In a triangle $ABC$, we denote $O$ and $G$ as the circumcenter and the centroid.

1) Identify the point $X$ such that $\overrightarrow{O X}=\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}$.
2) Deduce that if $H$ is the orthocenter of triangle $ABC$, the relation $\overrightarrow{O H}=3 \overrightarrow{O G}$ holds.
1) By the Chasles relation we have $\overrightarrow{A X}=\overrightarrow{O B}+\overrightarrow{O C}$. If we denote $A^{\prime}$ as the midpoint of $[B C]$, we also have $\overrightarrow{O B}+\overrightarrow{O C}=\overrightarrow{O A^{\prime}}+\overrightarrow{A^{\prime} B}+\overrightarrow{O A^{\prime}}+\overrightarrow{A^{\prime} C}=2 \overrightarrow{O A^{\prime}}$ because $\overrightarrow{A^{\prime} B}+\overrightarrow{A^{\prime} C}=\overrightarrow{0}$. We know that $\left(O A^{\prime}\right)$ is the perpendicular bisector of $[B C]$, so we have

$$
\overrightarrow{A X} \cdot \overrightarrow{B C}=2 \overrightarrow{O A^{\prime}} \cdot \overrightarrow{B C}=0
$$

Therefore, $X$ is on the altitude from $A$. Similarly, $X$ is on the altitudes from $B$ and $C$: $X$ is thus the orthocenter $H$ of triangle $A B C$.

2) From the previous question, we have $\overrightarrow{O H}=\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}$. By the Chasles relation, we have

$$
\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}=3 \overrightarrow{O G}+\overrightarrow{G A}+\overrightarrow{G B}+\overrightarrow{G C}
$$

From Exercise 1, we know that $\overrightarrow{G A}+\overrightarrow{G B}+\overrightarrow{G C}=\overrightarrow{0}$. We then obtain the desired equality.