Problem - 4273
Let $O$ and $H$ be the circumcenter and orthocenter of $\triangle{ABC}$ respectively. Show that $OH\parallel BC$ if and only if $\tan{B}\tan{C}=3$.
Let $M$ be the middle point of $BC$ and $D$ be the foot of altitude from $A$.
Let $R$ be the circumradius of $\triangle{ABC}$. Then $R=OC$. Meanwhile, $$\angle{MOC}=\frac{1}{2}\cdot\angle{BOC}=\angle{A}\implies OM= R\cos{A} = R(\cos{B}\cos{C}-\sin{B}\sin{C})$$
Meanwhile, by Laws of Sines, we have $$AD=AC\sin{C} = 2R\sin{B}\sin{C}$$
Now $$OH\parallel BC\Leftrightarrow AD=3OM \Leftrightarrow 2R\sin{B}\sin{C} = 3 R(\cos{B}\cos{C}-\sin{B}\sin{C})$$
Cancelling $R$ from the last relation and dividing both sides by $\cos{B}\cos{C}$ will give the desired result immediate.