Incenter-Related Configurations -Part (I)

by AlastorMoody, Nov 16, 2019, 7:15 AM

Let's Start-off from Really Basic Lemmas, 'coz this configuration is very interesting!!! Here we have the Incenter-Excenter Lemma:
[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(14cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -13.75752348035464, xmax = 22.325110003851645, ymin = -13.947881129874407, ymax = 7.4946674570515395; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen ttffqq = rgb(0.2,1,0); draw((-4.8127515399252045,5.283566240744204)--(-7.32,-3.13)--(3.64,-3.27)--cycle, linewidth(2) + rvwvcq); /* draw figures */ draw((-4.8127515399252045,5.283566240744204)--(-7.32,-3.13), linewidth(0.4) + rvwvcq); draw((-7.32,-3.13)--(3.64,-3.27), linewidth(0.4) + rvwvcq); draw((3.64,-3.27)--(-4.8127515399252045,5.283566240744204), linewidth(0.4) + rvwvcq); draw(circle((-1.8016037461609404,-0.1941218423136115), 6.250766174731304), linewidth(0.4)); draw(circle((-1.8814427861656493,-6.444378116967962), 6.368909394110248), linewidth(0.4) + linetype("4 4")); draw((-3.425794343555926,-0.26554447803803993)--(-7.32,-3.13), linewidth(0.4) + ttffqq); draw((-7.32,-3.13)--(-0.33709122877537645,-12.623211755897882), linewidth(0.4) + ttffqq); draw((-0.33709122877537645,-12.623211755897882)--(3.64,-3.27), linewidth(0.4) + ttffqq); draw((3.64,-3.27)--(-3.425794343555926,-0.26554447803803993), linewidth(0.4) + ttffqq); draw((-4.8127515399252045,5.283566240744204)--(-0.33709122877537645,-12.623211755897882), linewidth(0.4)); /* dots and labels */ dot((-4.8127515399252045,5.283566240744204),dotstyle); label("$A$", (-4.707289180217655,5.5722320664995575), NE * labelscalefactor); dot((-7.32,-3.13),dotstyle); label("$B$", (-7.191667223392514,-2.82733179375833), NE * labelscalefactor); dot((3.64,-3.27),dotstyle); label("$C$", (3.7514265382110312,-2.975211439185406), NE * labelscalefactor); dot((-3.425794343555926,-0.26554447803803993),dotstyle); label("$I$", (-3.317220513203151,0.04153332752693427), NE * labelscalefactor); dot((-0.33709122877537645,-12.623211755897882),dotstyle); label("$I_A$", (-0.21174795923457712,-12.321205030176577), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]
$I$ is the Incenter and $I_A$ is the $A-$Excenter. We can observe, $BICI_A$ is cyclic and $A,I,I_A$ are collinear. Let $M_A$ be midpoint of arc $BC$ not containing $A$ and $M_{BC}$ be midpoint of arc $BAC$. We get some more Trivial Results:
[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(14cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -13.75752348035465, xmax = 22.325110003851638, ymin = -13.947881129874407, ymax = 7.4946674570515395; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen ttffqq = rgb(0.2,1,0); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); draw((-4.8127515399252045,5.283566240744204)--(-7.32,-3.13)--(3.64,-3.27)--cycle, linewidth(2) + rvwvcq); /* draw figures */ draw((-4.8127515399252045,5.283566240744204)--(-7.32,-3.13), linewidth(0.4) + rvwvcq); draw((-7.32,-3.13)--(3.64,-3.27), linewidth(0.4) + rvwvcq); draw((3.64,-3.27)--(-4.8127515399252045,5.283566240744204), linewidth(0.4) + rvwvcq); draw(circle((-1.8016037461609404,-0.1941218423136115), 6.250766174731304), linewidth(0.4)); draw(circle((-1.8814427861656493,-6.444378116967962), 6.368909394110248), linewidth(0.4) + linetype("4 