If $AG^2 + BG^2 + CG^2 = 41$, then find $AB^2 + AC^2 + BC^2$.

Question

Let $G$ denote the centroid of triangle $ABC$. If $AG^2 + BG^2 + CG^2 = 41$, then find $AB^2 + AC^2 + BC^2$.
I don't know how to start. I feel like the Leibniz Relation (like the first identity) is similar, but I don't know.
So, if you drew it out, AG, BG, and CG are the lines that make the centroid. How is that related to AB, BC, and AC?

Answer 1

By the parallel axis theorem, the minimum of the moment of inertia $PA^2+PB^2+PC^2$ is attained at $P=G$. By computing such moment of inertia at $P=A$, $P=B$, $P=C$ and exploiting the PAT together with $G=\frac{A+B+C}{3}$, we have $$ AB^2+AC^2+BC^2=\color{red}{3}(AG^2+BG^2+CG^2).$$ This is also a straighforward consequence of Stewart's theorem for the length of the medians, or just the polarization identity.

Answer 2

1)$$ AG^2+BG^2+CG^2=\left(\frac23AE\right)^2+\left(\frac23BF\right)^2+\left(\frac23CD\right)^2=\frac49\left(AE^2+BF^2+CD^2\right)$$ 2) $$AE^2=\frac{2AB^2+2AC^2-BC^2}{4}$$ $$BF^2=\frac{2BA^2+2BC^2-AC^2}{4}$$ $$CD^2=\frac{2CA^2+2CB^2-AB^2}{4}$$ Then $$AE^2+BF^2+CD^2=\frac{3}4\left(AB^2+BC^2+CD^2\right)$$

3) Then $$AG^2+BG^2+CG^2=\frac49\cdot\frac34\left(AB^2+BC^2+CD^2\right)=\frac13\left(AB^2+BC^2+CD^2\right)$$

Answer:

$$AB^2+BC^2+CD^2=3\cdot41=123$$

Answer 3

HINT.-$$AG^2+BG^2-2AG\cdot BG\cos\angle{BGA}=AB^2\\BG^2+CG^2-2BG\cdot CG\cos\angle{BGC}=BC^2\\AG^2+CG^2-2AG\cdot CG\cos\angle{AGC}=AC^2$$

Use now the quite known formulas $$AG=2GE\\CG=2GD\\BG=2GF$$ to eliminate the angles whose sum is of $360^{\circ}$.