# Proving a relation given centroid of a triangle

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?

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.

• 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?
– user203867
Mar 5 '16 at 0:56
• 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.
– user203867
Mar 5 '16 at 1:03
• Yes, that's right. Mar 5 '16 at 1:03
• I think for tetrahedron $A_1A_2A_3A_4$, we would have $$\sum |XA_i|^2 = \sum |GA_i|^2 + 4 | XG|^2$$
– user203867
Mar 5 '16 at 1:05
• what do you think ?
– user203867
Mar 5 '16 at 1:05