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Math Help - average number of lines that touch a certain area

  1. #1
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    average number of lines that touch a certain area

    I'm thinking of the following problem. I'd appreciate if you share your thought about my solution,
    or correct it. There is no answer for the problem.

    Problem: There is a square with size of a^2 on 2 dimensional spaces. We imagine lines that touch the square. If we describe the line with x (length) and t (angle from horizon), then each line for the parameters x and t is associated with the probability f(x,t).

    Question is, what is the average number of lines that touch the square? Range of x : from b to c. Range of t: from 0 to 2*pi.


    I'm thinking following solution:
    For the fixed x and t, if the line should touch the square, one end of the line should be in the polygon with the size {sqrt(2)*a*x + a^2}.

    Thus the answer (for every x and t) is:
    int int_R {sqrt(2)*a*x +a^2}* f(x,t) dA ( int: integral )
    = int_b^c int_0^(2*pi) {(sqrt(2)*a*x +a^2}* f(x,t)* x dt dx.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by nicola View Post
    I'm thinking of the following problem. I'd appreciate if you share your thought about my solution,
    or correct it. There is no answer for the problem.

    Problem: There is a square with size of a^2 on 2 dimensional spaces. We imagine lines that touch the square. If we describe the line with x (length) and t (angle from horizon), then each line for the parameters x and t is associated with the probability f(x,t).

    Question is, what is the average number of lines that touch the square? Range of x : from b to c. Range of t: from 0 to 2*pi.


    I'm thinking following solution:
    For the fixed x and t, if the line should touch the square, one end of the line should be in the polygon with the size {sqrt(2)*a*x + a^2}.

    Thus the answer (for every x and t) is:
    int int_R {sqrt(2)*a*x +a^2}* f(x,t) dA ( int: integral )
    = int_b^c int_0^(2*pi) {(sqrt(2)*a*x +a^2}* f(x,t)* x dt dx.
    I think you need to be clearer about what you mean by touch the square, do you mean that one end of the line segment is on the boundary of the square, or that the line segment has some part of itself on or inside the square, ...

    CB
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  3. #3
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    Quote Originally Posted by CaptainBlack View Post
    I think you need to be clearer about what you mean by touch the square, do you mean that one end of the line segment is on the boundary of the square, or that the line segment has some part of itself on or inside the square, ...

    CB
    Thanks for the clarification.

    I meant that the line segment has some part of itself on or inside the square.
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by nicola View Post
    I'm thinking of the following problem. I'd appreciate if you share your thought about my solution,
    or correct it. There is no answer for the problem.

    Problem: There is a square with size of a^2 on 2 dimensional spaces. We imagine lines that touch the square. If we describe the line with x (length) and t (angle from horizon), then each line for the parameters x and t is associated with the probability f(x,t).

    Question is, what is the average number of lines that touch the square? Range of x : from b to c. Range of t: from 0 to 2*pi.


    I'm thinking following solution:
    For the fixed x and t, if the line should touch the square, one end of the line should be in the polygon with the size {sqrt(2)*a*x + a^2}.

    Thus the answer (for every x and t) is:
    int int_R {sqrt(2)*a*x +a^2}* f(x,t) dA ( int: integral )
    = int_b^c int_0^(2*pi) {(sqrt(2)*a*x +a^2}* f(x,t)* x dt dx.
    Without some constraint or distribution for the position of one end of the line the probability will be zero as there will be some finite region over which there is a non-zero probability of the line segment beeting a given square, but the rest of the plane is unbounded and so the required probability will be zero.

    CB
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  5. #5
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    Quote Originally Posted by CaptainBlack View Post
    Without some constraint or distribution for the position of one end of the line the probability will be zero as there will be some finite region over which there is a non-zero probability of the line segment beeting a given square, but the rest of the plane is unbounded and so the required probability will be zero.

    CB
    Sorry, I forgot to specify the constraint, and made the confusion. For simplicity, it can be assumed that ... for each unit area, N end points are placed, randomly uniformly. The probability a line (with x and t) exists follows the f(x,t) function.
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