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I tried an approach using vectors to solve this problem, but I wasn't able to find the answer. Any help would be great, thanks for all of your help in advance!

Problem: Let $G$ denote the centroid of triangle $ABC$. If $ AG^2+BG^2+CG^2 = 41 $ then find $ AB^2+AC^2+BC^2 $.

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By an identity your desired sum is three times the given sum. So your answer is $3 \cdot 41 = 123$.

To prove the identity, I suppose you would use the fact that the distance from a vertex to the centroid is $2/3$ the length of the corresponding median. You can use analytic geometry to look at the triangle with vertices $(0,0)$, $(p,0)$, and $(q,r)$. You can easily find the coordinates of the midpoints of the sides and of the centroid, so the lengths are easy to compare.

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  • $\begingroup$ Oh, thanks! I wasn't aware of that identity, thanks for linking to it. $\endgroup$
    – math-sd
    Sep 14 '14 at 22:42
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The Stewart's theorem gives that the squared length for a median is given by: $$ m_a^2 = \frac{2b^2+2c^2-a^2}{4}$$ and $AG=\frac{2}{3}m_a$, so: $$AG^2+BG^2+CG^2 = \frac{4}{9}\cdot\frac{3}{4}(AB^2+AC^2+BC^2)$$ and the answer is $3\cdot 41=123$.

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