Generalization of Lester's Theorem

by TelvCohl, May 2, 2023, 9:26 AM

In this short post, I present a proof to a generalization of Lester's theorem found by Dao Thanh Oai [1].

Preliminaries

First, recall the following well-known

Lemma 1 : Given a $ \triangle ABC $ and two points $ P, X. $ Then $ X $ and $ (P/X) $ are conjugate WRT any conic passing through $ A, B, C, P, $ where $ (P/X) $ is the cevian quotient of $ P, X $ WRT $ \triangle ABC. $ In particular, $ X(P/X) $ is tangent to the circumconic of $ \triangle ABC $ passing through $ P, X $ and the circumconic of $ \triangle ABC $ passing through $ P, (P/X). $

Corollary 1 : Given a $ \triangle ABC $ and two points $ U, V. $ Let $ W $ be the crosspoint of $ U, V $ WRT $ \triangle ABC $ and let $ X, Y, Z $ be the isogonal conjugate of $ U, V, W $ WRT $ \triangle ABC, $ respectively. Then $ X $ and $ Y $ are conjugate WRT any conic passing through $ A, B, C, Z. $

Lemma 2 : Given a $ \triangle ABC $ with orthocenter $ H $ and a pair of isogonal conjugate $ (P, Q) $ of $ \triangle ABC. $ Let $ \triangle Q_AQ_BQ_C $ be the pedal triangle of $ Q $ WRT $ \triangle ABC. $ Then the crosspoint of $ H, P $ WRT $ \triangle ABC $ is the anticomplement of $ Q $ WRT $ \triangle Q_AQ_BQ_C. $

Proof : Let $ \triangle DEF $ be the orthic triangle of $ \triangle ABC, $ $ T $ be the anticomplement of $ Q $ WRT $ \triangle Q_AQ_BQ_C $ and let $ X $ be the intersection of $ AP, EF. $ It suffices to prove that $ D, T, X $ are collinear. Let $ R $ be the reflection of $ Q $ in the midpoint of $ Q_BQ_C, $ then $ R \in AP $ and $ RT \stackrel{\parallel}{=} QQ_A, $ so note that $ ABCQ \stackrel{-}{\sim} AEFR $ we conclude that $$ \frac{XR}{XA} = \frac{QQ_A}{AD} = \frac{RT}{AD} \Longrightarrow T\in DX. \qquad \blacksquare $$
Lemma 3 : Given a rectangular hyperbola $ \mathcal{H} $ with antipode $ U, V. $ Let $ P, Q $ be two points such that $ P $ and $ Q $ are conjugate WRT $ \mathcal{H} $ and $ PQ $ is parallel to the tangents of $ \mathcal{H} $ at $ U, V. $ Then $ P, Q, U, V $ are concyclic.

Proof : Let $ W $ be the second intersection of $ PU $ with $ \mathcal{H}, $ then by Seydewitz-Staudt theorem we get $ W \in QV, $ so note that $ \mathcal{H} $ is the isogonal conjugate of the perpendicular bisector of $ UV $ WRT $ \triangle UVW $ we conclude that $ PQ $ and $ UV $ are antiparallel WRT $ \angle (PU, QV). $ i.e. $ P, Q, U, V $ are concyclic. $ \qquad \blacksquare $


Main result

Property : Given a $ \triangle ABC $ with Fermat point $ F_{1}, F_{2} $ and a point $ P $ lying on the Neuberg cubic of $ \triangle ABC. $ Let $ R $ be the perspector of $ \triangle ABC $ and the reflection triangle of $ P $ WRT $ \triangle ABC. $ Then $ F_{1}, F_{2}, P, R $ are concyclic.

Proof :

Let $ O, K $ be the circumcenter, symmedian point of $ \triangle ABC, $ respectively, $ Q $ be the isogonal conjugate of $ P $ WRT $ \triangle ABC $ and let $ U $ be the intersection of $ OK, PQ. $

Claim. $ O $ and $ U $ are conjugate WRT the circumconic $ \mathcal{C} $ of $ \triangle ABC $ passing through $ P, Q. $

Proof. Let $ G, H $ be the centroid, orthocenter of $ \triangle ABC, $ respectively and let $ V $ be the isogonal conjugate of $ U $ WRT $ \triangle ABC. $ By Corollary 1, it suffices to show that the crosspoint of $ H, V $ WRT $ \triangle ABC $ lies on $ PQ. $ Since $ G, K $ is the centroid of the pedal triangle of $ O, K $ WRT $ \triangle ABC, $ respectively, so the line connecting $ U $ and the centroid $ G_U $ of the pedal triangle of $ U $ WRT $ \triangle ABC $ is parallel to $ OG, $ hence from $ PQ \parallel OG $ we get $ G_U \in PQ $ and the claim is proved by Lemma 2. $ \qquad \square $

Let $ S $ be the isogonal conjugate of $ R $ WRT $ \triangle ABC. $ By New approach to Liang-Zelich Theorem (Sondat's theorem), $ R $ lies on $ PQ $ and the circum-rectangular hyperbola of $ \triangle ABC $ passing through $ Q, $ so $ S $ is the second intersection of $ OP $ with $ \mathcal{C} $ and hence from Claim. we get the crosspoint of $ Q, S $ WRT $ \triangle ABC $ lies on $ OK. $ By Corollary 1, $ P $ and $ R $ are conjugate WRT the Kiepert hyperbola $ \mathcal{K} $ of $ \triangle ABC. $ Finally, the tangents of $ \mathcal{K} $ at $ F_{1}, F_{2} $ are parallel to $ OH, PR $ (see Character of the Steiner line WRT Cevian triangle (Corollary 1)), so by Lemma 3 we conclude that $ F_{1}, F_{2}, P, R $ are concyclic. $ \qquad \blacksquare $

Bibliography

[1] $ \qquad \qquad $ Dao Thanh Oai, Three Conjectures in Euclidean Geometry, viXra:1507.0218, 2015.
Lester s theorem
Neuberg cubic
Cubic

New approach to Liang-Zelich Theorem

by TelvCohl, May 27, 2021, 10:22 AM

In this short post, I present a new proof to a result of Ivan Zelich and Xuming Liang [1].

Preliminaries

Sondat's theorem : Given two perspective and orthologic triangles $ \triangle A_1B_1C_1 $ and $ \triangle A_2B_2C_2 $ such that the perpendicular from $ A_1, B_1, C_1 $ to $ B_2C_2, C_2A_2, A_2B_2, $ resp. are concurrent at $ X_1 $ and the perpendicular from $ A_2, B_2, C_2 $ to $ B_1C_1, C_1A_1, A_1B_1, $ resp. are concurrent at $ X_2. $ Then their perspector $ P $ lies on $ X_1X_2 $ and the circum-rectangular hyperbola $ \mathcal{H}_1, \mathcal{H}_2 $ of $ \triangle A_1B_1C_1, \triangle A_2B_2C_2 $ passing through $ X_1, X_2, $ respectively.

Proof : Note that the second part is a particular case of a well-known result, so it suffices to prove that $ P \in X_1X_2. $ Let $ D_1\in\mathcal{H}_1 $ be the orthocenter of $ \triangle B_1X_1C_1, $ then $ D_1B_1 \parallel A_2B_2, D_1C_1 \parallel A_2C_2 $ $ \Longrightarrow $ $ A_1D_1 $ is parallel to the tangent of $ \mathcal{H}_2 $ at $ A_2, $ so $$ P(A_1, X_1; B_1, C_1) = D_1(A_1, X_1; B_1, C_1) = A_2(A_2, X_2; B_2, C_2) = P(A_2, X_2; B_2, C_2) \Longrightarrow P\in X_1X_2\ . \qquad \blacksquare $$
Lemma : Given a $ \triangle ABC $ and a pair $ \left (P, Q \right ) $ of isogonal conjugate WRT $ \triangle ABC. $ Let $ I, I_a, I_b, I_c $ be the incenter, A-excenter, B-excenter, C-excenter of $ \triangle ABC, $ resp. and let $ \mathcal{H} $ be a conic passing through $ I, I_a, I_b, I_c. $ Then $ Q $ lies on the polar of $ P $ WRT $ \mathcal{H}. $

Proof : Let $ U_a, V_a, U_b, V_b, U_c, V_c $ be the intersection of $ PQ $ with $ II_a, I_bI_c, II_b, I_cI_a, II_c, I_aI_b, $ respectively, then $$ (P, Q; U_a, V_a) = (P, Q; U_b, V_b) = (P, Q; U_c, V_c) = -1\ , $$so by Desargues involution theorem we know that the involution on $ PQ $ induced by the pencil of conics passing through $ I, I_a, I_b, I_c $ coincide with harmonic conjugate WRT $ P, Q. $ If $ U, V $ are the intersection of $ PQ $ with $ \mathcal{H}, $ then from the discussion above we get $ (P,Q;U,V) = -1 $ and hence $ Q $ lies on the polar of $ P $ WRT $ \mathcal{H}. $ $ \qquad \blacksquare $

Main result

Theorem : Given a $ \triangle ABC $ with circumcenter $ O $ and orthocenter $ H. $ Let $ (P, Q) $ be a pair of isogonal conjugate of $ \triangle ABC $ such that $ \{ P, Q \} \neq \{ O, H \} $ and let $ T $ be the intersection of $ PQ, OH. $ For $ t \in \mathbb{R} \cup \{ \infty \}, $ denote $ \triangle P_a^{t}P_b^{t}P_c^{t} $ as the image of the pedal triangle of $ P $ WRT $ \triangle ABC $ under the homothety $ (P, t). $ Then $ \triangle ABC $ and $ \triangle P_a^{t}P_b^{t}P_c^{t} $ are perspective if and only if $ t \in \left \{ 0\ ,\ \frac{2\ TO}{TH}\ ,\ \infty \right \}, $ where $ \frac{2\ TO}{TH} $ is undefined when $ T \equiv H. $

Proof :

Claim. If $ T \notin \{ O, H \}, $ then $ \triangle ABC $ and $ \triangle P_a^{t}P_b^{t}P_c^{t} $ are perspective for $ t = \frac{2\ TO}{TH}. $

Proof. Let $ S $ be the pole of $ OP $ WRT the circum-rectangular hyperbola $ \mathcal{H}_{P} $ of the excentral triangle of $ \triangle ABC $ passing through $ P $ and let $ V $ be the intersection of $ AS $ with $ PP_{a}^{1}, $ then it suffices to prove that $ \frac{PV}{PP_a^{1}} = \frac{2\ TO}{TH}. $ Let $ W, X, Y, Z $ be the intersection of $ OP $ with $ AH, HS, AS, BC, $ respectively. By Lemma we know that $ S \in PQ $ and $ SO, SP, SX, SY $ is the polar of $ X, P, O, Z $ WRT $ \mathcal{H}_{P}, $ respectively, so $ (Z,X;O,P) = (Y,O;X,P) $ and hence $$ \frac{AH}{2\ PP_a^{1}} = -\frac{OP}{XP} \cdot \frac{XY}{PY}\ . $$On the other hand, by Menelaus' theorem for $ \left ( \triangle XOH, \overline{TSP} \right ) $ and $ \left ( \triangle HXW, \overline{SYA} \right ) $ we get $ \frac{TO}{TH} = - \frac{SX}{HS} \cdot \frac{OP}{XP} $ and $ \frac{SX}{HS} = - \frac{WA}{AH} \cdot \frac{XY}{YW} , $ so $$ \frac{PV}{2\ PP_a^{1}} = \frac{PV}{AH}\ \cdot\ \frac{AH}{2\ PP_a^{1}} = \left ( \frac{WA}{AH} \cdot \frac{YP}{YW} \right ) \cdot \left ( -\frac{OP}{XP} \cdot \frac{XY}{PY} \right ) = \frac{TO}{TH}\ . \qquad \square$$
If $ AP_a^{t}, BP_b^{t}, CP_c^{t} $ are concurrent for some $ t \in \mathbb{R} \setminus \left \{ 0 \right \}, $ then by Sondat's theorem we know that their perspector lies on $ PQ $ and the circum-rectangular hyperbola of $ \triangle ABC $ passing through $ Q, $ so from Claim. we conclude that $ \triangle ABC $ and $ \triangle P_a^{t}P_b^{t}P_c^{t} $ are perspective if and only if $ t \in \left \{ 0\ ,\ \frac{2\ TO}{TH}\ ,\ \infty \right \}. \qquad \blacksquare $

Bibliography

[1] $ \qquad \qquad $ Ivan Zelich and Xuming Liang, GENERALISATIONS OF THE PROPERTIES OF THE NEUBERG CUBIC TO THE EULER PENCIL OF ISOPIVOTAL CUBICS, INTERNATIONAL JOURNAL OF GEOMETRY Vol. 4 (2015), No. 2, 5 - 25.
This post has been edited 1 time. Last edited by TelvCohl, May 28, 2021, 1:30 AM
Liang-Zelich Theorem
Isogonal conjugate

Investigate Orthocorrespondent using Inversion

by TelvCohl, Jun 16, 2020, 3:21 PM

Notation : Given a $ \triangle ABC, $ a point $ P $ and the orthocorrespondent $ T $ of $ P $ WRT $ \triangle ABC. $ For any point $ \cdot, $ denote $ \cdot^{*} $ as the image of $ \cdot $ under the inversion with center $ P $ that fixed $ \odot (ABC). $

Property 1 : Let $ Q $ be the isogonal conjugate of $ P $ WRT $ \triangle ABC.$ Then $ PT $ is parallel to the Euler line of the pedal triangle of $ Q $ WRT $ \triangle ABC. $

Proof : Let $ D, E, F $ be the intersection of $ BC, CA, AB $ with the orthotransversal of $ P $ WRT $ \triangle ABC, $ respectively$,$ then $ \triangle A^{*}B^{*}C^{*} $ is the circumcevian triangle of $ P $ WRT $ \triangle ABC $ and $ D^{*}, E^{*}, F^{*} $ is the projection of $ P $ on $ \widehat{P}_a\widehat{H}, \widehat{P}_b\widehat{H}, \widehat{P}_c\widehat{H}, $ respectively$,$ where $ \triangle \widehat{P}_a\widehat{P}_b\widehat{P}_c $ is the antipedal triangle of $ P $ WRT $ \triangle A^{*}B^{*}C^{*} $ and $ \widehat{H} $ is its orthocenter$.$ Now let $ \triangle T_aT_bT_c $ be the cevian triangle of $ T $ WRT $ \triangle ABC, $ then from $$ \widehat{P}_a(\widehat{P}_c, \widehat{P}_b; \widehat{H}, T_a^{*}) = P(B^{*}, C^{*}; D^{*}, T_a^{*}) = P(B,C; D,T_a) = -1 $$we know that the intersection of $ \widehat{P}_aT_a^{*} $ and $ \widehat{P}_b\widehat{P}_c $ lies on the orthic axis $ \widehat{\tau} $ of $ \triangle \widehat{P}_a\widehat{P}_b\widehat{P}_c, $ so $ \odot (PA^{*}T_a^{*}) $ passes through the projection of $ P $ on $ \widehat{\tau}. $ Analogously$,$ we can prove the projection of $ P $ on $ \widehat{\tau} $ lies on $ \odot (PB^{*}T_b^{*}), \odot (PC^{*}T_c^{*}), $ so $ T^{*} $ is the projection of $ P $ on $ \widehat{\tau}. $ Finally$,$ notice that $ \triangle \widehat{P}_a\widehat{P}_b\widehat{P}_c $ and the pedal triangle of $ Q $ WRT $ \triangle ABC $ are homothetic and $ \widehat{\tau} $ is perpendicular to the Euler line of $ \triangle \widehat{P}_a\widehat{P}_b\widehat{P}_c $ we conclude that $ PT $ is parallel to the Euler line of the pedal triangle of $ Q $ WRT $ \triangle ABC. $ $ \qquad \blacksquare $

