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...

)

What does "Minute" exactly mean \\\

by AlastorMoody, Apr 28, 2020, 7:52 PM

I came across this piece again i guess after a long period of time, this time the chills/vibes i got from it lasted for the whole day!! It felt so refreshed

It's chopin's minute waltz, but heyyy is about 2 minutes long lol! well minute signifies "small" lol (which i think most of us will not think of )

Master Blaster Jupiter

by AlastorMoody, Apr 26, 2020, 9:28 PM

So our friend, AoPS User Jupiter_Is_Big is a jupiter planet resident and has evoked a hope of development in interplanetary-travel, survival in ungodly places, time travel ,etc. among the smole brainy humans on earth. However

he not ready to share the *secret* behind those scientific marvels!!

Such a poor mean ~~kiddo~~ jumbo jupiter resident.

So i have undertaken the job of stalking jupiter_is_big in order to collect all the ~~useless/trash~~ hints he leaves on his posts, etc. that would give us a hint or idea for the development/improvements in scientific technology. I can sell these ideas to some big space ~~factory~~ institution for LOT of money (jupiter, if you are reading this, pls PM me the technology and i guarantee you $\left( e^{i\pi} +1 \right) \%$ of total amount of whatever money i earn in the auction

Don't worry, we'll somehow figure out how to convert it to your jupiter currency). So let's make a list of things that are beyond earth's tech capabilities and are proven to be possible by jupiter_is_big.

Some history: Jupiter_is_big made his first appearance on aops around april-may of 2019. Apparently, jupiter_is_big doesn't know about the existence of other heavenly/huge bodies or maybe he's teasing us earthlings by showing off how big his planet is :alien:

(we'll take a revenge, after we gather all their tech).

We know that it takes him about 3 earth hours to travel from Jupiter to Earth (so we should give up hopes of #AnywhereDoor as of now). Well that's pretty impressive. I am sure people there dont look like humans so we can also deduce he knows howtta change his appearance. He claims to land here on Earth with his spaceship which is apparently invisible to humans (it seems). Well he has a typical Indian accent, idk how he does that

and is exceptionally brilliant at studies (i wanna know how jupiterian parents compare with the asian parents, i know asian parents don't stand a chance lmao).

Well, now lets investigate some political features of jupiter. I found another alien, who brought Jupiterian Song to earth. It looks like some national anthem. Well so what are the odds of the existence of nations on jupiter? Or is the planet united? So in that case this becomes a Planetary Anthem. Well i need to confirm whether there exists nations on Jupiter from him sneakily. I'll continue further investigations, so stay tuned

omg, one of the best renditions of chopin's heroic polonaise

by AlastorMoody, Apr 26, 2020, 9:20 PM

So, i am gonna just put this video up 'coz i know people are lazy bums to search up. Anyway, let's begin commenting

i am sorry if i missed any other gr8 interpretations of this piece better than Evgeny Kissin did it. So, like there are out many people who play it flawlessly. But, the first few seconds in this interpretation are pUrE MegIC!! The dynamics is the beginning are literally played how it should be, eg, dynamics, articulations ,etc. It's so smooth and the tempo he maintained makes it really enjoyable!! This piece is really hard and technically demanding and idk how you keep up with that dynamics and flawless, emotional playing, so gr8 job!!

Another one which i usually listened to was horowitz'

considering that he was an oldie man

its really amazing how he kept up with the playing

This post has been edited 3 times. Last edited by AlastorMoody, Apr 26, 2020, 9:27 PM

Describe your feeling after reading this

by AlastorMoody, Apr 21, 2020, 9:45 AM

Here are some random lines

If one gazed up at the crystal blue sky and saw a shining rainbow, she would have been the mirror that cast it. Such was the girl Julliene.
She and the princess, peas in a pod, two beautiful twins. A melody of a brilliant and joyful time.

Today, again we head to the ball, our dresses a matching set. The sound of graceful footsteps rings in the night.

Symmetry...
Our laughter is unchanged from that of our childhood, our hands are, forever held.
I wish these days would continue forever. I pray, into those blue eyes.

Though we are twins. My sister is the princess; I am just a girl. The sense of inferiority I feel.

Even so, my sister always takes me by the hand and to the ball.

Symmetry...
On the stage stood two rose fragments, so alike one would confuse them.
They shone like glowing angels. A dazzling white.

Even for the lowly me, a wonderful prince with a white horse came riding to me.

Symmetry...
The melody we craft together beneath the moon colors the white rose blue. A tear of melancholy surrounds me, as I gaze up at this special night.

Symmetry...
Though I go to wed by his side, everyday I am thinking of you.
My dearest sister Lucienne. Let the both of us find happiness.

