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Greece — Selection Examination

Geometry Difficulty 7.2 National Olympiad, round 2 Prove it Greece

Let ABΓAB\Gamma be an acute angled triangle inscribed in the circle c(O,R)c(O, R) (with AB<AΓ<BΓAB < A\Gamma < B\Gamma) and let Δ,E,Z\Delta, E, Z be the touching points of the incircle of the triangle with the sides BΓB\Gamma, AΓA\Gamma, ABAB, respectively. The circumcircle of the triangle AEZAEZ (say, (c1)(c_1)) intersects the circle (c)(c) at point A′A'. The circumcircle of the triangle BΔZB\Delta Z (say, (c2)(c_2)) intersects the circle (c)(c) at point B′B'. The circumcircle of the triangle ΓΔE\Gamma\Delta E (say, (c3)(c_3)) intersects the circle (c)(c) at point Γ′\Gamma'. Prove that:

(α) The quadrilateral ΔEA′B′\Delta EA'B' is cyclic.

(β) The lines ΔA′\Delta A', EB′EB' and ZΓ′Z\Gamma' are concurrent.

Solution

From the inscribed in the circle c1c_1 quadrilateral AA′IZAA'IZ we have:
AA^′I=AZ^I=90∘=TAA^′I.(α) A\hat{A}'I = A\hat{Z}I = 90^\circ = T A\hat{A}'I \quad . (\alpha)
From the inscribed quadrilateral ΓΔIE\Gamma\Delta IE (since ΓI\Gamma I bisector), we have
Δ^1=Γ^2(1) \hat{\Delta}_1 = \frac{\hat{\Gamma}}{2} \qquad (1)
From the inscribed quadrilateral BΔIZB\Delta IZ (since BIBI bisector), we have:
Δ^2=B^2(2) \hat{\Delta}_2 = \frac{\hat{B}}{2} \qquad (2)
From the inscribed quadrilateral BΔIB′B\Delta IB' we have: Δ^3=B^1\hat{\Delta}_3 = \hat{B}_1.
From the inscribed quadrilateral BB′A′ABB'A'A we have B^1=A^1′=90∘−A^2′\hat{B}_1 = \hat{A}'_1 = 90^\circ - \hat{A}'_2.
Hence: Δ^3+A^2′=90∘\hat{\Delta}_3 + \hat{A}'_2 = 90^\circ.
From the inscribed quadrilateral AEIA′AEIA' we have: A^3′=A^2\hat{A}'_3 = \frac{\hat{A}}{2}
Hence:
Δ^1+Δ^2+Δ^3+Δ^2′+Δ^3′=180∘ \hat{\Delta}_1 + \hat{\Delta}_2 + \hat{\Delta}_3 + \hat{\Delta}'_2 + \hat{\Delta}'_3 = 180^\circ
Therefore the quadrilateral A′EΔB′A'Ε\Delta B' is cyclic. Similarly we prove that the quadrilaterals ΔZA′Γ′\Delta ZA'\Gamma' and ZEΓ′B′ΖΕ\Gamma'B' are cyclic.

Finally, we conclude that the lines ΔA′\Delta A', EB′EB' and ZΓ′Z\Gamma' are concurrent at the radical point of three circles.

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