Artzt Parabola

by MathPassionForever, Sep 27, 2019, 7:19 PM

I am extremely indebted to AlastorMoody, Supercali and math_pi_rate for this one. I admittedly have just shifted an entire thread to my blog post, but this was just too beautiful so........

$[asy] import graph; size(20cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(7); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -9.738376123141856, xmax = 12.707373326322468, ymin = -9.571364861640493, ymax = 6.319493433049379; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wwqqcc = rgb(0.4,0,0.8); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen zzwwqq = rgb(0.6,0.4,0); pen ffqqtt = rgb(1,0,0); pen qqccqq = rgb(0,0.8,0); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen ffqqff = rgb(1,0,1); pen wwccff = rgb(0.4,0.8,1); pen ffwwqq = rgb(1,0.4,0); draw((-0.75982,4.94852)--(-4.4106718182458415,-6.065689301708911)--(8.366593051504776,-5.3507159967228635)--cycle, linewidth(0.1) + rvwvcq); draw((-4.4106718182458415,-6.065689301708911)--(-2.585245909122921,-0.5585846508544554)--(0.6090703083147335,-0.3798413246079435)--cycle, linewidth(0.5) + zzwwqq); draw((8.366593051504776,-5.3507159967228635)--(3.8033865257523876,-0.2010979983614316)--(0.6090703083147335,-0.3798413246079435)--cycle, linewidth(0.5) + zzwwqq); /* draw figures */ draw((-0.75982,4.94852)--(-4.4106718182458415,-6.065689301708911), linewidth(0.8) + rvwvcq); draw((-4.4106718182458415,-6.065689301708911)--(8.366593051504776,-5.3507159967228635), linewidth(0.8) + rvwvcq); draw((8.366593051504776,-5.3507159967228635)--(-0.75982,4.94852), linewidth(0.8) + rvwvcq); draw((xmin, -6.460935923664841*xmin + 0.03937166648098678)--(xmax, -6.460935923664841*xmax + 0.03937166648098678), linewidth(0.5) + wwqqcc); /* line */ draw(circle((1.7705964655547837,-2.0024045945264013), 7.397185965483279), linewidth(0.5) + dtsfsf); draw((-4.4106718182458415,-6.065689301708911)--(-2.585245909122921,-0.5585846508544554), linewidth(0.5) + zzwwqq); draw((-2.585245909122921,-0.5585846508544554)--(0.6090703083147335,-0.3798413246079435), linewidth(0.5) + zzwwqq); draw((0.6090703083147335,-0.3798413246079435)--(-4.4106718182458415,-6.065689301708911), linewidth(0.5) + zzwwqq); draw((8.366593051504776,-5.3507159967228635)--(3.8033865257523876,-0.2010979983614316), linewidth(0.5) + zzwwqq); draw((3.8033865257523876,-0.2010979983614316)--(0.6090703083147335,-0.3798413246079435), linewidth(0.5) + zzwwqq); draw((0.6090703083147335,-0.3798413246079435)--(8.366593051504776,-5.3507159967228635), linewidth(0.5) + zzwwqq); draw((xmin, 0.2569064342526462*xmin + 1.384651407539936)--(xmax, 0.2569064342526462*xmax + 1.384651407539936), linewidth(0.5)); /* line */ real parabola1 (real x) {return x^2/2/3.5805800598820867;} draw(shift((-0.09542226795749895,-0.4882896694895455))*rotate(-165.59193348613547)*graph(parabola1,-21.48348035929252,21.48348035929252), linewidth(0.5) + ffqqtt); /* parabola construction */ draw(circle((1.5587528028867694,-0.3267002463918187), 2.2481451186557893), linewidth(0.4) + qqccqq); draw((-0.75982,4.94852)--(1.977960616629467,-5.708202649215887), linewidth(0.8) + dotted + wrwrwr); draw((xmin, -0.3902484892484888*xmin-2.0856656982162205)--(xmax, -0.3902484892484888*xmax-2.0856656982162205), linewidth(0.8) + dotted + wrwrwr); /* line */ draw((xmin, 0.8073184282603932*xmin-2.504872661830268)--(xmax, 0.8073184282603932*xmax-2.504872661830268), linewidth(0.8) + dotted + wrwrwr); /* line */ draw(circle((-0.5826055744380011,-0.44652354909410774), 2.0057731677920785), linewidth(0.4) + qqccqq); draw((-4.4106718182458415,-6.065689301708911)--(3.8033865257523876,-0.2010979983614316), linewidth(0.4) + linetype("4 4") + wrwrwr); draw((8.366593051504776,-5.3507159967228635)--(-2.585245909122921,-0.5585846508544554), linewidth(0.4) + linetype("4 4") + wrwrwr); draw((xmin, 1.6351611194359528*xmin + 1.1464697660785685)--(xmax, 1.6351611194359528*xmax + 1.1464697660785685), linewidth(0.4) + ffqqff); /* line */ draw((xmin, -0.8214199378593325*xmin + 1.5217703477385132)--(xmax, -0.8214199378593325*xmax + 1.5217703477385132), linewidth(0.4) + ffqqff); /* line */ fill(shift((-0.34509169785063354,-2.4630761093789717)) * scale(0.10583333333333333) * ((1,0)--(0,1)--(-1,0)--(0,-1)--cycle)); /* special point */ draw((xmin, 0.05595667870036119*xmin-0.41392287615624096)--(xmax, 0.05595667870036119*xmax-0.41392287615624096), linewidth(0.4) + wwccff); /* line *//* special point */ draw(circle((-4.066966825028381,-2.5001975640019505), 5.134319220144066), linewidth(0.6) + linetype("4 4") + ffwwqq); draw(shift((-0.15682087388725796,-5.82765793117621)) * scale(0.17638888888888887) * ((0,1)--(0,-1)^^(1,0)--(-1,0))); /* special point */ /* dots and labels */ dot((-0.75982,4.94852),linewidth(3pt) + dotstyle); label("$A$", (-0.6743597083187647,5.143247028072334), NE * labelscalefactor); dot((-4.4106718182458415,-6.065689301708911),linewidth(3pt) + dotstyle); label("$B$", (-4.318417198247642,-5.881180061712511), NE * labelscalefactor); dot((8.366593051504776,-5.3507159967228635),linewidth(3pt) + dotstyle); label("$C$", (8.458847671502976,-5.1662067567264645), NE * labelscalefactor); dot((1.977960616629467,-5.708202649215887),linewidth(3pt) + dotstyle); label("$M$", (2.070215236627668,-5.581352546718363), NE * labelscalefactor); dot((3.8033865257523876,-0.2010979983614316),linewidth(2pt) + dotstyle); label("$B'$", (3.892243981592106,-0.06913900182593982), SW * labelscalefactor); dot((-2.585245909122921,-0.5585846508544554),linewidth(2pt) + dotstyle); label("$C'$", (-2.496388453283203,-0.4150938268191881), NE * labelscalefactor); dot((0.6090703083147335,-0.3798413246079435),linewidth(2pt) + dotstyle); label("$Q$", (0.7094595916542265,-0.23058458682278904), NE * labelscalefactor); dot((0.35004888452167676,-2.2222717465639237),linewidth(2pt) + dotstyle); label("$D$", (0.43269573165962827,-2.07567698678678), N * labelscalefactor); dot((0.1461169215727915,1.4221897848451754),linewidth(2pt) + dotstyle); label("$M'$", (0.24818649166322942,1.568380503142102), NE * labelscalefactor); dot((-5.933120871784682,-0.13960551962021883),linewidth(2pt) + dotstyle); label("$D'_{1}$", (-5.840618428217931,0.000051963172709838995), NE * labelscalefactor); dot((6.225354714930267,2.983985089310572),linewidth(2pt) + dotstyle); label("$D'_{2}$", (6.313927756544841,3.1136453881119444), NE * labelscalefactor); dot((1.1640079116974356,-2.539918027229433),linewidth(2pt) + dotstyle); label("$R$", (1.262987311643423,-2.3985681567804784), SE * labelscalefactor); dot((0.9566005160561046,-1.7325914367347726),linewidth(2pt) + dotstyle); label("$P$", (1.0554144166474742,-1.5913402317962324), NE * labelscalefactor); dot((-1.3683749134976222,-1.0910436892842783),linewidth(2pt) + dotstyle); label("$S$", (-1.2740147383070608,-0.9455578918088355), NE * labelscalefactor); dot((2.8907133764192507,-0.8527192542889295),linewidth(2pt) + dotstyle); label("$T$", (2.992761436609662,-0.7149213418133367), SE * labelscalefactor); dot((1.953497694076105,1.886517534445795),linewidth(2pt) + dotstyle); label("$E$", (2.047151581628118,2.0296536031331), NE * labelscalefactor); dot((-2.121811664932106,0.839544338546558),linewidth(2pt) + dotstyle); label("$F$", (-2.035115353292206,0.9687254731538051), NE * labelscalefactor); dot((1.052053843145583,-2.1041404303096494),linewidth(2pt) + dotstyle); label("$X_{a}$", (1.1476690366456737,-1.9603587117890304), E * labelscalefactor); label("$H$", (-0.2592139183268673,-2.3293771917818287), SW*2); dot((-8.950368109446176,-0.9147557487064798),linewidth(2pt) + dotstyle); label("$K$", (-8.861957233158963,-0.7841123068119863), NE * labelscalefactor); dot((-0.5524558489253169,1.2427219453105143),linewidth(1pt) + dotstyle + invisible,UnFill(0)); label("$H_{A}$", (-0.46678681332281596,1.4299985731448028), N * labelscalefactor); label("$H_{a}$", (-0.0747046783304685,-5.604416201717912), N * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$
So here is the description of some of the objects:

$\Delta ABC$ is the reference triangle
$B',C'$ are the midpoints of sides $AB,AC$
$Q$ is the midpoint of $B'C'$ and $M$ is the midpoint of $BC$
$X_a$ is the $A$ -Humpty point
$H$ is the orthocenter of $\Delta ABC$ , $H_a$ is the foot of $A$ -altitude, $H_A$ is the orthocenter of $\Delta AB'C'$ and hence the
midpoint of $AH$ .
$D$ is the $A$ -Dumpty point
$M'$ is the reflection of $D$ about $B'C'$ and hence the midpoint of $AX_a$
$l$ is the line through $M'$ perpendicular to $AM$
$E,F$ are the intersections of $l$ with $AC,AB$ respectively
$D'_1,D'_2$ are the feet of perpendiculars from $B,C$ on $l$
$K$ is the intersection of $B'C'$ and $l$
$S$ and $T$ will be defined later.
$P,Q$ are $BD \cap AM$ and $CD \cap AM$ respectively.

Here's the first part:
EXISTENCE OF ARTZT PARABOLA: Prove that there exists a parabola tangent to $AB, AC, B'C'$ at $B,C,Q$ respectively. Prove that its focus is $D$ and its directrix is $l$ . It will be denoted by $\mathcal{P}$ from now on.

Supercali's pure angle chase proof:
(i) $B,R,D,M',E$ are cyclic and so are $C,P,D,M',F$
Proof: $\measuredangle M'EB = \measuredangle M'EA = \dfrac{\pi}{2} - \measuredangle EAM' = \dfrac{\pi}{2} - \measuredangle DBQ = \measuredangle M'DB$ and we have $B,D,M',E$ cyclic.
$\measuredangle M'RD = \measuredangle DAC + \measuredangle ACD = \measuredangle M'BQ + \measuredangle QBD = \measuredangle M'BD$ and we have $B,R,D,M'$ cyclic. Similarly for the other one.
(ii) $\measuredangle CDE = \measuredangle RDE = \measuredangle RM'E = \dfrac{\pi}{2}$ so it suffices to show $B,C$ lie on the parabola.
(iii) $\measuredangle ECD = \measuredangle EAM' = \measuredangle ECD'_2$ which makes $DED'_2C$ a kite $\implies ED=ED'_2$ and we're done!

