X(25) is the pole of the orthic axis WRT ABC

by AlastorMoody, Mar 28, 2019, 3:30 PM

Here's my attempt (together with Pluto1708 & Vrangr) at proving the following:

Problem: $X_{25}$ is the pole of the orthic axis WRT $\Delta ABC$

Here's the proof:
Lemma 1: Let $\ell_1 ,\ell_2$ be two $A-$ cevians in $\Delta ABC$ , such that, $\ell_2$ is the isogonal conjugate of $\ell_1$ WRT $\Delta ABC$ . Let $\Delta H_AH_BH_C$ be the orthic triangle WRT $\Delta ABC$ . Then, $\ell_1 ,\ell_2$ swap under the transformation sending $\Delta ABC \mapsto \Delta AH_BH_C$

Proof:
$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(13cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -10.00358821684574, xmax = 11.976486420685871, ymin = -5.837073883269276, ymax = 7.22485571690317; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); draw((-3.62,5.49)--(-7.14,-4.23)--(7.2,-4.45)--cycle, linewidth(0.8) + rvwvcq); draw((-3.769914548530757,-4.281702845141091)--(-5.547849924393275,0.16650532241402446)--(-0.6863764012673934,2.7949705571717094)--cycle, linewidth(2) + rvwvcq); /* draw figures */ draw((-3.62,5.49)--(-7.14,-4.23), linewidth(0.8) + rvwvcq); draw((-7.14,-4.23)--(7.2,-4.45), linewidth(0.8) + rvwvcq); draw((7.2,-4.45)--(-3.62,5.49), linewidth(0.8) + rvwvcq); draw((-3.769914548530757,-4.281702845141091)--(-5.547849924393275,0.16650532241402446), linewidth(0.8) + rvwvcq); draw((-5.547849924393275,0.16650532241402446)--(-0.6863764012673934,2.7949705571717094), linewidth(0.8) + rvwvcq); draw((-0.6863764012673934,2.7949705571717094)--(-3.769914548530757,-4.281702845141091), linewidth(0.8) + rvwvcq); draw((-3.62,5.49)--(xmax, -7.056981129354753*xmax-20.056271688264207), linewidth(0.4)); /* ray */ draw((xmin, -4.083586913433526*xmin-9.292584626629361)--(xmax, -4.083586913433526*xmax-9.292584626629361), linewidth(0.4) + linetype("4 4") + dtsfsf); /* line */ draw((-3.62,5.49)--(xmax, -2.831039625125939*xmax-4.758363442955894), linewidth(0.4)); /* ray */ /* dots and labels */ dot((-3.62,5.49),dotstyle); label("$A$", (-4.670717649051185,6.07180262116381), NE * labelscalefactor); dot((-7.14,-4.23),dotstyle); label("$B$", (-8.111860481648348,-5.044349879948465), NE * labelscalefactor); dot((7.2,-4.45),dotstyle); label("$C$", (8.102948677186449,-5.152448607674029), NE * labelscalefactor); dot((-3.769914548530757,-4.281702845141091),dotstyle); label("$H_A$", (-4.742783467534895,-5.098399243811247), NE * labelscalefactor); dot((-0.6863764012673934,2.7949705571717094),dotstyle); label("$H_B$", (0.13967573473647082,2.7928078801550034), NE * labelscalefactor); dot((-5.547849924393275,0.16650532241402446),dotstyle); label("$H_C$", (-6.7786428396997085,0.6308333256437019), NE * labelscalefactor); dot((-2.231970566004205,-4.305297522697286),dotstyle); label("$X_1$", (-2.86907218695843,-5.188481516915885), NE * labelscalefactor); dot((-3.62,5.49),dotstyle); dot((-0.14874596116747452,-4.33725773281333),dotstyle); label("$X_2$", (0.9504161926782105,-5.152448607674029), NE * labelscalefactor); dot((-3.0565156204606434,1.5135013669045863),dotstyle); label("$X_3$", (-4.256339192769851,1.8199193306249177), NE * labelscalefactor); dot((-2.350271460130228,1.8953481904753795),dotstyle); label("$X_4$", (-2.1664304567422557,2.9369395171224233), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

