Problems from Ukrainian Authors

Let $O_1$ be the center of circle $w_1$ and $X' = BP \cap AQ$ as on Fig. 26. Since $AP \perp BP$ and $BQ \perp AQ$, then $AP$ and $BQ$ are heights of the triangle $ABX'$, and $X$ is the orthocenter. Then $X'X$ is also a height in the triangle, therefore $X'X \perp AB$ and $\angle PX'X = 90^\circ - \angle ABX' = \angle PAB$. This implies that triangles $\triangle APB$ and $\triangle X'PX$ have the same angles, so there is a rotation and homothety that transform triangle $\triangle APB$ into $\triangle X'PX$. The center of transformations is at $P$, the rotation is by $90^\circ$ and the homothety coefficient is $\frac{XP}{BP} = \angle XBP = \frac{1}{2} \angle QO_1P$.

Denote by $M_1$ the midpoint of $XX'$. The segments $PO_1$ and $PM_1$ are medians of the triangles $\triangle APB$ and $\triangle X'PX$ respectively, therefore $PM_1 \perp PO_1$, that is, $PM_1$ is tangent to $w_1$. Similarly, we get that $QM_1$ is tangent to $w_1$ as well. Consequently, $M_1$ is the intersection of tangents to $w_1$ drawn from $P$ and $Q$ that are not dependent on the choice of points $A$ and $B$. Note that the length of the segment

$$XX' = AB \cdot \frac{1}{2} \angle QO_1P$$
does not depend on $A$ and $B$ as well, because the length of diameter $AB$ and the angle $\angle QO_1P$ are constant.

Analogously, we construct the point $M_2$ for the circle $w_2$. Assume that $T = XY \cap M_1M_2$ (the case when $XY \parallel M_1M_2$ will be considered further). We will show that point $T$ is fixed, which implies that all the lines $XY$ intersect at one point regardless of the choice of diameters $AB$ and $CD$.

Note that $M_1X$ and $M_2Y$ are parallel since both are perpendicular to $AB \parallel CD$, that is why
$$\frac{M_1T}{M_2T} = \frac{M_1X}{M_2Y}.$$
The points $M_1$ and $M_2$ as well as the ratio $\frac{M_1X}{M_2Y} = \frac{\frac{XX'}{2}}{\frac{YY'}{2}} = \frac{XX'}{YY'}$ are fixed, therefore $T$ is also fixed.

The lines $XY$ and $M_1M_2$ are parallel and do not coincide if and only if segments $M_1X$ and $M_2Y$ are equal and parallel. These properties hold as well with different choices of the diameters $AB$ and $CD$. Hence, if $XY$ and $M_1M_2$ are parallel it will imply that $XY$ is parallel to $M_1M_2$ for all other initializations of diameters $AB$ and $CD$.

The lines $XY$ and $M_1M_2$ coincide if and only if $AB \parallel CD \perp M_1M_2$. In this case one may consider another initialization of the diameters $AB$ and $CD$, that will determine in which of two cases considered above we are. Note, that the initial line $XY = M_1M_2$ satisfies both of the considered cases.