Problems from Ukrainian Authors

Let $H$ be the orthocenter of $\triangle ABC$, $A'$ be the antipode of $A$ in $(ABC)$, $S$ be the point, which is symmetric to $M$ with respect to $O$ (fig. 21), and $T$ be the intersection of $BC$ and $EF$. It's well-known that in this construction $H$ is the orthocenter of $\triangle AMT$, $BHCA'$ is parallelogram (then $M$ is the midpoint of $A'H$) and $AH = 2OM$.

As $SM = 2OM = AH$ and $AH \perp BC \perp SM$, we can obtain that $AHMS$ is parallelogram. Hence, $S$ is the antipode of $T$ in $(AMT)$ and $\angle AMH = 90^\circ - \angle MAT = 90^\circ - \angle MST = \angle STM$. As $OM \perp BC$ and $AO \perp EF$, we obtain that $\angle A'OM = \angle STM$. So, $\angle A'OM = \angle AMH$, hence $\triangle AA'M \sim \triangle AMO$ and then $AA' \cdot AO = AM^2 \Rightarrow \frac{AM}{AO} = \sqrt{2}$.