AnhTran2911

 

 

Proof : Let $ T $ $ \equiv $ $ AK $ $ \cap $ $ \odot (ABC) $ and $ T^* $ $ \equiv $ $ A^*K^* $ $ \cap $ $ \odot (A^*B^*C^*) $. Let the Lemoine axis of $ \triangle ABC $ cuts $ BC, $ $ CA, $ $ AB $ at $ D, $ $ E, $ $ F $, respectively and let the Lemoine axis of $ \triangle A^*B^*C^* $ cuts $ B^*C^*, $ $ C^*A^*, $ $ A^*B^* $ at $ D^*, $ $ E^*, $ $ F^* $, respectively. Let $ A_1 $ $ \equiv $ $ BC $ $ \cap $ $ B^*C^*, $ $ B_1 $ $ \equiv $ $ CA $ $ \cap $ $ C^*A^*, $ $ C_1 $ $ \equiv $ $ AB $ $ \cap $ $ A^*B^* $ and let $ O $ be the circumcenter of $ \triangle ABC $ ($\triangle A^*B^*C^* $). 

 

Clearly, $ A_1, $ $ B_1, $ $ C_1 $ lie on the polar $ \tau $ of $ P $ WRT $ \odot (O) $. Since $ ABTC $ and $ A^*B^*T^*C^* $ are harmonic quadrilateral, so $ T, $ $ P, $ $ T^* $ are collinear $ \Longrightarrow $ $ AK $ $ \cap $ $ A^*K^* $ $ \in $ $ \tau $. Similarly, we can prove $ BK $ $ \cap $ $ B^*K^* $ $ \in $ $ \tau $ and $ CK $ $ \cap $ $ C^*K^* $ $ \in $ $ \tau $. Since the tangent of $ \odot (O) $ through $ B, $ $ C $ and $ AK $ are concurrent, so $ D $ lies on the polar of $ AK $ $ \cap $ $ A^*K^* $ WRT $ \odot (O) $. Similarly, we can prove $ D^* $ lies on the polar of $ AK $ $ \cap $ $ A^*K^* $ WRT $ \odot (O) $ $ \Longrightarrow $ $ D, $ $ P, $ $ D^* $ are collinear. Analogously, we can prove $ P $ $ \in $ $ EE^* $ and $ P $ $ \in $ $ FF^* $, so from Desargue theorem ($ \triangle B_1EE^* $ and $ \triangle C_1FF^* $) we get $ \tau, $ $ EF, $ $ E^*F^* $ are concurrent, hence their pole $ P, $ $ K, $ $ K^* $ WRT $ \odot (O) $ are collinear.

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Lemma 2 : Let $ P $ be a point on the Kiepert hyperbola of $ \triangle ABC $. Let $ \triangle P_aP_bP_c $ be the pedal triangle of $ P $ WRT $ \triangle ABC $ and let $ \triangle XYZ $ be the circumcevian triangle of $ P $ WRT $ \triangle P_aP_bP_c $. Then $ P $ lies on the Kiepert hyperbola of $ \triangle XYZ $.

 

Proof : Let $ Q $ be the isogonal conjugate of $ P $ WRT $ \triangle ABC $. Let $ \triangle Q_aQ_bQ_c, $ $ \triangle Q_AQ_BQ_C $ be the pedal triangle, circumcevian triangle of $ Q $ WRT $ \triangle ABC $. Let $ R $ be the isogonal conjugate of $ Q $ WRT $ \triangle Q_AQ_BQ_C $. Since $ \triangle Q_AQ_BQ_C $ $ \cup $ $ R $ $ \sim $ $ \triangle Q_aQ_bQ_c $ $ \cup $ $ Q $ $ \cong $ $ \triangle XYZ $ $ \cup $ $ P $, so it suffices to prove $ Q $ lies on the Brocard axis of $ \triangle Q_AQ_BQ_C $. Let $ O $ be the circumcenter of $ \triangle ABC $. Let $ K, $ $ K_Q $ be the symmedian point of $ \triangle ABC, $ $ \triangle Q_AQ_BQ_C $, respectively. From Lemma 1 we get $ K, $ $ Q, $ $ K_Q $ are collinear, so notice $ Q $ lies on the Brocard axis $ OK $ of $ \triangle ABC $ we conclude that $ Q $ $ \in $ $ OK_Q $ (Brocard axis of $ \triangle Q_AQ_BQ_C $).

 

Remark : There is a stronger result of Lemma 2 : If $ P $ is the Kiepert perspector of $ \triangle ABC $ with angle $ \theta $, then $ P $ is the Kiepert perspector of $ \triangle XYZ $ with angle $ -\theta $ (but we don't need this stronger result in the proof). 

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Now we recall two well-known properties about conic as following :

 

Property 1 : Given a $ \triangle ABC $ and two points $ P, $ $ Q $. Let $ \triangle P_aP_bP_c $ be the anticevian triangle of $ P $ WRT $ \triangle ABC $ and let $ \triangle Q_aQ_bQ_c $ be the anticevian triangle of $ Q $ WRT $ \triangle ABC $. Then $ P, $ $ Q, $ $ P_a, $ $ P_b, $ $ P_c, $ $ Q_a, $ $ Q_b, $ $ Q_c $ lie on a conic. 

