50th Mathematical Olympiad in Ukraine, Fourth Round (March 23, 2010)

Denote by $X$ the second point of intersection of the circumcircles of $△ADQ$ and $BCQ$, $QE$ the ray which has the same direction as ray $CB$ (Fig.12).

Then $\angle EQB=\angle QBC$, $\angle EQA=\angle QAD$ $\Rightarrow \angle AQB=\angle QAD+\angle QBC=180^{\circ }-\angle QXD+180^{\circ }-\angle QXC=\angle CXD$. Moreover, $\angle AQB=\angle CPD$, because triangles $AQB$ and $CPD$ are symmetric with respect to the center of our parallelogram. Hence, points $D$, $P$, $X$, $C$ are cyclic. By analogy, $A$, $P$, $X$, $B$ also lie on the same circle, so all 4 circles pass through the common point $X$.

It is worth noting that this proof is valid for the specific configuration of given points only. For a complete solution, one has to consider at least three more configurations. However, it is possible to handle all these cases simultaneously using oriented angles. Indeed: $\angle (QB;QA)=\angle (QB;BC)+\angle (BC;QA)=\angle (QB;BC)+\angle (AD;QA)=\angle (QX;XC)+\angle (DX;XQ)=\angle (DX;XC)$. The rest coincides with what we have done before, because $\angle (QB;QA)=\angle (PD;PC)$ ($AQB$ and $CPD$ are symmetric with respect to the center of parallelogram). Hence, $D$, $P$, $X$, $C$ are cyclic as well as $A$, $P$, $X$, $B$. Therefore, all circles pass through $X$, and we are done.

Fig.12