Isogonal conjugate of Parry reflection point

by TelvCohl, Sep 7, 2018, 10:58 AM

In this short post, I'll prove two interesting properties of the isogonal conjugate of Parry reflection point. All properties about Parry reflection point used can be found in my previous post Generalization of Parry Reflection Point.

Property 1 : Given a $ \triangle ABC $ and a point $ P. $ Then the Euler line of $ \triangle ABC, \triangle PBC, \triangle PCA, \triangle PAB $ are parallel if and only if $ P $ is the isogonal conjugate of the Parry reflection point of $ \triangle ABC $ WRT $ \triangle ABC. $

Proof :

Lemma 1.1 : Given a quadrangle $ ABCD $ with Poncelet point $ T. $ Let $ \triangle D_aD_bD_c $ be the pedal triangle of $ D $ WRT $ \triangle ABC, $ $ \triangle D^*_aD^*_bD^*_c $ be the circumcevian triangle of $ D $ WRT $ \triangle D_aD_bD_c $ and let $ M_a, M_b, M_c $ be the midpoint of $ AD, BD, CD, $ respectively. Then $ T $ lies on $ D_a^*M_a, D_b^*M_b, D_c^*M_c. $

Proof : Since $ T $ lies on the pedal circle $ \odot (D_aD_bD_c) $ of $ D $ WRT $ \triangle ABC $ and the 9-point circle $ \odot (M_aM_bD_c) $ of $ \triangle ABD, $ so $ \measuredangle M_aTD_c = \measuredangle M_aM_bD_c = \measuredangle DBA = \measuredangle D^*_aD_aD_c = \measuredangle D^*_aTD_c $ $ \Longrightarrow $ $ T \in D_a^*M_a. $ $ \qquad \blacksquare $

Lemma 1.2 : Given a quadrangle $ BCUV $ such that $ \triangle UBC, \triangle VBC $ are equilateral triangle and two points $ P,Q. $ Then the Euler line of $ \triangle PBC, \triangle QBC $ are parallel if and only if $ B, C, P, Q, U, V $ lie on a conic.

Proof : First, from Collinearity of Kiepert perspectors (Property 4) we know that the line connecting $ U $ and the isogonal conjugate of $ U $ WRT $ \triangle QBC $ is parallel to the Euler line of $ \triangle QBC, $ so the Euler line of $ \triangle QBC $ is parallel to the Euler line of $ \triangle PBC $ if and only if the isogonal conjugate of $ U $ WRT $ \triangle QBC $ lies on the parallel $ \tau $ from $ U $ to the Euler line of $ \triangle PBC. $ i.e. the Euler line of $ \triangle PBC, \triangle QBC $ are parallel if and only if there is a point $ T \in \tau $ such that $ QB, QC $ is the reflection of $ BC $ in the bisector of $ \angle UBT, \angle UCT, $ respectively. Consider a point $ X $ moving on $ \tau $ and let $ \ell_b^{x}, \ell_c^{x} $ be the reflection of $ BC $ in the bisector of $ \angle UBX, \angle UCX, $ respectively, then $ \ell_b^{x} \mapsto \ell_c^{x} $ is a homography $ \Longrightarrow $ the locus of $ \ell_b^{x} \cap \ell_c^{x} $ is a conic $ \mathcal{C} $ passing through $ B, C, $ so the Euler line $ \triangle PBC, \triangle QBC $ are parallel if and only if $ Q \in \mathcal{C}. $ Finally, it's clear that $ P, U, V $ lie on $ \mathcal{C}, $ so the proof is completed. $ \qquad \blacksquare $

Back to the main problem :