Property 2 : Let $ \mathcal{C} $ be the inconic of $ \triangle ABC $ with foci $ P, Q $ and $ K $ be the perspector of $ \mathcal{C} $ WRT $ \triangle ABC. $ Then $ T \equiv (K/P), $ where $ (K/P) $ is the cevian quotient of $ P, K $ WRT $ \triangle ABC. $

Proof : Let $ \triangle P_aP_bP_c, \triangle P_AP_BP_C $ be the cevian triangle$,$ anticevian triangle of $ P $ WRT $ \triangle ABC, $ respectively$,$ then $ P_a^{*} $ is the projection of $ \widehat{P}_a $ on $ AP $ and from $ (A, P_a; P, P_A) = -1 $ we get $ P_A^{*} $ is the midpoint of $ A^{*}P_a^{*}, $ so $ P_A^{*} $ is the projection of $ P $ on the $ \widehat{P}_a $-midline of $ \triangle \widehat{P}_a\widehat{P}_b\widehat{P}_c. $ Now denote $ K_a, K_b, K_c $ as the point of tangency of $ \mathcal{C} $ with $ BC, CA, AB, $ respectively$,$ then $ PA, PK_a $ are isogonal conjugate WRT $ \angle BPC, $ so $ B^{*}C^{*} \parallel K_a^{*}P_a^{*} $ and hence $ \widehat{P}_aK_a^{*}, \widehat{P}_a P_a^{*} $ are isogonal conjugate WRT $ \angle \widehat{P}_a. $ i.e.$,$ $ \widehat{P}_aK_a^{*} $ is the tangent of $ \odot (\widehat{P}_a\widehat{P}_b\widehat{P}_c) $ at $ \widehat{P}_a, $ so note that the intersection of $ \widehat{P}_aK_a^{*} $ with the $ \widehat{P}_a $-midline of $ \triangle \widehat{P}_a\widehat{P}_b\widehat{P}_c $ lies on $ \widehat{\tau} $ we get $ T^{*} \in \odot (PP_A^{*}K_a^{*}) $ $ \Longrightarrow $ $ T\in P_AK_a. $ Similarly$,$ we can prove $ T \in P_BK_b, P_CK_c, $ so $ T \equiv (K/P). $ $ \qquad \blacksquare $

Property 3 : Let $ H $ be the orthocenter of $ \triangle ABC $ and $ R $ be the isogonal conjugate of $ P $ WRT $ \triangle P_AP_BP_C, $ then $ Q, R, T, (H/P) $ are collinear and $ (R,T;Q,(H/P)) = -1. $

Proof : For any point $ \cdot, $ let $ \lambda (\cdot) $ be the image of $ \cdot $ under the homothety with center $ P $ and factor $ 2. $ It suffices to prove $\lambda(R^{*})\lambda(Q^{*})\lambda(T^{*})\lambda((H/P)^{*}) $ is a harmonic quadrilateral and $ P $ lies on its circumcircle$.$ Let $ \widehat{O}, \widehat{N} $ be the circumcenter$,$ 9-point center of $ \triangle \widehat{P}_a\widehat{P}_b\widehat{P}_c, $ respectively and $ \Omega_{P} $ be the circle passing through $ P $ and orthogonal to $ \odot (\widehat{O}), \odot(\widehat{N}). $ Notice that the center of $ \Omega_{P} $ lies on the radical axis $ \widehat{\tau} $ of $ \odot (\widehat{O}), \odot(\widehat{N}), $ so $ \lambda(T^{*}) \in \Omega_{P}. $

Let $ X_a, X_b, X_c $ be the reflection of $ P $ in $ BC, CA, AB, $ respectively$,$ then $ X_a^{*},X_b^{*},X_c^{*} $ is the midpoint of $ \widehat{P}_aP, \widehat{P}_bP, \widehat{P}_cP, $ respectively and note that $ Q $ is the center of $ \odot (X_aX_bX_c) $ we get $ Q^{*} $ is the image of $ P $ under the inversion WRT $ \odot (X_a^{*}X_b^{*}X_c^{*}), $ so $ \lambda(Q^{*}) $ is the image of $ P $ under the inversion WRT $ \odot (\widehat{O}) $ $ \Longrightarrow $ $ \lambda(Q^{*}) \in \Omega_{P}. $ Since the medial triangle of $ \triangle \widehat{P}_a\widehat{P}_b\widehat{P}_c $ is the image of the pedal triangle of $ P $ WRT $ \triangle P_AP_BP_C $ under the inversion with center $ P $ that fixed $ \odot (ABC), $ so $ \lambda(R^{*}) $ is the image of $ P $ under the inversion WRT $ \odot (\widehat{N}) $ and hence it lies on $ \Omega_{P}. $ Finally$,$ since $ (H/P) $ is the perspector of $ \triangle P_AP_BP_C $ and $ \triangle X_aX_bX_c, $ so $ \lambda((H/P)^{*}) $ is the second intersection of $ \odot (P\widehat{P}_a\widehat{H}_a), \odot (P\widehat{P}_b\widehat{H}_b), \odot (P\widehat{P}_c\widehat{H}_c), $ where $ \triangle \widehat{H}_a\widehat{H}_b\widehat{H}_c $ is the orthic triangle of $ \triangle \widehat{P}_a\widehat{P}_b\widehat{P}_c, $ hence $ \lambda((H/P)^*) $ is the image of $ P $ under the inversion with center $ \widehat{H} $ that exchange $ \odot (\widehat{O}), \odot (\widehat{N}), $ so we conclude that $ \lambda((H/P)^*) \in \Omega_{P} $ and $$ (\lambda(Q^{*}), \lambda((H/P)^{*}); \lambda(R^{*}), \lambda(T^{*}))_{\Omega_{P}} = P(\lambda(Q^{*}), \lambda((H/P)^{*}); \lambda(R^{*}), \lambda(T^{*})) = (\widehat{O}, \widehat{H}; \widehat{N}, \infty) = -1. \qquad \blacksquare $$
This post has been edited 2 times. Last edited by TelvCohl, Jun 18, 2020, 4:11 PM
Orthocorrespondent

A cute problem about Radical Axis

by TelvCohl, Feb 29, 2020, 11:51 AM

Problem : Given a $ \triangle ABC $ with incenter $ I $ and a point $ P $ with pedal triangle $ \triangle P_aP_bP_c. $ Let $ A_1\in\overrightarrow{P_aB}, B_1\in\overrightarrow{P_bC}, C_1\in\overrightarrow{P_cA}, A_2\in\overrightarrow{P_aC}, B_2\in\overrightarrow{P_bA}, C_2\in\overrightarrow{P_cB} $ be the points such that $$ \overline{P_aA_1} = \overline{P_aA_2} = \overline{P_bB_1} = \overline{P_bB_2} = \overline{P_cC_1} = \overline{P_cC_2}. $$Prove that the orthocenter of $ \triangle DEF $ lies on the radical axis of $ \odot (A_1B_1C_1), \odot (A_2B_2C_2), $ where $ D, E, F $ is the projection of $ P $ on $ AI, BI, CI, $ respectively.

Proof :

Lemma : Given a $ \triangle ABC $ with circumcenter $ O $ and orthocenter $ H. $ Let $ P, Q $ be the isogonal conjugate WRT $ \triangle ABC $ and let $ \varrho $ be the circumradius of the pedal circle of $ P, Q $ WRT $ \triangle ABC. $ Then $$ 4\varrho^2 - HP^2-HQ^2=-\mathbf{Power}(H, \odot (O)). $$Proof : First, we prove the following

Claim. The radical axis of $ \odot (H, HP) $ and $ \odot (Q, 2\varrho) $ is the image of $ \odot (HQ) $ under the inversion with center $ H $ that swaps $ \odot (O) $ and the 9-point circle of $ \triangle ABC. $

Proof. Let $ T $ be the intersection of $ BC $ with the perpendicular from $ H $ to $ AP $ and $ D, U, V $ be the second intersection of $ \odot (O) $ with $ AH, AP, AQ, $ respectively, then $ HT $ is the Steiner line of $ V $ WRT $ \triangle ABC, $ so $ T \in DV $ and hence $ \triangle ADU \stackrel{-}{\sim} \triangle TKV, $ where $ K $ is the intersection of $ AQ, BC. $ Note that $ \frac{AP}{PU} = \frac{QK}{KV} $ (well-known), so $ \triangle DPU $ and $ \triangle QTV $ are inversely similar, too. Let $ P_a $ be the reflection of $ P $ in $ BC, $ then $ \measuredangle P_aHD = \measuredangle HDP = \measuredangle QTK, $ so $ HP_a \perp QT $ $ \Longrightarrow $ $ \odot (T, TP) $ passes through the intersection of $ AP, HP_a $ with $ \odot (H, HP), \odot (Q, 2\varrho), $ respectively, hence the intersection $ S $ of $ AP, HP_a $ lies on the radical axis of $ \odot (H, HP), \odot (Q, 2\varrho). $ It suffices to prove $ S $ lies on the line mentioned in Claim..

Let $ J, L $ be the intersection of $ AH, HS $ with $ BC, QT, $ respectively, then $ \measuredangle ASL = \measuredangle HTL = \measuredangle AJL $ $ \Longrightarrow $ $ A, J, L, S $ are concyclic, so $ AH \cdot HJ = LH \cdot HS. $ $ \qquad \color[rgb]{0.4,0.26,0.13} \blacksquare$

Now let $ R $ be the intersection of $ HQ $ with the radical axis $ \odot (H, HP), \odot (Q, 2\varrho), $ then by Claim. we conclude that
\begin{align*} 4\varrho^2 - HP^2 - HQ^2 &= QP_a^2 - HP^2 - HQ^2 \\ &= QR^2 - RH^2 - HQ^2\\ &= 2\ QH \cdot HR \\ &= -\mathbf{Power}(H, \odot (O)). \qquad\qquad\qquad \blacksquare \end{align*}Back to the main problem :

In this paragraph, $ i $ is a fixed number in $ \{ 1, 2 \}. $ Clearly, $ D $ is the midpoint of arc $ P_bP_c $ in $ \odot (AP) $ and $ \triangle DP_bB_i \stackrel{+}{\cong} \triangle DP_cC_i, $ so $ \triangle DP_bP_c \stackrel{+}{\sim} \triangle DB_iC_i $ and $ DB_i = DC_i. $ Similarly, $ E, F $ lie on the perpendicular bisector of $ C_iA_i, A_iB_i, $ respectively, so the circumcenter $ S_i $ of $ \triangle A_iB_iC_i $ is the orthology center of $ \triangle DEF $ WRT $ \triangle A_iB_iC_i. $ Let $ T_i $ be another orthology center of these triangles, then from $ \measuredangle B_iT_iC_i = \measuredangle FDE = \measuredangle CIB $ we get $ T_i $ lies on the circle $ \odot (D) $ with center $ D $ passing through $ B_i, C_i. $ Analogously, $ T_i $ lies on $ \odot (E), \odot (F), $ so $ \triangle A_i B_iC_i $ is the reflection triangle of $ T_i $ WRT $ \triangle DEF $ and hence $ T_i, S_i $ are isogonal conjugate WRT $ \triangle DEF. $

Let $ J $ be the orthocenter of $ \triangle DEF $ and let $ J_D $ be the projection of $ I $ on $ PP_a, $ then note that $ \triangle DEF $ and the circumcircle mid-arc triangle of $ \triangle ABC $ are inversely similar we get $ J_D\in DJ, $ so from $ J_DA_1 = J_D A_2 $ we get $ JT_1 = JT_2. $ Finally, by Lemma we conclude that $$ \mathbf{Power}(J, \odot (A_1B_1C_1))\ \stackrel{\text{Lemma}}{=}\ \mathbf{Power}(J, \odot (DEF))\ -\ JT_1^2 = \mathbf{Power}(J, \odot (DEF))\ -\ JT_2^2\ \stackrel{\text{Lemma}}{=}\ \mathbf{Power}(J, \odot (A_2B_2C_2)). \qquad\qquad\qquad \blacksquare $$Remark :

1. By the same argument we can prove the following

Corollary 1 : Given a $ \triangle ABC $ with orthocenter $ H $ and points $ U, V $ such that $ HU = HV. $ Then $ H $ lies on the radical axis of the circumcircle of reflection triangle of $ U, V $ WRT $ \triangle ABC, $ respectively.

2. Another interesting consequence of Lemma :

Corollary 2 : Given six points $ A, B, C, D, E, F $ lying on a conic that is not a rectangular hyperbola such that $ \triangle ABC $ and $ \triangle DEF $ share a common orthocenter $ T. $ Then $ T $ lies on the radical axis of $ \odot (ABC), \odot (DEF). $

3. Another interesting consequence of Lemma :

Corollary 3 : Given a $ \triangle ABC $ with orthocenter $ H $ and two pairs of isogonal conjugate $ (P, P^{*}), (Q, Q^{*}). $ Let $ \Omega_{P}, \Omega_{Q} $ be the pedal circle of $ (P, P^{*}), (Q, Q^{*}) $ WRT $ \triangle ABC, $ respectively. Then the radical axis of $ \Omega_{P}, \Omega_{Q} $ is the image of the Steiner line of $ PQP^{*}Q^{*} $ under the dilation $ (H, \tfrac{1}{2}). $
This post has been edited 2 times. Last edited by TelvCohl, Sep 1, 2024, 12:10 PM
Reason: Add Corollary 3
radical axis
Isogonal conjugate

Foci of Steiner inellipse and other triangle centers

by TelvCohl, Sep 12, 2018, 3:25 PM

Preliminaries

Lemma 1 : Given a $ \triangle ABC $ with isogonal conjugate $ (P,Q), (R,S). $ Then the Miquel point $ M $ of the quadrilateral $ PRQS $ lies on $ \odot (ABC). $ Furthermore, the Simson line of $ M $ WRT $ \triangle ABC $ is perpendicular to $ UV $ where $ U $ is the midpoint of $ PQ $ and $ V $ is the midpoint of $ RS. $

Proof :

Sublemma : Given a quadrilateral $ ABCD $ with its Miquel point $ M. $ Let $ \Phi $ be the composition of the transformation $ \mathbf{I} $ and $ \mathbf{S} $ where $ \mathbf{I} $ is the inversion with center $ M, $ power $ MA \cdot MC = MB \cdot MD $ and $ \mathbf{S} $ is the symmetry WRT the bisector of $ \angle AMC, \angle BMD. $ Then $ Y \stackrel{\Phi}{\leftrightarrow} Z $ for any isogonal conjugate $ (Y,Z) $ of the quadrilateral $ ABCD. $

Proof : Let $ \overline{Y} $ be the image of $ Y $ under $ \Phi, $ then it is clear that $ \triangle MAY \stackrel{+}{\sim} \triangle M\overline{Y}C $ and $ \triangle MDY \stackrel{+}{\sim} \triangle M\overline{Y}B, $ so $$ \measuredangle B\overline{Y}C = \measuredangle B\overline{Y}M + \measuredangle M\overline{Y}C = \measuredangle YDM + \measuredangle MAY = \measuredangle DYA + \measuredangle (AB,CD) = \measuredangle AZD = \measuredangle BZC. $$Similarly, we can prove $ \measuredangle C\overline{Y}D = \measuredangle CZD, \measuredangle D\overline{Y}A = \measuredangle DZA, \measuredangle A\overline{Y}B = \measuredangle AZB, $ so $ \overline{Y} $ coincide with $ Z. $ i.e. $ Y \stackrel{\Phi}{\leftrightarrow} Z. $ $ \qquad \blacksquare $

