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Math Help - Problem Solving (Paper)

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    Problem Solving (Paper)

    A rectangular sheet of paper is folded so that two diagonally opposite corners come together. If the crease formed is the same length as the longer side of the sheet, what is the ratio of the longer side of the sheet to the shorter side?
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    Quote Originally Posted by matgrl View Post
    A rectangular sheet of paper is folded so that two diagonally opposite corners come together. If the crease formed is the same length as the longer side of the sheet, what is the ratio of the longer side of the sheet to the shorter side?
    To be honest I don't think this is possible.
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    It is possible somehow..... i just dont know how
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    Please recheck the problem

    Quote Originally Posted by matgrl View Post
    It is possible somehow..... i just dont know how
    Crease is the hypotenuse and the two sides are the two legs of the triangle. The hypotenuse is always the longest side of any (right) triangle which is the case here.
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    MHF Contributor Unknown008's Avatar
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    If I understood well, this should be possible.

    If you fold a square diagonally, the crease (the line formed upon folding) is 1.41 times (that is square root of 2) the length of the square. And using a rectangle with a very long length as compared to the width makes the crease shorter and shorter.

    Except that I don't think that I can solve this problem... since wonderboy mentioned proof by induction in your other thread (which I don't know), I don't think that this problem is within my reach...



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    this is wonderful thank you
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    Quote Originally Posted by Unknown008 View Post
    If I understood well, this should be possible.

    If you fold a square diagonally, the crease (the line formed upon folding) is 1.41 times (that is square root of 2) the length of the square. And using a rectangle with a very long length as compared to the width makes the crease shorter and shorter.

    Except that I don't think that I can solve this problem... since wonderboy mentioned proof by induction in your other thread (which I don't know), I don't think that this problem is within my reach...



    The key here is that matgrl said, "...two diagonally opposite corners..." meaning the longest diagonal making two congruent right triangles. In this case the diagonal is longer than either leg so it's length can't equal either leg.
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    MHF Contributor Unknown008's Avatar
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    "...two diagonally opposite corners..."

    Yes, that's what I did. Let the vertices of the rectangle starting from te upper left, going clockwise be A, B, C and D.

    The points A and C will becomes toegther, as per my third diagram.

    "meaning the longest diagonal making two congruent right triangles."

    No, I don't think that this is what the question is looking for, or else, there is no solution as you yourself said.
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    Quote Originally Posted by Unknown008 View Post
    "...two diagonally opposite corners..."

    Yes, that's what I did. Let the vertices of the rectangle starting from te upper left, going clockwise be A, B, C and D.

    The points A and C will becomes toegther, as per my third diagram.

    "meaning the longest diagonal making two congruent right triangles."

    No, I don't think that this is what the question is looking for, or else, there is no solution as you yourself said.
    The dashed line in your second diagram doesn't cross the corners as the problem called for. That's why I suggested the way matgrl has the problem worded, there's no solution.
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    MHF Contributor Unknown008's Avatar
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    Well, I'm misunderstand the problem then... What I understand by it is that the rectangle is folded, so that two diagonally opposite corners (A and C in my previous explanation) meet at a point (or corners B and D meet at a point).

    I don't see how you see that the fold must pass through two diagonally opposite corners... the corners won't meet that way... =/
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    I'm sure this is what matgrl means

    Quote Originally Posted by Unknown008 View Post
    Well, I'm misunderstand the problem then... What I understand by it is that the rectangle is folded, so that two diagonally opposite corners (A and C in my previous explanation) meet at a point (or corners B and D meet at a point).

    I don't see how you see that the fold must pass through two diagonally opposite corners... the corners won't meet that way... =/
    Yes they do meet, just like you were going to diaper a baby by folding the cloth in half along the major diagonal and pinning the ends along the diagonal at the corners.
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    MHF Contributor Unknown008's Avatar
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    But... won't the corners be like this then?

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    Quote Originally Posted by Unknown008 View Post
    But... won't the corners be like this then?

    I was thinking about a square, sorry about that. I'm wondering what matgrl means precisely?
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    MHF Contributor Unknown008's Avatar
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    Yes, I mentioned a square because the crease is longer then, and gets shorter as the ratio of the length to the width increases, meaning that there must be a solution for a particular ratio.
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    My philosophy

    With more challenging problems, my belief is to start from what you already know that's most related to the problem to arrive at a solution so my suggestion to matgrl is to first look at the diagonal of the right triangle and see how it relates to the crease to see if you can derive a solution.

    Good luck to you.
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