idk:( :( :(

by AlastorMoody, Jul 5, 2020, 8:04 AM

(1) Making Inversion About Hyperbola Easier Using Inversion About Circle
Inversion is Pretty Cool (I'm talking about circle ofc)... how about performing Inversion About Conics? :maybe:

Ofcourse, I haven't discovered this, but I'll share though 'coz it's pretty Interesting!! (Thanks a lot to Vrangr :omighty:

)

Defining Inversion About Hyperbola
We'll First define the Polar of a Conic:
For a conic $\mathcal{C}$ and a point $A$ , The Polar of $A$ WRT $\mathcal{C}$ is the locus of all points such that, $-1=(X,Y;A,B)$ , where $X,Y$ $=$ $\ell_A$ $\cap$ $\mathcal{C}$ (where $\ell_A$ is a line passing through $A$ ).
Lemma: For a conic $\mathcal{C}$ and a point $O$ , Let $\ell$ be the Polar of $O$ WRT $\mathcal{C}$ . Let a line $m$ $\cap$ $\ell$ $=$ $L$ and Let $m$ $\cap$ $\mathcal{C}$ $=$ $A,B$ . Let $AA$ $\cap$ $BB$ $=$ $T$ and Let $OT$ $\cap$ $\ell$ $=$ $M$ . Then, $-1=(A,B;L,M)$ .
Proof: (By Vrangr) By LaHire's Theorem, $OT$ is polar of $L$ . Again By LaHire, $L$ lies on the Polar of $M$ $\implies$ $-1$ $=$ $(A,B;L,M)$ $\qquad \blacksquare$
When we Take $T$ to be the center of $\mathcal{C}$ , we get the following:

Corollary wrote:

Let $AB$ be a chord of Rectangular Hyperbola $\mathcal{H}$ . Let $M$ be midpoint of $AB$ and $L$ $=$ $AA \cap BB$ . If $K$ is center of $\mathcal{H}$ , Then $K,L,M$ collinear.

Surprisingly, the above Corollary is quite useful! Now, to simplify we can Define Inversion About Hyperbola with help of Circle Inversion and the Proof follows from the definition of Polars.

Take a Hyperbola $\mathcal{H}$ with $O$ as the center of $\mathcal{H}$ . Take Any Random Point $P$ in the plane and Suppose $\stackrel{\longrightarrow}{OP}$ $\cap$ $\mathcal{H}$ $=$ $R$ . Let $\omega$ be circle at $O$ with radius $OR$ . Then, the Image of $P$ after the Inversion about $\mathcal{H}$ coincides with the Image of $P$ after the Inversion about $\omega$ .

Some Well-Known Lemmas

Well-Known #1 wrote:

Let $X$ $\in$ $\mathcal{H}$ (Rectangular Hyperbola) with center $O$ . Then, reflection of $X$ over $O$ lies on $\mathcal{H}$

(This is Quite Well-Known and Easy to Prove,). We'll also need some basic lemmas regarding the center of rectangular hyperbola $\mathcal{H}$ . Check This BlogPost for the proofs.

From Now on, Let $\mathcal{H}$ be the Circum-Rectangular Hyperbola WRT $\Delta ABC$

Lemma #2 wrote:

Let $H$ be Orthocenter WRT $\Delta ABC$ . Then, $H$ $\in$ $\mathcal{H}$

Lemma #3 wrote:

Let $G$ be the Fourth-Intersection of $\mathcal{H}$ with $\odot (ABC)$ . Suppose, $K$ be the midpoint of $HG$ . Then, $K$ is the center of $\mathcal{H}$

Corollary#4 wrote:

$K$ lies on the Nine-Point Circle WRT $\Delta ABC$

Application (In our ~~Favourite~~ Comfortable Orthocenter Configuration)
$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(20cm); 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 = -9.872229745398423, xmax = 7.124774747659737, ymin = -5.9755366782192745, ymax = 4.125142221343976; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen ubqqys = rgb(0.29411764705882354,0,0.5098039215686274); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902); draw((-4.68,3.25)--(-6.62,-3.61)--(1.48,-3.59)--cycle, linewidth(0.4) + rvwvcq); /* draw figures */ draw((-4.68,3.25)--(-6.62,-3.61), linewidth(0.4) + rvwvcq); draw((-6.62,-3.61)--(1.48,-3.59), linewidth(0.4) + rvwvcq); draw((1.48,-3.59)--(-4.68,3.25), linewidth(0.4) + rvwvcq); draw(circle((-2.5762981745883096,-1.0492392917345015), 4.786336789010386), linewidth(0.4)); draw((-4.66740365082338,-1.851521416530997)--(-1.7090511261687018,3.6578726056864355), linewidth(0.4)); draw((-4.68,3.25)--(-4.663073659054051,-3.605168083108775), linewidth(0.4) + linetype("4 4") + ubqqys); draw((-6.62,-3.61)--(-2.1375183639556625,0.42685480672998877), linewidth(0.4) + linetype("4 4") + ubqqys); draw((-6.014935226432025,-1.4704410584142662)--(1.48,-3.59), linewidth(0.4) + linetype("4 4") + ubqqys); draw(circle((-3.6218509127058467,-1.4503803541327478), 2.3931683945051927), linewidth(0.4) + linetype("2 2")); pair hyperbolaLeft1 (real t) {return (2.054274217730499*(1+t^2)/(1-t^2),2.054274217730499*2*t/(1-t^2));} pair hyperbolaRight1 (real t) {return (2.054274217730499*(-1-t^2)/(1-t^2),2.054274217730499*(-2)*t/(1-t^2));} draw(shift((-3.1882273884960415,0.90317559457772))*rotate(-86.0207158012572)*graph(hyperbolaLeft1,-0.99,0.99), linewidth(0.4) + dtsfsf); draw(shift((-3.1882273884960415,0.90317559457772))*rotate(-86.0207158012572)*graph(hyperbolaRight1,-0.99,0.99), linewidth(0.4) + dtsfsf); /* hyperbola construction */ draw((-6.014935226432025,-1.4704410584142662)--(-2.1375183639556625,0.42685480672998877), linewidth(0.4)); draw((xmin, -7.284011802732607*xmin-22.319910333022797)--(xmax, -7.284011802732607*xmax-22.319910333022797), linewidth(0.4) + sexdts); /* line */ draw(circle((-3.188227388496041,0.9031755945777192), 2.0635395657672815), linewidth(0.4) + dtsfsf); draw((-4.88236010673622,-0.2750051523686752)--(-1.2392610068520262,0.2251449295569858), linewidth(0.4) + sexdts); draw(circle((-2.87911369424802,-1.348412202711139), 2.2727073909484585), linewidth(0.4) + ubqqys); draw(circle((-2.57,-3.6), 4.050012345660203), linewidth(0.4)); draw((-4.88236010673622,-0.2750051523686752)--(-2.57,-3.6), linewidth(0.4) + sexdts); draw((-2.57,-3.6)--(-1.2392610068520262,0.2251449295569858), linewidth(0.4) + sexdts); /* dots and labels */ dot((-4.68,3.25),dotstyle); label("$A$", (-4.619876717625532,3.386747763858663), NE * labelscalefactor); dot((-6.62,-3.61),dotstyle); label("$B$", (-6.570352643058436,-3.467781916948398), NE * labelscalefactor); dot((1.48,-3.59),dotstyle); label("$C$", (1.5380544183840634,-3.4538499460524488), NE * labelscalefactor); dot((-4.663073659054051,-3.605168083108775),dotstyle); label("$D$", (-4.605944746729583,-3.467781916948398), NE * labelscalefactor); dot((-2.1375183639556625,0.42685480672998877),dotstyle); label("$E$", (-2.0842580145627574,0.5724896428769022), NE * labelscalefactor); dot((-6.014935226432025,-1.4704410584142662),dotstyle); label("$F$", (-5.957345923636666,-1.3361903698681534), NE * labelscalefactor); dot((-4.66740365082338,-1.851521416530997),dotstyle); label("$H$", (-4.605944746729583,-1.712353584058785), NE * labelscalefactor); dot((-1.7090511261687018,3.6578726056864355),dotstyle); label("$G$", (-1.6523669167883288,3.7907749198411933), NE * labelscalefactor); dot((-3.188227388496041,0.9031755945777192),dotstyle); label("$K$", (-3.129155831758956,1.0461766533391788), NE * labelscalefactor); dot((-3.0608105567941233,-0.024930111405843853),dotstyle); label("$L$", (-3.003768093695412,0.1127346033105749), NE * labelscalefactor); dot((-2.57,-3.6),dotstyle); label("$M$", (-2.516149112337186,-3.4538499460524488), NE * labelscalefactor); dot((-2.9075628643510822,-1.1411881119024914),dotstyle); label("$N$", (-2.8505164138399697,-1.00182306836537), NE * labelscalefactor); dot((-4.88236010673622,-0.2750051523686752),dotstyle); label("$P$", (-4.8288562810647715,-0.1380408728165127), NE * labelscalefactor); dot((-1.2392610068520262,0.2251449295569858),dotstyle); label("$Q$", (-1.1786799063260522,0.3635100794376625), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
Keep everything same as we defined earlier. Let $\Delta DEF$ be the orthic triangle of $\Delta ABC$ . Applying Pascal on $ABBHCC$ $\implies$ $BB \cap CC$ lies on $EF$ . Suppose Let $BB \cap CC$ $=$ $L$ . Let $N$ $=$ $KM$ $\cap$ $\mathcal{H}$ . Let $\omega$ be the circle at $K$ with radius $KN$ .

