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Suppose we have a triangle $ABC$ with centroid $G$. Let $O$ be arbitrary point. Then, we have

$$ |OA|^2 + |OB|^2 + |OC|^2 = |GA|^2 + |GB|^2 + |GC|^2 + 3 |OG|^2 $$

My idea would be to put coordinates and write for instance $\vec{OA} = (a_1-O_1, a_2 - O_2)$ where $A = (a_1,a_2)$ and $O= (O_1,O_2)$ are points in the plane. So, by using dot product, we have

$$ | OA|^2 = \vec{OA} \cdot \vec{OA} = a_1^2 - 2 a_1 O_1 + O_1^2 + a_2^2 - 2 a_2 O_2 + O_2^2 $$

And we can compute also the other vectors as well, but then we will have a complicated equation. But, so far, am I on the right track?

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Note that, as vectors, $\overline{GA} +\overline{GB}+\overline{GC}=0$. I hope you know the following fact; if not, then look up a coordinate geometry approach to this kind of problem. Indeed, for every point X inside ABC, it is true that $\overline{XA} +\overline{XB}+\overline{XC} = 3\overline{XG}$. Now, the proof is trivial: just note that: $$ |XA|^2+|XB|^2+|XC|^2 = (XG+GA)^2 + (XG+GB)^2+ (XG+GC)^2 \\ = |GA|^2+|GB|^2 + |GC|^2 + 3|XG|^2 + XG.(GA+GB+GC) \\ = |GA|^2+|GB|^2 + |GC|^2 + 3|XG|^2 $$

You're done.

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  • $\begingroup$ Aston, thank you so much! Can we generalize this for the case of a tetrahedron? Say $ABCD$ is tetrahedron and $G$ its centroid. Can we have a similar result for this case? $\endgroup$
    – user203867
    Mar 5 '16 at 0:56
  • $\begingroup$ I understand. Also, I believe you have a small typo. $X$ does not need to be inside a triangle for the relations to work. Actually, for any $X$ we would have these relations. $\endgroup$
    – user203867
    Mar 5 '16 at 1:03
  • $\begingroup$ Yes, that's right. $\endgroup$ Mar 5 '16 at 1:03
  • $\begingroup$ I think for tetrahedron $A_1A_2A_3A_4$, we would have $$ \sum |XA_i|^2 = \sum |GA_i|^2 + 4 | XG|^2 $$ $\endgroup$
    – user203867
    Mar 5 '16 at 1:05
  • $\begingroup$ what do you think ? $\endgroup$
    – user203867
    Mar 5 '16 at 1:05

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