Basic Properties of Antigonal Conjugates with a taste of Rectangular Hyperbolas.

by amar_04, Aug 13, 2020, 11:26 AM

So as the title suggests, we will be discussing some basic facts and properties about Antigonal Conjuagtes. For those who don't know what Antigonal Conjugates are. Don't worry! I will be going through the very basics that is I will be defining Poncelet Point and other stuffs necessary. So without any delay we start.

Properties of Poncelet Point

$\textbf{Definition of Poncelet Point:-}$

$\bigstar$ Let $P$ be an arbitary point in the plane. Then the Point of Concurrency of the Nine Point Circles of $\Delta PBC,\Delta CPA,\Delta APB$ is the Poncelet Point of $\{A,B,C,P\}$. Sometimes it is referred to as the Euler-Poncelet Point too.

Throughout we will assume $P^*$ as the Poncelet Point of $\{A,B,C,P\}$. Here are some Interesting Properties of the Poncelet Point.

Property 1:- If $\epsilon$ is the Nine Point Circle of $\Delta ABC$ , then $P^*\in\epsilon$
Property 2:- If $\omega$ is the Pedal Circle of $P$ WRT $\Delta ABC$, then $P^*\in\omega$
Property 3:- If $\gamma$ is the Ceva Circle of $P$ WRT $\Delta ABC$, then $P^*\in\gamma$


Exercise:- Prove that the concurrency of the Nine Point Circles of $\Delta PBC,\Delta CPA,\Delta APB$ and Property 1 and Property 2 yourselves. Just simply angle chase.

We will come back to the Proof of Property 3 later. Before that let's move onto some Hyperbola, we will be resorting to that often!

$\textbf{Definition:-}$

$\bigstar$ What is an Equilateral or Rectangular Hyperbola?
The Hyperbolas which have perpendicular Asymptotes fall into the category of Rectangular Hyperbolas.

Basic Properties of Rectangular Hyperbolas

Lemma 1:- The Orthocenter $H$ of $\Delta ABC$ lies on the Rectangular Hyperbola $\mathcal H$ passing through $A,B,C$. The converse holds too.

We give an Elementary proof of this Lemma (rather non-standard). Let $X$ and $Y$ be the Infinite Line along two lines at Infinity $\infty_{\ell_1}$ and $\infty_{\ell_2}$ respectively such that $\infty_{\ell_1}\perp\infty_{\ell_2}$. Draw two lines through $A$ and $B$ respectively such that they are parallel to $\infty_{\ell}$ and draw two lines through $C$ and $H$ respectively which are parallel to $\infty_{\ell_2}$. Let $\overline{UV}$ be the Diagonal of the rectangle formed by those lines. Let $B'$ be the Projection of $B$ on $\overline{AC}$. The figure will look something like this. Now clearly by Coverse of Reims Theorem $B',U,V$ are collinear. But by Converse of Pascals Theorem on $AXBHYC$ we get $H$ lies on the Rectangular Hyperbola containing $\{A,B,C,X,Y\}$. So, $H\in\mathcal H$. Try the Converse yourself!

Lemma 2:- The Center of $\mathcal H$ lies on the Nine Point Circle of $\Delta ABC$

Let $\mathcal{H}$ be a fixed rectangular hyperbola through vertices of $\Delta ABC$. If $H$ is the orthocenter WRT $\Delta ABC$, then $H$ lies on $\mathcal{H}$. Let $E_A, E_B, E_C$ be the midpoints of $AH, BH, CH$. Let $\Delta H_AH_BH_C$ be the orthic triangle WRT $\Delta ABC$. $\implies$ $E_AE_BE_CH_AH_BH_C$ is the Nine-Point Circle WRT $\Delta ABC$. Let the tangent to $\mathcal{H}$ at $A,H$ ; $B,H$ and $C,H$ intersect at $K, M, N$, then $E_AK, E_BM, E_CN$ intersect at $P$ which is the center of $\mathcal{H}$. Consider the reflections of $H,A,B,C$ over $P$ to $Q, R, S, T$. Then, $Q, R, S, T$ lies on $\mathcal{H}$. Obviously $AHRQ$ is a parallelogram. Apply Pascal's Theorem on $HHRQQA$ $\implies$ $H-$tangent to $\mathcal{H}$ and the $Q-$tangent to $\mathcal{H}$ are parallel. Let $QS$ $\cap$ $AC$ $=$ $U$, $QR$ $\cap$ $BC$ $=$ $V$ and $QT$ $\cap$ $AB$ $=$ $W$. Then $\Delta VUW$ is the pedal triangle of $Q$ WRT $\Delta ABC$. Apply Pascal's Theorem on $SQRACB$ and $TCBARQ$ $\implies$ $VUW$ is the Simson Line $\implies$ $Q$ lies on $\odot (ABC)$ $\implies$ $P$ lies on Nine-Point Circle. $\qquad \blacksquare$

The above Proof of Lemma 2 was by @Alastor Moody here. Any way I give another Proof for Lemma 2 which will further prove that the Center of the Rectangular $\mathcal H$ passing through $P$ is the Poncelet Point $P^*$ of $\{A,B,C,P\}$.

