Coordinate geometry - centroid (SL LONEY exercise problem)

AI Thread Summary
The discussion revolves around proving two equations related to the centroid of triangle ABC using coordinate geometry. Participants express difficulty in solving the problems, particularly in applying the cosine rule and manipulating equations. Suggestions include using Pythagorean theorem relationships and evaluating the left-hand side of the first equation with specific triangle vertices. One participant eventually finds a simpler method to solve the problems after initial confusion. The conversation highlights the challenges and strategies in tackling coordinate geometry problems involving centroids.
matrixone
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Homework Statement



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##

Homework 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...

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 can't even get a start ...

Please help me with a hint ...
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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')?
 
I can't find any sir... : ( ...definitely , ΔABA' and ΔACA' are not similar ...
are you referring to the same cosine rule that i used ?
 
matrixone said:
I can't find any sir... : ( ...definitely , ΔABA' and ΔACA' are not similar ...
are you referring to the same cosine rule that i used ?
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?
 
haruspex said:
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?
triangle.png


BP2 + h2 = AB2
PC2 + h2 = AC2
x2 + h2 = AA'2
 
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matrixone said:
BP2 + h2 = AB2
PC2 + h2 = AC2
x2 + h2 = AA'2
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.
 
matrixone said:
i felt that co ordinate geometry approach will be cumbersome ...
so i started with
GA2+GB2=AB2+2.GA.GB.cos∠AGBGA2+GB2=AB2+2.GA.GB.cos∠AGBGA^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 can't even get a start ...

Please help me with a hint ...

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.
 
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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.
 
Buffu said:
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.

that was optimistic choice of vertices. I solved it that way (by comparing coefficients) and yes, not too cumbersome. thanks :)

haruspex said:
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.

I can't 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 can't see were this leads to the actual problem ...Or, were you thinking in some other way ?
 
  • #10
matrixone, how did you do the second one ? I tried but am stuck on cosines.
 
  • #11
matrixone said:
I can't see any mistake in the first one sir
Sorry, I thought I replied to this... No, you are right, I misread something.
 
  • #12
IssacNewton said:
matrixone, how did you do the second one ? I tried but am stuck on cosines.

I can solve this but my method is too primitive to do.
 
  • #13
matrixone said:
from last two equations i am able to get an expression for x (a big one
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.
 
  • #14
Buffu, what primitive method you are talking ?
 
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