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Coordinate geometry - centroid (SL LONEY exercise problem)

  1. Mar 4, 2017 #1
    1. The problem statement, all variables and given/known data

    If G be the centroid of ΔABC and O be any other point, prove that ,
    ## 3(GA^2 + GB^2 + GC^2)=BC^2+CA^2+AB^2##
    ##and,##
    ##OA^2 + OB^2 + OC^2 = GA^2.GB^2+GC^2+3GO^2##

    2. Relevant equations

    i m practising from S L LONEY coordinate geometry first chapter ... only the equation that i used to solve is mentioned in the chapter...

    3. The attempt at a solution

    i felt that co ordinate geometry approach will be cumbersome ...
    so i started with
    ##GA^2 + GB^2 = AB^2+2.GA.GB.cos∠AGB##
    and similar 2 more equations
    But i am stuck here !! :(

    And for the 2nd question i cant even get a start .....

    Please help me with a hint ...
     
  2. jcsd
  3. Mar 4, 2017 #2

    haruspex

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    If you have a triangle ABC and drop a median from A to A', the midpoint of BC, what is the relationship between the lengths AB, AC, AA' and BA'(=CA')?
     
  4. Mar 4, 2017 #3
    I can't find any sir...... : ( ...definitely , ΔABA' and ΔACA' are not similar .....
    are you referring to the same cosine rule that i used ?
     
  5. Mar 4, 2017 #4

    haruspex

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    No. It is a result I dimly recall from my own schooldays, so I thought you may have encountered it in your book.
    Drop a perpendicular from A to BC meeting it at P. Call AP length h and PA' length x. What three equations can you write relating these to AB, AC etc. using Pythagoras' Theorem?
     
  6. Mar 4, 2017 #5
    triangle.png

    BP2 + h2 = AB2
    PC2 + h2 = AC2
    x2 + h2 = AA'2
     
    Last edited: Mar 4, 2017
  7. Mar 4, 2017 #6

    haruspex

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    There's a mistake in the first one, probably just typed wrongly.
    You need to get rid of PC. What is the relationship between PC, x and A'C?
    Next manipulate the equations to eliminate x2 and h2.
     
  8. Mar 4, 2017 #7
    Pick a triangle of verticies ##(0,0), (a,0), (x,y)## then ##G = \left({a + x \over 3}, {y\over 3}\right)
    ##. Now you evaluate LHS of first equation and then RHS. It is not that cumbersome.

    For the second draw the figure and use cosine rule.
     
    Last edited: Mar 4, 2017
  9. Mar 4, 2017 #8
    upload_2017-3-5_3-5-57.png

    Use cosine rule on the vertex G of the red triangle. I am not sure it will work though.
     
  10. Mar 7, 2017 #9
    that was optimistic choice of vertices. I solved it that way (by comparing coefficients) and yes, not too cumbersome. thanx :)

    I cant see any mistake in the first one sir.. : ( ...and by changing PC = A'C - x , from last two equations i am able to get an expression for x (a big one) and then using that value to get expression for h seems even a time consuming approach .(Buffu's solution seems better).... And i cant see were this leads to the actual problem ....Or, were you thinking in some other way ?
     
  11. Mar 10, 2017 #10
    matrixone, how did you do the second one ? I tried but am stuck on cosines.
     
  12. Mar 10, 2017 #11

    haruspex

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    Sorry, I thought I replied to this.... No, you are right, I misread something.
     
  13. Mar 10, 2017 #12
    I can solve this but my method is too primitive to do.
     
  14. Mar 10, 2017 #13

    haruspex

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    No, you should get quite a simple relationship between AB, AC, A'B and AA'. Please post your working.
    It is a well-known theorem.

    Once you have that, the original problem is quite easy.
     
  15. Mar 10, 2017 #14
    Buffu, what primitive method you are talking ?
     
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