55rd Ukrainian National Mathematical Olympiad - Fourth Round

Let $M_1$ and $M_2$ be the middles of the diagonals (Fig. 26). Then $M_1{M}_2\parallel BC$. Consider circumcircles of $\angle AQC$ and $\angle BPD$. Let them intersect $BD$ and $AC$ in points $X$ and $Y$. Since $BD$ contains the bisector of $\angle AQC$, then point $X$ is the middle of the bigger arc $AC$ of the circumcircle of $\angle AQC$. Then $X$ is on the perpendicular bisector of $AC$. In the same way, $Y$ is on the perpendicular bisector of $BD$. So $X{M}_1{M}_{2}Y$ is inscribed, as $\angle X{M}_{1}Y=\angle X{M}_{2}Y=90^{\circ }$. Then $\angle M_{1}X{M}_2=\angle M_{1}Y{M}_2$. Also $\angle BXY=\angle C{M}_1{M}_2=\angle M_{1}CB$, so $XBCY$ is inscribed. Hence, $\angle BXC=\angle BYC$.

$$\begin{gathered} \angle AQC=180^{\circ }-2\angle M_{1}XC=180^{\circ }-2(\angle M_{1}X{M}_2+\angle M_{2}XC)= \\ =180^{\circ }-2(\angle M_{1}Y{M}_2+\angle M_{1}YB)=180^{\circ }-2\angle M_{2}YB=\angle BPD. \end{gathered}$$

Alternative solution.

Denote on $BD$ point $Q^{\prime }$ such that $\angle B{Q}^{\prime }C=\angle BPC$. Then $BC{Q}^{\prime }P$ is inscribed (Fig. 27). Hence $\angle Q^{\prime }PC=\angle Q^{\prime }BC=\angle Q^{\prime }DA$, so $AP{Q}^{\prime }D$ is also inscribed. Then $\angle CPD=\angle CP{Q}^{\prime }+\angle Q^{\prime }PD=\angle Q^{\prime }DA+\angle Q^{\prime }AD=\angle B{Q}^{\prime }A.$ So $\angle BPD=\angle C{Q}^{\prime }A$. Also $BD$ bisects $\angle C{Q}^{\prime }A$. Let us show that $Q^{\prime }=Q$. Assume these points are distinct. Because of the fact that