Back to the main problem :

Let $ \tau $ be the forth common tangent of the inconic of $ \triangle ABC $ with foci $ (P,Q), (R,S) $ and $ E, F $ be the intersection of $ \tau $ with $ CA, AB, $ respectively. Clearly, $ (P,Q), (R,S) $ are isogonal conjugate of the quadrilateral $ BCEF, $ so $ UV $ is the Newton line of the quadrilateral $ BCEF $ and hence the Miquel point $ \overline{M} $ of the quadrilateral $ BCEF $ is the pole of the Simson line of $ \triangle ABC $ with the direction $ \perp UV. $ Consider the mapping $ \Phi = \mathbf{S} \circ \mathbf{I} $ where $ \mathbf{I} $ is the inversion with center $ \overline{M}, $ power $ \overline{M}B \cdot \overline{M}E = \overline{M}C \cdot \overline{M}F $ and $ \mathbf{S} $ is the symmetry WRT the bisector of $ \angle B\overline{M}E, \angle C\overline{M}F, $ then by Sublemma we know that $ P \stackrel{\Phi}{\leftrightarrow} Q $ and $ R \stackrel{\Phi}{\leftrightarrow} S, $ so $ \overline{M}\equiv M. $ $ \qquad \blacksquare $

Lemma 2 : Given a $ \triangle ABC $ with isogonal conjugate $ (P,Q) $ and two points $ R, S $ lying on $ \odot (ABC) $ such that $ Q \in RS. $ Let $ D $ be the second intersection of $ AP $ with $ \odot (ABC) $ and $ X $ be the intersection of $ BC $ with $ DR. $ Then the Simson line of $ S $ WRT $ \triangle ABC $ is perpendicular to $ PX. $

Proof : Let $ J $ be the second intersection of $ AQ $ with $ \odot (ABC) $ and $ T $ be the intersection of $ AP, BC, $ then it is well-known that $ \frac{AQ}{QJ} = \frac{PT}{TD}, $ so notice $ \triangle AJR \stackrel{-}{\sim} \triangle XDT $ we get $ \triangle QJR \stackrel{-}{\sim} \triangle XDP $ and hence we conclude that $ \measuredangle ARS = \measuredangle (PX,BC). $ i.e. $ PX $ is perpendicular to the Simson line of $ S $ WRT $ \triangle ABC. $ $ \qquad \blacksquare $

Corollary 3 : Given a $ \triangle ABC $ with centroid $ G. $ Let $ R, S $ be two points lying on $ \odot (ABC) $ such that $ G \in RS. $ Then the Simson line of $ S $ WRT $ \triangle ABC $ is perpendicular to the tripolar of $ R $ WRT $ \triangle ABC. $

Proof : Let $ K $ be the symmedian point of $ \triangle ABC $ and $ \triangle K_aK_bK_c $ be the circumcevian triangle of $ K $ WRT $ \triangle ABC. $ By Lemma 2, the Simson line of $ S $ WRT $ \triangle ABC $ is perpendicular to $ \overline{DEF} $ where $ D, E, F $ is the intersection of $ RK_a, RK_b, RK_c $ with $ BC, CA, AB, $ respectively, but from $ R(A, K_a; B,C) $ $ = $ $ R(B, K_b; C, A) $ $ = $ $ R(C, K_c; A, B) $ $ = $ $ -1 $ we get $ D, E, F $ lie on the tripolar of $ R $ WRT $ \triangle ABC, $ so we conclude that the Simson line of $ S $ WRT $ \triangle ABC $ is perpendicular to the tripolar of $ R $ WRT $ \triangle ABC. $ $ \qquad \blacksquare $

Main result

Notation : Let $ U,V $ be the foci of the Steiner inellipse of $ \triangle ABC $ and $ G, O, N, K, F_1, F_2, S_1, S_2, T, S, E, P $ be the centroid, circumcenter, 9-point center, symmedian point, 1st Fermat point, 2nd Fermat point, 1st Isodynamic point, 2nd Isodynamic point, Tarry point, Steiner point, Euler reflection point, Parry point of $ \triangle ABC, $ respectively.

$ \bullet $ Recall that, by definition, $ T, S, E, P $ is the pole of the Simson line of $ \triangle ABC $ with the direction $ \perp OK, \parallel OK, \parallel OG, \perp GK, $ respectively.

Definition : Let $ \Omega $ be the composition of $ \mathbf{I} $ and $ \mathbf{S} $ where $ \mathbf{I} $ is the inversion with center $ G, $ power $ GU^2 = GV^2 $ and $ \mathbf{S} $ is the symmetry WRT the line $ UV. $

Property 1 : Let $ \triangle A_2B_2C_2 $ be the 2nd Brocard triangle of $ \triangle ABC. $ Then $ A \stackrel{\Omega}{\leftrightarrow} A_2, B \stackrel{\Omega}{\leftrightarrow} B_2, C \stackrel{\Omega}{\leftrightarrow} C_2. $

Proof : First, we prove that $ AU \cdot AV = AA_2 \cdot AG. $ Let $ M_a $ be the midpoint of $ BC $ and $ U_a, V_a $ be the second intersection of $ AU, AV $ with $ \odot (UBC), \odot (VBC), $ respectively. Since $ UM_a $ and $ VM_a $ are symmetric WRT $ BC, $ so the U-median of $ \triangle UBC $ passes through the reflection of $ V $ in $ BC $ which is the isogonal conjugate of $ A $ WRT $ \triangle UBC, $ hence $ AU $ is the U-symmedian of $ \triangle UBC $ and $ UBU_aC $ is a harmonic quadrilateral $ \Longrightarrow $ $ \overline{BM_aC} $ is the bisector of $ \angle UM_a U_a $ and $ V\in M_aU_a. $ Similarly, we can prove $ U $ lies on $ M_aV_a. $ Since $ AU \cdot AV_a = AB \cdot AC = AV \cdot AU_a, $ so $ UV \parallel U_aV_a $ $ \Longrightarrow $ the midpoint $ N_a $ of $ U_aV_a $ lies on $ AM_a $ and $ (A,M_a;G,N_a) = -1, $ hence from $ \frac{AM_a}{GM_a} = 3 $ we get $ \frac{AG}{AN_a} = \frac{1}{3} $ $ \Longrightarrow $ $ AU \cdot AV = \frac{AB \cdot AC}{3} = AA_2 \cdot AG. $

Since $ U, V $ are isogonal conjugate WRT $ \triangle ABC $ and $ AG $ is the A-median of $ \triangle AUV, $ so $ AA_2 $ is the A-symmedian of $ \triangle AUV, $ hence from $ AU \cdot AV = AA_2 \cdot AG $ we get $ AUA_2V $ is a harmonic quadrilateral $ \Longrightarrow $ $ \overline{UGV} $ is the bisector of $ \angle AGA_2 $ and $ GA \cdot GA_2 = GU^2 = GV^2. $ i.e. $ A \stackrel{\Omega}{\leftrightarrow} A_2. $ $ \qquad \blacksquare $

Corollary 1.1 : Let $ \triangle A_1B_1C_1 $ be the 1st Brocard triangle of $ \triangle ABC $ and let $ X \in \odot (ABC), X_1\in \odot (A_1B_1C_1) $ be the points such that $ \triangle ABC \cup X \stackrel{-}{\sim} \triangle A_1B_1C_1 \cup X_1. $ Then $ X \stackrel{\Omega}{\leftrightarrow}\overline{X_1}, X_1 \stackrel{\Omega}{\leftrightarrow} \overline{X} $ where $ \overline{X}, \overline{X_1} $ is the second intersection of $ GX, GX_1 $ with $ \odot (ABC), \odot (A_1B_1C_1), $ respectively.

Property 2 : $ E, G, T $ are collinear and $ E \leftrightarrow G $ under the inversion WRT the Brocard circle $ \odot (OK) $ of $ \triangle ABC. $

Proof : Let $ M $ be the midpoint of $ OK, $ then $$ \triangle ABC\ \cup\ G\ \cup\ O\ \cup\ T\ \stackrel{-}{\sim} \triangle A_1B_1C_1\ \cup\ G\ \cup\ M\ \cup\ O \Longrightarrow \triangle GOT \stackrel{-}{\sim} \triangle GMO, $$so $ \measuredangle OGT = \measuredangle OGM $ and hence $ G, M, T $ are collinear. Since the Simson line of $ S $ WRT $ \triangle ABC $ is parallel to $ OK, $ so $ \measuredangle GTS = \measuredangle GTO = \measuredangle MOG = \measuredangle (OK,OG) = \measuredangle ETS $ $ \Longrightarrow $ $ E, G, , T $ are collinear. Finally, from $ \measuredangle OEM = \measuredangle GTO = \measuredangle MOG $ we get $ \triangle MGO \stackrel{-}{\sim} \triangle MOE, $ so we conclude that $ MO^2 = ME \cdot MG. $ i.e. $ E \leftrightarrow G $ under the inversion WRT $ \odot (OK). $ $ \qquad \blacksquare $

Corollary 2.1 : $ E $ lies on the Parry circle $ \odot (GS_1S_2) $ of $ \triangle ABC. $

Proof : From Property 2 we get $ E \leftrightarrow G $ under the inversion WRT $ \odot (OK), $ but this mapping also exchanges $ S_1, S_2, $ so $ E, G, S_1, S_2 $ are concyclic. $ \qquad \blacksquare $

Property 3 : $ E \stackrel{\Omega}{\leftrightarrow} O,\ K \stackrel{\Omega}{\leftrightarrow} P,\ F_1 \stackrel{\Omega}{\leftrightarrow} F_2,\ S_1 \stackrel{\Omega}{\leftrightarrow} S_2. $

Proof : Since $ \triangle ABC \cup T \stackrel{-}{\sim} \triangle A_1B_1C_1 \cup O, $ so from Corollary 1.1 and Property 2 we get $ E \stackrel{\Omega}{\leftrightarrow} O. $ Since the Parry circle of $ \triangle ABC $ passes through $ E $ and is orthogonal to $ \odot (O), $ so its image under $ \Omega $ passes through $ O $ and the center $ \odot (OK) $ $ \Longrightarrow $ $ \Omega $ exchanges the Brocard axis and the Parry circle of $ \triangle ABC, $ hence note that $ S_1, S_2 $ lie on both of them we get $ S_1 \stackrel{\Omega}{\leftrightarrow} S_2. $ Note that $ G $ lies on $ F_1S_2, F_2S_1, $ $ K $ lies on $ F_1F_2, S_1S_2 $ and $ F_1S_1 \parallel F_2S_2 $ (see Collinearity of Kiepert perspectors (Application (1))), so $ F_1 \stackrel{\Omega}{\leftrightarrow} F_2. $ Furthermore, $ GK $ passes through the midpoint of $ F_1S_1 $ and the midpoint of $ F_2S_2, $ so by Lemma 1 we get $ P $ is the Miquel point of the quadrilateral $ F_1S_1F_2S_2 $ and hence we conclude that $ K \stackrel{\Omega}{\leftrightarrow} P. $ $ \qquad \blacksquare $

Corollary 3.1 : $ P $ lies on $ EK $ and $ GS. $

Proof : Since $ P, K $ lie on $ \odot (ABC), \odot (A_1B_1C_1), $ respectively and $ \triangle ABC \cup S \stackrel{-}{\sim} \triangle A_1B_1C_1 \cup K, $ so from Corollary 1.1 we get $ P \in GS. $ Since $ EP $ is the radical axis of $ \odot (O) $ and the Parry circle of $ \triangle ABC, $ so note that $ (O,K; S_1, S_2) = -1 $ we get $ K\in EP. $ $ \qquad \blacksquare $

Corollary 3.2 : $ E, N $ are isogonal conjugate WRT $ \triangle GF_1F_2. $

Proof : From Collinearity of Kiepert perspectors (Application (3)) we know that $ N $ is the second intersectioN of $ OG $ with $ \odot (OF_1F_2), $ so by Property 3 we get $ E, N $ are isogonal conjugate WRT $ \triangle GF_1F_2. $ $ \qquad \blacksquare $

Corollary 3.3 : $ EUOV, KUPV, F_1UF_2V, S_1US_2V $ are harmonic quadrilateral.

Property 4 : Let $ Q $ be the reflection of $ S $ in $ OG, $ then $ P,Q $ are the intersection of $ \odot (O) $ with $ \odot (GOK). $

Proof : Firom $ \measuredangle GPK = \measuredangle SPE = \measuredangle GOK $ we get $ G, O, K, P $ are concyclic, but $ \measuredangle QOG = \measuredangle QPS = \measuredangle QPG $ $ \Longrightarrow $ $ G, O, P, Q $ are concyclic, so $ G, O, K, P, Q $ lie on a circle. $ \qquad \blacksquare $

Property 5 : Let $ R $ be the reflection of $ E $ in $ OG, $ then $ R $ is the second intersection of $ \odot (O) $ with $ \odot (GF_1F_2). $

Proof : Since $ \measuredangle RET = \measuredangle (\perp OG, GE) = \measuredangle (\perp OK, OE), $ so the Simson line of $ R $ WRT $ \triangle ABC $ is parallel to $ OE $ $ \Longrightarrow $ $ \measuredangle GPR = \measuredangle SPR = \measuredangle EOK = \measuredangle EGO = \measuredangle OGR, $ hence we get $ \odot (GRV) $ is tangent to $ OG $ at $ G. $ On the other hand, from the proof of Collinearity of Kiepert perspectors (Application (3)) we get $ F_1F_2 $ passes through the midpoint $ J $ of $ GH $ and $ JF_1 \cdot JF_2 = JG^2, $ so $ \odot (GF_1F_2) $ is also tangent to $ OG $ at $ G, $ hence we conclude that $ G, F_1, F_2, P, R $ are concyclic. $ \qquad \blacksquare $

Property 6 : $ GK, OK, EK, F_1F_2 $ is the tripolar of $ S, E, Q, R $ WRT $ \triangle ABC, $ respectively.