Then According to our Corollary: $K,L,N,M$ is collinear. More Surprisingly, Let $PQ$ be Polar Chord of $M$ WRT $\omega$ . Then, $P,Q$ lie on $\odot (BFEC)$ . (Now this looks something familiar, isn't it?

That's because, when we take $G$ $\longrightarrow$ $A$ $\implies$ $P \longrightarrow F$ and $Q$ $\longrightarrow$ $E$ leading to the "Three Tangents Lemma". More Importantly, $N$ $\longrightarrow$ MidPoint of arc $FHE$ ! So Did we Generalise?

)

Well Anyways, To Prove: $P,Q \in \odot (BFEC)$ , Just Observe: $PL \times LQ=KL \times LM=FL \times LE$ . Hence, $BFPQEC$ is cyclic. $\qquad \blacksquare$

Wait!!!! It's not over yet :diablo:

(2) On some Special Hyperbolas

#1)A Hyperbola having (The Fourth Intersection with circumcircle coinciding with A)
In This BlogPost, we had proved that if $G$ is the Fourth Intersection of $\mathcal{H}$ $\cap$ $\odot (ABC)$ , then the Center of $\mathcal{H}$ is midpoint of $HG$ . But What if $G$ coincides with $A$ . Moreoever, when does that Happen?!! :flex:

Anyway We'll investigate that!

Let $\Delta DEF$ be orthic triangle WRT $\Delta ABC$ . Let $G$ be midpoint of arc $\widehat{FHE}$ . Suppose, $\mathcal{H}$ is the Circum-Rectangular Hyperbola WRT $\Delta ABC$ passing through $G$ . Now, from the configuration of "The Three Tangents Lemma", $MG$ is Perpendicular Bisector of $EF$ . Also, from our (Corollary), $MG$ $\cap$ $\odot (DEF)$ is the center of $\mathcal{H}$ . But simply, $MG$ $\cap$ $\odot (DEF)$ is midpoint of $AH$ $\implies$ The Fourth Intersection of $\mathcal{H}$ with $\odot (ABC)$ coincides with $A$ . $\qquad \blacksquare$

Hence, the Tangent to $\odot (ABC)$ at $A$ and the Tangent to $\mathcal{H}$ at $A$ are the same!! Cool :w00tb:

Let's Lean on to the Olympiad Side!! :wow:

(Problem: India TST Practice #2 2019 P1)

India TST 2019 Practice Test #2 P1 wrote:

Let the points $O$ and $H$ be the circumcenter and orthocenter of an acute angled triangle $ABC.$ Let $D$ be the midpoint of $BC.$ Let $E$ be the point on the angle bisector of $\angle BAC$ such that $AE\perp HE.$ Let $F$ be the point such that $AEHF$ is a rectangle. Prove that $D,E,F$ are collinear.

