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Vectors and points on a circle

  1. Apr 8, 2013 #1
    1. The problem statement, all variables and given/known data
    Let A, B, C, D be distinct points on a circle with centre O. If there exists non zero real numbers x and y such that ##|x\vec{OA}+y\vec{OB}|=|x\vec{OB}+y\vec{OC}|=|x\vec{OC}+y\vec{OD}|=|x \vec{OD}+y\vec{OA}|##, then which of the following is always true?
    A)ABCD is a trapezium
    B)ABCD is a rectangle
    C)ABCD is a rhombus
    D)ABCD is a square


    2. Relevant equations



    3. The attempt at a solution
    Honestly, I have no idea how to begin with this problem. Should I start by squaring the moduli? I don't know if that would help.

    Any help is appreciated. Thanks!
     
  2. jcsd
  3. Apr 8, 2013 #2
  4. Apr 8, 2013 #3

    ehild

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    Pranav, which statement is certainly true????


    ehild
     
  5. Apr 8, 2013 #4
    If I start by assuming x=y=1, it comes out that its a square. But how would I prove it for a general case?
     
  6. Apr 9, 2013 #5

    haruspex

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    I find it easiest to approach this geometrically.
    All that matters wrt x and y is the ratio. If yOA, yOB, yOC, yOD lie on a circle radius y around O, where do xOA etc. lie? Now, cheating a bit, let's take x as negative and look at the locations of |x|OA etc., so xOA+yOA = yOA-|x|OA, allowing us to look at the line joining the two points instead of a vector sum. Look at the triangle formed. What formula can you think of for |yOA-(|x|)OA|?
     
  7. Apr 9, 2013 #6

    haruspex

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    I find it easiest to approach this geometrically.
    All that matters wrt x and y is the ratio. If yOA, yOB, yOC, yOD lie on a circle radius y around O, where do xOA etc. lie? Now, cheating a bit, let's take x as negative and look at the locations of |x|OA etc., so xOA+yOA = yOA-|x|OA, allowing us to look at the line joining the two points instead of a vector sum. Look at the triangle formed. What formula can you think of for |yOA-(|x|)OA|?
     
  8. Apr 10, 2013 #7

    ehild

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    The magnitude of all vectors OA,OB,OC,OD are 1. Let the angles between two of them α,β,γ,δ.
    Find the magnitude of xOA+yOB, .... in terms of x,y,and the corresponding angles applying the law of cosines.

    ehild
     

    Attached Files:

  9. Apr 10, 2013 #8
    Sorry for the late reply.

    That's an excellent hint, ehild! Thanks a lot! :smile:

    It can be done without assuming the radius to be one.
    Referring to your attachment,
    [tex]|x\vec{OA}+y\vec{OB}|=x^2|\vec{OA}|^2+y^2|\vec{OB}|^2+2xy|\vec{OA}|| \vec{OB} |\cos \alpha[/tex]
    Similarly we can obtain expressions for ##|x\vec{OB}+y\vec{OC}|##, ##|x\vec{OC}+y\vec{OD}|## and ##|x\vec{OD}+y\vec{OA}|## and using the fact that ##|\vec{OA}|=|\vec{OB}|=|\vec{OC}|=|\vec{OD}|##, I get
    [tex]\cos \alpha=\cos \beta=\cos \gamma=\cos \delta[/tex]
    [tex]\Rightarrow \alpha=\beta=\gamma=\delta=\frac{\pi}{2}[/tex]

    This proves that these points form a square.

    Thanks ehild!
     
  10. Apr 10, 2013 #9

    ehild

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    You have to work a bit more on it. Two angles are not necessarily the same if their cosines are the same.
    And about the questions. Is it not true that ABCD is a rectangle? Or trapezium? Rhombus?

    ehild
     
  11. Apr 10, 2013 #10
    Yes, you are right, I need to work more on it.

    It is not rhombus because in a rhombus, the diagonals are not equal. In trapezium too, the diagonals are not equal.

    How would I show that it isn't a rectangle? :confused:

    I am not sure how would I do it but here's my take on it. Assuming that one of the angles between the diagonals is ##\alpha## and its cosine is ##\cos \alpha##, the other angle is ##\pi-\alpha## and its cosine ##-\cos \alpha## but this is contrary to the relation of cosines being equal, hence this isn't a rectangle.

    Is this correct?
     

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  12. Apr 10, 2013 #11

    haruspex

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    In my understanding of these terms, every square is also a rhombus and a rectangle and a trapezium. They are not mutually exclusive.
     
  13. Apr 11, 2013 #12

    ehild

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    and a quadrilateral, with a=b, and 90° angle between the diagonals.

    ehild
     
  14. Apr 11, 2013 #13
    Is my proof still incomplete?
     
  15. Apr 11, 2013 #14

    haruspex

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    I think the issue of cos α = cos β still needs some work. What are the possible solutions of that for each being in the range 0 to 2 pi?
     
  16. Apr 11, 2013 #15
    ##\beta=±\alpha## ?
     
  17. Apr 11, 2013 #16

    haruspex

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    That wouldn't allow both to be in the range 0 to 2 pi.
     
  18. Apr 11, 2013 #17
    This would be silly but is ##\alpha=\pi+\beta## or ##\alpha=\pi-\beta##?
     
  19. Apr 13, 2013 #18

    ehild

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    Well, if one of the angles is α1=(π-β), the other α2=π+β, and β≠π/2, what could be the other angles with the same cosine?


    ehild.
     
  20. Apr 13, 2013 #19

    ehild

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    Or β=2π-α...


    ehild
     
  21. Apr 13, 2013 #20
    Thanks but how should I complete my proof now?
     
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