Corollary 1: $\Delta QB'C, \Delta QC'B, \Delta ABC$ share an Artzt Parabola and hence a Dumpty point!
Proof 1: By degrees of freedom, a unique parabola is tangent to $AB,AC$ at $B,C$ . But since the parabola is tangent to $B'Q,B'C$ at $Q,C$ and $C'Q,C'B$ at $Q,B$ we're done!

And then math_pi_rate reminds us that Pascal might be a little more than a unit of pressure at times:

Supercali wrote:

Let $BB'$ meet the parabola at $T$ Then $CT$ passes through the centroid of $\triangle AB'C'$ .

Let $AQ \cap \mathcal{P}=Z$ . It is clear that $Z$ is the point at infinity on $AM$ since the two foci of any circumconic are isogonal.
Now, Let $B \infty_{AM} \cap MC'=L$ . It's easy to show that $AMBL,ACML$ and $AB'C'L$ are all parallelograms. This also implies that $L \in CQ$ and that $B'L$ bisects $AC'$ (i.e. $G' \in B'L$ ). Then, Pascal on $QZCCTQ$ and $QTBZCC$ gives that $CT \cap AM,B',L$ are collinear. As $B'L \cap AM=G'$ , so we get $G' \in CT$ , as desired.

Supercali wrote:

Let $X_a$ be the $A$ -Humpty point and let $H$ be the orthocentre. Then $Q,X_a$ , foot from $A$ onto $BC$ (say $A_H$ ), $K$ defined above and midpoint of $AH$ (say $H_A$ ) are concyclic.

Note that $A_H,H,M,X_a$ are concyclic. Then POP, and dividing both sides by half, easily gives that $A_H,X_a,Q,H_A$ are concyclic. Now, $\angle KQX_a=90^{\circ}+\angle H_AAX_a=90^{\circ}+\angle H_AX_aA=\angle KH_AX_a \Rightarrow K \in \odot (QH_AX_a)$

Supercali wrote:

Let $AT$ meet the parabola again at $T'$ . Then $BT'$ is parallel to $AC$ .

By Pascal on $BT'TCCB$ , it suffices to show that the line joining $AT \cap BC$ and $CT \cap AB$ is parallel to $AC$ . This is easy by some cross ratio chasing (to be added later).

Supercali wrote:

Let $L$ be any point on $BC$ and let $M,N$ be points on $AC,AB$ respectively such that $AMLN$ is a parallelogram. Then $MN$ is tangent to the parabola at a point $S$ .

Animate $S$ on the parabola $\mathcal{P}$ . One can easily show that $S \mapsto M,N$ are projective maps (Try to use the fact that focus of the the parabola lies on the circle passing through the point of intersection of the tangents at any two points, and the point where these tangents meet the tangent at the vertex). And, $M \mapsto M \infty_{AB} \cap BC$ and $N \mapsto N \infty_{AC} \cap BC$ are also projective. Thus it suffices to prove the problem for three positions of $S$ . Take $S=Q,B,C$ for this.

Supercali wrote:

$B'$ (midpoint of $AC$ ), $Q$ (midpoint of $B'C'$ ), $S$ , $B$ , $C$ , $N$ , $Q_B$ (intersection of $LN$ and $AQ$ ) lie on a conic.

Apply Pascal on $SQQCBB$ and $SSQCCB$ to get that $MC',SQ,BC$ are concurrent. Then, the converse of Pascal's Theorem on $B'QQ_BNBC$ and $NSQB'CB$ gives the desired conclusion.

