59th Ukrainian National Mathematical Olympiad

Let us denote by $P$ intersection point of $CF$ and $BE$. Then, $\angle CBE=60^{\circ }-\angle ABE=\angle FCA$. Analogously, $\angle FCB=\angle ABE$ (fig. 27). Then, $\angle FKE=\angle CKB=180^{\circ }-\angle CBE-\angle FCB=120^{\circ }$.

Thus, quadrilateral $AFPE$ is inscribed in circle $\gamma$, which point $D$ lies on. We draw a bisector of $\angle CAB$ until its intersection with $\gamma$ at some point $O$. Obviously, point $O$ is the middle of the smaller arc $EF$ of circle $\gamma$. Let us prove that $O$ is the center of equilateral $△ABC$. $△AFC=△BEC$ by the side and two adjacent angles. Hence, $FC=BE$. Naturally, $\angle OAFC=\angle BEO$ and $OF=OE$. Thus, $△OFC=△BEO$ by two pairs of sides and an angle between them, which yields $\angle PBO=\angle PCO$, i.e. quadrilateral $BPOC$ is inscribed, hence, $\angle BOC=120^{\circ }$. Therefore, point $O$ lies on the bisector of $\angle CAB$ in the same half-plane as point $P$ with respect to line $BC$, and $\angle BOC=120^{\circ }$. Obviously, then point $O$ is the center of $△ABC$.

Now, we prove that $O$ is the incenter of $△DXY$. Clearly, then this center does not depend on the points $E$ and $F$. Obviously, $\angle YDO=\angle XDO=30^{\circ }$, i.e. $DO$ is a bisector of $\angle XDY$. Suffices to show that $YO$ is a bisector of $\angle DYX$. $\angle YDO=\angle OBY=30^{\circ }$, hence, quadrilateral $DOYB$ is inscribed, and $DO=OB$, as radii. Therefore, $$\angle OYX=\angle ODB=\angle DBO=\angle DYO,$$ i.e. $YO$ is a bisector of $\angle DYX$.