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

Mesmerising Isogonals

by MathPassionForever, Sep 27, 2019, 2:59 PM

Finally what you expect from me: a GEOMETRY blog post!!
So I was having a conversation with Naruto.D.Luffy and we came up with this as a result of a problem's generalisation.
$[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 = -21.70393503651915, xmax = 16.96733784512575, ymin = -19.527249887479634, ymax = 8.639647115834748; /* image dimensions */ pen qqffff = rgb(0,1,1); pen ffxfqq = rgb(1,0.4980392156862745,0); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen ffqqff = rgb(1,0,1); draw((-8.204062972546984,5.370109264277287)--(-12.676912150369088,-11.22009495891742)--(11.029188492088062,-10.732147775882282)--cycle, linewidth(1.2) + qqffff); /* draw figures */ draw((-8.204062972546984,5.370109264277287)--(-12.676912150369088,-11.22009495891742), linewidth(1.2) + qqffff); draw((-12.676912150369088,-11.22009495891742)--(11.029188492088062,-10.732147775882282), linewidth(1.2) + qqffff); draw((11.029188492088062,-10.732147775882282)--(-8.204062972546984,5.370109264277287), linewidth(1.2) + qqffff); draw((xmin, -7.573770491803278*xmin-56.76558079009495)--(xmax, -7.573770491803278*xmax-56.76558079009495), linewidth(0.8)); /* line */ draw((xmin, -3.1740105245712344*xmin-20.669672954832006)--(xmax, -3.1740105245712344*xmax-20.669672954832006), linewidth(0.8) + green); /* line */ draw((xmin, -1.9247973507800558*xmin-10.421049410913895)--(xmax, -1.9247973507800558*xmax-10.421049410913895), linewidth(0.8)); /* line */ draw((xmin, -48.58333333333346*xmin-393.21061681863137)--(xmax, -48.58333333333346*xmax-393.21061681863137), linewidth(0.8) + ffxfqq); /* line */ draw((xmin, 0.13203463203463203*xmin-9.546303527808082)--(xmax, 0.13203463203463203*xmax-9.546303527808082), linewidth(0.8) + dotted + wrwrwr); /* line */ draw((xmin, 0.5195352121586896*xmin-4.633992715358346)--(xmax, 0.5195352121586896*xmax-4.633992715358346), linewidth(0.8) + dotted + wrwrwr); /* line */ draw((xmin, 0.5195352121586896*xmin-16.46219955905743)--(xmax, 0.5195352121586896*xmax-16.46219955905743), linewidth(0.8) + dotted + wrwrwr); /* line */ draw((xmin, 0.13203463203463203*xmin-12.188382620075727)--(xmax, 0.13203463203463203*xmax-12.188382620075727), linewidth(0.8) + dotted + wrwrwr); /* line */ draw(circle((-0.8238618291405168,-10.976121367399855), 5.340092852868624), linewidth(0.8) + ffqqff); draw((xmin, 1.194444444444445*xmin + 3.92177233180122)--(xmax, 1.194444444444445*xmax + 3.92177233180122), linewidth(0.8) + linetype("4 4") + wrwrwr); /* line */ draw((xmin, -0.2696078431372549*xmin-7.758592054976186)--(xmax, -0.2696078431372549*xmax-7.758592054976186), linewidth(0.8) + linetype("4 4") + wrwrwr); /* line */ draw(circle((-3.644468734917405,-4.7288267331171525), 3.29918604211756), linewidth(0.8) + blue); draw(circle((-4.669738921552697,-13.377517176814951), 5.680356398464501), linewidth(0.8) + red); draw((-6.9212894030477266,-4.345323344061346)--(-2.3675406461868462,-5.86401344726935), linewidth(0.4) + wrwrwr); draw((-4.598175673547878,-1.5704930560476382)--(-6.709460220477019,-5.949668956318167), linewidth(0.4) + wrwrwr); draw((xmin, 0.5138512998065294*xmin-1.5080719407995526)--(xmax, 0.5138512998065294*xmax-1.5080719407995526), linewidth(0.8) + linetype("2 2") + red); /* line */ draw((xmin, -2.2148760330578505*xmin-12.800873187314375)--(xmax, -2.2148760330578505*xmax-12.800873187314375), linewidth(0.8) + linetype("2 