Proof : By Connection between Inconic and Circumconic (Lemma 1), the tripolar of any point lying on $ \odot (O) $ WRT $ \triangle ABC $ passes through $ K, $ so by Corollary 3 we get $ GK, OK $ is the tripolar of $ S, E $ WRT $ \triangle ABC, $ respectively. Let $ \widehat{P} $ be the reflection of $ P $ in $ OG, $ then $ \measuredangle \widehat{P}PE = \measuredangle (\perp OG, EK) $ $ \Longrightarrow $ the Simson line of $ \widehat{P} $ WRT $ \triangle ABC $ is perpendicular to $ EK, $ so note that $ G \in \widehat{P}Q $ we get $ EK $ is the tripolar of $ Q $ WRT $ \triangle ABC $ by Corollary 3. Let $ \widehat{T} $ be the reflection of $ T $ in $ OG, $ then note that $ GE $ is the O-midline of $ \triangle O F_1F_2 $ we get $ \measuredangle \widehat{T}TE = \measuredangle (\perp OG, GE) = \measuredangle (OG, \perp F_1F_2), $ so the Simson line of $ \widehat{T} $ WRT $ \triangle ABC $ is perpendicular to $ F_1F_2 $ and hence by Corollary 3 again we conclude that $ F_1F_2 $ is the tripolar of $ R $ WRT $ \triangle ABC. $ $ \qquad \blacksquare $
Fermat points
Steiner inellipse

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
X1138

Collinearity of Kiepert perspectors

by TelvCohl, Sep 1, 2018, 7:38 AM

Preliminaries

Lemma 1 : Given a $ \triangle ABC $ with isogonal conjugate $ (P,P^*), (Q,Q^*). $ Then $ S $ $ \equiv $ $ PQ $ $ \cap $ $ P^*Q^* $ and $ T $ $ \equiv $ $ PQ^* $ $ \cap $ $ P^*Q $ are isogonal conjugate WRT $ \triangle ABC. $

Proof : Let $ X, Y $ be the intersection of $ AQ $ with $ PQ^*, P^*Q^*, $ respectively. Since $$ A(Q,Q^*;T,P) = (X,Q^*;T,P) \stackrel{Q}{=} (Y,Q^*;P^*,S) = A(Q,Q^*;P^*,S) = A(Q^*,Q;S,P^*), $$so $ (AS, AT) $ are isogonal conjugate WRT $ \angle A. $ Similarly, $ (BS, BT), (CS, CT) $ are isogonal conjugate WRT $ \angle B, \angle C, $ respectively, so we conclude that $ S, $ $ T $ are isogonal conjugate WRT $ \triangle ABC. $ $ \qquad \blacksquare $

Lemma 2 : Given a $ \triangle ABC $ and a point $ U $ lying on the perpendicular bisector of $ BC. $ Let $ V $ be the point such that $ \measuredangle ABU = \measuredangle VCB, \measuredangle ACU = \measuredangle VBC. $ Then $ A, U, V $ are collinear.

Proof : Let $ W $ be the reflection of $ A $ in the perpendicular bisector of $ BC, $ then $ \measuredangle UCW = \measuredangle VCB, \measuredangle UBW = \measuredangle VBC $ $ \Longrightarrow $ $ V,W $ are isogonal conjugate WRT $ \triangle BUC, $ so we conclude that $ A, U, V $ are collinear. $ \qquad \blacksquare $

Lemma 3 : Given a $ \triangle ABC $ with circumcircle $ \odot (O, R) $ and 9-point center $ N. $ Let $ P,Q $ be the isogonal conjugate WRT $ \triangle ABC $ and $ M $ be the midpoint of $ PQ. $ Then $ MN = \frac{OP \cdot OQ}{2R}. $

Proof : Let $ \triangle A_1B_1C_1 $ be the circumcevian triangle of $ P $ WRT $ \triangle ABC $ and $ A_2, B_2, C_2 $ be the reflection of $ A_1, B_1, C_1 $ in $ BC, CA, AB, $ respectively, then from Properties of Hagge circle (Property 1, Corollary 1.1, Property 3) we get $ H \in \odot (A_2B_2C_2), $ the reflection $ T $ of $ Q $ in $ N $ is the circumcenter of $ \triangle A_2B_2C_2 $ and $ \triangle A_1B_1C_1 \cup P \stackrel{-}{\sim} \triangle A_2B_2C_2 \cup P $ where $ H $ is the orthocenter of $ \triangle ABC, $ so $$ \frac{OP}{2MN} = \frac{OP}{TP} = \frac{\text{circumradius of }\triangle A_1B_1C_1}{\text{circumradius of }\triangle A_2B_2C_2} = \frac{R}{HT} = \frac{R}{OQ}. \qquad \blacksquare $$
Main result

Notation : Given a $ \triangle ABC $ with centroid $ G, $ circumcenter $ O, $ orthocenter $ H, $ 9-point center $ N, $ symmedian point $ K $ and $ \theta \in \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right ]. $ Let $ \triangle A_{\theta}B_{\theta}C_{\theta} $ be the Kiepert triangle of $ \triangle ABC $ with angle $ \theta, $ $ K_{\theta} $ be the perspector of $ \triangle ABC, \triangle A_{\theta}B_{\theta}C_{\theta}, $ and let $ A^*_{\theta}, B^*_{\theta}, C^*_{\theta}, K^*_{\theta} $ be the isogonal conjugate of $ A_{\theta}, B_{\theta}, C_{\theta}, K_{\theta} $ WRT $ \triangle ABC, $ respectively.

Property 1 : $ \color{blue} K \in K_{\theta}K_{-\theta}. $

Proof : Let $ \triangle T_aT_bT_c $ be the tangential triangle of $ \triangle ABC. $ Note that $ \measuredangle CBC_{-\theta} = \measuredangle T_bCB_{\theta}, \measuredangle BCB_{-\theta} = \measuredangle T_cBC_{\theta}, $ so we conclude that $$ B(C,T_b; B_{\theta},B_{-\theta}) = C(B,T_b; B_{\theta},B_{-\theta}) = B(C,T_c;C_{\theta},C_{-\theta}) = C(B,T_c;C_{\theta},C_{-\theta}). $$i.e. $ K \equiv BT_b \cap CT_c, K_{\theta} \equiv BB_{\theta} \cap CC_{\theta}, K_{-\theta} \equiv BB_{-\theta} \cap CC_{-\theta} $ are collinear. $ \qquad \blacksquare $

Corollary 1.1 : $ \color{blue} G \in K_{\theta}K^*_{-\theta}. $

Proof : From Property 1 we get $ K $ is the intersection of $ K_{\theta}K_{-\theta}, K^*_{\theta}K^*_{-\theta}, $ so by Lemma 1 we conclude that $ G $ lies on $ K_{\theta}K^*_{-\theta}, K_{-\theta}K^*_{\theta}. $ $ \qquad \blacksquare $

Property 2 : $ \color{blue} H \in A_{\theta}A^*_{\frac{\pi}{2}-\theta}, B_{\theta}B^*_{\frac{\pi}{2}-\theta}, C_{\theta}C^*_{\frac{\pi}{2}-\theta}, K_{\theta}K^*_{\frac{\pi}{2}-\theta}. $

Proof : Simple angle chasing yields $ \measuredangle HBA_{\theta} = \measuredangle A^*_{\frac{\pi}{2} - \theta}CB, \measuredangle HCA_{\theta} = \measuredangle A^*_{\frac{\pi}{2} - \theta}BC, $ so by Lemma 2 we get $ H $ lies on $ A_{\theta}A^*_{\frac{\pi}{2}-\theta}. $ Similarly, we can prove $ H \in B_{\theta}B^*_{\frac{\pi}{2}-\theta}, C_{\theta}C^*_{\frac{\pi}{2}-\theta}. $ By Desargues's theorem for $ \triangle BC_{\theta}C^*_{\frac{\pi}{2}-\theta} $ and $ \triangle CB_{\theta}B^*_{\frac{\pi}{2}-\theta} $ $ \Longrightarrow $ $ BC, B_{\theta}C_{\theta}, B^*_{\frac{\pi}{2}-\theta}C^*_{\frac{\pi}{2}-\theta} $ are concurrent, so by Desargues's theorem for $ \triangle BB_{\theta}B^*_{\frac{\pi}{2}-\theta} $ and $ \triangle CC_{\theta}C^*_{\frac{\pi}{2}-\theta} $ we conclude that $ H $ lies on $ K_{\theta}K^*_{\frac{\pi}{2}-\theta}. $ $ \qquad \blacksquare $

Corollary 2.1 : $ \color{blue} O \in A^*_{\theta}A^*_{\frac{\pi}{2}-\theta}, B^*_{\theta}B^*_{\frac{\pi}{2}-\theta}, C^*_{\theta}C^*_{\frac{\pi}{2}-\theta}, K_{\theta}K_{\frac{\pi}{2}-\theta}. $

Proof : From Property 2 we get $ H $ is the intersection of $ K_{\theta}K^*_{\frac{\pi}{2}-\theta}, K_{\frac{\pi}{2}-\theta}K^*_{\theta}, $ so by Lemma 1 $ \Longrightarrow $ $ O $ lies on $ K_{\theta}K_{\frac{\pi}{2}-\theta}. $ Analogously, we can prove $ O \in A^*_{\theta}A^*_{\frac{\pi}{2}-\theta}, B^*_{\theta}B^*_{\frac{\pi}{2}-\theta}, C^*_{\theta}C^*_{\frac{\pi}{2}-\theta}. $ $ \qquad \blacksquare $

Property 3 : Given $ \color{blue} \phi, \sigma \in \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right ]. $ Then $ \color{blue} K^*_{-(\phi + \sigma)} \in K_{\phi} K_{\sigma}. $

Proof : Note that for a fixed $ \tau \in \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right ] $ the mapping $ \mathbb{I}: K_{\theta} \mapsto K_{\tau - \theta} $ is an involution on the Kiepert hyperbola of $ \triangle ABC, $ so $ K_{\theta} K_{\tau - \theta} $ passes through the pole $ X $ of $ \mathbb{I} $ for all $ \theta \in \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right ]. $ Consider the case when $ \theta = 0 $ and $ \theta = \frac{\pi}{2} $ we conclude that $ X = K^*_{-\tau} $ by Corollary 1.1 and Property 2 $ \Longrightarrow $ $ K^*_{-\tau} \in K_{\theta} K_{\tau - \theta}. $ $ \qquad \blacksquare $

Property 4 : Let $ \color{blue} T $ be the intersection of $ \color{blue} OH $ with $ \color{blue} A_{\theta}A^*_{\theta}. $ Then $ \color{blue} \frac{OH}{OT} = 4\cos^2\theta-1. $

Proof : Let $ D $ be the intersection of $ AA^*_{\theta} $ with the perpendicular bisector of $ BC $ and let $ BD = CD = \eta, $ then from $ \measuredangle ADO = \measuredangle DAH = \measuredangle OAA_{\theta} $ we get $ OA^2 = OD \cdot OA_{\theta} $ $ \Longrightarrow $ $ D $ is the image of $ A_{\theta} $ under the inversion WRT $ \odot(O), $ so $ \eta = \frac{BC \cdot R}{2 \cdot OA_{\theta} \cdot \cos \theta}. $ On the other hand, by Lemma 2 we get $ \measuredangle A^*_{\theta}CD = \measuredangle ABC, \measuredangle A^*_{\theta}BD = \measuredangle ACB, $ so $$ \frac{HA^*_{\theta}}{A_{\frac{\pi}{2}-\theta}A^*_{\theta}} = \frac{AA^*_{\theta}}{DA^*_{\theta}} = \frac{2R \cdot \sin\theta}{\eta} = \frac{4 \cdot OA_{\theta} \cdot \sin \theta \cdot \cos \theta}{BC},$$hence by Menelaus' theorem for $ \triangle OHA_{\frac{\pi}{2}-\theta} $ and $ \overline{A^*_{\theta}A_{\theta}T} $ we conclude that $$ \frac{HT}{OT} = \frac{HA^*_{\theta}}{A_{\frac{\pi}{2}-\theta}A^*_{\theta}} \cdot \frac{A_{\frac{\pi}{2}-\theta}A_{\theta}}{OA_{\theta}} = -2\cos 2\theta = -2\left(2\cos^2\theta -1 \right ) \Longrightarrow \frac{OH}{OT} = 4\cos^2\theta-1. \qquad \blacksquare $$
Corollary 4.1 : $ \color{blue} A_{\theta}A^*_{\theta}, A_{-\theta}A^*_{-\theta}, B_{\theta}B^*_{\theta}, B_{-\theta}B^*_{-\theta}, C_{\theta}C^*_{\theta}, C_{-\theta}C^*_{-\theta}, K_{\theta}K^*_{\theta}, K_{-\theta}K^*_{-\theta} $ are concurrent on $ \color{blue} OH. $

Proof : From Property 4, $ A_{\theta}A^*_{\theta}, A_{-\theta}A^*_{-\theta}, B_{\theta}B^*_{\theta}, B_{-\theta}B^*_{-\theta}, C_{\theta}C^*_{\theta}, C_{-\theta}C^*_{-\theta} $ are concurrent at $ V. $ By Desargues's theorem for $ \triangle BC_{\theta}C^*_{\theta} $ and $ \triangle CB_{\theta}B^*_{\theta} $ we get $ BC, B_{\theta}C_{\theta}, B^*_{\theta}C^*_{\theta} $ are concurrent, so by Desargues's theorem for $ \triangle BB_{\theta}B^*_{\theta} $ and $ \triangle CC_{\theta}C^*_{\theta} $ we conclude that $ V $ lies on $ K_{\theta}K^*_{\theta}. $ Analogously, we can prove $ V \in K_{-\theta}K^*_{-\theta}. $ $ \qquad \blacksquare $

Property 5 : $ \color{blue} N \in K_{\theta}K_{\frac{\pi}{2}+\theta}. $

Proof : Let $ D_{-\theta}, E_{-\theta}, F_{-\theta} $ be the midpoint of $ AA_{-\theta}, BB_{-\theta}, CC_{-\theta}, $ respectively. First, note that $ AB_{\theta}A_{-\theta}C_{\theta}, BC_{\theta}B_{-\theta}A_{\theta}, CA_{\theta}C_{-\theta}B_{\theta} $ are parallelogram, so $ \triangle D_{-\theta}E_{-\theta}F_{-\theta} $ is the medial triangle of $ \triangle A_{\theta}B_{\theta}C_{\theta}.$ On the other hand, simple angle chasing yields the second intersection of $ AA_{\frac{\pi}{2}-\theta} $ with $ \odot (A_{\frac{\pi}{2}-\theta}BC) $ lies on $ \odot (B_{\theta}), \odot (C_{\theta}) $ where $ \odot (B_{\theta}), \odot (C_{\theta}) $ is the circle with center $ B_{\theta}, C_{\theta} $ and containing $ A, $ so $ AK_{\frac{\pi}{2}-\theta} \perp E_{\theta} F_{\theta}. $ Analogously, we can prove $ BK_{\frac{\pi}{2}-\theta} \perp F_{\theta} D_{\theta}, CK_{\frac{\pi}{2}-\theta} \perp D_{\theta} E_{\theta}, $ so by Sondat's theorem for $ \triangle ABC $ and $ \triangle D_{-\theta}E_{-\theta}F_{-\theta} $ we conclude that $ N \in K_{-\theta} K_{\frac{\pi}{2}-\theta}. $ $ \qquad \blacksquare $

Application

Notation : Let $ G, O, N, K, F_1, F_2, S_1, S_2 $ be the centroid, circumcenter, 9-point center, symmedian point, 1st Fermat point, 2nd Fermat point, 1st Isodynamic point, 2nd Isodynamic point of $ \triangle ABC, $ respectively.

Theorem (Basic properties of Fermat points and Isodynamic points) :

(1) $ \qquad $ $ G \in F_1S_2, F_2S_1. $ $ \qquad $ $ K \in F_1F_2, S_1S_2. $ $ \qquad $ $ \overline{GON} \parallel F_1S_1 \parallel F_2S_2. $ $ \qquad $ (2) $ \qquad $ $GK $ is the G-symmedian of $ \triangle GF_1F_2. $ $ \qquad $ (3) (Lester circle) $ O, N, F_1, F_2 $ are concyclic.

