55rd Ukrainian National Mathematical Olympiad - Fourth Round

Here is the solution for the location of points showed in the picture. For other cases, the location of the solution will be similar. Prove that for any selected point $D$ quadrilateral $XOYZ$ is inscribed. Then $\angle XZY=180^{\circ }-2\angle A$, so doesn't depend on $D$.

From the statement of the problem we have that circles $△BXD$ and $△DYC$ are tangent to lines $AB$ and $AC$. Denote these circles as $w_1$ and $w_2$, and their centers are $O_1$ and $O_2$. Let $K$ be the second point of intersection of these circles other than $D$ (see figure below).



Then $$\angle BKC=\angle BKD+\angle DKC=\angle B+\angle C=180^{\circ }-\angle A,$$ so $K$ is on the circumcircle $△ABC$. Also we have $$\angle XKY=\angle XKD+\angle DKY=\angle XBD+\angle DCY=180^{\circ }-\angle XOY,$$ so quadrilateral $XOYK$ is inscribed. Because $AZ$ is the diameter of the big circle, we have that $AB\perp BZ$, so $O_1$ is on line $BZ$. Similarly $O_2$ is on the line $CZ$. We prove that $O_1$, $O_2$ are on the circle $OXKY$. Indeed, $$\angle X{O}_{1}K=2\angle XBK=2(90^{\circ }-\frac{1}{2}\angle BOK)=180^{\circ }-\angle XOK,$$ Similarly for point $O_2$. Now we prove that point $Z$ is on circle $OXKY$, which will finish the solution.

For this purpose we prove that $\angle O_{1}ZC+\angle O_{1}K{O}_2=180^{\circ }$. Indeed, $$\angle O_{1}K{O}_2=\angle O_{1}KD+\angle DK{O}_2=90^{\circ }-\angle DBK+90^{\circ }-\angle DCK=180^{\circ }-\angle A=90^{\circ }-\angle BOC$$ $$\begin{array}{rl}& =\angle BZC=180^{\circ }-\angle O_{1}ZC\\ & \text{10th form}\\ & \text{10th form}\end{array}$$