This post has been edited 20 times. Last edited by MathPassionForever, Apr 1, 2020, 6:46 PM

geometry

Inparabola

Triangle Conics

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1 Comment

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Nice parabola,sadly no more blgpsts

by HoRI_DA_GRe8, Nov 8, 2021, 7:52 AM

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Cum'on post smth man
by Commander_Anta78, Dec 12, 2021, 8:55 AM
:spam: :spam: :spam: :spam: :spam: :spam: :spam: :spam::spam: :spam: :spam: :spam: :spam: :spam::spam: :spam: :spam: :spam:
by Project_Donkey_into_M4, Nov 8, 2021, 7:58 AM
Owners offline blogs dead,time for :spam: :spam: :spam: :spam: :spam: :spam: :spam: :spam: :spam: :spam: :spam: :spam: :spam::spam: :spam: :spam: :spam: :spam: :spam::spam: :spam: :spam: :spam: :spam: :spam:
by HoRI_DA_GRe8, Nov 8, 2021, 7:54 AM
？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？、？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？
by whatagreatday7, Dec 11, 2020, 2:22 PM
@3 below, that's an april fools prank ig?
by Synthetic_Potato, Apr 20, 2020, 9:03 AM
Kya yaar kuch bhi yaar
by MathPassionForever, Apr 1, 2020, 6:50 PM
Nice blog yaar geo pro
by Wizard_32, Apr 1, 2020, 6:02 PM
Lol okay, guess I will restart soon.
by MathPassionForever, Apr 1, 2020, 12:51 PM
No celebratory posts?
by Hexagrammum16, Mar 4, 2020, 6:02 AM
Nice blog, especially the Artzt Parabola post!
by PhysicsMonster_01, Jan 5, 2020, 1:09 PM
Uhh, looks like I'm a bit too stuck b/w JEE+Schoolwork+Oly prep. Perhaps I'll blog a bit after Mains. Sorry for no posts for now. And The algebraic to geometric ineq, oh man that'll take a LOT of time.
by MathPassionForever, Dec 4, 2019, 8:07 PM
New Post ??
by gamerrk1004, Nov 24, 2019, 6:43 AM
When are you releasing the article on Algebraic to Geometric inequalities?
by Math-wiz, Nov 5, 2019, 8:43 AM
HIGH LAVIL ... Not for me
by gamerrk1004, Oct 15, 2019, 12:40 PM
I found the properties actually!
I'd like to see some more geo posts
by Physicsknight, Oct 9, 2019, 7:57 PM
Chill, couldn't find that huge a compilation of properties anywhere
by MathPassionForever, Sep 28, 2019, 6:09 PM
Yayyy!!! Geo Posts!!!
I earlier thought that we made a groundbreaking discovery of the Dumpty Parabola, but bery sad to know that it was already known, namely Artzt Parabola
by AlastorMoody, Sep 27, 2019, 7:52 PM
Shocked!!!!
by Math-wiz, Sep 10, 2019, 6:18 AM
HURRAAYYYY!! #MPF Hai Lavil!! Next Post geo?!!
by AlastorMoody, Sep 9, 2019, 6:03 PM
yeeee this is nicer
by Hexagrammum16, Sep 5, 2019, 8:30 AM
Nice blog!
by Mathotsav, Sep 4, 2019, 10:08 AM
Okay, maybe after yet another inequality post.
by MathPassionForever, Sep 4, 2019, 8:03 AM
Get back to geo please
by Hexagrammum16, Sep 4, 2019, 6:36 AM
Shout!!!
Expecting some very interesting stuff here
by Naruto.D.Luffy, Sep 3, 2019, 6:25 PM
$\frac{1}{\cos{C}}$ blog..
by Mr.Chagol, Aug 28, 2019, 9:44 AM
Let me try this shout thing too.(Nothing much to say as of now,..well, I hope to see some interesting stuff here).
by Mathotsav, Aug 28, 2019, 7:15 AM
With that kind of support, you can expect nonzero blog posts soon

Also, those who know what I shall be posting, don't spoil for the others
by MathPassionForever, Aug 23, 2019, 5:07 PM
Even before you get to write I'm shouting
by Hexagrammum16, Aug 23, 2019, 4:47 PM
No posts yet I'll give a shout
by Pluto1708, Aug 23, 2019, 3:37 PM

29 shouts

What's in the name?

Artzt Parabola

by MathPassionForever, Sep 27, 2019, 7:19 PM

1 Comment