in a triangle $ABC, $ line joining the circumcenter and orthocenter is parallel to the side $AC,$
then relation between $\tan A, \tan B,\tan C$ is
as i know that $C$ (Circumcenter,) $G$ (Centroid,) $H$(Orthocenter) in $1:2$
could some help me to solve it, thanks
Let $O,H$ be the circumcenter, the orthocenter respectively.
From this answer, we have that $$\vec{OH}=\frac{\tan A\vec{OA}+\tan B\vec{OB}+\tan C\vec{OC}}{\tan A+\tan B+\tan C}$$ and that $$\vec{OO}=\vec 0=\frac{\sin2A\vec{OA}+\sin2B\vec{OB}+\sin2C\vec{OC}}{\sin2A+\sin2B+\sin2C}\iff \vec{OC}=-\frac{\sin2A}{\sin2C}\vec{OA}-\frac{\sin2B}{\sin2C}\vec{OB}$$ Here, let $$a=-\frac{\sin2A}{\sin2C},\qquad b=-\frac{\sin2B}{\sin 2C}$$
Now $OH$ is parallel to $AC$ iff there exists a real number $k$ such that $$\frac{\tan A\vec{OA}+\tan B\vec{OB}+\tan C(a\vec{OA}+b\vec{OB})}{\tan A+\tan B+\tan C}=k(a\vec{OA}+b\vec{OB}-\vec{OA})$$ to have $$\frac{\tan A+a\tan C}{\tan A+\tan B+\tan C}=k(a-1)\quad\text{and}\quad\frac{\tan B+b\tan C}{\tan A+\tan B+\tan C}=kb$$ Eliminating $k$ gives $$\frac{\tan A+a\tan C}{(a-1)(\tan A+\tan B+\tan C)}=\frac{\tan B+b\tan C}{b(\tan A+\tan B+\tan C)},$$ i.e. $$b(\tan A+\tan C)=(a-1)\tan B,$$ i.e. $$-\frac{\sin2B}{\sin2C}(\tan A+\tan C)=\left(-\frac{\sin2A}{\sin2C}-1\right)\tan B,$$ i.e. $$\frac{\sin2B}{\sin 2A+\sin2C}=\frac{\tan B}{\tan A+\tan C}\tag1$$
Here, note that $$\begin{align}\frac{\sin2B}{\sin 2A+\sin2C}&=\frac{2\cos B\sin B}{2\sin(A+C)\cos(A-C)}\\\\&=\frac{\cos(180^\circ-(A+C))}{\cos(A-C)}\\\\&=\frac{-\cos A\cos C+\sin A\sin C}{\cos A\cos C+\sin A\sin C}\\\\&=\frac{-1+\tan A\tan C}{1+\tan A\tan C}\tag1\end{align}$$ and that $$\begin{align}\tan A+\tan C&=\tan(A+C)(1-\tan A\tan C)\\\\&=-\tan B\ (1-\tan A\tan C)\tag3\end{align}$$
From $(1)(2)(3)$, we get $$\color{red}{\tan A\tan C=3}$$