Proof : (1) is a consequence of the results in the previous section. Let $ T $ be the intersection of $ GO, F_1F_2 $ and $ \triangle^{1}_{\mathbf{N}}, \triangle^{2}_{\mathbf{N}} $ be the 1st Napoleon triangle, 2nd Napoleon triangle of $ \triangle ABC, $ respectively. First, note that $ O $ lies on $ S_1S_2, $ so $ T $ is the reflection of $ O $ in $ G. $ Since $ G $ is the center of $ \triangle^{i}_{\mathbf{N}}, $ $ F_{j} $ lies on the circumcircle of $ \triangle^{i}_{\mathbf{N}} $ and $ O,F_i $ are isogonal conjugate of $ \triangle^{i}_{\mathbf{N}} $ for $ \{ i, j \} = \{ 1, 2 \}, $ so by Lemma 3 we get $$ \frac{KF_1}{KF_2} = \frac{TF_1}{TF_2} = \frac{GO \cdot GF_1}{GO \cdot GF_2} \cdot \frac{\text{circumradius of }\triangle^{2}_{\mathbf{N}}}{\text{circumradius of }\triangle^{1}_{\mathbf{N}}} = \left ( \frac{GF_1}{GF_2} \right )^2, $$hence $ GK $ is the G-symmedian of $ \triangle GF_1F_2 $ and (2) is proved. Finally, from the previous conclusion and $ (T,K; F_1, F_2) = -1 $ we get $ \overline{GON} $ is tangent to $ \odot (GF_1F_2) $ at $ G, $ so $ TO \cdot TN = {TG}^2 = TF_1 \cdot TF_2 $ and hence we conclude that $ O, N, F_1, F_2 $ are concyclic. i.e. (3) is proved. $ \qquad \blacksquare $

Remark : It is possible to prove (2) or (3) directly.

1. Proof of (2) without using (1) :

Since $ F_1, F_2 $ are antigonal conjugate WRT $ \triangle ABC, $ so the center of the Kiepert hyperbola $ \mathcal{K} $ of $ \triangle ABC $ is the midpoint of $ F_1F_2 $ $ \Longrightarrow $ the isogonal conjugate of $ \mathcal{K} $ WRT $ \triangle GF_1F_2 $ is the perpendicular bisector of $ F_1F_2, $ hence the tangent of $ \mathcal{K} $ at $ G $ is the G-symmedian of $ \triangle GF_1F_2. $ On the other hand, $ \mathcal{K} $ is the isotomic conjugate of $ GK $ WRT $ \triangle ABC, $ so $ GK $ is tangent to $ \mathcal{K} $ at $ G $ and hence $ GK $ is the G-symmedian of $ \triangle GF_1F_2. $ $ \qquad \blacksquare $

2. Proof of (3) without using (1) and (2) :

Lemma : Given a $ \triangle ABC $ and a circle $ \Omega $ passing through $ B $ and $ C. $ Let $ E, F $ be the intersection of $ \Omega $ with $ CA, AB, $ respectively and let $ Y \in BE, Z \in CF $ be the points such that $ \frac{BE}{EY} = \frac{CF}{FZ} = \kappa. $ Suppose that $ U, V $ is the intersection of the tangent of $ \odot (AYZ) $ at $ A $ with $ BC, YZ, $ respectively, then $ \frac{UA}{AV} = \kappa. $

Proof : Let $ B^* \in CA, C^* \in AB $ be the points such that $ \frac{BA}{AC^*} = \frac{CA}{AB^*} = \kappa $ and let $ A^* $ be the intersection of $ B^*Z, C^*Y. $ Clearly, $ \triangle ABC, \triangle A^*B^*C^* $ are homothetic and $ \triangle AEY \stackrel{-}{\sim} \triangle AFZ, $ so $$ \frac{VY}{VZ} = \left ( \frac{AY}{AZ} \right )^2 = \left ( \frac{AE}{AF} \right )^2 = \frac{AE}{AF} \cdot \frac{AB}{AC} = \frac{C^*Y}{B^*Z} \cdot \frac{A^*B^*}{A^*C^*}, $$hence by Menelaus' theorem we conclude that $ V \in B^*C^* $ $ \Longrightarrow $ $ \frac{UA}{AV} = \frac{\text{dist}(A,BC)}{\text{dist}(A,B^*C^*)} = \kappa. $ $ \qquad \blacksquare $

Back to the main problem :

Let $ \triangle N^{1}_aN^{1}_bN^{1}_c, \triangle N^{2}_aN^{2}_bN^{2}_c $ be the 1st Napoleon triangle, 2nd Napoleon triangle of $ \triangle ABC, $ respectively and let $ T^* $ be the intersection of $ F_1F_2 $ with the tangent of $ \odot (GF_1F_2) $ at $ G. $ Since $ F_1, F_2 $ is the reflection of $ N^{2}_a, N^{1}_a $ in $ GN^{1}_a, GN^{2}_a, $ respectively, so by Lemma for $ \triangle GN^{1}_aN^{2}_a $ we get the reflection $ \overline{T} $ of $ T^* $ in $ G $ lies on the perpendicular bisector of $ BC. $ Similarly, we can prove $ \overline{T} $ lies on the perpendicular bisector of $ CA, AB, $ so $ \overline{T} \equiv O $ and hence we conclude that $ T^*N \cdot T^*O = T^*G^2 = T^*F_1 \cdot T^*F_2 $ $ \Longrightarrow $ $ O, N, F_1, F_2 $ are concyclic. $ \qquad \blacksquare $
Kiepert perspector
Fermat points

Connection between Inconic and Circumconic

by TelvCohl, Aug 7, 2018, 2:57 PM

Notation : Given a $ \triangle ABC $ and a point $ T. $ Let $ \tau (T, \triangle ABC), $ $ \mathcal{C} (T, \triangle ABC), $ $ \mathcal{K} (T, \triangle ABC) $ be the tripolar of $ T,$ inconic of $ \triangle ABC $ with perspector $ T, $ circumconic of $ \triangle ABC $ with perspector $ T $ WRT $ \triangle ABC, $ respectively. When no confusion can be caused, we abbreviate $ \square (T, \triangle ABC) $ as $ \square_{T}. $

Property : Given a $ \triangle ABC $ with isogonal conjugate $ (P,Q), (R,S). $ Then $ \tau_R $ is tangent to the $ \mathcal{K}_{P} $ if and only if $ S\in \mathcal{C}_{Q}. $

Proof :

Lemma 1 : Given a $ \triangle ABC $ with isogonal conjugate $ (P,Q), (R,S). $ Then $ P\in\tau_{R} \Longleftrightarrow R\in\mathcal{K}_{P} \Longleftrightarrow S \in \tau_{Q}. $

Proof : First, when a line $\ell $ passing through $ P $ varies around $ P, $ the mapping $ \ell \cap AB \mapsto \ell \cap AC $ is a homography between $ AB $ and $ AC, $ so the mapping sending the harmonic conjugate of $ \ell \cap AB $ WRT $ (A,B) $ to the harmonic conjugate of $ \ell \cap AC $ WRT $ (A,C) $ is a homography, too $ \Longrightarrow $ the locus of the tripole of $ \ell $ WRT $ \triangle ABC $ is a circumconic of $ \triangle ABC, $ hence to prove the first part it suffices to prove $ P\in\tau_{R} $ when $ R \in \mathcal{K}_{P}. $ Let $ \triangle P^aP^bP^c $ be the anticevian triangle of $ P $ WRT $ \triangle ABC, $ then $ \mathcal{K}_{P} $ is tangent to $ P^bP^c, P^cP^a, P^aP^b $ at $ A, B, C, $ respectively. Let $ \triangle R_aR_bR_c $ be the cevian triangle of $ R $ WRT $ \triangle ABC $ and assume $ R\in\mathcal{K}_{P}, $ then $ P^a $ lie on the polar $ R_bR_c $ of $ R_a $ WRT $ \mathcal{K}_{P}, $ so note that $ \tau_{R} $ passes through the harmonic conjugate of $ R_b, R_c $ WRT $ (A,C), (A,B), $ respectively we get $ P\in\tau_{R}. $

For the second part, let $ X_P, X_Q $ be the intersection of $ BC $ with $ \tau_{P}, \tau_{Q}, $ respectively, then $ AX_P, AX_Q $ are isogonal conjugate WRT $ \angle A $ $ \Longrightarrow $ the isogonal conjugate of $ \tau_{Q} $ WRT $ \triangle ABC $ is tangent to $ P^bP^c $ at $ A. $ Similarly, we can prove the isogonal conjugate of $ \tau_{Q} $ WRT $ \triangle ABC $ is tangent to $ P^cP^a, P^aP^b $ at $ B,C, $ respectively, so $ \mathcal{K}_{P} $ is the isogonal conjugate of $ \tau_{Q} $ WRT $ \triangle ABC $ $ \qquad \blacksquare $

Lemma 2 : Given a $ \triangle ABC $ and a point $ P. $ Then the perspector of the circumconic of $ \triangle ABC $ with center $ P $ is the isotomcomplement of the anticomplement of $ P $ WRT $ \triangle ABC. $

Proof : Let $ Q $ be the anticomplement of $ P $ WRT $ \triangle ABC, $ $ R $ be the isotomcomplement of $ Q $ WRT $ \triangle ABC $ and $ A_P $ be the reflection of $ A $ in $ P. $ By Lemma 1, it suffices to prove that the isogonal conjugate $ A_P^* $ of $ A_P $ WRT $ \triangle ABC $ lies on $ \tau_S $ where $ S $ is the isogonal conjugate $ R $ WRT $ \triangle ABC. $ Let $ \triangle Q_aQ_bQ_c $ be the cevian triangle of $ Q $ WRT $ \triangle ABC, $ $ \triangle R^aR^bR^c $ be the anticevian triangle of $ R $ WRT $ \triangle ABC $ and $ B_s, C_s $ be the intersection of $ CA, AB $ with $ \tau_S, $ respectively. Note that $ \triangle Q_aQ_bQ_c, \triangle R^aR^bR^c$ are homothetic and $ BQ \parallel CA_P, CQ \parallel BA_P, $ so we conclude that $$ B(C,A_P^*;B_s,A) = B(A,A_P;R^a,C) = Q_c(A,C;Q_a,\parallel BC) = Q_b(B,A;Q_a,\parallel BC) = C(A,A_P;B, R^a) = C(B,A_P^*;A,C_s). $$i.e. $ A_P^*\in\tau_S. $ $ \qquad \blacksquare $

Corollary 2.1 : Given a $ \triangle ABC $ and points $ P, Q. $ Then $ \tau_Q $ is tangent to $ \mathcal{K}_{P} $ if and only if $ P \in \mathcal{C}_{Q}. $

Proof : Consider the homology taking $ Q $ to the centroid of $ \triangle ABC, $ then the desired result simplified follows from Lemma 2. $ \qquad \blacksquare $

Back to the main problem :

By Lemma 1, $ \mathcal{K}_{P}, \mathcal{K}_{S} $ is the isogonal conjugate of $ \tau_Q, \tau_R $ WRT $ \triangle ABC, $ respectively, so $ \tau_R $ is tangent to the $ \mathcal{K}_{P} $ if and only if $ \tau_Q $ is tangent to $ \mathcal{K}_{S}. $ But from Corollary 2.1 we know $ \tau_Q $ is tangent to $ \mathcal{K}_{S} $ if and only if $ S \in \mathcal{C}_{Q}, $ so we conclude that $ \tau_R $ is tangent to the $ \mathcal{K}_{P} $ if and only if $ S\in \mathcal{C}_{Q}. $ $ \qquad \blacksquare $
This post has been edited 2 times. Last edited by TelvCohl, Feb 14, 2020, 10:37 AM
circumconic
inconic

Ptolemy triangle and QA-Orthopole-circle

by TelvCohl, Feb 28, 2018, 11:08 AM

Notation

For two lines $ \tau, \eta $ and a points $ P, $ let $ \mathbb{T}_{\tau,\eta}(P) $ be the midpoint between the reflection of $ P $ in $ \tau $ and the reflection of $ P $ in $ \eta. $

Preliminaries

Lemma 1 : Given two lines $ \ell_1, \ell_2 $ and two points $ P,Q. $ Let $ X $ be the intersection of $ \ell_1, \ell_2. $ Then $ \triangle XPQ \stackrel{-}{\sim} \triangle X\mathbb{T}_{\ell_1,\ell_2} (P)\mathbb{T}_{\ell_1,\ell_2} (Q). $

Proof : Let $ P_i, Q_i $ be the projection of $ P,Q $ on $ \ell_i, $ respectively where $ i \in \{1,2 \} $ and let $ Y \equiv \mathbb{T}_{\ell_1,\ell_2} (P), Z \equiv \mathbb{T}_{\ell_1,\ell_2} (Q), $ then $ Y, Z $ is the orthocenter of $ \triangle XP_1P_2, \triangle XQ_1Q_2, $ respectively, so $ \frac{XP}{XQ} = \frac{XY}{XZ}. $ Since $ XY, XZ $ is the isogonal conjugate of $ XP, XQ $ WRT $ \angle (\ell_1, \ell_2), $ respectively, so we conclude that $ \triangle XPQ \stackrel{-}{\sim} \triangle XYZ. $ $ \qquad \blacksquare $

Corollary 1.1 : Given two lines $ \ell_1, \ell_2 $ and two points $ P,Q. $ Then $ PQ $ and $ \mathbb{T}_{\ell_1,\ell_2} (P)\mathbb{T}_{\ell_1,\ell_2} (Q) $ are antiparallel WRT $ \angle (\ell_1,\ell_2). $

Corollary 1.2 : Given two lines $ \ell_1, \ell_2 $ and three points $ P,Q, R. $ Then $ \triangle PQR \stackrel{-}{\sim} \triangle \mathbb{T}_{\ell_1,\ell_2} (P)\mathbb{T}_{\ell_1,\ell_2} (Q)\mathbb{T}_{\ell_1,\ell_2} (R). $

Corollary 1.3 : Given a $ \triangle ABC $ and a point $ P \in BC. $ Let $ E, F $ be the projection of $ B, C $ on $ CA, AB, $ respectively. Then $ \mathbb{T}_{AB,AC} (P) $ lies on $ EF. $

Lemma 2 : Given a quadrangle $ P_1P_2P_3P_4 $ with Euler-Poncelet point $ E $ and Isogonal center $ I. $ Then $ \mathbb{T}_{P_1P_2,P_3P_4} (I) = \mathbb{T}_{P_1P_3,P_2P_4} (I) = \mathbb{T}_{P_1P_4,P_2P_3} (I) = E. $

Proof : It suffices to prove $ \mathbb{T}_{P_1P_4, P_2P_3} (I) = E. $ Let $ O_4 $ be the circumcenter of $ \triangle P_1P_2P_3 $ and let the perpendicular from $ E $ to $ P_2P_3 $ cuts $ P_1P_4, IO_4 $ at $ U, V, $ respectively. By Miquel triangle and Isogonal center (Property 11), $ I $ is the image of the isogonal conjugate $ Q_4 $ of $ P_4 $ WRT $ \triangle P_1P_2P_3 $ under the inversion WRT $ \odot (O_4), $ so $ \measuredangle VIP_1 = \measuredangle Q_4P_1O_4 = \measuredangle (\perp P_2P_3, P_1P_4) = \measuredangle VUP_1 $ $ \Longrightarrow $ $ P_1, I, U, V $ are concyclic. Notice that $ E $ is the orthopole of $ O_4Q_4 $ WRT $ \triangle P_1P_2P_3 $ we get $ P_1V \perp O_4Q_4, $ so $ U $ is the projection of $ I $ on $ P_1P_4. $ Similarly, the intersection of the perpendicular from $ I,E $ to $ P_2P_3, P_1P_4, $ respectively lies on $ P_2P_3, $ so $ \mathbb{T}_{P_1P_4, P_2P_3} (I) = E. $ $ \qquad \blacksquare $

Lemma 3 : Given a quadrilateral $ P_1P_2P_3P_4 $ and a point $ T. $ Then the pedal circle of $ T $ WRT $ \triangle P_2P_3P_4, \triangle P_1P_3P_4, \triangle P_1P_2P_4, \triangle P_1P_2P_3 $ are concurrent.

