Problem
GEO-B3-M05-P001 Euler Line
In triangle \(ABC\), let \(O\), \(G\), \(H\) be the circumcenter, centroid, and orthocenter. Prove that \(O,G,H\) are collinear and \(OG:GH=1:2\).
C. Hint 1. Take \(O\) as the origin.
D. Hint 2. First prove that \(\vec h=\vec a+\vec b+\vec c\) is the orthocenter.
Let the position vectors of \(A,B,C\) be \(\vec a,\vec b,\vec c\), with \(|\vec a|=|\vec b|=|\vec c|=R\). Set \(\vec h=\vec a+\vec b+\vec c\). Then \((\vec h-\vec a)\cdot(\vec b-\vec c)=(\vec b+\vec c)\cdot(\vec b-\vec c)=0\), so \(AH\perp BC\). Similarly \(BH\perp CA\) and \(CH\perp AB\), hence \(H\) is the orthocenter.
The centroid has position vector \(\vec g=\frac{\vec a+\vec b+\vec c}{3}=\frac{\vec h}{3}\). Thus \(O,G,H\) are collinear, and from \(OG=\frac13 OH\), \(GH=\frac23 OH\) we get \(OG:GH=1:2\).
A core fact for all nine-point circle problems.