4")); draw((-3.425794343555926,-0.26554447803803993)--(-7.32,-3.13), linewidth(0.4) + ttffqq); draw((-7.32,-3.13)--(-0.33709122877537645,-12.623211755897882), linewidth(0.4) + ttffqq); draw((-0.33709122877537645,-12.623211755897882)--(3.64,-3.27), linewidth(0.4) + ttffqq); draw((3.64,-3.27)--(-3.425794343555926,-0.26554447803803993), linewidth(0.4) + ttffqq); draw((-4.8127515399252045,5.283566240744204)--(-0.33709122877537645,-12.623211755897882), linewidth(0.4)); draw((-7.32,-3.13)--(-1.7217647061562313,6.056134432340744), linewidth(0.4) + dtsfsf); draw((-1.7217647061562313,6.056134432340744)--(3.64,-3.27), linewidth(0.4) + dtsfsf); draw((-1.7217647061562313,6.056134432340744)--(-1.88144278616565,-6.444378116967967), linewidth(0.4)); /* dots and labels */ dot((-4.8127515399252045,5.283566240744204),dotstyle); label("$A$", (-4.707289180217664,5.5722320664995575), NE * labelscalefactor); dot((-7.32,-3.13),dotstyle); label("$B$", (-7.191667223392523,-2.82733179375833), NE * labelscalefactor); dot((3.64,-3.27),dotstyle); label("$C$", (3.751426538211023,-2.975211439185406), NE * labelscalefactor); dot((-3.425794343555926,-0.26554447803803993),dotstyle); label("$I$", (-3.3172205132031594,0.04153332752693427), NE * labelscalefactor); dot((-0.33709122877537645,-12.623211755897882),dotstyle); label("$I_A$", (-0.21174795923458553,-12.321205030176577), NE * labelscalefactor); dot((-1.88144278616565,-6.444378116967967),dotstyle); label("$M_A$", (-1.7496962716761648,-6.139835851324822), NE * labelscalefactor); dot((-1.7217647061562313,6.056134432340744),dotstyle); label("$M_{BC}$", (-1.60181662624909,6.34120622272035), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]
Just to mention: $M_A$ is the center of $\odot (BIC)$, $M_{BC}$ is the antipode of $M_A$ in $\odot (ABC)$ $\implies$ $M_AM_{BC}$ $\perp$ $BC$ and Lastly, $BM_{BC}$, $CM_{BC}$ are tangent to $\odot (BIC)$ at $B,C$. Cool!! I assume, these results are trivial and hence not proven here.

Now, that Tangency we received looks cool! How about invoking a Harmonic Quadrilateral?
[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(17cm); 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 = -17.107588967859602, xmax = 18.975044516346685, ymin = -13.77900257479671, ymax = 7.6635460121292684; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen ttffqq = rgb(0.2,1,0); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen ubqqys = rgb(0.29411764705882354,0,0.5098039215686274); draw((-4.8127515399252045,5.283566240744204)--(-7.32,-3.13)--(3.64,-3.27)--cycle, linewidth(2) + rvwvcq); /* draw figures */ draw((-4.8127515399252045,5.283566240744204)--(-7.32,-3.13), linewidth(0.4) + rvwvcq); draw((-7.32,-3.13)--(3.64,-3.27), linewidth(0.4) + rvwvcq); draw((3.64,-3.27)--(-4.8127515399252045,5.283566240744204), linewidth(0.4) + rvwvcq); draw(circle((-1.8016037461609404,-0.1941218423136115), 6.250766174731304), linewidth(0.4)); draw(circle((-1.8814427861656493,-6.444378116967962), 6.368909394110248), linewidth(0.4) + linetype("4 4")); draw((-3.425794343555926,-0.26554447803803993)--(-7.32,-3.13), linewidth(0.4) + ttffqq); draw((-7.32,-3.13)--(-0.33709122877537645,-12.623211755897882), linewidth(0.4) + ttffqq); draw((-0.33709122877537645,-12.623211755897882)--(3.64,-3.27), linewidth(0.4) + ttffqq); draw((3.64,-3.27)--(-3.425794343555926,-0.26554447803803993), linewidth(0.4) + ttffqq); draw((-4.8127515399252045,5.283566240744204)--(-0.33709122877537645,-12.623211755897882), linewidth(0.4)); draw((-7.32,-3.13)--(-1.7217647061562313,6.056134432340744), linewidth(0.4) + dtsfsf); draw((-1.7217647061562313,6.056134432340744)--(3.64,-3.27), linewidth(0.4) + dtsfsf); draw((-1.7217647061562313,6.056134432340744)--(-1.88144278616565,-6.444378116967967), linewidth(0.4)); draw((-1.7217647061562313,6.056134432340744)--(-6.321988149529278,-11.009962827199013), linewidth(0.4) + ubqqys); draw((-7.32,-3.13)--(-6.321988149529278,-11.