Yet another OH-symmetry

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Yet another OH-symmetry

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Source: All-Russian MO 2024 11.6

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9027 posts

#1 h Apr 22, 2024, 7:26 PM • 4 Y

Y by buratinogigle, GeoKing, Rounak_iitr, Funcshun840

Let $ABC$ be an acute non-isosceles triangle with circumcircle $\omega$ , circumcenter $O$ and orthocenter $H$ . We draw a line perpendicular to $AH$ through $O$ and a line perpendicular to $AO$ through $H$ . Prove that the points of intersection of these lines with sides $AB$ and $AC$ lie on a circle, which is tangent to $\omega$ .
Proposed by A. Kuznetsov

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#2 h Apr 22, 2024, 11:29 PM • 1 Y

Y by GeoKing

This happens for every pair of conjugate points.

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Om245

#3 h Apr 23, 2024, 4:41 AM • 2 Y

Y by GeoKing, Stuffybear

Let line through $H$ and $O$ meet $AB$ and $AC$ at $P,Q,D,E$ in respective order.

Let tangent at $A$ to $\omega$ meet $DE$ at $K$ and $R= PQ \cap BC$ . Let $T$ be reflection of $H$ above $BC$ .
Note that $KA \parallel RH$ . As $OK \perp AT$ , reflection of $A$ above $OK$ is $T$ . hence we have $\overline{K-R-T}$ .

by applying radical theorem on $(ADE),(DEPQ),(ABC)$ and $(PQBC),(ABC),(DEPQ)$ we have $K$ and $R$ lie on radical axis of $(ABC)$ and $(DEPQ)$ .
But $T$ lie on $(ABC)$ hence $T$ lie on $(DEPQ)$ with tangent to $(ABC)$ .

This post has been edited 2 times. Last edited by Om245, Apr 23, 2024, 12:52 PM
Reason: fast2

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#5 h Apr 28, 2024, 3:04 AM • 1 Y

Y by GeoKing

Here is another solution.

Let $E,F$ be the feet of altitudes from $B,C$ and $M,N$ be the midpoints of $AC,AB$ respectively.Let the intersections of $(O \perp AH)$ meet $AB,AC$ at $R,Q$ respectively and $(H \perp AO)$ meet $AB,AC$ at $T,S$ respectively.Let $AH$ meet $MN,RQ,BC,\omega$ at $I,J,K,L$ respectively.

See to the fact that $EFNM$ and $STRQ$ are homothetic,hence $RQTS$ is cyclic.The homothety at $A$ also sends $I$ to $J$ as well.Now since $I$ is the midpoint of $AK$ and $J$ is the midpoint of $AL$ ,the homothety sends $K$ to $L$ as well.Since $K$ also lies in the 9 pt circle, $RQTLS$ is cyclic.Finally as $A,L$ are reflections over $\overline{R-O-Q}$ we have,
$\measuredangle (RL,LL)=90-\measuredangle RLA-\measuredangle OLA=90-(\measuredangle LAR+\measuredangle OAL)=90-\measuredangle OAB=\measuredangle BCA=\measuredangle DMN=\measuredangle LQR$ Hence $LL$ is tangent to $(STRQ)$ as well ,hence done $\blacksquare$

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#6 h Jun 7, 2024, 8:03 PM • 1 Y

Y by tohill

Let $T$ be the reflection of $H$ across $BC$ . Let the line through $H$ and $O$ perpendicular to $AO$ and $AH$ intersect $AB$ and $AC$ at $H_B$ , $H_C$ , $O_B$ , and $O_C$ , respectively.

Claim: $H_BH_CO_BO_C$ is concyclic
Since $AO$ and $AH$ are isogonal, $H_BH_C$ and $O_BO_C$ are anti-parallel.
Claim: $H_BH_CT$ is tangent to the circumcircle of $ABC$
Notice that the tangent at $T$ to $w$ is a reflection of $H_BH_C$ across $BC$ so it must concur with $BC$ and $H_BH_C$ at some point $P$ . Notice that $BCH_BH_C$ is cyclic so $PT^2=PB\cdot PC=PH_B\cdot PH_C$ .
Claim: $O_BO_CT$ is tangent to the circumcircle of $ABC$
The circumcircle of $AO_BO_C$ is tangent to $w$ . Reflection across $O_BO_C$ yields the result.
Claim: $H_BH_CO_BO_CT$ is cyclic
Assume otherwise. Apply the Radical Axes Theorem on the circumcircles of $H_BH_CO_BO_C$ , $H_BH_CT$ , and $O_BO_CT$ yielding $H_BH_C$ , $O_BO_C$ , and the tangent at $T$ to $w$ concur. However the tangent at $T$ to $w$ and $H_BH_C$ intersect along $BC$ and $O_BO_C$ and $BC$ are strictly parallel.