Let $ S_1, S_2 $ be the 1st Isodynamic point, 2nd Isodynamic point of $ \triangle ABC, $ respectively, $ S_1^{a}, S_2^{a} $ be the second intersection of $ AS_1, AS_2 $ with $ \odot (S_1BC), \odot (S_2BC), $ respectively and $ Y_a, Z_a $ be the intersection of $ \{ BS_1^{a}, CS_2^{a} \}, $ $ \{ BS_2^{a}, CS_1^{a} \}, $ respectively. Simple angle chasing yields $ Y_a, Z_a $ lie on $ \odot (ABC) $ and $ \triangle AY_aZ_a $ is equilateral triangle, so the pole of $ BC, Y_aZ_a $ WRT $ \odot (ABC) $ lie on $ S_1^{a}S_2^{a}. $ Let $ \triangle T_aT_bT_c $ be the tangential triangle of $ \triangle ABC, $ then from Interesting Properties related to Four 9-point Centers (Lemma 4) we get the Euler reflection point $ E $ of $ \triangle ABC $ is the Poncelet point of $ T_aT_bT_cO $ where $ O $ is the circumcenter of $ \triangle ABC, $ so note that the pole of $ Y_aZ_a $ WRT $ \odot (ABC) $ is the image of $ A $ under the homothety $ (O,-2) $ we get $ S_1^{a}S_2^{a} $ passes through the reflection of $ O $ in $ E $ by Lemma 1.1. i.e. $ S_1^{a}S_2^{a} $ passes through the Parry reflection point $ P_{r} $ of $ \triangle ABC. $

Let $ R_1^{a}, R_2^{a}, Q $ be the isogonal conjugate of $ S_1^{a},S_2^{a}, P_{r} $ WRT $ \triangle ABC, $ respectively. Clearly, $ \triangle R_1^{a}BC, \triangle R_2^{a}BC $ are equilateral triangle, so $ Q $ lies on the circumconic $ \mathcal{C}_a $ of $ \triangle ABC $ passing through $ R_1^{a}, R_2^{a}. $ Analogously, we can construct $ \mathcal{C}_b, \mathcal{C}_c $ and prove $ Q $ lies on $ \mathcal{C}_b, \mathcal{C}_c, $ so by Lemma 1.2 and note that $ \mathcal{C}_a, \mathcal{C}_b, \mathcal{C}_c $ have at most four common points we conclude that the Euler line of $ \triangle ABC, \triangle PBC, \triangle PCA, \triangle PAB $ are parallel $ \Longleftrightarrow $ $ P \equiv Q. $ $ \qquad \blacksquare $

Property 2 : Given a $ \triangle ABC $ and a point $ P. $ Let $ \triangle DEF, \triangle XYZ $ be the pedal triangle, cevian triangle of $ P $ WRT $ \triangle ABC, $ respectively. Then $ \triangle DEF \sim \triangle XYZ $ if and only if $ P $ is the orthocenter of $ \triangle ABC $ or the isogonal conjugate of the Parry reflection point of $ \triangle ABC $ WRT $ \triangle ABC. $

Proof :

Lemma 2.1 : Given a $ \triangle ABC $ with isogonal conjugate $ P,Q. $ Let $ \triangle DEF $ be the cevian triangle of $ P $ WRT $ \triangle ABC $ and $ B_1, C_1 $ be the reflection of $ B, C $ in $ CA, AB, $ respectively. Then $ \measuredangle ERF = \measuredangle BAC + \measuredangle B_1QC_1 $ where $ R $ is the image of $ P $ under the inversion WRT the circumcircle of $ \triangle ABC. $