Before Proving Lemma 2 we start with a Theorem.

Theorem:- Let $P$ be a point on $\odot(ABC)$. Let $H_P$ be the Orthocenter of $\Delta ABC$ and define $H_A,H_B,H_C$ cyclically. Then the reflection $ABCP\overset{P^*}{\mapsto}H_AH_BH_CH_P$ where $P^*$ is the Poncelet Point of $\{A,B,C,P\}$.

It's fairly easy to see that $P^*$ is the midpoint of $PH_P$ because the Pedal Triangle degenerates into the Simson Line of $P$ and Midpoint of $PH_P\in\odot(X_5)$ where $\odot(X_5)$ is the Nine Point Circle of $\Delta ABC$. So, midpoint of $PH_P$ is $P^*$ $(\spadesuit)$ also it's easy to see that $ABCP$ and $H_AH_BH_CH_P$ with Homothetic Center at the midpoint of $PH_P$ which is $P^*$ from ($\spadesuit$). Now by Lemma 1 clearly $H_A,H_B,H_C,H_P$ lies on the Rectangular Hyperbola $\mathcal H$ passing through $P$ and from Theorem $P^*$ is the midpoint of $\mathcal H$!

So we proved Lemma 2.


Now we are ready to prove Property 3 mentioned above.

Lemma:- Given a $ \triangle ABC $ and two points $ P, $ $ Q $. Let $ \triangle P_aP_bP_c $ be the anticevian triangle of $ P $ WRT $ \triangle ABC $ and let $ \triangle Q_aQ_bQ_c $ be the anticevian triangle of $ Q $ WRT $ \triangle ABC $. Then $ P, $ $ Q, $ $ P_a, $ $ P_b, $ $ P_c, $ $ Q_a, $ $ Q_b, $ $ Q_c $ lie on a conic. (See Property 1 here.)

Let $P$ be the Point and $\Delta XYZ$ be the Cevian Triangle of $P$ WRT $\Delta ABC$. Let $\Delta DEF$ be the Excentral Triangle of $\Delta ABC$ and $H$ be the Incenter of $\Delta XYZ$. Then from Lemma we get that $\{A,B,C,D,E,F,P,H\}$ lies on a Conic and clearly from Lemma 1 we get that this is a Rectangular Hyperbola $\mathcal H$ and the Center of $\mathcal H$ lies on $\odot(XYZ)$ which is the Poncelet Point $P^*$ of $\{A,B,C,P\}$. Here is a diagram.