Solution: Add the Circum-Rectangular Hyperbola of $\Delta ABC$ passing through $E$ . Let $K$ be midpoint of $AH$ $\implies$ $K$ is center of $\mathcal{H}$ . Redefine $F$ as $KD$ $\cap$ $\mathcal{H}$ $=F$ . Hence, $AFHE$ is Parallelogram, but $\angle AEH$ $=$ $90^{\circ}$ $\implies$ $AEHF$ is Rectangle. $\qquad \blacksquare$
$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(16cm); 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 = -9.42510097180991, xmax = 8.97526050527123, ymin = -5.821934424538, ymax = 5.11270661717; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen ubqqys = rgb(0.29411764705882354,0,0.5098039215686274); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); draw((-4.44,3.89)--(-7,-3.89)--(4.36,-4.41)--cycle, linewidth(2) + rvwvcq); /* draw figures */ draw((-4.44,3.89)--(-7,-3.89), linewidth(0.4) + rvwvcq); draw((-7,-3.89)--(4.36,-4.41), linewidth(0.4) + rvwvcq); draw((4.36,-4.41)--(-4.44,3.89), linewidth(0.4) + rvwvcq); draw((-4.44,3.89)--(-1.32,-4.15), linewidth(0.4) + ubqqys); draw(circle((-1.1981438380595681,-1.4879115391474944), 6.279455700744047), linewidth(0.4)); pair hyperbolaLeft1 (real t) {return (2.4306337194779517*(1+t^2)/(1-t^2),2.4306337194779517*2*t/(1-t^2));} pair hyperbolaRight1 (real t) {return (2.4306337194779517*(-1-t^2)/(1-t^2),2.4306337194779517*(-2)*t/(1-t^2));} draw(shift((-4.561856161940435,1.2279115391474906))*rotate(-75.76944635286395)*graph(hyperbolaLeft1,-0.99,0.99), linewidth(0.4) + dtsfsf); draw(shift((-4.561856161940435,1.2279115391474906))*rotate(-75.76944635286395)*graph(hyperbolaRight1,-0.99,0.99), linewidth(0.4) + dtsfsf); /* hyperbola construction */ draw((-4.44,3.89)--(-4.683712323880865,-1.4341769217050127), linewidth(0.4) + ubqqys); draw((-4.561856161940432,1.227911539147494)--(-1.32,-4.15), linewidth(0.4)); draw((-4.44,3.89)--(-3.1860769586427318,-1.0543670331088528), linewidth(0.4)); draw((-5.93763536523813,3.5101901114038356)--(-4.561856161940432,1.227911539147494), linewidth(0.4)); draw((-4.683712323880865,-1.4341769217050127)--(-3.1860769586427318,-1.0543670331088528), linewidth(0.4) + ubqqys); draw((-5.93763536523813,3.5101901114038356)--(-4.44,3.89), linewidth(0.4) + ubqqys); /* dots and labels */ dot((-4.44,3.89),dotstyle); label("$A$", (-4.372542697365499,4.041865908257907), NE * labelscalefactor); dot((-7,-3.89),dotstyle); label("$B$", (-6.936527493352215,-3.740582060737028), NE * labelscalefactor); dot((4.36,-4.41),dotstyle); label("$C$", (4.4204169265183575,-4.253379019934369), NE * labelscalefactor); dot((-4.683712323880865,-1.4341769217050127),dotstyle); label("$H$", (-4.62894117696417,-1.2821731092909536), NE * labelscalefactor); dot((-1.32,-4.15),dotstyle); label("$D$", (-1.2655964151698307,-3.996980540335698), NE * labelscalefactor); dot((-3.1860769586427318,-1.0543670331088528),dotstyle); label("$E$", (-3.120714826383749,-0.9051165216458503), NE * labelscalefactor); dot((-4.561856161940432,1.227911539147494),dotstyle); label("$K$", (-4.508283068917736,1.3723052677305743), NE * labelscalefactor); dot((-5.93763536523813,3.5101901114038356),dotstyle); label("$F$", (-5.88076904794592,3.664809320612803), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
#2) MathPassionForever's Observation From my Misinterpretation of @bove Hyperbola
This Observation is Due to MathPassionForever. Here we deal with the Circum-Rectangular Hyperbola $\mathcal{H}$ WRT $\Delta ABC$ passign through the $A-$ Humpty Point. Let $K$ be the $A-$ Humpty Point.

Lemma #1 wrote:

If $\mathcal{H}$ is the Circum-Rectangular Hyperbola WRT $\Delta ABC$ passing through the $A-$ Humpty Point, say $K$ , Then on $\mathcal{H}$ , $-1=(B,C;A,K)$

Proof: It's well-known that $EF,HK,BC$ concur, say at $G$ ( $\Delta DEF$ is the Orthic Triangle) $-1=(B,C;D,G) \stackrel{H}{=} (B,C;A,K) \qquad \blacksquare$

Lemma #2 wrote:

If $\Delta DEF$ is the Orthic Triangle WRT $\Delta ABC$ and $\mathcal{H}$ is the Circum-Rectangular Hyperbola WRT $\Delta ABC$ passing through the $A-$ Humpty Point, namely $K$ , Then, $\odot (AEF)$ is Tangent to $\mathcal{H}$ at $A$ .

