Balkan Mathematical Olympiad

We first prove the following lemma. Lemma. Let $XYZT$ be a cyclic quadrilateral such that $XZ\perp YT$. Let $XZ\cap YT=O$, and let $V$ and $W$ be the orthogonal projections of $O$ to lines $XY$ and $ZT$, respectively. If $I$ lies in the middle of $YZ$, and $J$ lies in the middle of $XT$, then $IVJW$ is a cyclic deltoid. Proof. Let $Y^{\prime }$ and $Z^{\prime }$ be the points in the middle of $OY$ and $OZ$ respectively. Then $\overline{V{Y}^{\prime }}=\overline{O{Y}^{\prime }}=\overline{I{Z}^{\prime }}$ and $\overline{I{Y}^{\prime }}=\overline{O{Z}^{\prime }}=\overline{W{Z}^{\prime }}$, and also

$\angle V{Y}^{\prime }I=\angle V{Y}^{\prime }O+90^{\circ }=2\angle WZO+90^{\circ }=\angle W{Z}^{\prime }O+90^{\circ }=\angle I{Z}^{\prime }W.$ Hence, we have $\angle V{Y}^{\prime }I\cong \angle I{Z}^{\prime }W$, so $\overline{VI}=\overline{WI}$, and $$\angle VIW=90^{\circ }-\angle VI{Y}^{\prime }-\angle Z^{\prime }IW=90^{\circ }-\angle VI{Y}^{\prime }-\angle IV{Y}^{\prime }=\angle V{Y}^{\prime }O=2\angle XYT.$$ Analogously, we can show $\angle VJW=2\angle YXZ$, and therefore $$\angle VIW+\angle VJW=180^{\circ }.$$ The lemma is proved.

Let $Q{C}_1\cap H_{1}B=M_1$. As $\angle M_1{C}_{1}B=\angle AC{C}_1=\angle AB{H}_1$ (the last equality follows from the fact that $ACB{H}_1$ is cyclic), we have that $M_1$ lies in the middle of $B{H}_1$. Using the lemma, the quadrilayteral $PMR{M}_1$ is a cyclic deltoid, and the center of its circumscribed circle lies in the middle of $M{M}_1$. Hence, $Q$ also has to be on that circle, as $\angle MQ{M}_1=90^{\circ }$, and the middle of $M{M}_1$ is the circumcentrer of the triangle $PQR$.