IMO Team Selection Contest

Figure 32

As $OQ$ is the perpendicular bisector of $AD$, one has $\angle QAD=\angle ADQ=\angle BDC=\angle CBD$ (Fig. 32). Therefore $AQ\parallel BC$, whence $\angle QAC=180^{\circ }-\angle ACB=90^{\circ }$. From the cyclic quadrilateral $APCQ$ one also gets $\angle CPQ=\angle PQA=90^{\circ }$, i.e., $ACPQ$ is a rectangle.

As $\angle DOC=2\angle DAC$, one obtains $$\begin{array}{rl}\angle DOC& =2\angle BAC=2(90^{\circ }-\angle CBA)=180^{\circ }-2\angle CBA=\\ & =180^{\circ }-\angle CBD-\angle BDC=\angle DCB,\end{array}$$ which implies that isosceles triangles $BDC$ and $DCO$ are similar. Thus $\angle BDC=\angle DCO$, i.e., $OC\parallel AB$, whence $\angle QOC=90^{\circ }=\angle QAC$. So $O$ lies on the circle determined by $A$, $C$, $P$, $Q$. Therefore $$\angle ACQ=\angle AOQ=\frac{1}{2}\angle AOD=\frac{1}{2}\angle AOP=\frac{1}{2}\angle ACP.$$ Consequently, the diagonal of the rectangle $ACPQ$ bisects the angle of the rectangle, whence $ACPQ$ is a square.