Let $\ell_1 , \ell_2$ intersect $BC$ at $X_1,X_2$ and $H_BH_C$ at $X_3,X_4$ , then the following is trivially true:
$\left\{\begin{array}{cc} \Delta ABC \sim \Delta AH_BH_C \\\\ \Delta ABX_1 \sim \Delta AH_BX_4 \\\\ \Delta ACX_2 \sim \Delta AH_CX_3 \end{array}\right\| \Longrightarrow \Delta AX_1X_2 \sim \Delta AX_4X_3 \Longrightarrow \Delta AX_1X_2 \mapsto \Delta AX_4X_3 \text{ under } \Delta ABC \mapsto \Delta AH_BH_C$ Hence, under $\Delta ABC \mapsto \Delta AH_BH_C$ , we have, $AX_1 \mapsto AX_4$ and $AX_2 \mapsto AX_3$ $\qquad \blacksquare$

Using our (Lemma 1), we can prove the well-known result:
Application #1: $A-$ symmedian and $A-$ median in $\Delta ABC$ , swap places WRT $\Delta AH_BH_C$
Application #2: $A-$ altitude and the line passing through the circumcenter in $\Delta ABC$ , swap places WRT $\Delta AH_BH_C$

Lemma 2: The Isogonal conjugate of the Isotomic conjugate of the orthocenter of $\Delta ABC$ WRT $\Delta ABC$ WRT $\Delta ABC$ is the perspector of $\Delta ABC$ and the intouch triangle of the orthic triangle triangle of $\Delta ABC$

Proof:
$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -9.08183551693353, xmax = 12.526545681809704, ymin = -5.906325720866554, ymax = 6.934720483304593; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); draw((-3.407458545586132,5.667117878678224)--(-6.9274585455861315,-4.052882121321777)--(7.412541454413868,-4.272882121321778)--cycle, linewidth(0.8) + rvwvcq); draw((-3.557373094116889,-4.104584966462868)--(-5.335308469979409,0.34362320109224714)--(-0.4738349468535252,2.9720884358499324)--cycle, linewidth(2) + rvwvcq); /* draw figures */ draw((-3.407458545586132,5.667117878678224)--(-6.9274585455861315,-4.052882121321777), linewidth(0.8) + rvwvcq); draw((-6.9274585455861315,-4.052882121321777)--(7.412541454413868,-4.272882121321778), linewidth(0.8) + rvwvcq); draw((7.412541454413868,-4.272882121321778)--(-3.407458545586132,5.667117878678224), linewidth(0.8) + rvwvcq); draw((-3.557373094116889,-4.104584966462868)--(-5.335308469979409,0.34362320109224714), linewidth(0.8) + rvwvcq); draw((-5.335308469979409,0.34362320109224714)--(-0.4738349468535252,2.9720884358499324), linewidth(0.8) + rvwvcq); draw((-0.4738349468535252,2.9720884358499324)--(-3.557373094116889,-4.104584966462868), linewidth(0.8) + rvwvcq); draw((-3.557373094116889,-4.104584966462868)--(-3.407458545586132,5.667117878678224), linewidth(0.4)); draw((-6.9274585455861315,-4.052882121321777)--(-0.4738349468535252,2.9720884358499324), linewidth(0.4)); draw((7.412541454413868,-4.272882121321778)--(-5.335308469979409,0.34362320109224714), linewidth(0.4)); draw((-4.999608621192856,1.2706125562641997)--(7.412541454413868,-4.272882121321778), linewidth(0.4) + linetype("4 4")); draw((4.0424560029446255,-4.221179276180687)--(-3.407458545586132,5.667117878678224), linewidth(0.4) + linetype("4 4")); draw((-6.9274585455861315,-4.052882121321777)--(4.478917855681261,-1.5778526784934863), linewidth(0.4) + linetype("4 4")); draw((-3.4993304170009236,-0.3212577399040761)--(-4.1927993212353725,0.9613465237805017), linewidth(0.4) + linetype("4 4") + dtsfsf); draw((-3.4993304170009236,-0.3212577399040761)--(-4.853257344688113,-0.8624182455264973), linewidth(0.4) + linetype("4 4") + dtsfsf); draw((-3.4993304170009236,-0.3212577399040761)--(-2.1626414195993937,-0.9036968719922933), linewidth(0.4) + linetype("4 4") + dtsfsf); draw((-4.1927993212353725,0.9613465237805017)--(-3.407458545586132,5.667117878678224), linewidth(0.4) + linetype("4 4") + rvwvcq); draw((0.2884773901212642,-1.1686943120306272)--(-3.407458545586132,5.667117878678224), linewidth(0.4) + linetype("4 4") + rvwvcq); /* dots and labels */ dot((-3.407458545586132,5.667117878678224),dotstyle); label("$A$", (-4.4413470955640815,6.226248968591703), NE * labelscalefactor); dot((-6.9274585455861315,-4.052882121321777),dotstyle); label("$B$", (-7.877433941921613,-4.843618448797217), NE * labelscalefactor); dot((7.412541454413868,-4.272882121321778),dotstyle); label("$C$", (8.293428381400169,-4.967600963871973), NE * labelscalefactor); dot((-3.557373094116889,-4.104584966462868),dotstyle); label("$H_A$", (-4.370499944092792,-5.02073632747544), NE * labelscalefactor); dot((-0.4738349468535252,2.9720884358499324),dotstyle); label("$H_B$", (0.1460059622019004,3.0204153645158716), NE * labelscalefactor); dot((-5.335308469979409,0.34362320109224714),dotstyle); label("$H_C$", (-6.4604909124958265,-0.14999466382431517), NE * labelscalefactor); dot((-3.4993304170009236,-0.3212577399040761),dotstyle); label("$H$", (-3.05982764187394,-0.2916889667668933), NE * labelscalefactor); dot((4.0424560029446255,-4.221179276180687),dotstyle); label("$H_A'$", (4.556241141289659,-5.215565994021485), NE * labelscalefactor); dot((4.478917855681261,-1.5778526784934863),dotstyle); label("$H_B'$", (4.8573415350426385,-1.3898198145718743), NE * labelscalefactor); dot((-4.999608621192856,1.2706125562641997),dotstyle); label("$H_C'$", (-5.681172246311644,1.9754198803143574), NE * labelscalefactor); dot((2.3921050815288174,-2.030662832744015),dotstyle); label("$H'$", (2.4308265971509804,-1.6200730568535637), NE * labelscalefactor); dot((-4.1927993212353725,0.9613465237805017),dotstyle); label("$I_D$", (-4.512194247035371,1.5857605472222676), NE * labelscalefactor); dot((-2.1626414195993937,-0.9036968719922933),dotstyle); label("$I_F$", (-1.943985006201134,-1.5492259053822746), NE * labelscalefactor); dot((-4.853257344688113,-0.8624182455264973),dotstyle); label("$I_E$", (-5.450919004029954,-1.4075316024396964), NE * labelscalefactor); dot((0.2884773901212642,-1.1686943120306272),dotstyle); label("$O$", (0.5356652952939915,-0.9470251178763174), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
Define: $\Delta H_A'H_B'H_C'$ as the cevian triangle of $H'$ (The isotomic conjugate of orthocenter of $\Delta ABC$ WRT $\Delta ABC$ ) and Let $\Delta I_DI_EI_F$ be intouch triangle of the orthic triangle of $\Delta ABC$

