I played around using those formulas, and also analyzed the circle/square situation further. Apparently the triangle that is below the "circular segment" as depicted by wikipedia is an equilateral triangle, with sides of root3 over 3. Therefore theta = 60 degrees.

    By using the formula of wikipedia, and multiplying it by 4 to take into account the 4 circular regions out of the square, I ended up with

    <br />
A = \frac{\pi}{3} - 4(\frac{1}{6}(\frac{2\pi}{3} - \frac{\sqrt{3}}{2}))<br />

    Where A should be the answer to the question. But I get...

    A = \frac{-\pi}{9} + \frac{\sqrt{3}}{3}

    Something isn't right here ....

    Edit: I should have said this before, but pi/3 for the first term represents the area of the entire circle.
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  4. #4
    MHF Contributor Unknown008's Avatar
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    Area of a sector = \pi\left(\dfrac{\sqrt3}{3}\right)^2\cdot \dfrac{60}{360} = \dfrac{\pi}{18}

    Area of the triangle = \dfrac12 ab \sin c = \dfrac12 \left(\dfrac{\sqrt3}{3}\right)^2 \sin (60) = \dfrac{\sqrt3}{12}

    Area required = 4\left(\dfrac{\pi}{18} - \dfrac{\sqrt3}{12}\right)
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    Thanks, I forgot sin60 = 1/sqrt3
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