I Dare You To Take A Special Case! Part 1

For Euler line , $OH \parallel CB$ ,
$AH = 2\times OM = 2\times HP$ ,
$HP = \dfrac13 AP$ ,
$\tan B = \dfrac{AP}{PB}$ ,
$\angle C = 90^\circ - \angle HBP$ .

$\begin{aligned} \tan C &=& \cot (\angle HBP ) \\ &=& \dfrac{PB}{HP} \\ &=& \dfrac{AP / \tan B}{AP /3} \\ &=&\dfrac3{\tan B} \end{aligned}$ .

$\Rightarrow \tan B \tan C = \boxed3$ .

Solution using complex bash:

Let $B=R\\C=Re^{2iA}\\A=Re^{2i(A+B)}$ where $R$ is the circumradius of $\Delta ABC$ ,origin is the circumcentre and $A$ , $B$ and $C$ are used to denote vertices (in the argand plane) in the LHS and angles in the RHS. To see how this is true, draw a circle centred at origin and use central angle theorem .

Then, the centroid, $G=\dfrac{R}{3}(1+e^{2iA}+e^{2i(A+B)})$ $BC\parallel OG\iff \frac{G}{C-B} \in \mathbb{R}\\\iff 2\cos A+\cos (A+2B)=0\\\left(\text{on simplifying}\frac{G}{C-B}\text{and equating imaginary part to zero}\right)\\\iff 2\cos (B+C)+\cos (B-C)=0\\(\because A+B+C=\pi)\\\iff 3\cos B\cos C=\sin B\sin C\\ (\text{on expanding using addition formula})\\\iff\tan B\tan C=3$

$\implies \quad BC\parallel OG\iff \tan B\tan C=3$

Let $A=(1, 0,0),B=(0,1,0),C=(0,0,1)$ in areal coordinates.

The orthocentre and centroid have coordinates $H=(tan(A), tan(B), tan(C)), G=(1, 1,1)$

The line $BC$ has equation $x+0y+0z=0$ so all lines parallel to this have equation $(1+s)x+sy+sz=0$ for some $s$

The line GH (Euler Line) has equation $(tan(B) - tan(C)) x+(tan(C) - tan(A)) y+(tan(A) - tan(B)) z+0$

So $s+1=tan(B)-tan(C),s=tan(C)-tan(A)=tan(A)-tan(B)$ for some $s$ .

Adding together the second two equations we get $tan(C) - tan(B) =2s \Rightarrow - 2s=s+1 \Rightarrow s=-\frac{1}{3}$

Let $tan(A) =x$ giving $tan(B) =x+\frac{1}{3},tan(C)=x-\frac{1}{3}$

Using $tan(A) =tan(180-(B+C))=-tan(B+C)$ we get:

$x=-\frac{x+\frac{1}{3}+x-\frac{1}{3}}{1-(x+\frac{1}{3})(x-\frac{1}{3})} \Rightarrow x=\frac{2x}{x^2-\frac{10}{9}} \Rightarrow x^3-\frac{28}{9} x=0$

But $x=tan(A) \neq 0$ so $x^2=\frac{28}{9}$

$tan(B)tan(C)=(x+\frac{1}{3})(x-\frac{1}{3}) =x^2-\frac{1}{9}=\frac{28}{9}-\frac{1}{9}=3$