Proof : Let $ H_{ij} $ be the projection of $ T $ on $ P_iP_j $ where $ i<j $ and $ i, j \in \{ 1,2,3,4 \}. $ Consider an inversion with center $ T $ and denote the image of $ \Box $ as $ {\Box}^*, $ then $ H^*_{ij}, H^*_{ik}, H^*_{il} $ are collinear where $ \{ i, j, k, l \} = \{ 1,2,3,4 \}, $ so $ \odot (H^*_{23}H^*_{24}H^*_{34}), \odot (H^*_{13}H^*_{14}H^*_{34}), \odot (H^*_{12}H^*_{14}H^*_{24}), \odot (H^*_{12}H^*_{13}H^*_{23}) $ are concurrent $ \Longrightarrow $ $ \odot (H_{23}H_{24}H_{34}), \odot (H_{13}H_{14}H_{34}), \odot (H_{12}H_{14}H_{24}), \odot (H_{12}H_{13}H_{23}) $ are concurrent. $ \qquad \blacksquare $

Lemma 4 : Given a $ \triangle ABC $ with orthocenter $ H $ and a point $ P. $ Let $ Y, Z $ be the reflection of $ P $ in $ CA, AB, $ respectively. Then the anti-Steiner point $ T $ of $ HP $ WRT $ \triangle ABC $ lies on $ \odot (AYZ). $

Proof : Let $ E, F $ be the reflection of $ H $ in $ CA, AB, $ respectively, then $ T $ lies on $ EY, FZ $ and $ E, F $ lie on $ \odot (ABC), $ so $ \measuredangle YAZ = \measuredangle EAF = \measuredangle ETF = \measuredangle YTZ $ $ \Longrightarrow $ $ T \in \odot (AYZ). $ $ \qquad \blacksquare $

Main result

In this section, I'll give a brief introduction to Ptolemy triangle, QA-Orthopole-circle and related transformation WRT a quadrangle. All symbols in this section bear the same meaning.

Notation

Given a quadrangle $ P_1P_2P_3P_4 $ with Euler-Poncelet point $ E, $ Isogonal center $ I $ and a point $ T. $ Let $ T_X \equiv \mathbb{T}_{P_1P_4, P_2P_3} (T), T_Y \equiv \mathbb{T}_{P_1P_3, P_2P_4} (T), T_Z \equiv \mathbb{T}_{P_1P_2, P_3P_4} (T). $

Property 1 : $ \triangle T_XT_YT_Z $ is the image of the pedal triangle of $ P_i $ WRT $ \triangle P_jP_kP_l $ under the spiral similarity with center $ E, $ ratio $ \frac{IT}{IP_i} $ and angle $ \measuredangle TIP_i $ where $ \{ i,j,k,l \} = \{ 1,2,3,4 \}. $

Proof : It suffices to prove the case when $ i=4. $ Let $ \triangle A_4B_4C_4 $ be the pedal triangle of $ P_4 $ WRT $ \triangle P_1P_2P_3. $ From Lemma 2 $ \Longrightarrow $ $ \mathbb{T}_{P_1P_4, P_2P_3} (I) = E, $ so combining Corollary 1.2 for lines $ P_1P_4, P_2P_3 $ and points $ P_4, I, T $ we get $ \triangle ITP_4 \stackrel{-}{\sim} \triangle ET_XA_4. $ Similarly, we can prove $ \triangle ITP_4 \stackrel{-}{\sim} \triangle ET_YB_4, $ $ \triangle ITP_4 \stackrel{-}{\sim} \triangle ET_ZC_4, $ so $ \triangle T_XT_YT_Z $ is the image of $ \triangle A_4B_4C_4 $ under the spiral similarity with center $ E, $ ratio $ \frac{IT}{IP_4} $ and angle $ \measuredangle TIP_4. $ $ \qquad \blacksquare $

Corollary 2 : $ E $ lies on $ \odot (T_XT_YT_Z). $

Corollary 3 : Let $ O_T $ be the circumcenter of $ \triangle T_XT_YT_Z. $ Then the mapping $ T \mapsto O_T $ is linear.

Proof : Let $ M_4 $ be the circumcenter $ \triangle A_4B_4C_4, $ then from Property 1 we get $ \triangle ITP_4 \stackrel{-}{\sim} \triangle EO_TM_4, $ so the mapping $ T \mapsto O_T $ is linear. $ \qquad \blacksquare $

Remark : $ \triangle T_XT_YT_Z $ is called the Ptolemy triangle of $ T $ and the mapping $ T \mapsto O_T $ is called the QA-Orthopole transformation WRT the quadrangle $ P_1P_2P_3P_4. $

Property 4 : Let $ \varrho $ be a line passing through $ T, $ $ R_{ij} $ be the intersection of $ \varrho $ with $ P_iP_j $ and $ H_{kl} $ be the projection of $ R_{ij} $ on $ P_kP_l $ where $ \{ i,j,k,l \} = \{ 1,2,3,4 \}. $ Then $ H_{12}H_{34}, H_{13}H_{24}, H_{14}H_{23} $ are concurrent on $ \odot (O_T). $

Proof : From Corollary 1.3 $\Longrightarrow $ $ T_Y \in H_{13}H_{24}, $ $ T_Z \in H_{12}H_{34}, $ so note that $ \{ \varrho, H_{13}H_{24} \}, \{ \varrho, H_{12}H_{34}\} $ are antiparallel WRT $ \angle (P_1P_3, P_2P_4), \angle (P_1P_2, P_3P_4), $ respectively we get $$ \measuredangle (H_{13}H_{24}, H_{14}H_{23}) = \measuredangle (H_{13}H_{24}, P_2P_4) + \measuredangle (P_2P_4, P_3P_4) + \measuredangle (P_3P_4, H_{12}H_{34}) = \measuredangle (P_1P_3, \varrho) + \measuredangle (P_2P_4, P_3P_4) + \measuredangle (\varrho, P_1P_2) = \measuredangle (P_1P_3, P_1P_2) + \measuredangle (P_2P_4, P_3P_4) = \measuredangle T_YT_XT_Z, $$hence the intersection of $ H_{13}H_{24}, H_{12}H_{34} $ lies on $ \odot (O_T). $ Similarly, the intersection of $ \{ H_{12}H_{34}, H_{14}H_{23} \}, \{ H_{14}H_{23}, H_{13}H_{24}\} $ lies on $ \odot (O_T), $ so $ H_{12}H_{34}, H_{13}H_{24}, H_{14}H_{23}, \odot (O_T) $ are concurrent at $ T_{\varrho}. $ $ \qquad \blacksquare $

Remark : The mapping $ \varrho \mapsto T_{\varrho} $ is called the Quang Duong’s transformation WRT the quadrangle $ P_1P_2P_3P_4. $

Property 5 : Let $ O_i $ be the circumcenter of $ \triangle P_jP_kP_l $ and $ V_i $ be the orthopole of $ O_iT $ WRT $ \triangle P_jP_kP_l $ where $ \{ i,j,k,l \} = \{ 1,2,3,4 \}. $ Then $ V_i $ lies on $ \odot (O_T) $ for any $ i \in \{ 1,2,3,4 \}. $

Proof : Let $ U $ be the midpoint of $ P_2P_3, $ $ Q $ be the projection of $ T $ on $ P_2P_3 $ and $ E_1, E_4 $ be the Poncelet point of $ P_2P_3P_4T, P_1P_2P_3T, $ respectively, then $ E, U, E_1, V_1 $ lie on the 9-point circle of $ \triangle P_2P_3P_4 $ and $ E, U, E_4, V_4 $ lie on the 9-point circle of $ \triangle P_1P_2P_3. $ From Lemma 3 $ \Longrightarrow $ the pedal circle of $ T $ WRT $ \triangle P_2P_3P_4, \triangle P_1P_3P_4, \triangle P_1P_2P_4, \triangle P_1P_2P_3 $ are concurrent at $ R, $ so note that $ Q, U, E_1, E_4 $ lie on the 9-point circle of $ \triangle TP_2P_3 $ we get $$ \measuredangle EV_1R + \measuredangle RV_4E = \measuredangle EV_1E_1 + \measuredangle E_1V_1R + \measuredangle RV_4E_4 + \measuredangle E_4V_4E = \measuredangle EUE_1 + \measuredangle E_1QR + \measuredangle RQE_4 + \measuredangle E_4UE = \measuredangle E_1QE_4 + \measuredangle E_4UE_1 = 0, $$hence $ E, U, V_1, V_4 $ are concyclic. Similarly, we can prove $ V_2, V_3 $ lie on $ \odot (EUV_4), $ so $ E, U, V_1, V_2, V_3, V_4 $ lie on a circle $ \odot (O_T^*). $

Let $ N_i $ be the 9-point center of $ \triangle P_jP_kP_l $ where $ \{ i,j,k,l \} = \{ 1,2,3,4 \}. $ Since $ O_1O_4, O_1T $ is the Steiner line of $ U, V_1 $ WRT the medial triangle of $ \triangle P_2P_3P_4, $ respectively, so $ \measuredangle O_T^*N_1N_4 = \measuredangle V_1EU = \measuredangle O_4O_1T. $ Similarly, we can prove $ \measuredangle O_T^*N_4N_1 = \measuredangle O_1O_4T, $ so $ \triangle O_T^*N_1N_4 \stackrel{-}{\sim} \triangle TO_1O_4. $ Analogously, we can prove $ \triangle O_T^*N_2N_4 \stackrel{-}{\sim} \triangle TO_2O_4, $ $ \triangle O_T^*N_3N_4 \stackrel{-}{\sim} \triangle TO_3O_4, $ so $ \triangle N_1N_2N_3 \cup O_T^* \stackrel{-}{\sim} \triangle O_1O_2O_3 \cup T. $

On the other hand, from Interesting Properties related to Four 9-point Centers (Property 1 and Corollary 11) we get $ \triangle N_1N_2N_3 \cup M_4 \cup E \stackrel{-}{\sim} \triangle O_1O_2O_3 \cup P_4 \cup I $ and $ E, I $ is the image of $ M_4, P_4 $ under the inversion WRT $ \odot (N_1N_2N_3), \odot (O_1O_2O_3), $ so combining $ \triangle EO_TM_4 \stackrel{-}{\sim} \triangle ITP_4, $ by Property 1, we conclude that $ \triangle N_1N_2N_3 \cup O_T \stackrel{-}{\sim} \triangle O_1O_2O_3 \cup T $ $ \Longrightarrow $ $ O_T \equiv O_T^* $ and $ \odot (O_T) \equiv \odot (O_T^*). $ $ \qquad \blacksquare $

Remark : $ \odot (O_T) $ is called the QA-Orthopole-circle of $ T $ WRT the quadrangle $ P_1P_2P_3P_4. $

Property 6 : The image of $ \varrho $ under the QA-Orthopole transformation WRT $ P_1P_2P_3P_4 $ is the perpendicular bisector of $ ET_{\varrho}. $

Proof : Let $ \gamma $ be the perpendicular bisector of $ ET_{\varrho}, $ then note that $ ET_X $ and $ IT $ are antiparallel WRT $ \angle (P_1P_4, P_2P_3), $ by Corollary 1.1 and Lemma 2, we get $ \measuredangle (EO_T, \gamma) = \measuredangle (ET_X, T_{\varrho}T_X) = \measuredangle (\varrho, IT), $ so $ \gamma $ is the image of $ \varrho $ under the QA-Orthopole transformation WRT $ P_1P_2P_3P_4. $ $ \qquad \blacksquare $

Let $ \triangle XYZ $ be the diagonal triangle of the quadrangle $ P_1P_2P_3P_4 $ where $ X, Y, Z $ is the intersection of $ \{ P_1P_4, P_2P_3 \}, \{ P_1P_3, P_2P_4 \}, \{ P_1P_2, P_3P_4 \}, $ respectively, then from New proof of the property of Poncelet point and Corollary 2 we know $ E $ is one of the intersection of $ \odot (O_T) $ and $ \odot (XYZ). $ The following property gives the character of another intersection of these two circles.

Property 7 : Let $ H^{\mathbf{D}} $ be the orthocenter of $ \triangle XYZ. $ Then the anti-Steiner point $ S $ of $ H^{\mathbf{D}}T $ WRT $ \triangle XYZ $ lies on $ \odot (O_T). $

Proof : Let $ X^T, Y^T. Z^T $ be the reflection of $ T $ in $ YZ, ZX, XY, $ respectively, then $ T_X, T_Y, T_Z $ lie on $ \odot (XY^TZ^T), \odot (YZ^TX^T), \odot (ZX^TY^T), $ respectively (well-known). On the other hand, from Lemma 4 we know $ S $ lies on $ \odot (XY^TZ^T), \odot (YZ^TX^T), \odot (ZX^TY^T), $ so $$ \measuredangle T_YST_Z = \measuredangle T_YSX^T + \measuredangle X^TST_Z = \measuredangle T_YYX^T + \measuredangle X^TZT_Z. $$Note that $ \{ YT, YT_Y \}, \{ ZT, ZT_Z \} $ are isogonal conjugate WRT $ \angle (T_1T_3, T_2T_4), \angle (T_1T_2, T_3T_4), $ respectively, so we get $$ \measuredangle T_YST_Z = \measuredangle T_YYP_1 + \measuredangle P_1YX^T + \measuredangle X^TZP_1 + \measuredangle P_1ZT_Z = \measuredangle P_4YT + \measuredangle P_1YX^T + \measuredangle X^TZP_1 + \measuredangle TZP_4 = \measuredangle P_4YZ + \measuredangle P_1YZ + \measuredangle YZP_1 + \measuredangle YZP_4 = \measuredangle P_2P_4P_3 + \measuredangle P_3P_1P_2, $$hence by Property 1 we conclude that $ \measuredangle T_YST_Z = \measuredangle T_YT_XT_Z $ $ \Longrightarrow $ $ S \in \odot (O_T). $ $ \qquad \blacksquare $
isogonal center
Ptolemy triangle
QA-Orthopole-circle

Miquel triangle and Isogonal center

by TelvCohl, Feb 3, 2018, 1:14 PM

Main result

In this section, I'll give a brief introduction to Miquel triangle, Isogonal center and Gergonne-Steiner point of a quadrangle. All symbols in this section bear the same meaning.