009962827199013), linewidth(0.4) + ubqqys); draw((-6.321988149529278,-11.009962827199013)--(3.64,-3.27), linewidth(0.4) + ubqqys); draw((-14.51837862789293,-3.03804990803786)--(-3.425794343555926,-0.26554447803803993), linewidth(0.4) + ttffqq); draw((-14.51837862789293,-3.03804990803786)--(-6.321988149529278,-11.009962827199013), linewidth(0.4) + ttffqq); draw((-14.51837862789293,-3.03804990803786)--(-7.32,-3.13), linewidth(0.4) + rvwvcq); draw((-14.51837862789293,-3.03804990803786)--(-1.88144278616565,-6.444378116967967), linewidth(0.4) + linetype("4 4")); /* dots and labels */ dot((-4.8127515399252045,5.283566240744204),dotstyle); label("$A$", (-4.685698751985307,5.593230976150208), NE * labelscalefactor); dot((-7.32,-3.13),dotstyle); label("$B$", (-7.199652724245581,-2.835908813193107), NE * labelscalefactor); dot((3.64,-3.27),dotstyle); label("$C$", (3.7730169664433797,-2.983788458620183), NE * labelscalefactor); dot((-3.425794343555926,-0.26554447803803993),dotstyle); label("$I$", (-3.2956300849708025,0.03295630809216165), NE * labelscalefactor); dot((-0.33709122877537645,-12.623211755897882),dotstyle); label("$I_A$", (-0.21973346008764366,-12.329782049611367), NE * labelscalefactor); dot((-1.88144278616565,-6.444378116967967),dotstyle); label("$M_A$", (-1.757681772529223,-6.148412870759603), NE * labelscalefactor); dot((-1.7217647061562313,6.056134432340744),dotstyle); label("$M_{BC}$", (-1.609802127102148,6.362205132371002), NE * labelscalefactor); dot((-4.873891246542604,-5.637753652618531),dotstyle); label("$T$", (-4.744850610156137,-5.349862785453394), NE * labelscalefactor); dot((-6.321988149529278,-11.009962827199013),dotstyle); label("$J$", (-6.1940711353414715,-10.703105949913535), NE * labelscalefactor); dot((-14.51837862789293,-3.03804990803786),dotstyle); label("$K$", (-14.386603492001424,-2.7471810259368614), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]
Let $\stackrel{\longrightarrow}{IM_{BC}}$ $\cap$ $\odot (ABC)$, $\odot (BIC)$ $=$ $T,J$ respectively. It's a Well-Known Lemma in the Mixtilinear Configuration that, $T$ is the $A-$ Mixtilinear Point. Also, Using the Antipode fact, $\implies$ $\angle M_{BC}TM_A$ $=$ $90^{\circ}$ $\implies$ $T$ is midpoint of $IJ$. It follows from La-Hire's Theorem, $K$ which is the intersection of tangents at $I,J$ to $\odot (BIC)$, lies on $BC$. Details Lastly, Also note $\implies$ $K,T,M_A$ are collinear. Anyway, one can use harmonic quadrilateral instead of La-Hire's Theorem. Let's make things interesting!! :diablo: We'll merge two different configurations in a big picture now. Add the Incircle in the diagram.
[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(18cm); 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 = -17.107588967859602, xmax = 18.975044516346685, ymin = -13.779002574796706, ymax = 7.6635460121292684; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen ttffqq = rgb(0.2,1,0); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen ubqqys = rgb(0.29411764705882354,0,0.5098039215686274); pen qqqqcc = rgb(0,0,0.8); draw((-4.8127515399252045,5.283566240744204)--(-7.32,-3.13)--(3.64,-3.27)--cycle, linewidth(2) + rvwvcq); /* draw figures */ draw((-4.8127515399252045,5.283566240744204)--(-7.32,-3.13), linewidth(0.4) + rvwvcq); draw((-7.32,-3.13)--(3.64,-3.27), linewidth(0.4) + rvwvcq); draw((3.64,-3.27)--(-4.8127515399252045,5.283566240744204), linewidth(0.4) + rvwvcq); draw(circle((-1.8016037461609404,-0.1941218423136115), 6.250766174731304), linewidth(0.4)); draw(circle((-1.8814427861656493,-6.444378116967962), 6.368909394110248), linewidth(0.4) + linetype("4 