Problems from Ukrainian Authors

Let $H$ be the orthocenter of $△ABC$, $A^{\prime }$ 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 $△AMT$, $BHC{A}^{\prime }$ is parallelogram (then $M$ is the midpoint of $A^{\prime }H$) 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 $\angle A^{\prime }OM=\angle STM$. So, $\angle A^{\prime }OM=\angle AMH$, hence $△A{A}^{\prime }M\sim △AMO$ and then $A{A}^{\prime }\cdot AO=A{M}^2\Rightarrow \frac{AM}{AO}=\sqrt{2}$.