2") + red); /* line */ /* dots and labels */ dot((-8.204062972546984,5.370109264277287),linewidth(4pt) + dotstyle); label("$A$", (-8.04141391153527,5.695407386300712), NE * labelscalefactor); dot((-12.676912150369088,-11.22009495891742),linewidth(4pt) + dotstyle); label("$B$", (-12.514263089357375,-10.894796836893994), NW * labelscalefactor); dot((11.029188492088062,-10.732147775882282),linewidth(4pt) + dotstyle); label("$C$", (11.191837553099775,-10.406849653858856), NE * labelscalefactor); dot((-5.784885220120563,-12.952187811476929),linewidth(3pt) + dotstyle); label("$P$", (-5.642340261612507,-12.724598773275764), NE * labelscalefactor); dot((-2.3675406461868462,-5.86401344726935),linewidth(3pt) + dotstyle); label("$Q$", (-2.186047715113608,-5.608702354013328), NE * labelscalefactor); dot((2.471492725556317,-15.178172061536873),linewidth(3pt) + dotstyle); label("$R$", (2.6527618499848495,-14.920361096933886), E * labelscalefactor); dot((-6.1277538820978,-10.355379256829654),linewidth(3pt) + dotstyle); label("$S$", (-5.967638383635932,-10.122213797088358), NE * labelscalefactor); dot((-0.8238618291405126,-10.976121367399852),linewidth(3pt) + dotstyle); label("$M$", (-0.6408816355023359,-10.732147775882282), SW*2); dot((-7.8646224521743635,-11.121042683825934),linewidth(3pt) + dotstyle); label("$I$", (-7.7161157895118455,-10.894796836893994), NE * labelscalefactor); dot((-7.978106032039922,-5.60763209535758),linewidth(3pt) + dotstyle); label("$H$", (-7.797440320017702,-5.3647287624957585), W*2); dot((-1.6085513735097214,-7.32491398858876),linewidth(3pt) + dotstyle); label("$R'$", (-1.4541269405609003,-7.072543903118744), E * labelscalefactor); dot((-4.598175673547878,-1.5704930560476382),linewidth(3pt) + dotstyle); label("$Q'$", (-4.42247230402466,-1.339164502455867), N * labelscalefactor); dot((-6.9212894030477266,-4.345323344061346),linewidth(3pt) + dotstyle); label("$S'$", (-6.740221423441568,-4.104198539654985), W); dot((-6.709460220477019,-5.949668956318167),linewidth(3pt) + dotstyle); label("$P'$", (-6.536910097176928,-5.690026884519185), NE * labelscalefactor); dot((0.8592501594336901,-12.0749318414492),linewidth(3pt) + dotstyle); label("$W$", (1.026271239867721,-11.830028937711344), NE * labelscalefactor); dot((-4.979841699456322,-19.049402672901163),linewidth(3pt) + dotstyle); label("$Z$", (-4.829094956553942,-18.823938561214995), NE * labelscalefactor); dot((-0.4252879624658302,-9.602456267441015),linewidth(3pt) + dotstyle); label("$Y$", (-0.27492124822598196,-9.349630757282723), NE * labelscalefactor); dot((-6.441321998897964,-7.980486306638234),linewidth(3pt) + dotstyle); label("$X$", (-6.292936505659358,-7.723140147165594), W * labelscalefactor); dot((-6.0724568386395825,-4.628411780353551),linewidth(3pt) + dotstyle); label("$T$", (-5.926976118383004,-4.388834396425482), NE * labelscalefactor); dot((-4.138486506330636,-3.634638611309332),linewidth(3pt) + dotstyle); label("$H_{A}$", (-3.9751873862424496,-3.372277765102277), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy]$
Problem: Perpendiculars $BP,CS$ are dropped on line $l$ through $A$ and perpendiculars $BQ,CR$ are dropped on lime $m$ , which is the isogonal conjugate of $l$ with respect to $\angle A$ such that $P,S$ lie on $l$ , $Q,R$ lie on $m$ . $M$ is the midpoint of $BC$ and $I$ is the foot of $A$ -altitude on $BC$ . Then prove that:
(i) $PQRS$ is cyclic with circumcenter $M$ .
(ii) $MIPR$ and $MIQS$ are cyclic.

