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Math Help - Easy but very tricky geometry problem!

  1. #1
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    Easy but very tricky geometry problem!

    The image is self explanatory. The problem is really of geometry and not one in the probability area(since the probability matter of the problem is trivial).

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    Re: Easy but very tricky geometry problem!

    Surely there has to be some restriction on the way in which the insect moves
    Edit: nevermind, I missread the question.
    Last edited by Shakarri; August 14th 2013 at 03:43 PM.
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    Re: Easy but very tricky geometry problem!

    Quote Originally Posted by Shakarri View Post
    Surely there has to be some restriction on the way in which the insect moves
    Edit: nevermind, I missread the question.
    The probability is(not the solution but a hint)
    Spoiler:
    just the "area of ABCDEFG polygon" / "area of AHJK rectangle" so the real problem is to find these areas....
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    Re: Easy but very tricky geometry problem!

    Quote Originally Posted by ChessTal View Post
    The probability is(not the solution but a hint)
    Spoiler:
    just the "area of ABCDEFG polygon" / "area of AHJK rectangle" so the real problem is to find these areas....
    Spoiler:
    I thought that was the trick to it but you'd have to assume that the time is long enough for initial conditions to be unimportant (I'm thinking of an insect which moves at a finite speed between points, I suppose an insect that can jump between spots would be modeled as instantly moving between points).
    I still have no idea how to get the area with only 3 measurements or how the inequalities help
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  5. #5
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    Re: Easy but very tricky geometry problem!

    Quote Originally Posted by Shakarri View Post
    Spoiler:
    I thought that was the trick to it but you'd have to assume that the time is long enough for initial conditions to be unimportant (I'm thinking of an insect which moves at a finite speed between points, I suppose an insect that can jump between spots would be modeled as instantly moving between points).
    I still have no idea how to get the area with only 3 measurements or how the inequalities help
    Well:
    Spoiler:
    Well it is a very beautiful and relatively unknown feature of compass and straightedge that when you have a field that consists of straight lines and with one right angle(90) and by knowing the one side that forms the right angle then you can find its area no matter how complicated the shape is with just 1(yes one) measurement!!
    E.g for this:


    We can find its area by using compass and straightedge as many times as we want and with just one measurement with a ruler.

    Hint: For the construction/proof of that you will have to think that a trapezoid's diagonals create 2 equal area triangles.
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  6. #6
    Super Member ebaines's Avatar
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    Re: Easy but very tricky geometry problem!

    Similar to the previous post, I have a way to solve this with only two measurements, not three, and without using the 7cm length information. It involves breaking the polygon into five adjacent trapezoids, each with vertical sides, and using the base of the rectangle as its base, and with slanted top edges, and then combining two adjacent trapezoids into a single trapezoid of equal area. This combining of trapezoids requires use of the compass and straightedge - the process is to find the midpoint of the upper edge of both adjacent trapezoids and then draw a straight line through those two points, creating a single trapezoid with combined width of the original two trapezoids, which now has the same area as the original two. Repeat this process four times and you end up with a single trapezoid with base length equal to the original large rectangle. The only trick is that for the right-most trapezoid, which has the triangle area cut out of the bottom, you can move that triangle to an equivalent area at the top - simply copy its height using the compass to extend the upper portion down an equal amount. After all five sections have been converted into one large trapezoid the last step is to convert that trapezoid into a rectangle - simply find the midpoint of its top slanted edge and draw a horizontal line through it. So now the original polygon has been converted into a rectangle with width equal to the large rectangle. Use the ruler to measure the height of the inner rectangle that has been converted from the original polygon (call this height x) and the height of the large rectangle (call this y), and the probability of the bug being in the original polygon is the same as the probability of it being in the equivalent rectangle, which is x/y. So again - only two measurements are needed.
    Last edited by ebaines; August 17th 2013 at 04:48 AM.
    Thanks from Shakarri
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    Re: Easy but very tricky geometry problem!

    Quote Originally Posted by ebaines View Post
    Similar to the previous post, I have a way to solve this with only two measurements, not three, and without using the 7cm length information. It involves breaking the polygon into five adjacent trapezoids, each with vertical sides, and using the base of the rectangle as its base, and with slanted top edges, and then combining two adjacent trapezoids into a single trapezoid of equal area. This combining of trapezoids requires use of the compass and straightedge - the process is to find the midpoint of the upper edge of both adjacent trapezoids and then draw a straight line through those two points, creating a single trapezoid with combined width of the original two trapezoids, which now has the same area as the original two. Repeat this process four times and you end up with a single trapezoid with base length equal to the original large rectangle. The only trick is that for the right-most trapezoid, which has the triangle area cut out of the bottom, you can move that triangle to an equivalent area at the top - simply copy its height using the compass to extend the upper portion down an equal amount. After all five sections have been converted into one large trapezoid the last step is to convert that trapezoid into a rectangle - simply find the midpoint of its top slanted edge and draw a horizontal line through it. So now the original polygon has been converted into a rectangle with width equal to the large rectangle. Use the ruler to measure the height of the inner rectangle that has been converted from the original polygon (call this height x) and the height of the large rectangle (call this y), and the probability of the bug being in the original polygon is the same as the probability of it being in the equivalent rectangle, which is x/y. So again - only two measurements are needed.
    Interesting but can you elaborate more into how exactly to do what you are saying? I mean can you draw an image or be more specific on your description?
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  8. #8
    Super Member ebaines's Avatar
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    Re: Easy but very tricky geometry problem!

    I hope the attached helps explain it. I break the original polygon into 4 trapezoids and a triangle, labelled A-E. In each step I combine a trapezoid with its neighbor, keeping the same total area. After I get a trapezod for ABCD I convert it to a rectangle of same area. Then for triange E I first convert it to a right angle trianangle of equal area as E, and then combine with rectangle ABCD to create trapezoid ABCDE. Finally I turn that into a rectangle. The percentage of the area of the original polygon to the overall reactangle is then x/y.

    Attachment 29011
    Attached Thumbnails Attached Thumbnails Easy but very tricky geometry problem!-trapezoids.jpg  
    Last edited by ebaines; August 17th 2013 at 02:40 PM.
    Thanks from ChessTal
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