Notation

Given a quadrangle $ P_1P_2P_3P_4. $ Let $ M_X, M_Y, M_Z $ be the Miquel point of the complete quadrilateral $ \{ P_1P_2, P_1P_3, P_2P_4, P_3P_4 \}, $ $ \{ P_1P_2, P_2P_3, P_1P_4, P_3P_4 \}, $ $ \{ P_1P_3, P_2P_3, P_1P_4, P_2P_4 \}, $ respectively and let $ X, Y, Z $ be the intersection of $ \{ P_1P_4, P_2P_3 \}, $ $ \{ P_1P_3, P_2P_4 \}, $ $ \{ P_1P_2, P_3P_4 \}, $ respectively. Consider the transformation $ \Psi_X $ defined by inversion with center $ M_X, $ power $ M_XM_Y \cdot M_XM_Z $ followed by reflection in the angle bisector of $ \angle M_YM_XM_Z. $ For two points $ J,K, $ we say $ \{ J,K \} \in \mathbb{X} $ if $ J, K $ is the image of $ K, J $ under $ \Psi_X. $ Similarly, we can define $ \Psi_Y,\Psi_Z, \mathbb{Y}, \mathbb{Z}. $ Note that $ \{ J,K \} \in \mathbb{X} $ $ \Longleftrightarrow $ $ \triangle M_XM_YJ \stackrel{+}{\sim} \triangle M_XKM_Z. $

Property 1 : $ \{ P_1, P_4 \}, \{ P_2, P_3 \}, \{ Y,Z \} \in \mathbb{X}, $ $ \{ P_1, P_3 \}, \{ P_2, P_4 \}, \{ X,Z \} \in \mathbb{Y}, $ $ \{ P_1, P_2 \}, \{ P_3, P_4 \}, \{ X,Y \} \in \mathbb{Z}. $

Proof : Since $ M_X $ is the Miquel point of the complete quadrilateral $ \{ P_1P_2, P_1P_3, P_2P_4, P_3P_4 \}, $ so we get $ M_XP_1 \cdot M_XP_4 = M_XP_2 \cdot M_XP_3 = M_XY \cdot M_XZ = \varsigma $ and $ \{ M_XP_1, M_XP_4 \}, \{ M_XP_2, M_XP_3 \} $ are isogonal conjugate WRT $ \angle YM_XZ. $ Consider the transformation $ \Phi_X $ define by inversion with center $ M_X, $ power $ \varsigma $ followed by reflection in the angle bisector of $ \angle YM_XZ, $ then $ \odot (P_2P_3Z) \mapsto \odot (P_2P_3Y) \odot (P_1P_4Z) \mapsto \odot (P_1P_4Y) $ under $ \Phi_X, $ so we conclude that $ M_Z $ is the image of $ M_Y $ under $ \Phi_X $ $ \Longrightarrow $ $ \Phi_X = \Psi_X. $ i.e. $ \{ P_1, P_4 \}, \{ P_2, P_3 \}, \{ Y,Z \} \in \mathbb{X}. $ $ \qquad \blacksquare $

Property 2 : The isogonal conjugate $ T_i $ of $ P_i $ WRT $ \triangle M_XM_YM_Z $ is the isogonal conjugate of the complement of $ P_i $ WRT $ \triangle P_jP_kP_l $ WRT $ \triangle P_jP_kP_l $ where $ \{ i,j,k,l \} = \{ 1,2,3,4\}. $

Proof : It suffices to prove the case when $ i=4. $ Let $ M_{ij} $ be the midpoint of $ P_iP_j $ where $ i, j \in \{ 1,2,3,4\} $ and let $ P^*_4 $ be the reflection of $ P_4 $ in $ M_{23}. $ From $ \triangle M_XP_1P_2 \stackrel{+}{\sim} \triangle M_XP_3P_4 $ we get $ \frac{P_2P_4^*}{M_XP_3} = \frac{P_3P_4}{M_XP_3} = \frac{P_1P_2}{M_XP_1} $ and $ \measuredangle (P_1P_2, P_2P_4^*) = \measuredangle (P_1P_2, P_3P_4) = \measuredangle (M_XP_1, M_XP_3), $ so $ \triangle P_1P_2P_4^* \stackrel{+}{\sim} \triangle P_1M_XP_3 $ $ \Longrightarrow $ $ \{ P_1M_X, P_1P_4^* \} $ are isogonal conjugate WRT $ \angle P_1. $ i.e. $ P_1M_X $ passes through the isogonal conjugate of the complement of $ P_4 $ WRT $ \triangle P_1P_2P_3 $ WRT $ \triangle P_1P_2P_3. $ Similarly, $ P_2M_Y, P_3M_Z $ pass through the isogonal conjugate of the complement of $ P_4 $ WRT $ \triangle P_1P_2P_3 $ WRT $ \triangle P_1P_2P_3. $ On the other hand, from Property 1 we know $ \{ P_1, P_4 \} \in \mathbb{X}, $ so $ T_4 $ lies on $ P_1M_X. $ Similarly, $ T_4 $ lies on $ P_2M_Y, $ $ P_3M_Z, $ so we conclude that $ T_4 $ is the isogonal conjugate of the complement of $ P_4 $ WRT $ \triangle P_1P_2P_3 $ WRT $ \triangle P_1P_2P_3. $ $ \qquad \blacksquare $

Property 3 : $ \{ T_1,T_4 \}, \{ T_2, T_3 \} \in \mathbb{X}, \{ T_1, T_3 \}, \{ T_2,T_4 \} \in \mathbb{Y}, \{ T_1, T_2 \}, \{ T_3,T_4 \} \in \mathbb{Z} $ and $ P_1T_1 \parallel P_2T_2 \parallel P_3T_3 \parallel P_4T_4. $

Proof : From Property 1 we know $ \{ P_1, P_4 \} \in \mathbb{X}, $ so $ \triangle M_XM_YP_1 \stackrel{+}{\sim} \triangle M_XP_4M_Z $ $ \Longrightarrow $ $ \measuredangle T_4P_1M_Y = \measuredangle M_XM_ZP_4 = \measuredangle T_4M_ZM_Y, $ so $ T_4 \in \odot (P_1M_YM_Z). $ Similarly, we can prove $ T_1, T_2, T_3 $ is the second intersection of $ P_4M_X, P_4M_Y, P_4M_Z $ with $ \odot (P_4M_YM_Z), $ $ \odot (P_4M_ZM_X), $ $ \odot (P_4M_XM_Y), $ respectively, so $ \triangle M_XM_ZT_2 \stackrel{+}{\sim} \triangle M_XT_3M_Y $ and $ \{ T_1,T_4 \}, \{ T_2, T_3 \} \in \mathbb{X}. $ From $ \{ P_1,P_4 \}, \{ T_1, T_4 \} \in \mathbb{X} $ $ \Longrightarrow $ $ \frac{M_XP_1}{M_XT_4} = \frac{M_XT_1}{M_XP_4}, $ so $ P_1T_1 $ and $ P_4T_4 $ are parallel. Analogously, we can prove $ \{ T_1, T_3 \}, \{ T_2,T_4 \} \in \mathbb{Y}, \{ T_1, T_2 \}, \{ T_3,T_4 \} \in \mathbb{Z} $ and $ P_2T_2 \parallel P_4T_4, $ $ P_3T_3 \parallel P_4T_4, $ so we conclude that $ P_1T_1 \parallel P_2T_2 \parallel P_3T_3 \parallel P_4T_4. $ $ \qquad \blacksquare $

Property 4 : $ \odot (P_1P_4M_X), \odot (P_2P_3M_X), \odot (P_1P_3M_Y), \odot (P_2P_4M_Y), \odot (P_1P_2M_Z), \odot (P_3P_4M_Z) $ are concurrent at $ I $ and $ \{ I,X \} \in \mathbb{X}, $ $ \{ I,Y \} \in \mathbb{Y}, $ $ \{ I,Z \} \in \mathbb{Z}. $

Proof : Let $ I $ be the second intersection of $ \odot (P_1P_2M_Z), \odot (P_1P_3M_Y). $ Notice $ \odot (P_1P_2M_Z), \odot (P_1P_3M_Y) $ is the image of $ \odot (P_3P_4M_Y), \odot (P_2P_4M_Z) $ under $ \Psi_X, $ respectively, so $ \{ I,X \} \in \mathbb{X}. $ Combining $ \{ P_2, P_3 \} \in \mathbb{X} $ we get $ \triangle M_XP_2I \stackrel{+}{\sim} \triangle M_XXP_3 $ $ \Longrightarrow $ $ \measuredangle P_2IM_X = \measuredangle P_2P_3M_X $ $ \Longrightarrow $ $ I \in \odot (P_2P_3M_X). $ Similarly, by $ \{ P_1, P_4 \} \in \mathbb{X} $ we get $ \triangle M_XP_1I \stackrel{+}{\sim} \triangle M_XXP_4, $ so $ \measuredangle P_1IM_X = \measuredangle P_1P_4M_X $ and $ I \in \odot (P_1P_4M_X). $ Analogously, we can prove $ I $ lies on $ \odot (P_2P_4M_Y), $ $ \odot (P_3P_4M_Z), $ so we conclude that $ \odot (P_1P_4M_X), \odot (P_2P_3M_X), \odot (P_1P_3M_Y), \odot (P_2P_4M_Y), \odot (P_1P_2M_Z), \odot (P_3P_4M_Z) $ are concurrent at $ I $ and $ \{ I,X \} \in \mathbb{X}.$ $ \qquad \blacksquare $

Corollary 5 : $ IP_i $ is the common tangent of $\odot (P_iM_XX), \odot (P_iM_YY), \odot (P_iM_ZZ) $ where $ i \in \{ 1,2,3,4 \}. $

Proof : It suffices to prove the case when $ i=4. $ From $ I \in \odot (P_1P_4M_X) $ and $ \triangle M_XP_1I \stackrel{+}{\sim} \triangle M_XXP_4 $ $ \Longrightarrow $ $ \measuredangle IP_4M_X = \measuredangle IP_1M_X = \measuredangle P_4XM_X, $ so $ IP_4 $ is tangent to $ \odot (P_4M_XX). $ Similarly, we can prove $ IP_4 $ is tangent to $ \odot (P_4M_YY), \odot (P_4M_ZZ), $ so $ IP_4 $ is the common tangent of $\odot (P_4M_XX), \odot (P_4M_YY), \odot (P_4M_ZZ). $ $ \qquad \blacksquare $

Corollary 6 : $ M_XX, M_YY, M_ZZ $ are concurrent at $ G. $

Proof : From $ \{ I,X \} \in \mathbb{X} $ $ \Longrightarrow $ $ \{ M_XI, M_XX \} $ are isogonal conjugate WRT $ \angle M_YM_XM_Z, $ so the isogonal conjugate $ G $ of $ I $ WRT $ \triangle M_XM_YM_Z $ lies on $ M_XX. $ Similarly, we can prove $ G \in M_YY, $ $ G \in M_ZZ. $ $ \qquad \blacksquare $

Remark : $ G $ is called the Gergonne-Steiner point and $ I $ is called the Isogonal center of the quadrangle $ P_1P_2P_3P_4. $

Property 7 : $ G $ lies on $ \odot (XM_YM_Z), \odot (YM_ZM_X), \odot (ZM_XM_Y). $

Proof : From the proof of Property 3 $ \Longrightarrow $ $ T_4 $ lies on $ \odot (P_2M_ZM_X), $ $ \odot (P_3M_XM_Y), $ so $ \measuredangle M_YM_XM_Z = \measuredangle M_YM_XT_4 + \measuredangle T_4M_XM_Z = \measuredangle M_YP_3T_4 + \measuredangle T_4P_2M_Z. $ On the other hand, since $ I $ lies on $ \odot (P_1P_2M_Z) $ and $ \odot (P_1P_3M_Y), $ so $ \measuredangle M_ZIM_Y = \measuredangle M_ZIP_1 + \measuredangle P_1IM_Y = \measuredangle M_ZP_2P_1 + \measuredangle P_1P_3M_Y, $ hence keep in mind that $ G,I $ are isogonal conjugate WRT $ \triangle M_XM_YM_Z $ we conclude that $$ \measuredangle M_YGM_Z = \measuredangle M_YM_XM_Z + \measuredangle M_ZIM_Y = \measuredangle M_YP_3T_4 + \measuredangle T_4P_2M_Z + \measuredangle M_ZP_2P_1 + \measuredangle P_1P_3M_Y = \measuredangle P_1P_3M_Z + \measuredangle M_YP_2P_1 = \measuredangle P_1XM_Z + \measuredangle M_YXP_1 = \measuredangle M_YXM_Z. $$i.e. $ G $ lies on $ \odot (XM_YM_Z). $ $ \qquad \blacksquare $

Property 8 : $ G $ is the complement of the antigonal conjugate of $ P_i $ WRT $ \triangle P_jP_kP_l $ WRT $ \triangle P_jP_kP_l $ where $ \{ i,j,k,l \} = \{ 1,2,3,4\}. $

Proof : It suffices to prove the case when $ i=4. $ Since $ M_Y $ is the center of the spiral similarity of $ \overline{P_1P_4M_{14}} \mapsto \overline{P_2P_3M_{23}}, $ so $ \measuredangle M_{14}M_YM_{23} = \measuredangle P_1M_YP_2 = \measuredangle M_{14}XM_{23} $ $ \Longrightarrow $ $ M_Y $ lies on $ \odot (XM_{14}M_{23}). $ Analogously, we can prove $ M_Z \in \odot (XM_{14}M_{23}), $ so $ M_{14} $ and $ M_{23} $ lie on $ \odot (XM_YM_Z). $ Similarly, we can prove $ M_{13}, M_{24} \in \odot (YM_ZM_X), $ $ M_{12}, M_{34} \in \odot (ZM_XM_Y), $ so $$ \measuredangle M_{12}GM_{13} = \measuredangle M_{12}GM_X + \measuredangle M_XGM_{13} = \measuredangle M_{12}M_{34}M_X + \measuredangle M_XM_{24}M_{13} = \measuredangle P_2P_4M_X + \measuredangle M_XP_4P_3 = \measuredangle P_2P_4P_3, $$hence the anticomplement $ G^*_4 $ of $ G $ WRT $ \triangle P_1P_2P_3 $ lies on the reflection of $ \odot (P_2P_3P_4) $ in $ P_2P_3. $ Analogously, we can prove $ G^*_4 $ lies on the reflection $ \odot (P_1P_3P_4), \odot (P_1P_2P_4) $ in $ P_1P_3, $ $ P_1P_2, $ respectively, so we conclude that $ G^*_4 $ is the antigonal conjugate of $ P_4 $ WRT $ \triangle P_1P_2P_3. $ i.e. $ G $ is the complement of the antigonal conjugate of $ P_4 $ WRT $ \triangle P_1P_2P_3 $ WRT $ \triangle P_1P_2P_3. $ $ \qquad \blacksquare $

Corollary 9 : $ G $ lies on $ \odot (M_{12}M_{13}M_{14}), \odot (M_{12}M_{23}M_{24}), \odot (M_{13}M_{23}M_{34}), \odot (M_{14}M_{24}M_{34}). $

Property 10 : Let $ S_i $ be the Miquel associate of $ P_i $ WRT $ \triangle P_jP_kP_l $ where $ \{ i,j,k,l \} = \{ 1,2,3,4\}, $ then $ I $ lies on $ \odot (P_2S_1Z), $ $ \odot (P_3S_1Y), $ $ \odot (P_4S_1X), $ $ \odot (P_1S_2Z), $ $ \odot (P_3S_2X), $ $ \odot (P_4S_2Y), $ $ \odot (P_1S_3Y), $ $ \odot (P_2S_3X), $ $ \odot (P_4S_3Z), $ $ \odot (P_1S_4X), $ $ \odot (P_2S_4Y), $ $ \odot (P_3S_4Z). $