Proof:

And this is the question from where we got this:

Elnino2k wrote:

Given the circle $(ABC)$ and $BC$ fixed, $A$ moves on $(ABC)$ . The circles with diameter $AB$ and $AC$ intersect at $K$ . Let $E,F$ be the intersection of $AD$ with $(AB),(AC)$ respectively where $D$ is midpoint of $BC$ . Suppose that $(KDF)$ intersects $(AB)$ at $N$ and $(KDE)$ intersects $(AC)$ at $M$ .
Prove that $MN$ passes through a fixed point as $A$ moves.

Solution:

Edit:
Supercali and BOBTHEGR8 helped me study this configuration even more.

Let $P',Q',R',S'$ be intersections of $BH,CH$ with $l,m$ as shown in the diagram. Then:
1) $P'Q'R'S'$ and $PQRS$ have similicenter $A$ .
2) Let $T=P'Q' \cap R'S'$ . Then $H,T,H_A$ are collinear, where the last point is the $A$ -Humpty point.

This post has been edited 25 times. Last edited by MathPassionForever, Apr 1, 2020, 6:43 PM

Cyclic to symmetric conversion

by MathPassionForever, Sep 4, 2019, 7:54 AM

A comparatively much smaller post (as of now), as I don't have much information on this as of now.

We can clearly see that the SD theorems are strong. But then, some inequalities are cyclic, sigh. So the basic objective of this post is to serve as a collection of inequalities which can make cyclic inequalities symmetric (and these results need to be strong themselves,for our inequality shouldn't become false after their use

)
Here is the first one:
$a^2b + b^2c + c^2a + abc \leq \dfrac{4}{27}(a+b+c)^3 \ \forall a,b,c \in \mathbb{R}^+$
Proof: This inequality is obviously cyclic. So we may work with just 2 cases: $a\geq b \geq c$ and $a \leq b \leq c$
In either case, we have:
$\begin{align*} c(b-a)(b-c) &\leq 0\\ \iff b^2c - bc^2 + c^2a - abc &\leq 0\\ \iff a^2b + b^2c + c^2a + abc &\leq 2abc + a^2b + bc^2 = b(a+c)^2\\ \iff a^2b + b^2c + c^2a + abc &\leq \dfrac{1}{2} (2b)(a+c)(a+c)\\ \overset{AM-GM}{\iff} a^2b + b^2c + c^2a + abc &\leq \dfrac{1}{2 \times 27} \left(\dfrac{2b+2a+2c}{3}\right)^3\\ &= \dfrac{4}{27}(a+b+c)^3 \end{align*}$
Luckily, I have a question to post :-)

Question: For nonnegative reals $a,b,c$ , prove that
$(a^2 + b^2 + c^2)^2 \geq (\sqrt{2}-1)(a+b+c)(\sqrt{2}(a^3+b^3+c^3)+a^2b + b^2c + c^2a)$

Solution

And in case someone doesn't find it that sharp, we have the famous and super strong Vasc inequality here!
$(a^2 + b^2 + c^2)^2 \geq 3(a^3b + b^3c + c^3a) \ \forall a,b,c \in \mathbb{R}$ I shall take leave here. Bye, see you soon!

