Austrian Mathematical Olympiad

Let $M$ and $O$ denote the circumcenters of the triangles $DEF$ and $ABC$, respectively. We will use the well-known facts that the points $H$, $M$, $S$ and $O$ lie on the Euler line of triangle $ABC$ in this order, and that $HM:MS:SO=3:1:2$.

Since $S$ is in the interior of $ABC$, it is inside the circumcircle of $ABC$. However, since $ABC$ is obtuse, the orthocenter $H$ is outside the circumcircle. This implies that the circle with diameter $HS$ intersects the circumcircle of $ABC$ in two points.

Let $X$ be one of these intersection points. We will prove that $MX:OX=1:2$. Let $N$ be the midpoint of $HS$. Using Thales' theorem, we get that $HN$, $XN$ and $SN$ have the same length, and $HN:NM:MS:SO=2:1:1:2$. This implies that $MN:XN=XN:ON$.

Therefore, the triangles $XNM$ and $ONX$ are similar with ratio $1:2=MN:XN$. Therefore, we have $MX:OX=1:2$ as desired.

Figure 4: Figure for Problem 4