[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(25cm); 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 = -15.842401898734185, xmax = 22.075934810126604, ymin = -10.103860759493674, ymax = 11.24605316455697; /* image dimensions */ /* draw figures */ draw((xmin, 4.568714416151212*xmin + 21.662856469515518)--(xmax, 4.568714416151212*xmax + 21.662856469515518), linewidth(0.8)); /* line */ draw((xmin, 0.007788161993769443*xmin-4.744906542056075)--(xmax, 0.007788161993769443*xmax-4.744906542056075), linewidth(0.8)); /* line */ draw((xmin, -1.0716650680340856*xmin + 2.8652387296403026)--(xmax, -1.0716650680340856*xmax + 2.8652387296403026), linewidth(0.8)); /* line */ draw((xmin, -3.5116785471619987*xmin-5.266561761690379)--(xmax, -3.5116785471619987*xmax-5.266561761690379), linewidth(0.4)); /* line */ draw((xmin, 0.625*xmin-1.17125)--(xmax, 0.625*xmax-1.17125), linewidth(0.4)); /* line */ draw((xmin, -0.3606965174129354*xmin-2.147089552238806)--(xmax, -0.3606965174129354*xmax-2.147089552238806), linewidth(0.4)); /* line */ draw((xmin, -0.9272185087388412*xmin-4.88349319466219)--(xmax, -0.9272185087388412*xmax-4.88349319466219), linewidth(0.4)); /* line */ draw((xmin, 2.0028278005435047*xmin-4.449201843921155)--(xmax, 2.0028278005435047*xmax-4.449201843921155), linewidth(0.4)); /* line */ draw((xmin, 0.09994520887466238*xmin + 0.07789326743154854)--(xmax, 0.09994520887466238*xmax + 0.07789326743154854), linewidth(0.4)); /* line */ draw((xmin, -5.494449467187799*xmin-5.560447990839757)--(xmax, -5.494449467187799*xmax-5.560447990839757), linewidth(0.4)); /* line */ draw((xmin, 2.9774754756368607*xmin + 13.976884698854816)--(xmax, 2.9774754756368607*xmax + 13.976884698854816), linewidth(0.4)); /* line */ draw((xmin, -0.3358549913114254*xmin-2.0271004899458607)--(xmax, -0.3358549913114254*xmax-2.0271004899458607), linewidth(0.4)); /* line */ draw((xmin, -1.4502647138281781*xmin + 3.765954629369643)--(xmax, -1.4502647138281781*xmax + 3.765954629369643), linewidth(0.4)); /* line */ draw((xmin, 0.689529291077047*xmin-1.324769843763852)--(xmax, 0.689529291077047*xmax-1.324769843763852), linewidth(0.4)); /* line */ pair hyperbolaLeft1 (real t) {return (4.1547889871044115*(1+t^2)/(1-t^2),4.1547889871044115*2*t/(1-t^2));} pair hyperbolaRight1 (real t) {return (4.1547889871044115*(-1-t^2)/(1-t^2),4.1547889871044115*(-2)*t/(1-t^2));} draw(shift((-3.1406908410689613,2.08904391317063))*rotate(-75.32413781366773)*graph(hyperbolaLeft1,-0.99,0.99), linewidth(0.8)); draw(shift((-3.1406908410689613,2.08904391317063))*rotate(-75.32413781366773)*graph(hyperbolaRight1,-0.99,0.99), linewidth(0.8)); /* hyperbola construction */ draw((xmin, 0.1820018558677955*xmin-4.719084595081908)--(xmax, 0.1820018558677955*xmax-4.719084595081908), linewidth(0.4)); /* line */ /* dots and labels */ dot((-3.332686708860754,6.436762658227854),dotstyle); label("$A$", (-4.111068987341767,6.6591575949367146), NE * labelscalefactor); dot((-5.79,-4.79),dotstyle); label("$B$", (-6.140422784810124,-4.432789873417722), NE * labelscalefactor); dot((7.05,-4.69),dotstyle); label("$C$", (7.147674683544319,-4.4049905063291135), NE * labelscalefactor); dot((-0.99,-1.79),dotstyle); label("$P$", (-0.8863424050632832,-1.5138563291139224), NE * labelscalefactor); dot((-0.14821996135870136,-4.746060903125846),linewidth(4pt) + dotstyle); label("$X$", (0.2812310126582368,-5.26677088607595), NE * labelscalefactor); dot((2.3790722197854612,0.31567013736591315),linewidth(4pt) + dotstyle); label("$Y$", (2.4773810126582387,0.5432968354430405), NE * labelscalefactor); dot((-4.830180794957348,-0.40486016102284644),linewidth(4pt) + dotstyle); label("$Z$", (-5.139645569620249,0.18190506329114164), NE * labelscalefactor); dot((5.198317102109108,-3.772981235108749),linewidth(4pt) + dotstyle); label("$F$", (5.312916455696216,-3.543210126582278), NE * labelscalefactor); dot((-6.687943381658956,-5.936302702482577),linewidth(4pt) + dotstyle); label("$E$", (-7.252397468354428,-5.739360126582279), NE * labelscalefactor); dot((-2.306126744694759,7.110448872815898),linewidth(4pt) + dotstyle); label("$D$", (-2.1929126582278413,7.659934810126589), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy]

Phew! So we have learned about some basic and elementary properties of Poncelet Point and Conics. Now it's time for us to work with Antigonal Conjuagtes! We define $\widehat{P}$ as the Antigonal Conjugate of $P$ WRT $\Delta ABC$ throughout.

Properties of Antigonal Conjugtes

$\bigstar$ If $P$ is an aribary point then $\widehat{P}$ is the reflection of $P$ over $P^*$ where $P^*$ is the Poncelet Point of $\{A,B,C,P\}$.

Now we will talk about some well known Properties of Antigonal Conjugates.
Lemma 1:- $\measuredangle BPC=\pi-\measuredangle B\widehat PC$ and similarly others.

Lemma 2:- $\widehat P$ is the Isogonal Conjuagte of the Inverse of the Isogonal Conjugate of a Point $P$ WRT $\Delta ABC$

Lemma 3:- $\widehat P$ lies on the Reflection Circle of $P$ WRT $\Delta ABC$.