4")); draw((-3.425794343555926,-0.26554447803803993)--(-7.32,-3.13), linewidth(0.4) + ttffqq); draw((-7.32,-3.13)--(-0.33709122877537645,-12.623211755897882), linewidth(0.4) + ttffqq); draw((-0.33709122877537645,-12.623211755897882)--(3.64,-3.27), linewidth(0.4) + ttffqq); draw((3.64,-3.27)--(-3.425794343555926,-0.26554447803803993), linewidth(0.4) + ttffqq); draw((-4.8127515399252045,5.283566240744204)--(-0.33709122877537645,-12.623211755897882), linewidth(0.4)); draw((-7.32,-3.13)--(-1.7217647061562313,6.056134432340744), linewidth(0.4) + dtsfsf); draw((-1.7217647061562313,6.056134432340744)--(3.64,-3.27), linewidth(0.4) + dtsfsf); draw((-1.7217647061562313,6.056134432340744)--(-1.88144278616565,-6.444378116967967), linewidth(0.4)); draw((-1.7217647061562313,6.056134432340744)--(-6.321988149529278,-11.009962827199013), linewidth(0.4) + ubqqys); draw((-7.32,-3.13)--(-6.321988149529278,-11.009962827199013), linewidth(0.4) + ubqqys); draw((-6.321988149529278,-11.009962827199013)--(3.64,-3.27), linewidth(0.4) + ubqqys); draw((-14.51837862789293,-3.03804990803786)--(-3.425794343555926,-0.26554447803803993), linewidth(0.4) + ttffqq); draw((-14.51837862789293,-3.03804990803786)--(-6.321988149529278,-11.009962827199013), linewidth(0.4) + ttffqq); draw((-14.51837862789293,-3.03804990803786)--(-7.32,-3.13), linewidth(0.4) + rvwvcq); draw((-14.51837862789293,-3.03804990803786)--(-1.88144278616565,-6.444378116967967), linewidth(0.4) + linetype("4 4")); draw(circle((-3.425794343555926,-0.26554447803803993), 2.913961301965583), linewidth(0.4) + qqqqcc); /* dots and labels */ dot((-4.8127515399252045,5.283566240744204),dotstyle); label("$A$", (-4.685698751985307,5.593230976150209), NE * labelscalefactor); dot((-7.32,-3.13),dotstyle); label("$B$", (-7.199652724245581,-2.835908813193105), NE * labelscalefactor); dot((3.64,-3.27),dotstyle); label("$C$", (3.7730169664433797,-2.9837884586201806), NE * labelscalefactor); dot((-3.425794343555926,-0.26554447803803993),dotstyle); label("$I$", (-3.2956300849708025,0.03295630809216332), NE * labelscalefactor); dot((-0.33709122877537645,-12.623211755897882),dotstyle); label("$I_A$", (-0.21973346008764366,-12.329782049611364), NE * labelscalefactor); dot((-1.88144278616565,-6.444378116967967),dotstyle); label("$M_A$", (-1.757681772529223,-6.148412870759601), NE * labelscalefactor); dot((-1.7217647061562313,6.056134432340744),dotstyle); label("$M_{BC}$", (-1.609802127102148,6.3622051323710025), NE * labelscalefactor); dot((-4.873891246542604,-5.637753652618531),dotstyle); label("$T$", (-4.744850610156137,-5.349862785453392), NE * labelscalefactor); dot((-6.321988149529278,-11.009962827199013),dotstyle); label("$J$", (-6.1940711353414715,-10.703105949913532), NE * labelscalefactor); dot((-14.51837862789293,-3.03804990803786),dotstyle); label("$K$", (-14.386603492001424,-2.7471810259368596), NE * labelscalefactor); dot((-3.4630134406164252,-3.1792680764884764),dotstyle); label("$D$", (-3.3547819431416324,-2.895060671363935), NE * labelscalefactor); dot((-1.3531342778462379,1.7826866420666359),dotstyle); label("$E$", (-1.2253150489917533,2.073695414985808), NE * labelscalefactor); dot((-6.218395061190416,0.5666524344894496),dotstyle); label("$F$", (-6.105343348085226,0.8610823224837871), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]
As usual, Let $D,E,F$ be the $A,B,C-$Intouch Points. Let $\Psi _{\odot (BIC)}$ denote Inversion around $\odot (BIC)$. Let $\odot (AIFE)$ $\cap$ $\odot (ABC)$ $=$ $L$. Clearly, $L$ lies on $IA'$ where $A'$ is $A-$antipode in $\odot (ABC)$. Radical Axes Theorem on $\odot (AFIE)$, $\odot (BIC)$ and $\odot (ABC)$ $\implies$ $K$ $\in$ $LA$.