On each their own paths, the roses bloom magnificent. (The two bloom magnificent)

lmao these are the lyrics of song :rotfl:

Seems like a fairy tale scenario but really, the emotions invoked by the song + meaningful lyrics is something that I cannot really describe

Brooklyn Tech Spam lmao

by AlastorMoody, Apr 10, 2020, 5:50 AM

So, I was listening to Performances of Chopin's pieces yesterday and I somehow reached this (Piano Concerto #1)

WTHHHHHHH!!!! OMGGG

Read the comment section lmao :rotfl:

Apparently, it looks like 6000 kids from some "brooklyn tech" were told to watch this 44 minute video as a "virtual field trip" (wtf lmao :rotfl:

) by their principal. Kids were spamming the whole commment section lolol. I am damm lucky to having witness this event lol yesterday

Notice wrote:

Dear Brooklyn Tech Students and Parents:

In the interest of creating a remote learning culture that educates the whole child, I will be providing lessons for all Tech students this Thursday and Friday, April 9th and 10th. I am excited to share some enrichment activities with our students in the form of two "virtual field trips."

Teachers will not post lessons, take class attendance, or require assignments on either day since I am providing students with their lessons. Additionally, no teacher office hours will be available during these two days.

Thursday's Field Trip & Reflection:

All Tech students will watch two classical music concerts and write a short reflection piece.

Performance #1: Listen to this concert featuring Olga Scheps playing Chopin's "Piano Concerto Number 1 in E Minor".

Performance # 2: Listen to this concert featuring the extremely gifted Alice Sara Ott performing Beethoven's "Piano Concerto Number 3".

After you view both you are to write a reflection (~300-400 word paragraph) addressing the following questions:

What resonated with you the most from the performances?
Compare and contrast the two performances and the featured artists.

Post your attendance and submit your reflection for Thursday's attendance using this link.

Friday's Field trip and Reflection:

On Friday, all Tech students will take virtual tours of two museums: the National Museum of American History and the Salvador Dali Museum, culminating with a short written reflection.

Museum # 1: The virtual tour through the incredible Salvador Dali Museum is found here.

Museum # 2: The virtual tour of the National Museum of American History is found here.

After you have visited both museums write a reflection based on these questions:

What did you enjoy most about each museum?
If you could visit (or revisit) one of these museums in person, which one would it be and why?

Post your attendance and submit your reflection for Friday's field trip using this link.

Be sure you have submitted your attendance and reflections by 9:30 am on Monday, April 13th.

Finally, you have probably heard that the Regents examinations in June have been cancelled. All students enrolled in Regents courses will obtain credit towards graduation by passing the course. A Regents examination does not need to be taken in the future for a course enrolled in this year. Furthermore, if you have previously passed a Regents course but not the examination you will not have to take the examination for graduation.

Be well,

David Newman
Principal

This post has been edited 2 times. Last edited by AlastorMoody, Apr 16, 2020, 1:26 PM

Piano is one of the best inventions!

by AlastorMoody, Apr 9, 2020, 9:11 AM

So, recently I have got a lot of useless time (thnx corona) with me and suddenly my interest started diverting again towards piano. I didnt listen to those beautiful piano pieces for, i think, almost 2 years.

Today, I finally went and listened to this (Schumann's Traumerei). OMG!! Gooseebumpsssss... This piece was kinda special for me when I started learning it long back (The reason for speciality is ... a secret XD). This piece kinda reminds me of those joyful and stress-free days during the childhood and also makes me really feel happy + gives me hope to do atleast something in my life. (d0nt judge me, I felt like writing abt this (>_<) )

btw- Holy crap! Horowitz rocksss!!! See this (Liszt's Hungarian Rhapsody). I am damm sure that many people might recall this piece from Tom & Jerry. Well, the most special thing about this interpretation is that horowitz's interpretation is really full of emotions. The tempo, modulation, swings, melody ,etc are awesome, infact way different than any other i think out there. Although he messed up the Friska part which I think is the best part of the piece, but nvm.

This post has been edited 5 times. Last edited by AlastorMoody, Apr 9, 2020, 9:57 AM

#Make(AOT)OP2TheNationalAnthem !!!

by AlastorMoody, Apr 8, 2020, 8:24 AM

Well, I wouldn't say how good Attack on Titan is, because almost everyone knows this. But lemme point out one thing

See this

THIS IS HOW A FREAKKIIIINNNNG NATIONAL ANTHEM should be!!! Such a majestic feel and every line of the song brings out the Patriotism and the enthusiasm to give endless sacrifices!!! Damm, it. This is how a National Anthem should be!!! Anyway, the anime is also good, so those who haven't watched it are missing an important part of their Life

This post has been edited 3 times. Last edited by AlastorMoody, Apr 8, 2020, 8:27 AM

Well, It's Not The End!!!!

by AlastorMoody, Apr 8, 2020, 8:23 AM

I guess, I should have written this long before (like longgg before), but THIS IS NOT THE END!!

I thought I wouldn't qualify INMO, but did, so yay! So the last post was not the final good bye! Hence, Bring on SPAMMINGGGG

This is THE END

by AlastorMoody, Jan 18, 2020, 5:57 AM

I guess, This is the end of my MO career. I think I'll stop this here. I enjoyed a lot, solving problems, trashtalking and sh*t-posting on forums for these 2 years. I'll really miss AoPS, which is a really wonderful place where I spent my whole 2 years. I think people need to move ahead and I guess I'll do the same

Thank you everyone, I had a really nice time. See you later, if we ever meet. Good Bye

Submit

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!!!!