Lemma 3:- $P,\widehat P$ are Antipodal Points WRT the Rectangular Hyperbola $\mathcal H$ passing through $\{A,B,C,P\}$.

Lemma 5:- Let $\Delta M_AM_BM_C$ be the Medial Triangle of $\Delta ABC$. Let $\Delta M_A^*M_B^*M_C^*$ be the triangle formed by the reflection of $P$ about $M_A,M_B,M_C$ respectively. Then $\{A,B,C,M_A^*,M_B^*,M_C^*,\widehat P\}$ lies on a Conic $\mathcal C$.

Lemma 6:- The Rectangular Hyperbola $\mathcal H$ passing through $\{A,B,C,P,\widehat P\}$ is the Isogonal Conjugate of the Perendicular Bisector of $P\widehat P$ WRT $\Delta AP\widehat P,\Delta BP\widehat P,\Delta CP\widehat P$.

Lemma 7:- Let $K,\widehat K$ be the Isogonal Conjugates of $P,\widehat P$ WRT $\Delta ABC$ respectively. Then $K,\widehat K$ are Inverses WRT $\Delta ABC$.

Try angle chasing (Lemma 1 , Lemma 2 and Lemma 7) and Lemma 3 is trivial from Lemma 2. So I will just prove Lemma 5 and Lemma 6.


Proof of Lemma 5

[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(25cm); 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 = -13.75, xmax = 15.41, ymin = -7.87, ymax = 7.47; /* image dimensions */ pen qqttzz = rgb(0,0.2,0.6); /* draw figures */ draw((xmin, 4.53913043478261*xmin + 11.813130434782611)--(xmax, 4.53913043478261*xmax + 11.813130434782611), linewidth(0.8) + qqttzz); /* line */ draw((xmin, 0.011382113821138186*xmin-4.169821138211383)--(xmax, 0.011382113821138186*xmax-4.169821138211383), linewidth(0.8) + qqttzz); /* line */ draw((xmin, -1.03*xmin + 4.9631)--(xmax, -1.03*xmax + 4.9631), linewidth(0.8) + qqttzz); /* line */ pair hyperbolaLeft1 (real t) {return (3.7648470037776307*(1+t^2)/(1-t^2),3.7648470037776307*2*t/(1-t^2));} pair hyperbolaRight1 (real t) {return (3.7648470037776307*(-1-t^2)/(1-t^2),3.7648470037776307*(-2)*t/(1-t^2));} draw(shift((-1.207373023347141,2.157029928327833))*rotate(-74.02819642656856)*graph(hyperbolaLeft1,-0.99,0.99), linewidth(0.8)); draw(shift((-1.207373023347141,2.157029928327833))*rotate(-74.02819642656856)*graph(hyperbolaRight1,-0.99,0.99), linewidth(0.8)); /* hyperbola construction */ draw((2.21,-1.45)--(-1.23,6.23), linewidth(0.4)); draw((2.21,-1.45)--(-3.53,-4.21), linewidth(0.4)); draw((2.21,-1.45)--(8.77,-4.07), linewidth(0.4)); draw(shift((1.555,-0.875))*rotate(150.97180357343146)*xscale(4.373670789935975)*yscale(2.8581171359992825)*unitcircle, linewidth(0.4) + red); draw(shift((0.9,-0.3))*rotate(-29.028196426568545)*xscale(8.74734157987195)*yscale(5.716234271998564)*unitcircle, linewidth(0.4) + red); draw((2.21,-1.45)--(5.33,3.61), linewidth(0.4)); draw((2.21,-1.45)--(-6.97,3.47), linewidth(0.4)); draw((2.21,-1.45)--(-4.624746046694282,5.764059856655666), linewidth(0.4)); draw((2.21,-1.45)--(3.03,-6.83), linewidth(0.4)); /* dots and labels */ dot((-1.23,6.23),dotstyle); label("$A$", (-1.15,6.43), NE * labelscalefactor); dot((-3.53,-4.21),dotstyle); label("$B$", (-3.45,-4.01), NE * labelscalefactor); dot((8.77,-4.07),dotstyle); label("$C$", (8.85,-3.87), NE * labelscalefactor); dot((-1.1376080880630828,-1.8872894058864897),linewidth(4pt) + dotstyle); label("$H$", (-1.05,-1.73), NE * labelscalefactor); dot((2.21,-1.45),dotstyle); label("$P$", (2.29,-1.25), NE * labelscalefactor); dot((0.49,2.39),linewidth(4pt) + dotstyle); label("$X_A$", (0.57,2.55), NE * labelscalefactor); dot((5.49,-2.76),linewidth(4pt) + dotstyle); label("$X_C$", (5.57,-2.61), NE * labelscalefactor); dot((3.77,1.08),linewidth(4pt) + dotstyle); label("$M_B$", (3.85,1.23), NE * labelscalefactor); dot((-2.38,1.01),linewidth(4pt) + dotstyle); label("$M_C$", (-2.31,1.17), NE * labelscalefactor); dot((-0.66,-2.83),linewidth(4pt) + dotstyle); label("$X_B$", (-0.59,-2.67), NE * labelscalefactor); dot((2.62,-4.14),linewidth(4pt) + dotstyle); label("$M_A$", (2.69,-3.99), NE * labelscalefactor); dot((-1.207373023347141,2.157029928327833),linewidth(4pt) + dotstyle); label("$P^*$", (-1.13,2.31), NE * labelscalefactor); dot((5.33,3.61),dotstyle); label("$M_B^*$", (5.41,3.81), NE * labelscalefactor); dot((3.03,-6.83),dotstyle); label("$M_A^*$", (3.11,-6.63), NE * labelscalefactor); dot((-6.97,3.47),dotstyle); label("$M_C^*$", (-6.89,3.67), NE * labelscalefactor); dot((-4.624746046694282,5.764059856655666),dotstyle); label("$\widehat P$", (-4.55,5.97), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy]