[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(18cm); 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 = -17.107588967859602, xmax = 18.975044516346685, ymin = -13.77900257479671, ymax = 7.6635460121292684; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen ttffqq = rgb(0.2,1,0); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen ubqqys = rgb(0.29411764705882354,0,0.5098039215686274); pen qqqqcc = rgb(0,0,0.8); draw((-4.8127515399252045,5.283566240744204)--(-7.32,-3.13)--(3.64,-3.27)--cycle, linewidth(2) + rvwvcq); /* draw figures */ draw((-4.8127515399252045,5.283566240744204)--(-7.32,-3.13), linewidth(0.4) + rvwvcq); draw((-7.32,-3.13)--(3.64,-3.27), linewidth(0.4) + rvwvcq); draw((3.64,-3.27)--(-4.8127515399252045,5.283566240744204), linewidth(0.4) + rvwvcq); draw(circle((-1.8016037461609404,-0.1941218423136115), 6.250766174731304), linewidth(0.4)); draw(circle((-1.8814427861656493,-6.444378116967962), 6.368909394110248), linewidth(0.4) + linetype("4 4")); draw((-3.425794343555926,-0.26554447803803993)--(-7.32,-3.13), linewidth(0.4) + ttffqq); draw((-7.32,-3.13)--(-0.33709122877537645,-12.623211755897882), linewidth(0.4) + ttffqq); draw((-0.33709122877537645,-12.623211755897882)--(3.64,-3.27), linewidth(0.4) + ttffqq); draw((3.64,-3.27)--(-3.425794343555926,-0.26554447803803993), linewidth(0.4) + ttffqq); draw((-4.8127515399252045,5.283566240744204)--(-0.33709122877537645,-12.623211755897882), linewidth(0.4)); draw((-7.32,-3.13)--(-1.7217647061562313,6.056134432340744), linewidth(0.4) + dtsfsf); draw((-1.7217647061562313,6.056134432340744)--(3.64,-3.27), linewidth(0.4) + dtsfsf); draw((-1.7217647061562313,6.056134432340744)--(-1.88144278616565,-6.444378116967967), linewidth(0.4)); draw((-1.7217647061562313,6.056134432340744)--(-6.321988149529278,-11.009962827199013), linewidth(0.4) + ubqqys); draw((-7.32,-3.13)--(-6.321988149529278,-11.009962827199013), linewidth(0.4) + ubqqys); draw((-6.321988149529278,-11.009962827199013)--(3.64,-3.27), linewidth(0.4) + ubqqys); draw((-14.51837862789293,-3.03804990803786)--(-3.425794343555926,-0.26554447803803993), linewidth(0.4) + ttffqq); draw((-14.51837862789293,-3.03804990803786)--(-6.321988149529278,-11.009962827199013), linewidth(0.4) + ttffqq); draw((-14.51837862789293,-3.03804990803786)--(-7.32,-3.13), linewidth(0.4) + rvwvcq); draw((-14.51837862789293,-3.03804990803786)--(-1.88144278616565,-6.444378116967967), linewidth(0.4) + linetype("4 4")); draw(circle((-3.425794343555926,-0.26554447803803993), 2.913961301965583), linewidth(0.4) + qqqqcc); draw(circle((-4.119272941740566,2.5090108813530825), 2.8599073426364945), linewidth(0.4)); draw((-6.755455378061811,3.6178896713466084)--(-1.88144278616565,-6.444378116967967), linewidth(0.4) + linetype("4 4")); draw((-14.51837862789293,-3.03804990803786)--(-4.8127515399252045,5.283566240744204), linewidth(0.4) + ttffqq); draw((-6.755455378061811,3.6178896713466084)--(1.209544047603321,-5.67180992537143), linewidth(0.4) + dtsfsf); /* dots and labels */ dot((-4.8127515399252045,5.283566240744204),dotstyle); label("$A$", (-4.685698751985307,5.593230976150208), NE * labelscalefactor); dot((-7.32,-3.13),dotstyle); label("$B$", (-7.199652724245581,-2.835908813193107), NE * labelscalefactor); dot((3.64,-3.27),dotstyle); label("$C$", (3.7730169664433797,-2.983788458620183), NE * labelscalefactor); dot((-3.425794343555926,-0.26554447803803993),dotstyle); label("$I$", (-3.2956300849708025,0.03295630809216165), NE * labelscalefactor); dot((-0.33709122877537645,-12.623211755897882),dotstyle); label("$I_A$", (-0.21973346008764366,-12.329782049611367), NE * labelscalefactor); dot((-1.88144278616565,-6.444378116967967),dotstyle); label("$M_A$", (-1.757681772529223,-6.148412870759603), NE * labelscalefactor); dot((-1.7217647061562313,6.056134432340744),dotstyle); label("$M_{BC}$", (-1.609802127102148,6.362205132371002), NE * labelscalefactor); dot((-4.873891246542604,-5.637753652618531),dotstyle); label("$T$", (-4.744850610156137,-5.349862785453394), NE * labelscalefactor); dot((-6.321988149529278,-11.009962827199013),dotstyle); label("$J$", (-6.1940711353414715,-10.703105949913535), NE * labelscalefactor); dot((-14.51837862789293,-3.03804990803786),dotstyle); label("$K$", (-14.386603492001424,-2.7471810259368614), NE * labelscalefactor); dot((-3.4630134406164252,-3.1792680764884764),dotstyle); label("$D$", (-3.3547819431416324,-2.8950606713639373), NE * labelscalefactor); dot((-1.3531342778462379,1.7826866420666359),dotstyle); label("$E$", (-1.2253150489917533,2.0736954149858065), NE * labelscalefactor); dot((-6.218395061190416,0.5666524344894496),dotstyle); label("$F$", (-6.105343348085226,0.8610823224837856), NE * labelscalefactor); dot((-6.755455378061811,3.6178896713466084),dotstyle); label("$L$", (-6.637710071622696,3.907403018281545), NE * labelscalefactor); dot((1.209544047603321,-5.67180992537143),dotstyle); label("$A'$", (1.3182148523539357,-5.379438714538809), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]
Observe, $\Psi _{\odot (BIC)} (\odot (ABC))$ $=$ $BC$. Also By Shooting Lemma, $\Psi _{\odot (BIC)} (AI ~ \cap ~ BC)$ $=$ $A$. Let $AI \cap BC =X$ (For now). So Notice, $\Delta IDX$ $\stackrel{\Psi _{\odot (BIC)}}{ \mapsto} $ $\Delta ILA$ $\implies$ $ALDX$ cyclic (or) $L,D,M_A$ collinear. Another way to notice this is using angle chasing.
[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 = -17.069946876296346, xmax = 19.012686607909938, ymin = -13.319231313559806, ymax = 8.123317273366181; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen ttffqq = rgb(0.2,1,0); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen ubqqys = rgb(0.29411764705882354,0,0.5098039215686274); pen qqqqcc = rgb(0,0,0.8); draw((-4.8127515399252045,5.283566240744204)--(-7.32,-3.13)--(3.64,-3.27)--cycle, linewidth(2) + rvwvcq); /* draw figures */ draw((-4.8127515399252045,5.283566240744204)--(-7.32,-3.13), linewidth(0.4) + rvwvcq); draw((-7.32,-3.13)--(3.64,-3.27), linewidth(0.4) + rvwvcq); draw((3.64,-3.27)--(-4.8127515399252045,5.283566240744204), linewidth(0.4) + rvwvcq); draw(circle((-1.8016037461609404,-0.1941218423136115), 6.250766174731304), linewidth(0.4)); draw(circle((-1.8814427861656493,-6.444378116967962), 6.368909394110248), linewidth(0.4) + linetype("4 4")); draw((-3.425794343555926,-0.26554447803803993)--(-7.32,-3.13), linewidth(0.4) + ttffqq); draw((-7.32,-3.13)--(-0.33709122877537645,-12.623211755897882), linewidth(0.4) + ttffqq); draw((-0.33709122877537645,-12.623211755897882)--(3.64,-3.27), linewidth(0.4) + ttffqq); draw((3.64,-3.27)--(-3.425794343555926,-0.26554447803803993), linewidth(0.4) + ttffqq); draw((-4.8127515399252045,5.283566240744204)--(-0.33709122877537645,-12.623211755897882), linewidth(0.4)); draw((-7.32,-3.13)--(-1.7217647061562313,6.056134432340744), linewidth(0.4) + dtsfsf); draw((-1.7217647061562313,6.056134432340744)--(3.64,-3.27), linewidth(0.4) + dtsfsf); draw((-1.7217647061562313,6.056134432340744)--(-1.88144278616565,-6.444378116967967), linewidth(0.4)); draw((-1.7217647061562313,6.056134432340744)--(-6.321988149529278,-11.009962827199013), linewidth(0.4) + ubqqys); draw((-7.32,-3.13)--(-6.321988149529278,-11.009962827199013), linewidth(0.4) + ubqqys); draw((-6.321988149529278,-11.009962827199013)--(3.64,-3.27), linewidth(0.4) + ubqqys); draw((-14.51837862789293,-3.03804990803786)--(-3.425794343555926,-0.26554447803803993), linewidth(0.4) + ttffqq); draw((-14.51837862789293,-3.03804990803786)--(-6.321988149529278,-11.009962827199013), linewidth(0.4) + ttffqq); draw((-14.51837862789293,-3.03804990803786)--(-7.32,-3.13), linewidth(0.4) + rvwvcq); draw((-14.51837862789293,-3.03804990803786)--(-1.88144278616565,-6.444378116967967), linewidth(0.4) + linetype("4 