 

Property 2 : Given a $ \triangle ABC $ and a point $ P $. Let $ I, $ $ I_a, $ $ I_b, $ $ I_c $ be the incenter, A-excenter, B-excenter, C-excenter of $ \triangle ABC $, respectively. Let $ \mathcal{H} $ be a conic passing through $ I, $ $ I_a, $ $ I_b, $ $ I_c $. Then the polar of $ P $ WRT $ \mathcal{H} $ passes through $ P^* $ where $ P^* $ is the isogonal conjugate of $ P $ WRT $ \triangle ABC $.

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From Property 1 and Property 2 we get the following lemma :

 

Lemma 3 : Let $ \mathcal{H} $ be a circum-rectangular hyperbola of $ \triangle ABC $ and let $ P, $ $ Q $ be the points on $ \mathcal{H} $. Let $ \triangle DEF $ be the cevian triangle of $ Q $ WRT $ \triangle ABC $ and let $ P^* $ be the isogonal conjugate of $ P $ WRT $ \triangle DEF $. Then $ PP^* $ is tangent to $ \mathcal{H} $.

 

Proof : Let $ I, $ $ I_a, $ $ I_b, $ $ I_c $ be the incenter, A-excenter, B-excenter, C-excenter of $ \triangle DEF $, respectively. From Property 1 we get $ A, $ $ B, $ $ C, $ $ Q, $ $ I, $ $ I_a, $ $ I_b, $ $ I_c $ lie on a conic $ \mathcal{C} $, but notice $ I $ is the orthocenter of $ \triangle I_aI_bI_c $ we get $ \mathcal{C} $ is a rectangular hyperbola $ \Longrightarrow $ $ \mathcal{C} $ $ \equiv $ $ \mathcal{H} $, so from Property 2 we conclude that $ P^* $ lies on the polar of $ P $ WRT $ \mathcal{H} $. i.e. $ PP^* $ is tangent to $ \mathcal{H} $

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Let $ H $ be the orthocenter of $ \triangle ABC $. Let $ \triangle XYZ $ be the anticevian triangle of $ P $ WRT $ \triangle ABC $ and $ J $ be the circumcenter of $ \triangle XYZ $. Let $ \triangle A_1B_1C_1 $ be the pedal triangle of $ P $ WRT $ \triangle ABC $. Perform the Inversion with center $ P $ and denote $ V^* $ as the image of $ V $ ($ V $ is an arbitrary point). Obviously, $ \triangle A^*B^*C^* $ is the pedal triangle of $ P $ WRT $ \triangle A_1^*B_1^*C_1^* $ and $ \triangle X^*Y^*Z^* $ is the pedal triangle of $ P $ WRT the medial triangle $ \triangle A_2^*B_2^*C_2^* $ of $ \triangle A_1^*B_1^*C_1^* $, so $ PJ $ $ \equiv $ $ PJ^* $ passes through the isogonal conjugate $ Q^* $ of $ P $ WRT $ \triangle A_2^*B_2^*C_2^* $.

 

Let $ \triangle DEF $ be the anticomplementary triangle of $ \triangle ABC $. Let $ \triangle A_3^*B_3^*C_3^* $ be the cevian triangle of $ P $ WRT $ \triangle A_1^*B_1^*C_1^* $. From $ EF $ $ \parallel $ $ BC $ and $ A $ $ \in $ $ EF $ $ \Longrightarrow $ the image of the line $ EF $ under the Inversion is the circle with diameter $ PA_3^* $. Similarly, we can prove $ \odot (PB_3^*), $ $ \odot (PC_3^*) $ is the image of the line $ FD, $ $ DE $ under the Inversion, respectively, so $ \triangle D^*E^*F^* $ is the pedal triangle of $ P $ WRT $ \triangle A_3^*B_3^*C_3^* $. Since $ H $ is the circumcenter of $ \triangle DEF $, so we get $ PH $ $ \equiv $ $ PH^* $ passes through the isogonal conjugate $ R^* $ of $ P $ WRT $ \triangle A_3^*B_3^*C_3^* $.

 

Let $ G^* $ $ \equiv $ $ A_1^*A_2^* $ $ \cap $ $ B_1^*B_2^* $ $ \cap $ $ C_1^*C_2^* $ be the Centroid of $ \triangle A_1^*B_1^*C_1^* $. From Lemma 2 we know $ P $ lies on the Kiepert hyperbola of $ \triangle A_1^*B_1^*C_1^* $, so $ A_1^*, $ $ B_1^*, $ $ C_1^*, $ $ G^*, $ $ P $ lie on a rectangular hyperbola $ \mathcal{K} $, hence from Lemma 3 we conclude that $ P, $ $ Q^*, $ $ R^* $ lie on the tangent of $ \mathcal{K} $ through $ P $ $ \Longrightarrow $ $ P, $ $ J^*, $ $ H^* $ are collinear $ \Longrightarrow $ $ P, $ $ J, $ $ H $ are collinear.