NMO Selection Tests for the Balkan and International Mathematical Olympiads

Let the lines $CX$ and $C^{\prime }{X}^{\prime }$ meet at point $Q$. The line $CX$ meets $\ell$ at $D$, and $\gamma$ a second time at $Y$; similarly, the line $C^{\prime }{X}^{\prime }$ meets $\ell$ at $D^{\prime }$, and ${\gamma }^{\prime }$ a second time at $Y^{\prime }$. Notice that the cross-ratios $(CDXY)$ and $(C^{\prime }{D}^{\prime }{X}^{\prime }{Y}^{\prime })$ are both harmonic, for $C$ and $C^{\prime }$ are the poles of $\ell$ relative to $\gamma$ and ${\gamma }^{\prime }$, respectively. Since the lines $C{C}^{\prime }$, $D{D}^{\prime }$ and $X{X}^{\prime }$ all pass through $P$, so does the line $Y{Y}^{\prime }$. Consequently, $(QCXD)=(Q{C}^{\prime }{X}^{\prime }{D}^{\prime })$ and $(QCYD)=(Q{C}^{\prime }{Y}^{\prime }{D}^{\prime })$. Let $\varrho$ and ${\varrho }^{\prime }$ be the powers of $Q$ relative to $\gamma$ and ${\gamma }^{\prime }$, respectively, let $\omega$ be the power of $C$ relative to $\gamma$, and let ${\omega }^{\prime }$ be the power of $C^{\prime }$ relative to ${\gamma }^{\prime }$. Multiply the last two equalities involving cross-ratios and apply Menelaus' theorem to triangle $QC{C}^{\prime }$ and transversal $PD{D}^{\prime }$ to get $$\frac{\varrho }{{\varrho }^{\prime }}=\frac{\omega }{{\omega }^{\prime }}{\left(\frac{C^{\prime }{D}^{\prime }}{Q{D}^{\prime }}\cdot \frac{QD}{CD}\right)}^2=\frac{\omega }{{\omega }^{\prime }}{\left(\frac{P{C}^{\prime }}{PC}\right)}^2=\text{constant}$$ and infer that $Q$ lies on a fixed circle of the pencil of circles generated by $\gamma$ and ${\gamma }^{\prime }$.