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 $P^{\prime }$, such that $OE\cdot O{P}^{\prime }=O{A}^2$. Let $\angle AOB=2\alpha$, $\angle BOC=2\beta$, $\angle COD=2\gamma$, $\angle OEA=x$. Then the triangles $O{P}^{\prime }B$, $OEB$ are similar by the angle and two adjacent sides $\Rightarrow \angle OEB=\angle OB{P}^{\prime }$. The triangles $OAE$, $OA{P}^{\prime }$ are also similar $\Rightarrow \angle OEA=\angle OA{P}^{\prime }$. Hence the points $O$, $A$, $B$, $P^{\prime }$ lie on the same circle. Analogously, the points $D$, $C$, $P^{\prime }$ lie on the same circle since $\angle OED=\angle OD{P}^{\prime }=\angle OC{P}^{\prime }$. Therefore, we have: $$\angle P^{\prime }BC=\angle OBC-\angle OB{P}^{\prime }=90^{\circ }-\beta -x;\angle OED=\angle AED-\angle AEO=180^{\circ }-2\beta -\gamma -\alpha -x;$$ $$\angle EOC=\angle OCD-\angle OED=90^{\circ }-\gamma -(180^{\circ }-2\beta -\gamma -\alpha -x)=2\beta +\alpha +x-90^{\circ };$$ $$\angle P^{\prime }DA=\angle EDA-\angle ED{P}^{\prime }=\angle EDA-\angle EOC=\alpha +\beta -(2\beta +\alpha +x-90^{\circ })=90^{\circ }-\beta -x.$$

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

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 $OM\cdot O{M}_1=O{A}^2$.

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