$O$ is the Bevan point WRT $\Delta H_AH_BH_C$ , hence $AI_D$ is the isotomic conjugate of $AO$ WRT $\Delta AH_BH_C$ and from (Lemma 1) it follows that $AI_D$ is the isotomic conjugate of the $A-$ altitude in $\Delta AH_BH_C$ WRT $\Delta AH_BH_C$ and hence, $AI_D$ becomes the Isogonal conjugate of the Isotomic conjugate of $A-$ altitude in $\Delta ABC$ WRT $\Delta ABC$ WRT $\Delta ABC$ , similarly, it can be shown that, $AI_D,BI_E,CI_F$ concur at $H^*$ (The Isogonal conjugate of the Isotomic conjugate of the orthocenter of $\Delta ABC$ WRT $\Delta ABC$ WRT $\Delta ABC$ )

Since, $HH_A \perp I_EI_F$ and $HH_A \perp I_EI_F$ $\implies$ $I_EI_F || BC$ , hence $\Delta I_DI_EI_F$ and $\Delta ABC$ are homothetical triangles about $H^*$ $\qquad \blacksquare$

Result #3: $X_{25}$ lies on the Euler Line WRT $\Delta ABC$
Proof: Trivial, to see that $H^* \equiv X_{25}$ is the ex-similicenter (Post#4) of $\odot (I_DI_EI_F)$ and $\odot (ABC)$ and now very easy to see that, $X_{25}$ lies on $OH$ (Euler Line) WRT $\Delta ABC$ $\qquad \blacksquare$

