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Math Help - any more possible arrangement of AC>AB and BD<BC

  1. #1
    rcs
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    any more possible arrangement of AC>AB and BD<BC

    Can anybody help me on this problem please.

    For any points ABCD are arranged along a line such that AC>AB and BD<BC. Draw a picture with the four points in plane. Is there more than one order possible?


    Thanks a lot
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    Have you tried drawing a line with A, B, C such that AB < AC and then trying to insert D so that BD < BC? I know it's just a restatement of the problem, but I don't see any difficulty in this problem other than trying a few cases.
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    Quote Originally Posted by rcs View Post
    For any points ABCD are arranged along a line such that AC>AB and BD<BC. Draw a picture with the four points in plane. Is there more than one order possible?
    You need to explain the notation. What does AB stand for?
    Is it simply the length of line segment \overline{AB}?
    Or is it a signed length? If we arrange A-B-C, B is between A & C, then using just length we have AB<AC.

    Do you see what we need to know about the notation?
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    rcs
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    i think it is about the segment sir
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    If it is simply length of a line segment then A-B-D-C will give AB<AC~\&~BD<BC.
    Can you find another?
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    rcs
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    DC<DA ... is that correct sir
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    Quote Originally Posted by rcs View Post
    DC<DA ... is that correct sir
    Why do you think that works?
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    rcs
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    DC is shorter than DA ... am i correct sir ?
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    Hello, rcs!

    \text{Points }A,B,C,D\text{ are arranged along a line}
    . . \text{such that: } AC>AB\text{ and }BD<BC.

    \text{Is there more than one order possible?}

    Using emakarov's suggestion, I found 12.
    The following six and their mirror-images.

    A - B - D - C

    A - D - B - - \,C

    D - A - B - - - - \,C

    B - A - D - C

    B - D - A - - - - \,C

    D - B - A - - \,C


    Did I miss any?

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  10. #10
    rcs
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    oh my God! u mean their mirror images are the opposite arrangement sir soroban?
    like A-B-D-C its mirror is C-D-B-A ? am i right sir?

    thanks alot
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    Quote Originally Posted by Soroban View Post
    [size=3]Using emakarov's suggestion, I found 12.
    The following six and their mirror-images.
    A - B - D - C
    A - D - B - - \,C
    This is precisely why I asked for clarification as to exactly what is involved here. I assumed that this question has to do with ordering collinear points using some form of Hilbertís betweenness axioms. If that is correct then there is only one possible order: A-B-D-C.
    There is no notion of a mirror image in the axioms.
    Short of something like Ed Moiseís ruler axiom spacing makes no difference.

    To be clear on what this means, given P-Q-R we know that PQ<PR.
    BUT there is no way to compare PQ~\&~QR, without knowing the full context of the question.
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  12. #12
    rcs
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    sir plato, what do mirror images meant for?
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  13. #13
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    Quote Originally Posted by rcs View Post
    what do mirror images meant for?
    @rcs
    Unless you post the exact statement of this problem along with all definitions and relevant axioms, it is pointless to continue this thread.
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