If $x,y,z$ are perpendicular distances from circumcenter on the sides $BC,AC$ and $AB$ respectively. In need find $k$ such that
$$\frac ax+\frac by+\frac cz=\frac{abc}{kxyz}$$
(Lowercase letters denote opposite sides like $BC=a$)
I have added up area of all small triangles and couldn't make it through
Note that $a=2R\sin A$ and $x=R\cos A$, so that $\dfrac ax=2\tan A$. Using the identity $\tan A+\tan B+\tan C=\tan A\tan B\tan C$, $\boxed{k=4}$.