Proof : It suffices to prove $ I $ lies on $ \odot (P_1S_4X). $ Consider an inversion with center $ I $ and denote the image of $ \Box $ as $ {\Box}^{\mathbb{I}}, $ then by Property 4 $ \Longrightarrow $ $ \triangle M_X^{\mathbb{I}}M_Y^{\mathbb{I}}M_Z^{\mathbb{I}} $ is the diagonal triangle of the quadrangle $ P_1^{\mathbb{I}}P_2^{\mathbb{I}}P_3^{\mathbb{I}}P_4^{\mathbb{I}}. $ Since $ X^{\mathbb{I}} $ lies on $ \odot (P_2^{\mathbb{I}}P_4^{\mathbb{I}}M_Z^{\mathbb{I}}), $ $ \odot (P_3^{\mathbb{I}}P_4^{\mathbb{I}}M_Y^{\mathbb{I}}), $ so $ X^{\mathbb{I}} $ is the Miquel point of the complete quadrilateral $ \{ P_1^{\mathbb{I}}P_2^{\mathbb{I}}, P_1^{\mathbb{I}}P_3^{\mathbb{I}}, P_2^{\mathbb{I}}P_4^{\mathbb{I}}, P_3^{\mathbb{I}}P_4^{\mathbb{I}} \}. $ Analogously, $ Y^{\mathbb{I}}, Z^{\mathbb{I}} $ is the Miquel point of the complete quadrilateral $ \{ P_1^{\mathbb{I}}P_2^{\mathbb{I}}, P_2^{\mathbb{I}}P_3^{\mathbb{I}}, P_1^{\mathbb{I}}P_4^{\mathbb{I}}, P_3^{\mathbb{I}}P_4^{\mathbb{I}} \}, $ $ \{ P_1^{\mathbb{I}}P_3^{\mathbb{I}}, P_2^{\mathbb{I}}P_3^{\mathbb{I}}, P_1^{\mathbb{I}}P_4^{\mathbb{I}}, P_2^{\mathbb{I}}P_4^{\mathbb{I}} \}, $ respectively. Since $ S_4^{\mathbb{I}} $ lies on $ \odot (P_1^{\mathbb{I}}Y^{\mathbb{I}}Z^{\mathbb{I}}), $ $ \odot (P_2^{\mathbb{I}}Z^{\mathbb{I}}X^{\mathbb{I}}), $ $ \odot (P_3^{\mathbb{I}}X^{\mathbb{I}}Y^{\mathbb{I}}), $ so $ S_4^{\mathbb{I}} \in P_1^{\mathbb{I}}X^{\mathbb{I}}, $ by the proof of Property 3, hence we conclude that $ I $ lies on $ \odot (P_1S_4X). $ $ \qquad \blacksquare $

Property 11 : $ I $ is the image of the isogonal conjugate of $ P_i $ WRT $ \triangle P_jP_kP_l $ under the inversion WRT $ \odot (P_jP_kP_l) $ where $ \{ i,j,k,l \} = \{ 1,2,3,4\}. $

Proof : It suffices to prove the case when $ i=4. $ Let $ I^* $ be the image of the isogonal conjugate of $ P_4 $ WRT $ \triangle P_1P_2P_3 $ under the inversion WRT $ \odot (P_1P_2P_3). $ Since $ I $ lies on $ \odot (P_2P_3M_X), $ so $$ \measuredangle P_2IP_3 = \measuredangle P_2M_XP_3 = \measuredangle P_2M_XP_4 + \measuredangle P_4M_XP_3 = \measuredangle P_2ZP_3 + \measuredangle P_2YP_3 = \measuredangle P_2P_1P_3 + \measuredangle P_2P_4P_3 = \measuredangle P_2I^*P_3. $$Similarly, we can prove $ \measuredangle P_3IP_1 = \measuredangle P_3I^*P_1, $ $ \measuredangle P_1IP_2 = \measuredangle P_1I^*P_2, $ so we conclude that $ I \equiv I^*. $ i.e. $ I $ is the image of the isogonal conjugate of $ P_4 $ WRT $ \triangle P_1P_2P_3 $ under the inversion WRT $ \odot (P_1P_2P_3). $ $ \qquad \blacksquare $

Property 12 : $ S_i $ is the cevian quotient $ (P_i / T_i) $ of $ P_i, T_i $ WRT $ \triangle P_jP_kP_l $ where $ \{ i,j,k,l \} = \{ 1,2,3,4\}. $

Proof : It suffices to prove the case when $ i=4. $ Let $ \widehat{P}_4 $ be the isotomic conjugate of $ P_4 $ WRT $ \triangle P_1P_2P_3, $ $ \triangle \widehat{X}_4\widehat{Y}_4\widehat{Z}_4 $ be the cevian triangle of $ \widehat{P}_4 $ WRT $ \triangle P_1P_2P_3, $ and $ R_1, R_2, R_3 $ be the second intersection of $ \odot (P_1P_2P_3) $ with $ \odot (P_1YZ), $ $ \odot (P_2ZX), $ $ \odot (P_3XY), $ respectively. Since $ R_1 $ is the center of the spiral similarity of $ P_2Z \mapsto P_3Y, $ so we get $ \frac{R_1P_2}{R_1P_3} = \frac{P_2Z}{P_3Y} = \frac{P_1\widehat{Z}_4}{P_1\widehat{Y}_4} $ $ \Longrightarrow $ $ \triangle R_1P_2P_3 \stackrel{+}{\sim} \triangle P_1\widehat{Z}_4\widehat{Y}_4, $ hence $ P_1R_1 $ is tangent to $ \odot (P_1\widehat{Y}_4\widehat{Z}_4) $ at $ P_1. $ Similarly, we can prove $ P_2R_2, P_3R_3 $ is tangent to $ \odot (P_2\widehat{Z}_4\widehat{X}_4), \odot (P_3\widehat{X}_4\widehat{Y}_4) $ at $ P_2, $ $ P_3, $ respectively. Let $ \triangle T_4^1T_4^2T_4^3 $ be the triangle formed by $ P_1R_1, P_2R_2, P_3R_3 $ and $ V_4^1, V_4^2, V_4^3 $ be the isogonal conjugate of $ T_4^1, T_4^2, T_4^3 $ WRT $ \triangle P_1P_2P_3, $ respectively. Note that $ \triangle \widehat{X}_4\widehat{Y}_4\widehat{Z}_4 $ and $ \triangle V_4^1V_4^2V_4^3 $ are homothetic, so $ \triangle V_4^1V_4^2V_4^3 $ is the anticevian triangle of the isotomcomplement of $ \widehat{P}_4 $ WRT $ \triangle P_1P_2P_3 $ WRT $ \triangle P_1P_2P_3, $ hence by Property 2 we conclude that $ \triangle T_4^1T_4^2T_4^3 $ is the anticevian triangle of $ T_4 $ WRT $ \triangle P_1P_2P_3 $ $ \Longrightarrow $ $ S_4, $ the perspector of $ \triangle XYZ $ and $ \triangle T_4^1T_4^2T_4^3, $ is the cevian quotient $ (P_4 / T_4 ) $ of $ P_4, T_4 $ WRT $ \triangle P_1P_2P_3. $ $ \qquad \blacksquare $

Property 13 : $ I $ lies on $ S_iT_i $ where $ i \in \{ 1,2,3,4 \}. $

Proof : It suffices to prove the case when $ i=4. $ Let $ \triangle Q_1Q_2Q_3 $ be the circumcevian triangle of $ T_4 $ WRT $ \triangle P_1P_2P_3, $ then $ P_4^1Q_1 \cdot P_4^1P_1 = P_4^1X \cdot P_4^1S_4, $ so $ Q_1 $ lies on $ \odot (P_1S_4X). $ Similarly, we can prove $ Q_2 \in \odot (P_2S_4Y), $ $ Q_3 \in \odot (P_3S_4Z), $ so from $ P_1T_4 \cdot Q_1T_4 = P_2T_4 \cdot Q_2T_4 =P_3T_4 \cdot Q_3T_4 $ we conclude that $ T_4 $ lies on the radical axis of $ \odot (P_1S_4X), \odot (P_2S_4Y), \odot (P_3S_4Z). $ i.e. $ I $ lies on $ S_4T_4. $ $ \qquad \blacksquare $

Property 14 : $ T_1 $ lies on $ XS_4, YS_3, ZS_2. $ $ T_2 $ lies on $ XS_3, YS_4, ZS_1. $ $ T_3 $ lies on $ XS_2, YS_1, ZS_4. $ $ T_4 $ lies on $ XS_1, YS_2, ZS_3. $

Proof : It suffices to prove $ T_4 $ lies on $ XS_1. $ Since $ \odot (IP_2Z), \odot (IP_3Y) $ is the image of $ \odot (P_3XY), \odot (P_2ZX) $ under $ \Psi_X, $ respectively, so $ \{ S_1, S_4 \} \in \mathbb{X}. $ Combining Property 13 we get $ M_X, S_4, T_4, X $ are concyclic. Similarly, we can prove $ M_X, S_1, T_1, X $ are concyclic, so note that $ \{ T_1, T_4 \} \in \mathbb{X}, $ by Property 3, we conclude that $ \measuredangle M_XXT_4 = \measuredangle M_XS_4T_4 = \measuredangle M_XT_1S_1 = \measuredangle M_XXS_1 $ $ \Longrightarrow $ $ T_4\in XS_1. $ $ \qquad \blacksquare $

Application

Problem 1 (Generalization of Musselman's theorem) : Given a $ \triangle ABC $ with isogonal conjugate $ P,Q. $ Let $ D, E, F $ be the second intersection of $ AQ, BQ, CQ $ with $ \odot (BQC), $ $ \odot (CQA), $ $ \odot (AQB), $ respectively. Then the circles $ \odot (APD), \odot (BPE), \odot (CPF) $ are coaxial.

Proof : Simple angle chasing yields $ \triangle ABF \stackrel{+}{\sim} \triangle AEC, $ $ \triangle ABD \stackrel{+}{\sim} \triangle APC, $ so we get $ \frac{AE}{AP} = \frac{AD}{AF} $ and $ \measuredangle PAE = \measuredangle FAD $ $ \Longrightarrow $ $ \triangle AEP \stackrel{+}{\sim} \triangle ADF, $ hence $ A $ is the center of the spiral similarity of $ EP \mapsto DF $ $ \Longrightarrow $ $ A $ is the Miquel point of the complete quadrilateral $ \{ DE, DF, PE, PF\}. $ Similarly, we can prove $ B,C $ is the Miquel point of the complete quadrilateral $ \{ EF, ED, PF, PD\}, $ $ \{ FD, FE, PD, PE\}, $ respectively, so by Property 4 we conclude that $ \odot (APD), \odot (BPE), \odot (CPF) $ are coaxial. $ \qquad \blacksquare $

Remark : Musselman's theorem is the case when $ P, Q $ is the circumcenter, orthocenter of $ \triangle ABC, $ respectively.

Problem 2 (2012 IMO shortlist G8) : Let $ \triangle ABC $ be a triangle with circumcircle $ \omega $ and $ \ell $ be a line without common points with $ \omega. $ Denote by $ P $ the foot of the perpendicular from the center of $ \omega $ to $ \ell. $ The side-lines $ BC, CA, AB $ intersect $ \ell $ at the points $ X, Y, Z $ different from $ P. $ Prove that $ \odot (APX), \odot (BPY), \odot (CPZ) $ have a common point different from $ P $ or are mutually tangent at $ P. $

Proof :

Lemma : Given a $ \triangle ABC $ with isogonal conjugate $ P,Q. $ Let $ \triangle DEF $ be the cevian triangle of $ P $ WRT $ \triangle ABC $ and $ X\in BC $ be the point s.t. $ AX $ is tangent to $ \odot (AEF) $ at $ A. $ Then $ X $ lies on the polar of $ Q $ WRT $ \odot (ABC). $

Proof : Let $ BP, CP $ cuts $ \odot (AEF) $ again at $ P_B, P_C, $ respectively and $ \triangle Q_aQ_bQ_c $ be the circumcevian triangle of $ Q $ WRT $ \triangle ABC. $ From Pascal's theorem for $ EP_BP_CFAA $ we get $ BC, P_BP_C $ and the tangent of $ \odot (AEF) $ at $ A $ are concurrent, so $ X $ lies on $ P_BP_C. $ Since $ \measuredangle BAQ_b = \measuredangle BEC = \measuredangle P_BAX, $ so combining $ \measuredangle ABQ_b = \measuredangle P_BBX $ we get $ Q_b $ is the isogonal conjugate of $ P_B $ WRT $ \triangle AXB $ $ \Longrightarrow $ $ XQ_b $ is the isogonal conjugate of $ P_BP_C $ WRT $ \angle AXD. $ Similarly, we can prove $ XQ_c $ is the isogonal conjugate of $ P_BP_C $ WRT $ \angle AXD, $ so we conclude that $ X \in Q_bQ_c $ $ \Longrightarrow $ $ X $ lies on the polar of $ Q $ WRT $ \odot (ABC). $ $ \qquad \square $

Back to the main proof :

Let $ R $ be the isogonal conjugate of the image of $ P $ under the inversion WRT $ \omega $ WRT $ \triangle ABC $ and $ T $ be the Miquel associate of $ R $ WRT $ \triangle ABC, $ it suffices to prove $ T $ lies on $ \odot (APX), \odot (BPY), \odot (CPZ). $ Let $ \triangle DEF $ be the cevian triangle of $ R $ WRT $ \triangle ABC, $ then by Lemma $ \Longrightarrow $ $ X $ lies on the tangent of $ \odot (AEF) $ at $ A, $ so $ \measuredangle TDX = \measuredangle TFA = \measuredangle TAX $ $ \Longrightarrow $ $ A, D, T, X $ are concyclic. Note that $ P $ is the Isogonal center of the quadrangle $ ABCR, $ so by Property 10 we conclude that $ A, D, P, T, X $ are concyclic. i.e. $ T \in \odot (APX). $ $ \qquad \blacksquare $

Remark : The condition $ \ell \cap \omega = \varnothing $ is redundant.

Problem 3 : Given a $ \triangle ABC $ with isogonal conjugate $ P,Q. $ Let $ \triangle DEF $ be the antipedal triangle of $ P $ WRT $ \triangle ABC $ and $ R, S $ be the image of $ Q, P $ under the inversion WRT $ \odot (ABC), $ $ \odot (DEF), $ respectively. Then $ P, R, S $ are collinear and $ R $ is the midpoint of $ PS. $

Proof : Let $ \triangle XYZ $ be the cevian triangle of $ P $ WRT $ \triangle ABC $ and $ U, V, W $ be the Miquel point of the complete quadrilateral $ \{ AB, AC, PB, PC\}, $ $ \{ BC, BA, PC, PA\}, $ $ \{ CA, CB, PA, PB\}, $ respectively. Consider an inversion with center $ P $ and denote the image of $ \Box $ as $ {\Box}^*, $ then $ X^*, Y^*, Z^* $ is the second intersection of $ A^*P^*, B^*P^*, C^*P^* $ with $ \odot (B^*PC^*), $ $ \odot (C^*PA^*), $ $ \odot (A^*PB^*), $ respectively and $ U^*, V^*, W^* $ is the intersection of $ \{ B^*Z^*, C^*Y^* \}, $ $ \{ C^*X^*, A^*Z^* \}, $ $ \{ A^*Y^*, B^*X^* \}, $ respectively. Furthermore, by Property 4 we get $ R^* $ lies on $ A^*U^*, B^*V^*, C^*W^*. $

Note that $ \{ U^*, X^* \}, \{ V^*, Y^* \}, \{ W^*, Z^* \} $ are isogonal conjugate WRT $ \triangle A^*B^*C^*, $ so $ R^* $ is the isogonal conjugate of $ P $ WRT $ \triangle A^*B^*C^*. $ On the other hand, $ \triangle D^*E^*F^* $ is the pedal triangle of $ P $ WRT $ \triangle A^*B^*C^*, $ so $ S^* $ is the center of $ \odot (D^*E^*F^*) $ $ \Longrightarrow $ $ S^* $ is the midpoint of $ PR^*, $ hence we conclude that $ P, R, S $ are collinear and $ R $ is the midpoint of $ PS. $ $ \qquad \blacksquare $

Remark : See Interesting Properties related to Four 9-point Centers (Property 4) for another proof to this problem.
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