Proof: We know the tangen to $\odot (AEF)$ at $A$ is parallel to $BC$ . From the (Corollary) mentioned in (1) $\implies -1=(A,K;B,C) \stackrel{A}{=} (AA ~ \cap ~ BC, AK ~ \cap ~ BC ~ ; B,C)=(AA ~ \cap ~ BC, M;B,C)$ Where $M$ is the midpoint of $BC$ . Hence, $\odot (AEF)$ is tangent to $\mathcal{H}$ $\qquad \blacksquare$

#3) Darij Grinberg's Medial Hyperbola
This cute name was given By Darij Grinberg during 2004~2005... ok enough history

Suppose, $A-$ antipode in $\odot (ABC)$ $=$ $A'$ . then The Circum-Rectangular Hyperbola passing through $A'$ is the A-Medial Hyperbola. To make everything look better, the $A-$ Medial Hyperbola is the Isogonal Conjugate of the Perpendicular bisector of $\overline{BC}$ . This property appeared recently in enhanced's post: Easy Conic

Anyway,

enhanced wrote:

In a $\Delta ABC$ let $\mathcal{H}$ be the isogonal conjugate of the perpendicular bisector of $BC$ then show that $A-$ symmedian is tangent to $\mathcal{H}$ at $A$

AlastorMoody wrote:

Under Isogonal conjugation on $\Delta ABC$ , the midpoint of $BC$ maps to $A$ . Now the $A-$ median and perpendicular bisector of $BC$ intersect exactly at one point, which is midpoint of $BC$ and since, isogonal conjugation preserves intersection $\implies$ $A-$ symmedian is tangent to $\mathcal{H}$ at $A$

Similarly, for:

WizardMath wrote:

A line passing through none of the vertices of the triangle cuts $BC$ at $K$ , then the isogonal of $AK$ is the tangent at $A$ to the circumconic $\mathcal{H}$ of $\triangle ABC$ that is the isogonal conjugate of the line.

Let the line be $\ell$ , now, $AK \cap \ell=K$ and since, $K \mapsto A$ $\implies$ isogonal of $AK$ is tangent to $\mathcal{H}$ at $A$

The $\mathcal{H}$ mentioned is nothing but the $A-$ Medial Hyperbola. Now, we also know, $A-$ symmedian is tangent to $\mathcal{H}$

So, suppose if $\mathcal{H}$ is the Isogonal Conjugate of Perpendicular Bisector of $\overline{BC}$ . Then, we wanna show $M$ is the center of $\mathcal{H}$ . Let $A'$ be $A-$ Antipode in $\odot (ABC)$ . Since, $OM$ $\cap$ $AH$ $=$ $\infty_{AH}$ $\implies$ $\mathcal{I}(\infty_{BC})$ $=$ $A'$ $\in$ $\mathcal{H}$ $\implies$ $M$ is the center of $\mathcal{H}$ $\qquad \blacksquare$

