Problems from Ukrainian Authors

Let $H$ and $O$ be the orthocentre and the centre of the circumcircle of $△ABC$ respectively, and $E$ be the centre of $\omega$. Then, as you know $E$ be the centre of $HO$. Let $M$ be the midpoint of $BC$. Let the circle $\Omega$ intersect $BC$ for the second time at point $X$. Since $\angle CAB=60^{\circ }$, then $\angle BHC=\angle BOC=120^{\circ }$, so points $B,H,O$ and $C$ are on the same circle (fig. 23). Let $T$ be the point of intersection of the lines $UV$ and $BC$, then, according to the degree of the point, $TD\cdot TX=TU\cdot TV=TB\cdot TC=TH\cdot TO$. Thus, the points lie on the same circle, and since $\angle HDX=90^{\circ }$, then $\angle HOX=90^{\circ }$, whence $XO\perp HO$. And remembering that $OU=OV$, we get that $XO$ bisects chord $UV$ of the circle $\Omega$, from where $XO$ passes through the centre of the circle $\Omega$, and hence $XO$ passes through the point diametrically opposite to the point $X$ on the circle $\Omega$. However $\angle XDA=90^{\circ }$, hence $Z$ is the point diametrically opposite to $X$ on the circle $\omega$. So $XO$ passes through $Z$. In the right $△BOM$, $\angle BOM=\frac{1}{2}\angle BOC=\angle BAC=60^{\circ }$, so

$AH=BO=AO$. Thus, the condition $\angle BAC=60^{\circ }$ implies that $AH=AO$, and therefore $AE\perp HO$, with $AE$ being the bisector $\angle HAO$, and hence $AE$ is also the bisector of $\angle CAB$. The condition $\angle BAC=60^{\circ }$ implies that the radius of the circle $\omega$ is $\frac{1}{2}AO=\frac{1}{2}AH=OM$. Since $\angle HOZ=90^{\circ }$ the condition $AH=AO$ implies that $A$ is the midpoint of the segment $HZ$. Denote by $W$ the point that is symmetric to $O$ with respect to $M$. Then $W$ is also symmetric to $O$ with respect to $BC$, since $OM\perp BC$, so $BW=WC$ and $\angle BWC=120^{\circ }$, whence $W$ is the midpoint of the smaller arc $BC$ of the circumcircle $△ABC$, and therefore $W$ lies on $AE$. Moreover, since $OA=OW$ and $OE\perp AW$, then $E$ is the midpoint of the segment $AW$. In the trapezoid $ZOWH$, $A$ and $M$ are the midpoints of the bases $ZH$ and $OW$, so, by a well-known fact, the point of intersection of its diagonals $ZW\cap HO$ lies on $AM$, but $AM\cap HO=G$ is the centroid of $△ABC$, so $G$ lies on $ZW$. Let us prove the collinearity of the points $Z,K$ and $W$, which implies that $G,K,Z$ and $W$ lie on the same line. To do this, let $K^{\prime }$ be the projection of point $X$ onto $ZW$. Moreover, since $OA=OW$, and $OE\perp AW$, $E$ is the midpoint of the segment $AW$. In the trapezoid $ZOWH$, $A$ and $M$ are the midpoints of the bases $ZH$ and $OW$, so, by a well-known fact, the point of intersection of its diagonals $ZW\cap HO$ lies on $AM$, but $AM\cap HO=G$ is the centroid of $△ABC$, so $G$ lies on $ZW$. Let us prove the collinearity of the points $Z,K$ and $W$, which implies that $G,K,Z$ and $W$ lie on the same line. To do this, let $K^{\prime }$ be the projection of point $X$ onto $ZW$. Then, since, $\angle X{K}^{\prime }Z=\angle XDZ=90^{\circ }$, the points $X,Z,D$ to $K^{\prime }$ are cyclic, and therefore $K^{\prime }$ lies on the circle $\Omega$. However $K^{\prime }$ lies on $ZW$, so to prove the collinearity of the points $Z,K$ and $W$, it is enough to show that $K^{\prime }$ lies on the circle $\omega$, which implies that $K^{\prime }$ is the second intersection of the circles $\Omega$ and $\omega$, i.e. $K^{\prime }=K$, and therefore $K$ like $K^{\prime }$ will lie on the line $ZW$. We have that, $\angle W{K}^{\prime }X=\angle WMX=90^{\circ }$, so the points $X,M,K^{\prime }$ and $W$ lie on the same circle, and hence $\angle M{K}^{\prime }X=\angle MWX=\angle MOX$ (this equality is obtained from the axial symmetry of points $O$ and $W$ with respect to line $BC$). However, the two lines $ZD$ and $OM$ are perpendicular to $BC$, so they are parallel, so $\angle MOX=\angle DZX$, i.e. $\angle M{K}^{\prime }X=\angle DZX$. Since the points $X,Z,K^{\prime }$ and $D$ are cyclic, then $\angle D{K}^{\prime }X=180^{\circ }-\angle DZX$. Therefore $$\angle D{K}^{\prime }M=\angle D{K}^{\prime }X-\angle M{K}^{\prime }X=180^{\circ }-\angle DZX-\angle DZX=180^{\circ }-2\angle DZX.$$

$$\angle D{K}^{\prime }M=180^{\circ }-2\angle DZX=180^{\circ }-2\angle DAE=180^{\circ }-\angle DAO.$$ Let $N$ be the midpoint of $AH$, then, as we know, $N$ lies on $\omega$, and $\angle MDN$ is right, so $MN$ is the diameter of the circle $\omega$, so the points $M,E$ and $N$ are collinear. Moreover, since $AN=\frac{1}{2}AH=OM$ and $AH\parallel OM$, then $AOMN$ is a parallelogram, hence $MN\parallel AO$, and therefore $\angle DAO=\angle DNM$. Thus, as we proved above, $\angle D{K}^{\prime }M=180^{\circ }-\angle DAO=180^{\circ }-\angle DNM$, and hence $\angle D{K}^{\prime }M+\angle DNM=180^{\circ }$, so the point $K^{\prime }$ lies on the circumcircle $\Delta DNM$ – $\omega$, which proves the collinearity of points $G,K,Z$ and $W$. In particular points $G,K$ and $W$ lie on the same line. Let $P$ be a point symmetric to $M$ with respect to $HO$. Then, since $A$ and $W$ are also symmetric with respect to $HO$, then $APMW$ is an isosceles trapezoid, so the point of intersection of its diagonals $PW\cap AM$ lies on $HO$. However $G=HO\cap AM$, so $T$ lies on $GW$, which passes through $K$. So $S$ is the second intersection of $KE$ and the circle $\omega$. Then $\angle KSM=\angle KTM$, and from the isosceles trapezoid $APMW$, $\angle KTM=\angle EAM$, so $\angle EAM=\angle ESM$, from which points $A,S,E$ and $M$ lie on the same circle, but $ES=EM$, from which $AE$ is the bisector of $\angle SAM$, that is $AS$ is the symmedian $\Delta ABC$, which was the proof.