Track / Stage 5 / 207 of 400 #807 of 1964

Problem 807

AIME late

Geometry Difficulty 5.4 Prove it

Prove: The circumcenter $O$ , centroid $G$ , and orthocenter $H$ of $\triangle ABC$ are collinear, and $O G: G H=$ $1: 2$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

14. $\overrightarrow{O H}=\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}$ , and $\overrightarrow{O G}=\frac{1}{3}(\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C})=\frac{1}{3} \overrightarrow{O H}=\frac{1}{3}(\overrightarrow{O G}+\overrightarrow{G H})$ , thus $\overrightarrow{O G}=\frac{1}{2} \overrightarrow{G H}$ . Therefore, $O$ , $G$ , $H$ are collinear, and $|O G|:|G H|=1: 2$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.