Let $\{X_A.X_B,X_C\}$ be the midpoints of $AP,BP,CP$ respectively. Then clearly $\{M_A,M_B,M_C,X_A,X_B,X_C\}$ lies on a Conic. By Converse of Pascal's Theorem on $M_AX_BM_CX_AM_BX_C$ and also it's easy to notice that the Conic formed is the Locus of the Center of those Hyperbolas $\mathcal T$ passing through $\{A,B,C,T\}$ (Why?) $(\spadesuit)$. So Clearly the Center of $\mathcal H$ which is $P^*$ from Lemma 2 must lie on this Conic. Now take a Homothety at $P$ mapping $\Delta M_AM_BM_C$ to $\Delta M_A^*M_B^*M_C^*$. We get the desired result. $\blacksquare$

Remark:- Try using $(\spadesuit)$ to prove that the Locus of the Centers of the Hyperbolas $\mathcal T$ passing through $\{A,B,C,G\}$ where $G$ is the Centroid of $\Delta ABC$. Is the Steiner Inellipse of $\Delta ABC$.

Proof of Lemma 6:-

Notice that from Lemma 1 the Orthocenter $H$ of $\Delta AP\widehat P$ lies on $\mathcal H$ and the reflection of $A$ over $P^*$ lies on $\mathcal H$. Let this point be $T$. So the Isogonal Conjugte of $\mathcal H$ is the Line joining the Isogonal Conjugates of $H,T$ WRT $\Delta AP\widehat P$ which is the Perpendicular bisector of $P\widehat P$. Similary for $\Delta BP\widehat P$ and $\Delta CP\widehat P$. $\blacksquare$

Bonus:- Try finding more properties!

We will end our discussion after solving two Problems which involves Antigonal Conjugates.
ISL 2018 G4 wrote:
A point $T$ is chosen inside a triangle $ABC$. Let $A_1$, $B_1$, and $C_1$ be the reflections of $T$ in $BC$, $CA$, and $AB$, respectively. Let $\Omega$ be the circumcircle of the triangle $A_1B_1C_1$. The lines $A_1T$, $B_1T$, and $C_1T$ meet $\Omega$ again at $A_2$, $B_2$, and $C_2$, respectively. Prove that the lines $AA_2$, $BB_2$, and $CC_2$ are concurrent on $\Omega$.

Take a Negative Homothety at $T$ with a scale factor of $-2$. So the problem transforms to this one.

Proposition:- $ABC$ be a triangle with a point $T$ in it's plane and let $A_1B_1C_1$ be the $T-\text{Pedal Triangle}$ WRT $\triangle ABC$. Let $\{A_1,B_1,C_1\}\cap\odot(A_1B_1C_1)=\{X,Y,Z\}$ and let $\{A_2,B_2,C_2\}$ be the midpoints of $\{PA,PB,PC\}$ respectively. Then $XA_2,YB_2ZC_2$ intersect on $\odot(A_1B_1C_1)$


Proof:- Let $\odot(A_1B_1C_1)\cap BC=\{A_1,D\}$ and let $K$ be the Poncelet Point of $\{ABCT\}$. We claim that $XA_2,YB_2,ZC_2$ concur at the Poncelet Point of $\{ABCT\}$. Notice that $\angle DKX=90^\circ$ and similarly we get that $\angle DKA_2=\angle DKZ+\angle ZKA_2=\angle ATC+90^\circ-\angle ATC=90^\circ\implies\overline{A_2-X-K}$. Similarly we get that $\overline{B_2-Y-K}$ and $\overline{C_2-Z-K}$. So, $XA_2,YB_2,ZC_2$ concur at the Poncelet Point of $\{ABCT\}$. So now transform back.

