Suppose in a triangle $ABC$ , A point $P$ exists such that the euler lines of $\triangle ABC$ and $\triangle PBC$ coincide. Suppose in a triangle $ABC$ , $AB=3$ and $AC=4$. A point $P$ exists other than $A$ and not lying on $BC$ such that the euler lines of $\triangle ABC$ and $\triangle PBC$ coincide.What can be the possible lengths of $BC$?
So, we can say they share the same centroid and we have $\frac{HG}{OG}=\frac{2}{1}$.
But what more can we infer?
 A: Observe first of all that the circumcenter of $PBC$ must lie both on the perpendicular bisector of $BC$ and on the Euler line of $ABC$: it follows that the circumcenter of $PBC$ is the same as the circumcenter $O$ of $ABC$, that is $P$ lies on the circumcircle of $ABC$.
On the other hand the centroid $G'$ of $PBC$ cannot be the same as the centroid $G$ of $ABC$, for that would entail $P=A$. If $M$ is the midpoint of $BC$, we have that $M$, $G'$, $P$ are aligned and $MP/MG'=MA/MG=3$. But $G'$ also belongs to the Euler line of $ABC$: it follows that $P$ lies on a line $r$, parallel to the Euler line of $ABC$ and passing through $A$.
Point $P$ is then the other intersection (different from $A$) of line $r$ with the circumcircle of $ABC$. This construction fails only if $r$ is tangent to the circumcircle of $ABC$, that is if line $OG$ is perpendicular to $AO$. One also has to discard the case when $OG$ is parallel to $AC$, for in that case $P=C$.
Those exceptions arise for some well defined values of $\theta=\angle BAC$. The length of $BC$ can then take any value between $1$ and $7$, with the exception of those corresponding to the forbidden values of $\theta$.
To find the excluded values of $\theta$ it is convenient to refer our points to a coordinate system. We can set for instance
$$
A=(0,0);\quad
C=(4,0);\quad
B=(3\cos\theta,3\sin\theta).
$$
From that we easily get
$$
G=\left(\cos\theta+{4\over3}, \sin\theta\right);\quad
O=\left(2,{3/2-2\cos\theta\over \sin\theta}\right).
$$
The condition $AO\perp OG$ gives two excluded values:
$$
\cos\theta={36\pm\sqrt{46}\over50},
$$
while the condition $OG\parallel AC$ gives only one:
$$
\cos\theta=1-{\sqrt{2}\over2}.
$$
From the cosine law it is then immediate to find the forbidden values of $BC$.

