Vectors - Prove the following relation about the centroid

AI Thread Summary
The discussion focuses on proving the relation PA² + PB² + PO² = GA² + GB² + GO² + 3(PG²) for a triangle OAB with centroid G. Participants suggest treating point P as the origin for a more symmetrical approach, which simplifies the proof. One user successfully derives a related equation involving the vectors of points A, B, and C. However, there is uncertainty regarding the angles between the vectors, which complicates the proof process. The conversation emphasizes the importance of starting with the more complex parts of the equation for simplification.
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Homework Statement


http://img147.imageshack.us/img147/733/vectorsba4.png
Given is the triangle OAB and a variable point P. G is the centroid. Prove that:
PA2+PB2+PO2=GA2+GB2+GO2+3 (PG2)


The Attempt at a Solution


I treat O as the origin.
Vectors are denoted in bold.
Position vectors of A, B, P are a,b and p respectively.
G=a+b/3
How on Earth am I going to prove that:
9(|p-a|2+|p-b|2+|p|2)=|b-2a|2+|a-2b|2+|a+b|2+3|a+b-3p|2

I mean is there any criteria to prove this equation in modulus of vectors when I don't know the angles between any of them?
 
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ritwik06 said:
Given is the triangle OAB and a variable point P. G is the centroid. Prove that:
PA2+PB2+PO2=GA2+GB2+GO2+3 (PG2)

I treat O as the origin.
Vectors are denoted in bold.
Position vectors of A, B, P are a,b and p respectively.
G=a+b/3

Hi ritwik06! :smile:

No, it'll be easier ('cos it's more symmetrical) if you treat P as the origin, and use g = (1/3)(a + b + c) :wink:

(when the answer is symmetrical, always try to keep the proof symmetrical! :wink:)
 


tiny-tim said:
Hi ritwik06! :smile:

No, it'll be easier ('cos it's more symmetrical) if you treat P as the origin, and use g = (1/3)(a + b + c) :wink:

(when the answer is symmetrical, always try to keep the proof symmetrical! :wink:)


Thanks a lot tim.
I have done exactly that. And I get:
a2+b2+c2=1/9(|a+b-2c|2+|b+c-2a|2+|a+c-2b|2+3|a+b+c|2)

But the fact still remains that I don't know many angle such as th one made by b+c-2a.?? I think it is 0.
 
No … start with the complicated part of the equation, and try to simplify it, not the other way round!

In other words, start with (g - a)2 + (g - b)2 + (g - c)2 :smile:
 
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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