We get that $AA_2,BB_2,CC_2$ are concurrent at $\odot(ABC)$. The Point of Concurrency is the $\text{Antigonal Conjugate}$ of $\{ABCT\}$. $\blacksquare$

Now here is a harder problem proposed by @buratinogigle. It is also the P127 of the famous GeoZebra Marathon. We will extensively use our Lemmas in this problem. Here goes the problem and my Solution.
Buratinogigle wrote:
Let $ABC$ be a triangle and $P_1,P_2$ are two antigonal conjugate points. $Q_1,Q_2$ are isogonal conjugate of $P_1,P_2$. Reflection of line $P_2Q_1$ thourgh $BC$ cuts line $P_1Q_2$ at $X$ ,Prove that $\measuredangle BXC=\measuredangle BP_1C$

[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(25cm); 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 = -26.408453486015034, xmax = 27.107031745977462, ymin = -14.562765893491065, ymax = 9.596373113535464; /* image dimensions */ pen qqttzz = rgb(0,0.2,0.6); /* draw figures */ draw((xmin, 4.008403361344537*xmin + 15.841512605042016)--(xmax, 4.008403361344537*xmax + 15.841512605042016), linewidth(0.8)); /* line */ draw((xmin, 0.01754385964912279*xmin-3.873333333333334)--(xmax, 0.01754385964912279*xmax-3.873333333333334), linewidth(0.8)); /* line */ draw((xmin, -0.9173228346456693*xmin + 3.231653543307087)--(xmax, -0.9173228346456693*xmax + 3.231653543307087), linewidth(0.8)); /* line */ draw((xmin, 0.23694018521389304*xmin + 1.7294420252666707)--(xmax, 0.23694018521389304*xmax + 1.7294420252666707), linewidth(0.4)); /* line */ draw((xmin, -0.20017694148807477*xmin-9.43332043214197)--(xmax, -0.20017694148807477*xmax-9.43332043214197), linewidth(0.4)); /* line */ draw((xmin, -0.7783864084613183*xmin-1.7142370359195227)--(xmax, -0.7783864084613183*xmax-1.7142370359195227), linewidth(0.4)); /* line */ draw(circle((1.5095621309330547,-14.085041463184131), 12.004720584501898), linewidth(0.8) + red); pair hyperbolaLeft1 (real t) {return (1.0923863821057558*(1+t^2)/(1-t^2),1.0923863821057558*2*t/(1-t^2));} pair hyperbolaRight1 (real t) {return (1.0923863821057558*(-1-t^2)/(1-t^2),1.0923863821057558*(-2)*t/(1-t^2));} draw(shift((-4.13148719458663,-1.2365818293833253))*rotate(-147.28342646627334)*graph(hyperbolaLeft1,-0.99,0.99), linewidth(0.4) + qqttzz); draw(shift((-4.13148719458663,-1.2365818293833253))*rotate(-147.28342646627334)*graph(hyperbolaRight1,-0.99,0.99), linewidth(0.4) + qqttzz); /* hyperbola construction */ draw((xmin, -0.6000748066769415*xmin + 0.3198024459780313)--(xmax, -0.6000748066769415*xmax + 0.3198024459780313), linewidth(0.4)); /* line */ draw((xmin, -0.19002617507558908*xmin-2.021672538344399)--(xmax, -0.19002617507558908*xmax-2.021672538344399), linewidth(0.4)); /* line */ /* dots and labels */ dot((-2.56,5.58),dotstyle); label("$A$", (-2.438303832368012,5.879582497069844), NE * labelscalefactor); dot((-4.94,-3.96),dotstyle); label("$B$", (-4.800670749613119,-3.6328816230370813), NE * labelscalefactor); dot((7.6,-3.74),dotstyle); label("$C$", (7.735623024567583,-3.4123940440942055), NE * labelscalefactor); dot((0.5225293705791998,-2.120966796000219),dotstyle); label("$P_1$", (0.5540275961424571,-1.428005833608324), NE * labelscalefactor); dot((-8.785503759752462,-0.35219686276643136),dotstyle); label("$P_2$", (-8.674952493895095,-0.04208390882453335), NE * labelscalefactor); dot((-1.6841270382794864,1.330404652893004),linewidth(4pt) + dotstyle); label("$Q_1$", (-1.5563535165965052,1.5958238204654007), NE * labelscalefactor); dot((-11.407218944492547,7.164987148815941),linewidth(4pt) + dotstyle); label("$Q_2$", (-11.289305215646348,7.422995549669975), NE * labelscalefactor); dot((13.349977537776379,-12.105678104588545),linewidth(4pt) + dotstyle); label("$X$", (13.468300077082377,-11.85391849505002), NE * labelscalefactor); dot((-3.391695916011611,0.925812966737675),linewidth(4pt) + dotstyle); label("$V$", (-3.2572576970129825,1.1863468881429173), NE * labelscalefactor); dot((5.710237283754192,-3.106767088150229),linewidth(4pt) + dotstyle); label("$U$", (5.845729490771498,-2.845425983955382), NE * labelscalefactor); dot((1.2702179534540083,-0.44242334687846196),linewidth(4pt) + dotstyle); label("$O$", (1.4044796863506959,-0.19957503664087317), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy]

