Problem
GEO-B2-M09-P012 Euler Line with Numbers
In triangle \(A(0,0)\), \(B(6,0)\), \(C(2,4)\), find the circumcenter \(O\), the orthocenter \(H\), and the centroid \(G\). Prove that \(O,G,H\) are collinear and \(OG:GH=1:2\).
First find \(O\) as the intersection of perpendicular bisectors. For \(H\), use altitudes.
The perpendicular bisector of \(AB\) has equation \(x=3\). From \(OA=OC\), we get \(9+y^2=(3-2)^2+(y-4)^2\), so \(y=1\). Thus \(O(3,1)\). By the orthocenter formula with \(p=6,q=2,r=4\), \(H(2,2)\). The centroid is \(G\left(\frac{0+6+2}{3},\frac{0+0+4}{3}\right)=\left(\frac83,\frac43\right)\). Vector \(\overrightarrow{OH}=(-1,1)\), and \(\overrightarrow{OG}=\left(-\frac13,\frac13\right)=\frac13\overrightarrow{OH}\). Therefore the points are collinear and \(OG:GH=1:2\).
A numerical preparation for the general Euler line problem.