Problems from Ukrainian Authors

Let $R$ be the projection of $B$ onto $AC$ (fig. 23). Then $P,Q,R$ are the projections of $B$ onto the sides of $△AOC$. Let $B_1$ be the isogonal conjugate of $B$ with respect to this triangle. Since $\angle BAC=\angle BOC=120^{\circ }$, we have $\angle B_{1}AO=\angle B_{1}OA$. Then the projection of $B_1$ onto $AC$ is $M$. Then the points $P,Q,R$, and $M$ are concyclic. Thus, it suffices to prove $\angle QRA=\angle QPR$. We will use the cyclicity of $ARBQ,OQBP,$ and $CPBR$. Indeed, $$\angle RPQ=\angle RPC-\angle QPC=\angle RBC-\angle QBO=(90^{\circ }-\angle BCA)-(90^{\circ }-\angle BOA)=\angle BCA,$$ $$\angle QRA=\angle ABQ=90^{\circ }-\angle BAO=\angle BCA,$$ completing the proof.

Fig. 23