Problems from Ukrainian Authors

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

Denote by $M_1$ the midpoint of $X{X}^{\prime }$. The segments $P{O}_1$ and $P{M}_1$ are medians of the triangles $△APB$ and $△X^{\prime }PX$ respectively, therefore $P{M}_1\perp P{O}_1$, that is, $P{M}_1$ is tangent to $w_1$. Similarly, we get that $Q{M}_1$ 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

$$X{X}^{\prime }=AB\cdot \frac{1}{2}\angle Q{O}_{1}P$$ does not depend on $A$ and $B$ as well, because the length of diameter $AB$ and the angle $\angle Q{O}_{1}P$ are constant.

Analogously, we construct the point $M_2$ for the circle $w_2$. Assume that $T=XY\cap M_1{M}_2$ (the case when $XY\parallel M_1{M}_2$ 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 $M_{1}X$ and $M_{2}Y$ are parallel since both are perpendicular to $AB\parallel CD$, that is why $$\frac{M_{1}T}{M_{2}T}=\frac{M_{1}X}{M_{2}Y}.$$ The points $M_1$ and $M_2$ as well as the ratio $\frac{M_{1}X}{M_{2}Y}=\frac{\frac{X{X}^{\prime }}{2}}{\frac{Y{Y}^{\prime }}{2}}=\frac{X{X}^{\prime }}{Y{Y}^{\prime }}$ are fixed, therefore $T$ is also fixed.

The lines $XY$ and $M_1{M}_2$ are parallel and do not coincide if and only if segments $M_{1}X$ and $M_{2}Y$ are equal and parallel. These properties hold as well with different choices of the diameters $AB$ and $CD$. Hence, if $XY$ and $M_1{M}_2$ are parallel it will imply that $XY$ is parallel to $M_1{M}_2$ for all other initializations of diameters $AB$ and $CD$.

The lines $XY$ and $M_1{M}_2$ coincide if and only if $AB\parallel CD\perp M_1{M}_2$. 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 $XY=M_1{M}_2$ satisfies both of the considered cases.