4")); draw(circle((-3.425794343555926,-0.26554447803803993), 2.913961301965583), linewidth(0.4) + qqqqcc); draw(circle((-4.119272941740566,2.5090108813530825), 2.8599073426364945), linewidth(0.4)); draw((-6.755455378061811,3.6178896713466084)--(-1.88144278616565,-6.444378116967967), linewidth(0.4) + linetype("4 4")); draw((-14.51837862789293,-3.03804990803786)--(-4.8127515399252045,5.283566240744204), linewidth(0.4) + ttffqq); draw((-6.755455378061811,3.6178896713466084)--(1.209544047603321,-5.67180992537143), linewidth(0.4) + dtsfsf); draw(circle((-3.409081408517464,1.0428395792587513), 4.2224520981067), linewidth(0.4) + ubqqys); draw((-6.755455378061811,3.6178896713466084)--(-1.7217647061562313,6.056134432340744), linewidth(0.4) + ubqqys); draw(circle((-8.97208648572443,-1.6517971930379418), 5.716909402476542), linewidth(0.4) + linetype("4 4")); draw(circle((-9.665565083909069,1.122758166353172), 6.392348795598595), linewidth(0.4) + linetype("4 4")); draw((-4.8127515399252045,5.283566240744204)--(-4.92061560412313,-3.160649070750248), linewidth(0.4)); /* dots and labels */ dot((-4.8127515399252045,5.283566240744204),dotstyle); label("$A$", (-4.707208518592882,5.579787372020479), NE * labelscalefactor); dot((-7.32,-3.13),dotstyle); label("$B$", (-7.191586561767741,-2.819776488237425), NE * labelscalefactor); dot((3.64,-3.27),dotstyle); label("$C$", (3.7515071998358045,-2.967656133664501), NE * labelscalefactor); dot((-3.425794343555926,-0.26554447803803993),dotstyle); label("$I$", (-3.3171398515783777,0.019512703962429642), NE * labelscalefactor); dot((-0.33709122877537645,-12.623211755897882),dotstyle); label("$I_A$", (-0.21166729760980382,-12.31364972465569), NE * labelscalefactor); dot((-1.88144278616565,-6.444378116967967),dotstyle); label("$M_A$", (-1.7496156100513833,-6.161856474889338), NE * labelscalefactor); dot((-1.7217647061562313,6.056134432340744),dotstyle); label("$M_{BC}$", (-1.6017359646243083,6.348761528241273), NE * labelscalefactor); dot((-4.873891246542604,-5.637753652618531),dotstyle); label("$T$", (-4.766360376763712,-5.333730460497714), NE * labelscalefactor); dot((-6.321988149529278,-11.009962827199013),dotstyle); label("$J$", (-6.215580901949046,-10.716549554043272), NE * labelscalefactor); dot((-14.51837862789293,-3.03804990803786),dotstyle); label("$K$", (-14.408113258608997,-2.7310487009811797), NE * labelscalefactor); dot((-3.4630134406164252,-3.1792680764884764),dotstyle); label("$D$", (-3.3467157806637924,-2.8789283464082556), NE * labelscalefactor); dot((-1.3531342778462379,1.7826866420666359),dotstyle); label("$E$", (-1.2468248155993285,2.0898277399414904), NE * labelscalefactor); dot((-6.218395061190416,0.5666524344894496),dotstyle); label("$F$", (-6.097277185607386,0.847638718354054), NE * labelscalefactor); dot((-6.755455378061811,3.6178896713466084),dotstyle); label("$L$", (-6.629643909144856,3.92353534323723), NE * labelscalefactor); dot((1.209544047603321,-5.67180992537143),dotstyle); label("$A'$", (1.3262810148317756,-5.363306389583129), NE * labelscalefactor); dot((-3.3551493764184945,5.2649472350059785),dotstyle); label("$G$", (-3.2284120643221326,5.550211442935064), NE * labelscalefactor); dot((-4.92061560412313,-3.160649070750248),dotstyle); label("$H_A$", (-4.795936305849127,-2.8789283464082556), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]
So suppose, someone hates inversion. So, here's a way to do using angle chasing. Let $H_A$ be foot from $A$ to $BC$. Easy to observe, $AIH_AK$ and $KLIDT$ is cyclic. Our goal should be to prove, $IM_A$ tangent to $\odot (KLIDT)$.
$$\angle DIM_A=\angle H_AAI=\angle DKI $$Hence, We must have, $\Psi _{\odot (BIC)} (L) = D$. By Converse of Reim's Theorem, If $LM_{BC}$ $\cap$ $\odot (AFIE)$ $=$ $G$ $\implies$ $G$ lies on $DI$. Further angle chasing, should imply $\odot (DLG)$ tangent to $BC$.
