Isogonal Conjugate lies on Euler line.

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Isogonal Conjugate lies on Euler line.

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#1 h Jul 25, 2017, 10:50 AM • 2 Y

Y by Adventure10, Mango247

Let the cevian triangle of the isotomic conjugate of the circumcenter of $\triangle ABC$ be $\triangle XYZ$ and let the orthocenter of $\triangle ABC$ be $H$ . Then prove that the isogonal conjugate of $H$ wrt $\triangle XYZ$ lies on the Euler line $\mathcal{L}_\mathrm{E}$ of $\triangle ABC$ .

This post has been edited 2 times. Last edited by WizardMath, Jul 25, 2017, 1:16 PM

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#3 h Jul 29, 2017, 12:34 AM • 6 Y

Y by WizardMath, zed1969, Akatsuki1010, enhanced, AlastorMoody, Adventure10

Lemma : Given a $\triangle ABC$ with circumcenter $O,$ orthic triangle $\triangle H_aH_bH_c,$ tangential triangle $\triangle XYZ.$ Let $V$ be the isogonal conjugate (WRT $\triangle ABC$ ) of the isotomic conjugate of $O$ WRT $\triangle ABC$ and $\triangle M_XM_YM_Z$ be the medial triangle of $\triangle XYZ.$ Then $V$ is the homothetic center of $\triangle H_aH_bH_c,$ $\triangle M_XM_YM_Z.$

Proof : Let $O_X$ be the midpoint of $OX$ and $E,F$ be the intersection of $AC,AB$ with $OZ,OY,$ respectively, then $E,F$ lie on the X -midline $M_YM_Z$ of $\triangle XYZ.$ Let $\triangle I_XI_YI_Z$ be the excentral triangle of $\triangle M_XM_YM_Z,$ then it's clear that $\triangle AEF \cup O$ and $\triangle I_XM_YM_Z\cup O_X$ are homothetic, so $AI_X,$ $EF,$ $OO_X$ are concurrent.

On the other hand, $E,$ $F$ lie on the circle $\odot (O_X)$ with center $O_X$ and diameter $OX,$ so $BC$ and $EF$ are antiparallel WRT $\angle A$ and $\triangle BOC$ $\stackrel{-}{\sim}$ $\triangle FO_XE,$ $\Longrightarrow$ $V$ lies on $AI_X.$ Similarly, we can prove $V$ lies on $BI_Y,$ $CI_Z,$ so $V$ is the homothetic center of $\triangle ABC \cup \triangle H_aH_bH_c,$ $\triangle I_XI_YI_Z \cup \triangle M_XM_YM_Z.$ $\qquad \blacksquare$

Corollary : Given a $\triangle ABC$ with orthocenter $H,$ circumcenter $O.$ Let $O^*$ be the isotomic conjugate of $O$ WRT $\triangle ABC$ and $\mathcal{H}$ be the circum-rectangular hyperbola of $\triangle ABC$ passing through $O^*.$ Then $OH$ is a tangent of $\mathcal{H}.$

Proof : Let $\triangle H_aH_bH_c$ be the orthic triangle of $\triangle ABC$ and $T$ be the Nagel point of $\triangle H_aH_bH_c.$ Since $T$ is the crosspoint of $O$ and $H$ WRT $\triangle ABC,$ so $OT$ is tangent to the Jerabek hyperbola $\mathcal{J}$ of $\triangle ABC$ at $O.$ On the other hand, by Lemma we get $OT$ passes through the isogonal conjugate of $O^*$ WRT $\triangle ABC,$ so $OH$ is tangent to $\mathcal{H}$ at $H.$ $\qquad \blacksquare$
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Back to the main problem :

Let $O^*$ be the isotomic conjugate of the circumcenter of $\triangle ABC$ WRT $\triangle ABC$ and let $\mathcal{H}$ be the circum-rectangular hyperbola of $\triangle ABC$ passing through $O^*.$ Notice the tangent of $\mathcal{H}$ at $H$ passes through the isogonal conjugate $W$ of $H$ WRT $\triangle XYZ,$ so by Corollary we conclude that $W$ lies on the Euler line of $\triangle ABC.$ $\qquad \blacksquare$

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#4 h Jul 29, 2017, 4:59 AM • 4 Y

Y by WizardMath, AlastorMoody, Adventure10, Mango247

Triangle $ABC$ , given a point $T$ , define it's isotomic conjugate $T'$ , and isogonal conjugate of $T'$ be $T^*$
Let $O, G, H$ be circumcenter, centroid, orthocenter respectively.
It's sufficient to proof that $O^*O$ tangent to the Jerabek hyperbola $J$ of $ABC$

First of all, noticed that $H'$ lies on $J$ , $H', G, G^*$ are collinear, $OG^* \parallel HH'$ and the image of isotomic conjugate of $J$ is $HH'$

Cause the image of a line after isotomic and isogonal transform is still a line, $O^*, G^*, H^*$ are collinear. $G^*(G^*, H'; O, H) = (G, H^*; H, O)$ so $O^*G^*$ is tangent to $J$ . Let the center of $J$ be $X_{125}$ , the midpoint of $OG^*$ be $S$ , We want to proof $O^*, X_{125}, S$ are collinear, it's well-known that this line pass through $G$ and Tarry point (the fourth intersection of $(ABC)$ and Kiepert hyperbola of $ABC$ .)
Let the isotomic conjugate of $G^*$ be $K'$ , $G^*$ lies on $J$ so $K'$ lies on $HH'$ . Since both $(O, O'), (G^*, K')$ are isotomic conjugate, so $OK'$ pass through the isotomic conjugate of the infinitely point on $OG^*$ , consider the image of this line after isotomic and isogonal conjugate, it will lead to the fact that $O^*, X_{125}, S$ are collinear, as desired.
Remark. Or you can just proof that isogonal conjugate of isotomic conjugate of ninepoint center wrt $ABC$ , is the centroid of orthic triangle wrt $ABC$

This post has been edited 1 time. Last edited by houkai, Jul 29, 2017, 5:01 AM

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#5 h Feb 14, 2023, 8:12 PM

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The point is $X(14188)$ .
A Construction is as follows. (This construction is not mentioned in ETC).
Let $H=X(4)$ -Orthocenter of $ABC$ . Parallel fron $H$ to $BC$ intersect the $AC, AB$ at $Ab, Ac$ resp.
Define $BA, Bc, Ca, Cb$ cyclically. $H_A$ : Inverse of $H$ respect to circle $(ABcCb)$ . Define $H_B, H_C$ cyclically.

* $O'$ : Circumcenter of $H_AH_BH_C$ lies on Euler line of $ABC$ . $O'=X(14118)$ .
Figure:
https://geometry-diary.blogspot.com/2023/02/1904-construction-for-x14118.html

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