# Centroid of Triangle problem

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 $AB^2+AC^2+BC^2$.

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.
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 $3\cdot 41=123$.