Some Observations: $H^*$ $\equiv$ $X_{25}$ is the Isogonal Mittenpunkt WRT $\Delta H_AH_BH_C$ and Since, $X_6$ (Symmedian Point WRT $\Delta ABC$ ) is the Mittenpunkt WRT $\Delta H_AH_BH_C$ , hence, $X_6$ and $X_{25}$ are isogonal conjugates WRT $\Delta H_AH_BH_C$

Result #4: $X_{25}$ is the homothetic perspector of $\Delta ABC$ and its tangential triangle
Proof: $X_{25}$ is the homothetic center of $\odot (ABC)$ and $\odot (I_DI_EI_F)$ , Let $\Delta DEF$ be the tangential triangle WRT $\Delta ABC$ , using the fact that $\Delta H_AH_BH_C$ is the tangential triangle WRT $\Delta I_DI_EI_F$ , hence, $X_{25}$ is the homothetic center of $\odot (DEF)$ and $\odot (H_AH_BH_C)$ $\implies$ $X_{25}$ is the perspector of $\Delta DEF$ and $\Delta H_AH_BH_C$ $\qquad \blacksquare$

Result #5: $X_{25}$ is the Pole of Orthic Axis WRT $\Delta ABC$ (Proof by Vrangr communicated to me via Pluto1708)
Proof: We'll make use of La-Hire's Theorem repeatedly, If $H_BH_C\cap BC=T$ , then, $H_A$ lies on the polar of $T$ and now obviously $D$ also lies on this polar and so the polar of $T$ is $H_AD$ .Similarly if $S=H_AH_B\cap AB$ , then, polar of $S$ is $FH_C$ $\implies$ ,therefore, pole of $ST$ is $FH_C\cap DH_A=X_{25}$ $\qquad \blacksquare$

This post has been edited 4 times. Last edited by AlastorMoody, Mar 28, 2019, 3:51 PM

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Great work dude this point is really awesome!!

Edit (AlastorMoody): Thanks :thumbup:

wouldn't have been possible without your help

This post has been edited 1 time. Last edited by AlastorMoody, Mar 29, 2019, 5:21 AM

by Pluto1708, Mar 28, 2019, 8:58 PM

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X(25) yayyyy. It's the only center I know >12.

Edit (AlastorMoody): LOL

though X(20) is also an innocent triangle center

This post has been edited 1 time. Last edited by AlastorMoody, Apr 7, 2019, 9:11 AM

by Hexagrammum16, Apr 5, 2019, 4:04 PM

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what a goat, u used to be friends with my brother
by bookstuffthanks, Jul 31, 2024, 12:05 PM
hello fellow moody!!
by crazyeyemoody907, Oct 31, 2023, 1:55 AM
@below I wish I started earlier / didn't have to do JEE and leave oly way before I could study conics and projective stuff which I really wanted to study . Huh, life really sucks when u are forced due to peer pressure to read sh_t u dont want to read
by kamatadu, Jan 3, 2023, 1:25 PM
Lots of good stuffs here.
by amar_04, Dec 30, 2022, 2:31 PM
But even if he went to jee he could continue with this.

Doing JEE(and completely leaving oly) seems like a insult to the oly math he knows
by HoRI_DA_GRe8, Feb 11, 2022, 2:11 PM
Ohhh did he go for JEE? Good for him, bad for us :sadge:. Hmmm so that is the reason why he is inactive
Btw @below finally everyone falls to the monopoly of JEE Coz IIT's are the best in India.
by BVKRB-, Feb 1, 2022, 12:57 PM
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Thanks dude!
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Amazing work bro!!
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@below Thanks a lot
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Nice Blog, loved it.
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@below Thanks buddy!
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by PhysicsMonster_01, Jan 29, 2019, 9:21 AM
@below I am noob right now I hope to develop myself within a year or so
by AlastorMoody, Jan 28, 2019, 7:24 PM
Nice blog. But the properties are a bit 'basic'.
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Nice!!!!!!!!
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Math is fun!
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120 shouts

Magic of Hogwarts

X(25) is the pole of the orthic axis WRT ABC

by AlastorMoody, Mar 28, 2019, 3:30 PM

2 Comments