Let $H$ and $O$ be the orthocenter and the circumcenter of the triangle $ABC$. Line $OH$ intersects the sides $AB, AC$ at points $X, Y$ correspondingly, so that $H$ belongs to the segment $OX$. It turned out that $XH = HO = OY$. Find $\angle BAC$.
(Proposed by Oleksii Masalitin)
Let $BD$, $CE$ be the altitudes and $M$, $N$ the midpoints of $AC$ and $AB$ respectively. Due to the Thales theorem $EH=ON/2$ and $DH=2OM$. At the same time it is known that $ON= CH/2$ and $OM=BH/2$. So $BH=HD$ and $CH=4HE$. Now we use the similarity of triangles $BHE$ and $CHD$ and find out that $BH/HE=CH/HD$. It means that $HE= BH/2$ and $\angle EBH=30^\circ$. Thus, $\angle BAC=60^\circ$.
By the way, the property $XH=OY$ is enough for the result. The argumentation could be the following.
Triangles $AXH$ and $AOY$ are equal because $XH=OY$, $\angle XAH=\angle YAO$ and these triangles have the same length of the altitude from $A$. Thus, $R=AH$. At the same time $R=\frac{a}{2\sin\alpha}$ and $AH=\frac{a}{\tan \alpha}$ and as a consequence $\cos \alpha=\frac{1}{2}$.
