Problems of Ukrainian Authors

Let $E$, $F$ be the points of intersection of the lines $AB$, $CD$ and $AD$, $BC$ respectively, and $O$ be the center of the circumscribed circle of the quadrilateral $ABCD$ (Fig. 50). On the half-line $OE$ we choose the point $P'$, such that $OE \cdot OP' = OA^2$. Let $\angle AOB = 2\alpha$, $\angle BOC = 2\beta$, $\angle COD = 2\gamma$, $\angle OEA = x$. Then the triangles $OP'B$, $OEB$ are similar by the angle and two adjacent sides $\Rightarrow \angle OEB = \angle OBP'$. The triangles $OAE$, $OAP'$ are also similar $\Rightarrow \angle OEA = \angle OAP'$. Hence the points $O$, $A$, $B$, $P'$ lie on the same circle. Analogously, the points $D$, $C$, $P'$ lie on the same circle since $\angle OED = \angle ODP' = \angle OCP'$. Therefore, we have:
$$\angle P'BC = \angle OBC - \angle OBP' = 90^\circ - \beta - x; \quad \angle OED = \angle AED - \angle AEO = 180^\circ - 2\beta - \gamma - \alpha - x;$$
$$\angle EOC = \angle OCD - \angle OED = 90^\circ - \gamma - (180^\circ - 2\beta - \gamma - \alpha - x) = 2\beta + \alpha + x - 90^\circ;$$
$$\angle P'DA = \angle EDA - \angle EDP' = \angle EDA - \angle EOC = \alpha + \beta - (2\beta + \alpha + x - 90^\circ) = 90^\circ - \beta - x.$$

Similarly, $\angle P'CB = \angle P'AD$. So, the points $A$, $P'$, $C$, $F$ lie on the same circle, and the points $B$, $P'$, $D$, $F$ lie on the same circle. The same is true for the point $P' \Rightarrow P = P'$.

Let $w$ be the circle that touches the lines $AB$, $CD$ at points $M$, $N$ respectively and touches the circumscribed circle of the triangle $OAB$ at a point $M_1$. By a known lemma the points $M$, $M_1$, $O$ are collinear and $OM \cdot OM_1 = OA^2$.

Apply the inversion with the center $O$ and radius $OA$ (Fig. 51). Then the circumscribed circles of the triangles $OAB$, $OCD$ will be mapped to the lines $AB$, $CD$ respectively, the point $M$ will be mapped to the point $M_1$, and so the circle $w$ will be mapped to $w$ itself.

So $w$ touches the image of the line $CD$, that is the circumscribed circle of the triangle $OCD$. Thus $w$ is the circle that we were looking for.

Fig. 50

Fig. 51