Romanian Mathematical Olympiad

We will prove that $YF$ is the Euler line of the triangle $ABX$, hence it contains the centroid of that triangle.

Because $BFA$ is a right isosceles triangle, it follows that $BF=AF$. In the right triangle $BFC$ we have $\angle CBF=30^{\circ }$, hence $FC=\frac{BC}{2}$, thus $FC=CX$, and a short computation shows that $\angle BXF=30^{\circ }$. We deduce that $BF=FX$, hence $AF=BF=FX$, making $F$ the circumcenter of $ABX$.

Let $Y^{\prime }$ be the orthocenter of $ABX$. Applying Sylvester's relation yields $\overline{F{Y}^{\prime }}=\overline{FA}+\overline{FB}+\overline{FX}=\overline{FB}+\overline{BY}=\overline{FY}$. It follows that $Y^{\prime }=Y$, hence $Y$ is the orthocenter of $ABX$. We conclude that $FY$ is the Euler line of $ABX$ and since the centroid lies between the circumcenter and the orthocenter, we obtain the desired result.