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

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New approach to Liang-Zelich Theorem

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

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