This post has been edited 5 times. Last edited by MathPassionForever, Sep 5, 2019, 8:32 PM

Inequality

inequalities

Symmetric Polynomial Inequalities in Three Variables

by MathPassionForever, Sep 3, 2019, 1:30 PM

Hey guys! I shall present here, contrary to what you would normally expect me to, some strong INEQUALITY STUFF! Here we go:

We'll start off with with something which has REALLY ugly looks, but get along with it, it gets MUCH nicer in the end.
Define $3u=a+b+c, 3v^2 = ab+bc+ca, w^3 = abc$
Theorem 1: Given $u,v^2, w^3 \in \mathbb{R}$ , there exist corresponding $a,b,c \in \mathbb{R}$ iff:
$\begin{align*} &(i)u^2 \geq v^2\\ &(ii)w \in\left[3uv^2 - 2u^3 - 2\sqrt{(u^2 - v^2)^3}, 3uv^2 - 2u^3 + 2\sqrt{(u^2-v^2)^3}\right] \end{align*}$ Proof:
Define $f(t) = t^3 - 3ut^2 + 3v^2t - w^3$ . Clearly the roots of this equation are $a,b,c$ .
Lemma 1: $a,b,c \in \mathbb{R} \iff (a-b)(b-c)(c-a) \in \mathbb{R}$
One way is clear. The other way is the one we need to focus on.
Let $a \notin \ \mathbb{R}$ and let it denote the complex number $z$ . Since the coefficients of $f(t)$ are real, there must be another nonreal root, WLOG $b$ , which is $\bar{z}$ . Plug into our expression to get $(z-\bar{z})(\bar{z}-c)(c-z)= (\bar{z} - z)|z-c|^2 \notin \mathbb{R}$ .

After our lemma, the problem looks like this:
Given $u,v^2,w^3 \in \mathbb{R}$ , there exist corresponding $a,b,c \in \mathbb{R}$ iff $\{(a-b)(b-c)(c-a)\}^2 \geq 0$
I'll skip some computational pain for you and present right away that thing thing becomes $4(u^2-v^2)^3 - 27(w^3 - (3u^2v - 2u^3))^2 \geq 0$
Condition (i) clearly is necessary here (It is of course necessary, but here we learn why it is part of criteria that make it sufficient). And Condition (ii) is mere rewriting of what I just presented!

Theorem 2: $a,b,c \geq 0 \iff u,v,w \geq 0$
Proof: One direction is obvious, the other direction is what I shall prove i.e. if at least one of $a,b,c$ is negative, then so is the case for $u,v,w$ .
The case when all three are negative gives us negative $u,w$ .
The case where one is negative gives negative $w$ .
Now we assume that exactly two are negative, say $b,c$ . Also assume that $u$ is nonnegative i.e. $a \geq -(b+c)$ .Now, $v= ab+bc+ca = bc + (b+c)a \leq bc - (b+c)^2 \leq 0$ , and we're done.

Okay, this was quite boring so far. But, just as I said, stay with me, please.
Hey, have you noticed that in most positive real inequalities, equality cases are where three, or more generally speaking, two variables are equal, or one of them is zero? Well, the following theorem explains why!

Tej's theroem: There are three parts of this theorem. I'll write all the three statements, but present the proof to just one, for each of them has nearly the same proof. Well, let me get straight to the point. That's an exercise for the readers.
(i) Fix $u,v^2$ and let there exist a value of $w^3$ such that there exist corresponding positive reals $a,b,c$ . Then $w^3$ has both a global maximum and a global minimum. The former is attained only when two of $a,b,c$ are equal to each other and the latter is attained only when two of $a,b,c$ are equal or one of $a,b,c$ is zero.
(ii) Fix $u,w^3$ and let there exist a value of $v^2$ such that there exist corresponding positive reals $a,b,c$ . Then $v^2$ has both a global maximum and a global minimum. The former and latter are both attained only when two of $a,b,c$ are equal.
(iii) Fix $v^2,w^3$ and let there exist a value of $u$ such that there exist corresponding positive reals $a,b,c$ . Then $u$ has both a global maximum and a global minimum. The former and latter are both attained only when two of $a,b,c$ are equal.