Proof : Inversion with center $A,$ power $AB \cdot AC$ followed by reflection in the bisector of $ \angle A, $ labeling inverse points with $*.$ Clearly, $ B^{*} = C, C^{*} = B, B_1^{*} = C_1, C_1^{*} = B_1 $ and $ P^{*}, Q^{*} $ is the second intersection of $ AQ, AP $ with $ \odot (QBC), \odot (PBC), $ respectively, so $$ \left\{\begin{array}{cc} \triangle ABP \stackrel{+}{\sim} \triangle AP^{*}C \Longrightarrow \measuredangle Q^{*}CB = \measuredangle APB = \measuredangle F^{*}CP^{*} \\\\ \triangle AFP \stackrel{+}{\sim} \triangle AP^{*}F^{*} \Longrightarrow \measuredangle Q^{*}BC = \measuredangle APF = \measuredangle CF^{*}P^{*} \end{array}\right\| \Longrightarrow \triangle Q^*BC \stackrel{-}{\sim} \triangle P^*F^*C, $$hence note that $ P^* $ and $ R^* $ are symmetric WRT $ BC $ we get $ \frac{CQ^*}{CB_1} = \frac{CR^*}{CF^*}. $ On the other hand, simple angle chasing yields $ \measuredangle Q^*CB_1 = \measuredangle R^*CF^*, $ so $ \triangle Q^*B_1C \stackrel{+}{\sim} \triangle R^*F^*C. $ Similarly, $ \triangle Q^*C_1B \stackrel{+}{\sim} \triangle R^*E^*B, $ so \begin{align*} \measuredangle BAC + \measuredangle B_1QC_1 &=\measuredangle BAC + \measuredangle B_1QA + \measuredangle AQC_1 \\ &=\measuredangle BAC + \measuredangle Q^*C_1A + \measuredangle AB_1Q^* \\ &=\measuredangle BAC + \measuredangle Q^*C_1B + \measuredangle BC_1A + \measuredangle CB_1Q^* + \measuredangle AB_1C \\ &= \measuredangle R^*E^*A + \measuredangle AF^*R^* \\ &= \measuredangle ERA + \measuredangle ARF \\ &= \measuredangle ERF. \qquad \qquad \qquad \blacksquare \\ \end{align*}Back to the main problem :

Let $ R $ be the image of $ P $ under the inversion WRT $ \odot (ABC) $ and $ M $ be the Miquel point of $ X,Y,Z $ WRT $ \triangle ABC, $ then note that the pedal triangle of $ R $ is inversely similar to $ \triangle DEF $ we get $$ \triangle DEF \sim \triangle XYZ \Longleftrightarrow \triangle DEF \stackrel{+}{\sim} \triangle XYZ \text{ or } \triangle DEF \stackrel{-}{\sim} \triangle XYZ \Longleftrightarrow M \equiv P \text{ or } M \equiv R. $$Let $ \triangle T_aT_bT_c $ be the tangential triangle of $ \triangle ABC $ and let $ A_1, B_1, C_1 $ be the reflection of $ A, B, C $ in $ BC, CA, AB, $ respectively. Since $ \triangle T_aBC_1 $ and $ \triangle T_aCB_1 $ are directly congruent, so by Lemma 2.1 we get $$ R \in \odot (AYZ) \Longleftrightarrow \measuredangle B_1QC_1 + \measuredangle BAC = \measuredangle CAB \Longleftrightarrow \measuredangle B_1QC_1 = \measuredangle B_1T_aC_1 \Longleftrightarrow Q \in \odot (T_aB_1C_1) $$where $ Q $ is the isogonal conjugate of $ P $ WRT $\triangle ABC. $ Similarly, we can prove $ R \in \odot (BZX) \Longleftrightarrow Q \in \odot (T_bC_1A_1) $ and $ R \in \odot (CXY) \Longleftrightarrow Q \in \odot (T_cA_1B_1). $

Let $ P_r $ be the Parry reflection point of $ \triangle ABC, $ then $ A_1P_r, B_1P_r, C_1P_r $ is parallel to the reflection of the Euler line of $ \triangle ABC $ in $ BC, CA, AB, $ respectively, so $ P_r \in \odot (T_aB_1C_1), \odot (T_bC_1A_1), \odot (T_cA_1B_1) $ and hence we get $ \triangle DEF \stackrel{-}{\sim} \triangle XTZ $ if and only if $ Q \equiv P_r. $ On the other hand, it's clearly that $ M \equiv P $ if and only if $ P $ is the orthocenter $ H $ of $ \triangle ABC, $ so we conclude that $ \triangle DEF \sim \triangle XYZ $ $ \Longleftrightarrow $ $ P \equiv H $ or $ Q \equiv P_r. $ $ \qquad \blacksquare $
Parry reflection point
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Parry reflection point
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