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.

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