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Math Help - Square

  1. #1
    MHF Contributor alexmahone's Avatar
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    Square

    What is the probability that a point chosen at random within a square is closer to the centre than to the boundary?
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  2. #2
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    Quote Originally Posted by alexmahone View Post
    What is the probability that a point chosen at random within a square is closer to the centre than to the boundary?
    This boils down to a problem in geometry .... Find the area of a square that is closer to the centre than the sides.
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  3. #3
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by mr fantastic View Post
    This boils down to a problem in geometry .... Find the area of a square that is closer to the centre than the sides.
    You're probably referring to a square of side length \frac{a}{2} located within the original square of side length a.

    However, the corner of the inner square is \frac{a}{2\sqrt{2}} units away from the centre whereas it is only \frac{a}{4} units away from the boundary.

    So I don't think the above approach is correct.
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  4. #4
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    Quote Originally Posted by alexmahone View Post
    You're probably referring to a square of side length \frac{a}{2} located within the original square of side length a.

    However, the corner of the inner square is \frac{a}{2\sqrt{2}} units away from the centre whereas it is only \frac{a}{4} units away from the boundary.

    So I don't think the above approach is correct.
    No I didn't mean that at all. Let me re-word it:

    Find the area inside a square that is closer to the centre of the square than the sides of the square.
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  5. #5
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by mr fantastic View Post
    No I didn't mean that at all. Let me re-word it:

    Find the area inside a square that is closer to the centre of the square than the sides of the square.
    I get a "square" of length \frac{a}{2} with circular corners of radius \frac{a}{4}. Is that right?
    Last edited by alexmahone; January 23rd 2009 at 05:05 AM.
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  6. #6
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    Quote Originally Posted by alexmahone View Post
    I get a "square" of length \frac{a}{2} with circular corners of radius \frac{a}{4}. Is that right?
    No, that is not correct. Recall that a parabola is a set of points equally distance from a point and a line. So in this question the area is the intersection of the interiors of four parabolas. See the graph below.
    Attached Thumbnails Attached Thumbnails Square-area.gif  
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  7. #7
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    Quote Originally Posted by Plato View Post
    No, that is not correct. Recall that a parabola is a set of points equally distance from a point and a line. So in this question the area is the intersection of the interiors of four parabolas. See the graph below.
    I will expand slightly on what Plato has said. Think about this:

    \sqrt{x^2 + y^2} = 1 - y \Rightarrow \, ....

    \sqrt{x^2 + y^2} = 1 - x \Rightarrow \, ....

    Since these parabolas are inverses of each other, they intersect on the line y = x \, ....

    By symmetry, the required area is four times the required area in one quadrant of the square.
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  8. #8
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    Quote Originally Posted by mr fantastic View Post
    By symmetry, the required area is four times the required area in one quadrant of the square.
    Or eight a fairly easy integral.
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