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Geometry Difficulty 8.5 Shortlist Prove it Romania

Let ℓ\ell be a line, and let γ\gamma and γ′\gamma' be two circles. The line ℓ\ell meets γ\gamma at points AA and BB, and γ′\gamma' at points A′A' and B′B'. The tangents to γ\gamma at AA and BB meet at point CC, and the tangents to γ′\gamma' at A′A' and B′B' meet at point C′C'. The lines ℓ\ell and CC′CC' meet at point PP. Let λ\lambda be a variable line through PP and let XX be one of the points where λ\lambda meets γ\gamma, and X′X' be one of the points where λ\lambda meets γ′\gamma'. Prove that the point of intersection of the lines CXCX and C′X′C'X' lies on a fixed circle.

Solution

Let the lines CXCX and C′X′C'X' meet at point QQ. The line CXCX meets ℓ\ell at DD, and γ\gamma a second time at YY; similarly, the line C′X′C'X' meets ℓ\ell at D′D', and γ′\gamma' a second time at Y′Y'. Notice that the cross-ratios (CDXY)(CDXY) and (C′D′X′Y′)(C'D'X'Y') are both harmonic, for CC and C′C' are the poles of ℓ\ell relative to γ\gamma and γ′\gamma', respectively. Since the lines CC′CC', DD′DD' and XX′XX' all pass through PP, so does the line YY′YY'. Consequently, (QCXD)=(QC′X′D′)(QCXD) = (QC'X'D') and (QCYD)=(QC′Y′D′)(QCYD) = (QC'Y'D'). Let ϱ\varrho and ϱ′\varrho' be the powers of QQ relative to γ\gamma and γ′\gamma', respectively, let ω\omega be the power of CC relative to γ\gamma, and let ω′\omega' be the power of C′C' relative to γ′\gamma'. Multiply the last two equalities involving cross-ratios and apply Menelaus' theorem to triangle QCC′QCC' and transversal PDD′PDD' to get
ϱϱ′=ωω′(C′D′QD′⋅QDCD)2=ωω′(PC′PC)2=constant \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 QQ lies on a fixed circle of the pencil of circles generated by γ\gamma and γ′\gamma'.

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