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mcmp

#7 h Nov 6, 2024, 8:50 AM • 1 Y

Y by Rounak_iitr

Russian geo :love:

:love:

Define the points as defined in the diagram below.
$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10) + blue; defaultpen(dps); /* default pen style */ pen dotstyle = blue; /* point style */ real xmin = -20.61703458825209, xmax = 24.072148158469165, ymin = -12.505052373958343, ymax = 12.073998136738606; /* image dimensions */ pen qqwuqq = rgb(0,0.39215686274509803,0); pen ccqqqq = rgb(0.8,0,0); pen ffvvqq = rgb(1,0.3333333333333333,0); draw((-4.2029584148481405,9.073871310695187)--(-8.416490391756525,-5.400249020690629)--(8.480028787723565,-5.299916203052611)--cycle, linewidth(0.5) + ccqqqq); /* draw figures */ draw(circle((0,0), 10), linewidth(0.5) + qqwuqq); draw((-4.2029584148481405,9.073871310695187)--(-8.416490391756525,-5.400249020690629), linewidth(0.5) + ccqqqq); draw((-8.416490391756525,-5.400249020690629)--(8.480028787723565,-5.299916203052611), linewidth(0.5) + ccqqqq); draw((8.480028787723565,-5.299916203052611)--(-4.2029584148481405,9.073871310695187), linewidth(0.5) + ccqqqq); draw((-7.813236953859962,-3.3279821200739916)--(2.517715274359453,1.4572480329930906), linewidth(0.5) + ffvvqq); draw((-6.856286714825485,-0.04071315265146354)--(3.7837209451897897,0.022468023120875688), linewidth(0.5) + ffvvqq); draw(circle((-1.516277314066464,-3.378154929795888), 6.297159519658339), linewidth(0.5) + qqwuqq); draw((-4.2029584148481405,9.073871310695187)--(0,0), linewidth(0.5) + ffvvqq); draw((-7.813236953859962,-3.3279821200739916)--(-4.094903148316446,-9.123144644578867), linewidth(0.5) + ffvvqq); draw((-6.856286714825485,-0.04071315265146354)--(-4.094903148316446,-9.123144644578867), linewidth(0.5) + ffvvqq); draw((-9.999682587261418,0.0796753037068001)--(-7.813236953859962,-3.3279821200739916), linewidth(0.5) + ffvvqq); draw((-4.094903148316446,-9.123144644578867)--(6.406051134376832,7.678704894951316), linewidth(0.5) + ffvvqq); draw((-9.999682587261418,0.0796753037068001)--(8.480028787723565,-5.299916203052611), linewidth(0.5) + ffvvqq); draw((-8.416490391756525,-5.400249020690629)--(6.406051134376832,7.678704894951316), linewidth(0.5) + ffvvqq); draw((-4.2029584148481405,9.073871310695187)--(-4.094903148316446,-9.123144644578867), linewidth(0.5) + ffvvqq); draw((-7.069551303071259,-0.7733093046706251)--(-4.117161583598773,-5.37471927881346), linewidth(0.5) + ffvvqq); draw((-4.117161583598773,-5.37471927881346)--(1.1333155577478653,3.0262054909516327), linewidth(0.5) + ffvvqq); draw((-7.069551303071259,-0.7733093046706251)--(1.1333155577478653,3.0262054909516327), linewidth(0.5) + ffvvqq); /* dots and labels */ dot((-4.2029584148481405,9.073871310695187),dotstyle); label("$A$", (-4.068071602356874,9.420577911152003), NE * labelscalefactor); dot((-8.416490391756525,-5.400249020690629),dotstyle); label("$B$", (-8.292595908882868,-5.068493057511681), NE * labelscalefactor); dot((8.480028787723565,-5.299916203052611),dotstyle); label("$C$", (8.605501317221107,-4.963752785449052), NE * labelscalefactor); dot((0,0),dotstyle); label("$O$", (0.1564527041691194,0.34308766572415295), NE * labelscalefactor); dot((-4.139420018881101,-1.6262939130480514),dotstyle); label("$H$", (-3.998244754315122,-1.262929839236159), NE * labelscalefactor); dot((-6.856286714825485,-0.04071315265146354),dotstyle); label("$B'$", (-6.721491827943448,0.3081742417032766), NE * labelscalefactor); dot((3.7837209451897897,0.022468023120875688),dotstyle); label("$C'$", (3.927102498423725,0.37800108974502933), NE * labelscalefactor); dot((-7.813236953859962,-3.3279821200739916),dotstyle); label("$B''$", (-7.6641542765071,-2.9736876162591), NE * labelscalefactor); dot((2.517715274359453,1.4572480329930906),dotstyle); label("$C''$", (2.67021923367219,1.8094514746009596), NE * labelscalefactor); dot((-4.094903148316446,-9.123144644578867),dotstyle); label("$D$", (-3.9633313302942463,-8.769316003724574), NE * labelscalefactor); dot((6.406051134376832,7.678704894951316),dotstyle); label("$E$", (6.545609299989423,8.02404095031695), NE * labelscalefactor); dot((-9.999682587261418,0.0796753037068001),dotstyle); label("$F$", (-9.863699989822287,0.41291451376590566), NE * labelscalefactor); dot((-7.069551303071259,-0.7733093046706251),dotstyle); label("$F'$", (-6.930972372068704,-0.4250076627351267), NE * labelscalefactor); dot((1.1333155577478653,3.0262054909516327),dotstyle); label("$E'$", (1.2736822728371509,3.380555555540395), NE * labelscalefactor); dot((-4.117161583598773,-5.37471927881346),dotstyle); label("$D'$", (-3.9633313302942463,-5.033579633490804), NE * labelscalefactor); dot((-4.152508370055177,0.5778460866718097),dotstyle); label("$P$", (-3.998244754315122,0.9366158740790509), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
I’m going to first show that $D-B’’-F$ collinear, and same for $D-C’’-E$ . However note $B’’HC’’$ is just $F’PE’$ dilated from $A$ with factor $\frac{AP}{AH}$ , so assuming $\frac{AP}{AH}=\frac{AD’}{AD}$ , $\measuredangle ADF=\measuredangle ACF=\measuredangle E’CH=\measuredangle E’D’H=\measuredangle HD’F’=\measuredangle ADB’’$ , so the ratio is all we need to prove to finish the collinearity. However the ratio is equivalent to $\frac{AP}{AD’}=\frac{AH}{AD}=\frac{PH}{D’D}=-\frac{PH}{D’H}$ , which is actually true since $(AH;PD’)=-1$ .

