67th NMO Selection Tests for BMO and IMO

Letting $DF$ and $DG$ meet $\gamma$ again at $R$ and $S$, respectively, we claim that $R$ and $S$ both lie on the line $H^{\prime }{I}^{\prime }$.





Notice that $\angle H^{\prime }CQ=\angle SDC=\angle SRC$ and $\angle QC{I}^{\prime }=\angle CDR=\angle CSR$ to deduce that $(C{H}^{\prime }Q,RCQ)$ and $(C{I}^{\prime }Q,SCQ)$ are pairs of similar triangles, so $Q{H}^{\prime }\cdot QR=Q{C}^2=Q{I}^{\prime }\cdot QS$. Finally, invert from $Q$ with radius $QC$ (Figure 2) and notice that $R,C$ and $S$ are mapped to $H^{\prime },C$ and $I^{\prime }$, respectively, to infer that $\gamma$, the circumcircle of the triangle $RCS$, is mapped to $\omega$, the circumcircle of the triangle $H^{\prime }C{I}^{\prime }$. Since $P$ and $C$ both lie on either circle, and the latter is fixed under the inversion, so is the former. Consequently, $Q{P}^2=Q{C}^2$, i.e., $QP=QC$.