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Romania — NMO Selection Tests for the Balkan and International Mathematical Olympiads

Geometry Difficulty 8.5 Shortlist Prove it Romania

Let $\ell$ be a line, and let $\gamma$ and $\gamma'$ be two circles. The line $\ell$ meets $\gamma$ at points $A$ and $B$ , and $\gamma'$ at points $A'$ and $B'$ . The tangents to $\gamma$ at $A$ and $B$ meet at point $C$ , and the tangents to $\gamma'$ at $A'$ and $B'$ meet at point $C'$ . The lines $\ell$ and $CC'$ meet at point $P$ . Let $\lambda$ be a variable line through $P$ and let $X$ be one of the points where $\lambda$ meets $\gamma$ , and $X'$ be one of the points where $\lambda$ meets $\gamma'$ . Prove that the point of intersection of the lines $CX$ and $C'X'$ lies on a fixed circle.

Solution

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

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