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Math Help - help Geometry E.C.

  1. #1
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    help Geometry E.C.


    please help
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  2. #2
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    background

    my teacher said that they were relatively simple problems and that if i couldn't get them, i was thinking too hard
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  3. #3
    Grand Panjandrum
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    For the second problem the area of overlap is independent of the
    relative orientation of the two squares. So orient them so that finding
    the area of overlap is easy, and that area is the area of overlap for
    any other orientation.

    Below is the hint text I produced when this was the Geometry Problem
    of the week on The Math Forum at Drexel (green square is the darker
    square in your diagram, and the blue the lighter square:

    Also I have now found the diagram that goes with this problem and I
    have attached it.

    =================================================

    I hope you will find the following of some help
    in solving this problem:

    Hint text for "Squares on Squares"

    1. The first thing you could do with this problem
    is to produce a labelled diagram. So copy the
    diagram from the problem and follow the instructions
    below

    2. Draw a horizontal line from P to the right
    hand edge of the green square, and label the
    point where the line meets the square u. Also
    label the point on the right hand edge of the
    green square above u, where it meets the edge
    of the blue square v.

    3. Draw a vertical line from P to the top
    edge of the green square and label the
    point where the line meets the square w.
    Also label the point on the top edge of the
    green square to the left of w, where it meets
    the edge of the blue square x.

    4. Now look at triangles Puv and Pwx. Do they
    have the same area? If so, can you explain why
    the areas are the same. What does this tell
    us about the area of the overlap?

    RonL
    Attached Thumbnails Attached Thumbnails help Geometry E.C.-squares.jpg  
    Last edited by CaptainBlack; January 25th 2006 at 05:10 AM.
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  4. #4
    Grand Panjandrum
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    Problem of the Week 25

    Part 1.

    With N lines, each line cuts N-1 other lines, which apparently
    gives us N.(N-1) points, but we have counted each point twice
    in this argument so there are N.(N-1)/2 points of intersection.

    Reality check:

    Two lines cut each other at 1 point, and 2*1/2=1 so that's OK

    Three lines cut each other at 3 points and 3*2/2=3 so that's OK

    Four lines cut each other at 6 points, and 4*3/2=6 so that's OK

    when N=10 our formula gives 10*9/2=45 points of intersection.

    RonL
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  5. #5
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    Quote Originally Posted by kkirk909
    my teacher said that they were relatively simple problems and that if i couldn't get them, i was thinking too hard
    Hello,

    seems to me as if your teacher likes to make some jokes.
    I've thought about your 2nd post:
    When the square on top is situated as I've drawn in my 1st drawing you have exactly a quarter of the square as overlayed area.
    If the suare on top rotates a little bit further on you only have to show, that the triangle T1 which you "cut" from quarter of the square has the same value for the area as the triangle T2 (use cogruent rule(?) ASA on both triangles.

    That means the overlayed area doesn't change it's value for the area: It has always a quarter of the square which means 64 square inches.

    Bye
    Attached Thumbnails Attached Thumbnails help Geometry E.C.-qudfl1.gif   help Geometry E.C.-qudfl2.gif  
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  6. #6
    Grand Panjandrum
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    Problem of the Week 25

    Part 2.

    Suppose that we know how many regions R(n-1) that n-1 lines satisfying
    the conditions of the problem devide the plane into.

    Now add a new line starting from a long way away, as it comes in it divides
    the region that it was being drawn through into 2 parts when it meets the
    first of the existing lines, thus increasing the region count by 1.

    Similarly when it reaches the second line it adds one more region to the
    region count, and so on untill it has cut the last of the existing lines. At this
    point the region count has increased by n-1. As the line
    moves off towards infinity it divides the last region into 2.

    Thus R(n)=R(n-1)+n, which we may use to evaluate R(10),
    or we can expand this and find the closed form for R(n-1) .

    R(1)=2,
    R(2)=2+2=4,
    R(3)=4+3=7,
    R(4)=7+4=11,
    R(5)=11+5=16,
    R(6)=16+6=22,
    R(7)=22+7=29,
    R(8)=29+8=37,
    R(9)=37+9=46,
    R(10)=46+10=56.

    RonL
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