I shall prove only the former.
We know that $4(u^2-v^2)^3 - 27(w^3 - (3u^2v - 2u^3))^2 \geq 0$ . Now substitute $x=w^3$ and write it in the form of a polynomial of $x$ , i.e. $P(x)=Ax^2 + Bx + C$ . While $B$ and $C$ can be hideous, what is important is that $A=-1$ .So as $x$ tends to infinity, $P(x)$ is definitely negative. And we assumed that there exists a value of $x$ for which $P(x)$ is positive. This means that $P(x)$ has a positive root. Let $\alpha$ be the largest root and $\beta$ be the smallest positive root. Equality is clearly achieved iff the polynomial is zero i.e. $(a-b)(b-c)(c-a)=0$ or $x=0$ , i.e. $abc=0$ . Take your time to realise this.

Aha! this powerful theorem gives us the nuke we were waiting for! * Insert devious laughter *

SD-3,4,5 theorem(For all nonnegative $n\leq 5$ tho).
To verify if a symmetric homogeneous 3 variable polynomial $f(a,b,c)$ of degree $n\leq 5$ is nonnegative for nonnegative $a,b,c$ , we just need to verify the cases:
(i) $f(a,1,1) \geq 0$
(ii) $f(0,b,c) \geq 0$
Told you we'd get to nice stuff :-)

Proof:
Every polynomial function in 3 variables of degree $\leq 5$ can be written as $A(u,v^2)w^3 + B(u,v^2)$ . Fixing $u,v^2$ we maximise or minimise $w^3$ to find extrema of function. And extrema are achieved when two variables are equal or one is zero, and we're done!

Based on similar logic, just that you need to focus on the Tej's theorem reasoning,to find this:
Let $a,b,c$ be sides of a possibly degenerate triangle instead. Then we need to verify this just for:
(i) $f(x,1,1) \geq 0 \forall x \in [0,2]$
(ii) $f(k+l,k,l) \geq 0$

We also have a non-homogenous variant!
Non-homogenous variant: Let $f(a,b,c)$ satisfy all criteria that it did earlier, except homogeneity. We can still check for just the cases:
(i) $f(a,b,b) \geq 0$
(ii) $f(0,b,c) \geq 0$

Some crucial facts that help you do manipulations as per your need:
(i)If the degree is $\leq 3$ , you can make it a polynomial in $v^2$ instead and check that $f(x,1,1) \geq 0$ only from the second part of Tej's theorem!
(ii) We used the fact that $A(u,v^2)w^3 + B(u,v^2)$ has $A,B$ as functions. But they can be ANY functions, not just polynomials!

Edit: Hey guys, back with part of what I promised!
SD-6 Theorem: Let $f(a,b,c)$ be degree 6 instead (and follows all other criteria as mentioned earlier). Then, it can be written as
$f(a,b,c) = Aw^6 + B(u,v^2)w^3 + C(u,v^2)$ After noting that $A$ is a constant, see that there are two cases:
(i) $A \leq 0$ : Just check if $f(a,1,1) \geq 0$ and $f(0,b,c) \geq 0$ . The same logic as used in Tej's theorem.
(ii) Um..... this is the bad part. For $a>0$ , we need to prove
$f(a,b,c) \geq A\left(w^3 + Ku^3 + Luv^2 + M\dfrac{v^4}{u}\right)^2$ So yeah, we need to find $K,L,M$ ourselves. The point of doing this is to make $A = 0$ to apply (i) now.

Adding problems soon!

Dec 27:
Edit: Ok maybe not so soon

April 1:
Edit: Can we act like I never said this?

This post has been edited 14 times. Last edited by MathPassionForever, Apr 1, 2020, 12:54 PM

Inequality

inequalities

Submit

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