We state Two Lemmas.

$\textbf{LEMMA 1:-}$ (Well known) Consider two arbitrary points $P,Q$ in the plain of $\triangle ABC$. Let $P',Q'$ be the isogonal conjugate of $P,Q$ WRT $\triangle ABC$ respectively. Let $PQ\cap P'Q'=S, P'Q\cap PQ'=T$ prove that $T$ is isogonal conjugate of $S$ WRT $\triangle ABC$.

Let $T'$ denote the isogonal conjugate of $T$ WRT $\triangle ABC.$ Then $A(T,P,Q,Q')=A(T',P',Q',Q)$ $\Longrightarrow$ $AT'$ is the image of $AT$ under the involution defined by $\{AP,AP'\},\{AQ,AQ'\}.$ But by dual of Desargues involution theorem for the complete quadrilateral $PQP'Q',$ it follows that $AS$ is the image of $AT$ under the involution defined by $\{AP,AP'\},\{AQ,AQ\}$ $\Longrightarrow$ $AT' \equiv AS$ and cyclically we get $BT' \equiv BS$ $\Longrightarrow$ $T' \equiv S.$ (Proof by @Luis Gonzalez)

$\textbf{LEMMA 2:-}$ (Well known) $ABC$ be a triangle then $\forall$ points $X$ on the Isogonal Conjugate of the Perpendicular Bisector of $BC$. $\measuredangle XBA=\measuredangle ACX$.

Let $X^*$ be the Isogonal Conjuagte of $X$ WRT $\Delta ABC$. Then $\measuredangle XBA=\measuredangle CBX^*=\measuredangle X^*CB=\measuredangle ACX$.

____________________________________________________________________________________________
Coming back to the Problem. Let $\mathcal H$ bee the Rectangular Hyperbola passing through $\{A,B,C,P_1,P_2\}$. From Lemma 3 we get $\{P_1,P_2\}$ are Antipodal Points WRT the Rectangular Hyperbola $\mathcal H$ passing through $\{A,B,C,P_1,P_2\}$. Let $\overline{Q_1P_2}\cap\overline{Q_2P_1}=V$ and $\overline{Q_1Q_2}\cap\overline{P_1P_2}=U$. Now from $\textbf{LEMMA 1}$ we get that $\{U,V\}$ are Isogonal Conjugates. From Lemma 6 we get that $\mathcal H$ is the perpendicular bisector of $P\widehat P$ WRT $\Delta CP\widehat P$ and from $\textbf{LEMMA 2}$ we get that $\measuredangle CP_1V=\measuredangle VP_2C(\bigstar)$. Let $\tau$ be the reflection of $\overline{Q_1P_2}$ over $\overline{BC}$.Let $P_2^*$ be the reflection of $P_2$ over $\overline{BC}$ and clearly $P_2^*\in\tau$ and $P_2^*\in\odot(BP_1C)$. Now combining with $(\bigstar)$ we get $-\measuredangle CP_1V=\measuredangle CP_2V=-\measuredangle CDX$. So $X\in\odot(P_1CP_2^*)\equiv\odot(BP_1C)$. $\blacksquare$

I hope this will help y'all. If you have any better suggestions and problems write in the Comment box below. Goodbye!
This post has been edited 11 times. Last edited by amar_04, Aug 14, 2020, 7:11 PM
Poncelet point
Antigonal conjugates
rectangular hyperbola
conics
geometry
Lemmas
involution