Now for the actual proof. Note it’s clear that $(B’C’D)$ is tangent to $(ABC)$ at $D$ since it’s the reflection of $(AB’C’)$ over $B’C’$ (clear) hence it suffices to show that $B’’,C’’\in(B’C’D)$ . Notice that $\triangle DEF\sim\triangle DB’’C’’$ , so it’s clear that $(DB’’C’’)$ is the homothetic copy of $(ABC)$ scaled from $D$ so it’s also tangent at $D$ . Finally $B’B’’C’’C’$ cyclic is clear. Now if they aren’t the same circle then using radax on the three previous circles gives the tangent at $D$ , $B’C’$ and $B’’C’’$ concur which is nonsense, done.

This post has been edited 1 time. Last edited by mcmp, Nov 6, 2024, 8:52 AM

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#8 h Mar 24, 2025, 2:07 AM • 1 Y

Y by Funcshun840

cesarlozada wrote:

This happens for every pair of conjugate points.

That's not true, I was thinking about it on my day off. Certainly I discarded this idea after drawing it in Geogebra but I was sure I'd be able to find a generalization. So $H,O,I,I_a,I_bI_b,I_c$ lies on that locus easily, based on stablished problem and mixtilinear incircles-excicrcles. So it's a suspicious isopivotal cubic, where we need a pivot, supposing my pair of isogonal conjugates points were $A$ and $O_a$ , where $O_a \equiv BC \cap AO$ "we'd obtain" as the circle tangent as $\odot (ABC)$ it does makes sense, may be the M'Cay Cubic. So I got a generalization what I solved using Pascal so many times(I'll post my solution later) it really took me some time

General Problem

Let $ABC$ be an acute non-isosceles triangle with circumcircle $\omega$ , let $P,Q$ be two pair of isogonal conjugates. We draw a line perpendicular to $AP$ through $Q$ and a line perpendicular to $AQ$ through $P$ . Prove that the points of intersection of these lines with sides $AB$ and $AC$ lie on a circle, which is tangent to $\omega$ if only if $PQ$ passes through the center of $\omega$ .

This post has been edited 1 time. Last edited by drmzjoseph, Mar 24, 2025, 2:26 AM
Reason: PQ instead P,Q

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