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(20cm); 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 = -14.05658698813138, xmax = 16.710851987372646, ymin = -12.475466754866938, ymax = 5.808462144510303; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen ubqqys = rgb(0.29411764705882354,0,0.5098039215686274); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); draw((-7.084221324561282,4.063766311689345)--(-8.8,-2.17)--(0.14,-2.21)--cycle, linewidth(2) + rvwvcq); /* draw figures */ draw((-7.084221324561282,4.063766311689345)--(-8.8,-2.17), linewidth(0.4) + rvwvcq); draw((-8.8,-2.17)--(0.14,-2.21), linewidth(0.4) + rvwvcq); draw((0.14,-2.21)--(-7.084221324561282,4.063766311689345), linewidth(0.4) + rvwvcq); draw(circle((-4.320424822444672,-0.04994781638420633), 4.955927262644594), linewidth(0.4)); draw((-7.103371679671937,-0.2163380555422419)--(-1.5566283203280638,-4.163661944457758), linewidth(0.4) + ubqqys); draw((-1.5566283203280638,-4.163661944457758)--(0.14,-2.21), linewidth(0.4)); draw((-1.5566283203280638,-4.163661944457758)--(-8.8,-2.17), linewidth(0.4)); pair hyperbolaLeft1 (real t) {return (3.349203943246869*(1+t^2)/(1-t^2),3.349203943246869*2*t/(1-t^2));} pair hyperbolaRight1 (real t) {return (3.349203943246869*(-1-t^2)/(1-t^2),3.349203943246869*(-2)*t/(1-t^2));} draw(shift((-4.33,-2.19))*rotate(-28.180598996547513)*graph(hyperbolaLeft1,-0.99,0.99), linewidth(0.4) + dtsfsf); draw(shift((-4.33,-2.19))*rotate(-28.180598996547513)*graph(hyperbolaRight1,-0.99,0.99), linewidth(0.4) + dtsfsf); /* hyperbola construction */ draw((-7.084221324561282,4.063766311689345)--(-1.5566283203280638,-4.163661944457758), linewidth(0.4) + ubqqys); draw((-7.084221324561282,4.063766311689345)--(-4.371774682264501,-11.526641486115832), linewidth(0.4) + ubqqys); draw((-4.371774682264501,-11.526641486115832)--(-1.5566283203280638,-4.163661944457758), linewidth(0.4) + ubqqys); draw((-4.33,-2.19)--(-4.371774682264501,-11.526641486115832), linewidth(0.4) + ubqqys); draw((-7.103371679671937,-0.2163380555422419)--(-7.084221324561282,4.063766311689345), linewidth(0.4) + ubqqys); /* dots and labels */ dot((-7.084221324561282,4.063766311689345),dotstyle); label("$A$", (-6.995207551130457,4.3205286202851205), NE * labelscalefactor); dot((-8.8,-2.17),dotstyle); label("$B$", (-8.71011398583068,-1.9086168116406432), NE * labelscalefactor); dot((0.14,-2.21),dotstyle); label("$C$", (0.2427063717954904,-1.9590552361906495), NE * labelscalefactor); dot((-4.33,-2.19),dotstyle); label("$M$", (-4.221094200880094,-1.9338360239156465), NE * labelscalefactor); dot((-7.103371679671937,-0.2163380555422419),dotstyle); label("$H$", (-6.995207551130457,0.033262533534594727), NE * labelscalefactor); dot((-1.5566283203280638,-4.163661944457758),dotstyle); label("$A'$", (-1.3208847892547142,-4.607072525065974), NE * labelscalefactor); dot((-4.371774682264501,-11.526641486115832),dotstyle); label("$K$", (-4.2715326254301,-11.26494456566679), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
The Fact that $A'K$ is tangent to $\mathcal{H}$ can be proved using the Main Powerful Corollary that we proved earlier. To write it out, the tangent at $A'$ should intersect the $A-$ symmedian such that the intersection also lies on $MO$ $\implies$ the intersection clearly if $K$ . $\qquad \blacksquare$

(Just too much in a Single Post I guess...

)

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Nice post!!

by GeoMetrix, Jul 19, 2020, 11:58 AM

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Alastor #proyaargodyaar
by GeoMetrix, Apr 24, 2020, 5:41 AM
@below why do you look so jobless
by AlastorMoody, Apr 16, 2020, 7:45 AM
e
eeeeeeeeeeeeeee
by gatorpokemon, Apr 15, 2020, 12:59 AM
Thanks a lot guyss and @below congrats for INMO merit
by AlastorMoody, Mar 19, 2020, 3:25 PM
Ahh! Now u belong to one of those kinds of ppl congrats!
by RAMUGAUSS, Mar 5, 2020, 7:28 PM
Congo for IMOTC!
by PhysicsMonster_01, Mar 3, 2020, 1:55 PM
Congrats alastor for IMOTC.
by GeoMetrix, Mar 3, 2020, 11:49 AM
im going to say something
by bingo2019, Feb 19, 2020, 1:43 AM
Probably because of JEE @below.
by gamerrk1004, Jan 29, 2020, 3:56 PM
Why are you living Olympiads Now ?
You have 2 more years in your high School ...
by a_simple_guy, Jan 20, 2020, 6:05 AM
contrib!!!
by Kagebaka, Jan 18, 2020, 9:48 PM
Hello
by A-student, Dec 19, 2019, 1:21 PM
hii.............................
by GeoMetrix, Nov 26, 2019, 12:50 PM
First shout man
by GAUSSIANGAUSS, Nov 16, 2019, 3:36 PM

14 shouts

Well, I am Dumb!!!!

idk:( :( :(

by AlastorMoody, Jul 5, 2020, 8:04 AM

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