Comment

15 Comments

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First Comment:Very Nice! :D

Amar~ Thanks!
This post has been edited 1 time. Last edited by amar_04, Aug 13, 2020, 11:53 AM

by Functional_equation, Aug 13, 2020, 11:37 AM

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Hau so pro?? BTW nice blogpost

Amar~ IDK. :P. Thanks! :)
This post has been edited 1 time. Last edited by amar_04, Aug 13, 2020, 11:54 AM

by GeoMetrix, Aug 13, 2020, 11:49 AM

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what? $\qquad$

Amar~ :huuh:
This post has been edited 2 times. Last edited by amar_04, Aug 13, 2020, 1:33 PM

by mueller.25, Aug 13, 2020, 12:46 PM

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Wooow Amar! How did you know that the P127 I sent to the Geozebra marathon belonged to Buratinogigle? :o :thumbup:.

Amar~ Lol I used the Search Function to know the source. :P
This post has been edited 1 time. Last edited by amar_04, Aug 13, 2020, 1:34 PM

by Functional_equation, Aug 13, 2020, 12:54 PM

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Another Proof(Luis Gonzáles's Proof) :)

Amar~ He is a Legend. :omighty:
This post has been edited 1 time. Last edited by amar_04, Aug 14, 2020, 7:20 PM

by Functional_equation, Aug 13, 2020, 3:44 PM

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Lemma 2 = Lemma 7 in Section 2.

Amar~ Oh! Yes. Anyway I am leaving as it is. :P
This post has been edited 1 time. Last edited by amar_04, Aug 14, 2020, 7:18 PM

by Synthetic_Potato, Aug 14, 2020, 8:05 AM

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Over the top for me (I bogged down somewhere in the middle, well, rather at the beggining),. but I appreciate how much effort you put in it!! (impressed emogi is here).

Amar ~ Thanks :P
This post has been edited 2 times. Last edited by amar_04, Aug 14, 2020, 7:19 PM

by zuss77, Aug 14, 2020, 4:54 PM

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Absolute joy to read Amar! :D ;)

Amar ~ I am glad that you enjoyed it! :)
This post has been edited 1 time. Last edited by amar_04, Aug 16, 2020, 9:54 AM

by cmsgr8er, Aug 15, 2020, 9:26 PM

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Cool post. Great work !! :D

Amar ~ Thanks! :)
This post has been edited 1 time. Last edited by amar_04, Aug 16, 2020, 8:08 PM

by lilavati_2005, Aug 16, 2020, 6:38 PM

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Really $\textbf{amazing}$. your posts are getting more interesting every time. :thumbup:

Amar ~ Lol Thanks! :rotfl:
This post has been edited 2 times. Last edited by amar_04, Aug 17, 2020, 11:26 AM

by Gaussian_cyber, Aug 17, 2020, 11:23 AM

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very nice post :)

Amar ~ Thanks!
This post has been edited 1 time. Last edited by amar_04, Sep 9, 2020, 10:07 AM

by Aritra12, Sep 7, 2020, 11:34 AM

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In the proof of lemma 2, Why $E_AK, E_BM, E_CN$ pass through $P$?

Amar~ Maybe you can check alastor's blogpost :P
This post has been edited 1 time. Last edited by amar_04, May 17, 2021, 6:47 AM

by shalomrav, Mar 4, 2021, 10:41 PM

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For Lemma 1 about rectangular hyperbolas, here's something I found.

Isogonally conjugate the hyperbola in $\triangle ABC$, it goes to a line. Let the points at infinity of the hyperbola be $P_{infty 1}$ and $P_{\infty 2}$. Their isogonal conjugates are $P_1$ and $P_2$ on $(ABC)$. Because of the rectangular condition, $P_1,P_2$ are antipodes on $(ABC)$, so the isogonal conjuate of the hyperbola, $\overline{P_1P_2}$, goes through the circumcenter $O$. Un-isogonally-conjugating we get Lemma 1.

Also do you have anything on why isogonally conjugating a line gives a circumconic? I've never seen this properly explained, so I'm not fully sure why it's true.

AMAR~ check PM ;)
This post has been edited 1 time. Last edited by amar_04, May 17, 2021, 6:48 AM

by i3435, Apr 5, 2021, 2:23 PM

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nice blogpost ! I remember how frequently Alastor had been using the hyperbola stuff)

Amar~ Lol yeah
This post has been edited 1 time. Last edited by amar_04, May 17, 2021, 6:46 AM

by Kamran011, Apr 20, 2021, 6:17 AM

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interesting!

by kn07, Sep 9, 2023, 5:36 PM

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