Which implies, $\odot (DLG)$ is also tangen to $\odot (ABC)$ at $L$.

[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(21cm); 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 = -20.310021989506694, xmax = 16.515608576902242, ymin = -13.697015176967454, ymax = 8.18706856126748; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen ttffqq = rgb(0.2,1,0); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen ubqqys = rgb(0.29411764705882354,0,0.5098039215686274); pen qqqqcc = rgb(0,0,0.8); draw((-4.8127515399252045,5.283566240744204)--(-7.32,-3.13)--(3.64,-3.27)--cycle, linewidth(2) + rvwvcq); /* draw figures */ draw((-4.8127515399252045,5.283566240744204)--(-7.32,-3.13), linewidth(0.4) + rvwvcq); draw((-7.32,-3.13)--(3.64,-3.27), linewidth(0.4) + rvwvcq); draw((3.64,-3.27)--(-4.8127515399252045,5.283566240744204), linewidth(0.4) + rvwvcq); draw(circle((-1.8016037461609404,-0.1941218423136115), 6.250766174731304), linewidth(0.4)); draw(circle((-1.8814427861656493,-6.444378116967962), 6.368909394110248), linewidth(0.4) + linetype("4 4")); draw((-3.425794343555926,-0.26554447803803993)--(-7.32,-3.13), linewidth(0.4) + ttffqq); draw((-7.32,-3.13)--(-0.33709122877537645,-12.623211755897882), linewidth(0.4) + ttffqq); draw((-0.33709122877537645,-12.623211755897882)--(3.64,-3.27), linewidth(0.4) + ttffqq); draw((3.64,-3.27)--(-3.425794343555926,-0.26554447803803993), linewidth(0.4) + ttffqq); draw((-4.8127515399252045,5.283566240744204)--(-0.33709122877537645,-12.623211755897882), linewidth(0.4)); draw((-7.32,-3.13)--(-1.7217647061562313,6.056134432340744), linewidth(0.4) + dtsfsf); draw((-1.7217647061562313,6.056134432340744)--(3.64,-3.27), linewidth(0.4) + dtsfsf); draw((-1.7217647061562313,6.056134432340744)--(-1.88144278616565,-6.444378116967967), linewidth(0.4)); draw((-1.7217647061562313,6.056134432340744)--(-6.321988149529278,-11.009962827199013), linewidth(0.4) + ubqqys); draw((-7.32,-3.13)--(-6.321988149529278,-11.009962827199013), linewidth(0.4) + ubqqys); draw((-6.321988149529278,-11.009962827199013)--(3.64,-3.27), linewidth(0.4) + ubqqys); draw((-14.51837862789293,-3.03804990803786)--(-3.425794343555926,-0.26554447803803993), linewidth(0.4) + ttffqq); draw((-14.51837862789293,-3.03804990803786)--(-6.321988149529278,-11.009962827199013), linewidth(0.4) + ttffqq); draw((-14.51837862789293,-3.03804990803786)--(-7.32,-3.13), linewidth(0.4) + rvwvcq); draw((-14.51837862789293,-3.03804990803786)--(-1.88144278616565,-6.444378116967967), linewidth(0.4) + linetype("4 4")); draw(circle((-3.425794343555926,-0.26554447803803993), 2.913961301965583), linewidth(0.4) + qqqqcc); draw(circle((-4.119272941740566,2.5090108813530825), 2.8599073426364945), linewidth(0.4)); draw((-6.755455378061811,3.6178896713466084)--(-1.88144278616565,-6.444378116967967), linewidth(0.4) + linetype("4 4")); draw((-14.51837862789293,-3.03804990803786)--(-4.8127515399252045,5.283566240744204), linewidth(0.4) + ttffqq); draw((-6.755455378061811,3.6178896713466084)--(1.209544047603321,-5.67180992537143), linewidth(0.4) + dtsfsf); draw(circle((-3.409081408517464,1.0428395792587513), 4.2224520981067), linewidth(0.4) + ubqqys); draw((-6.755455378061811,3.6178896713466084)--(-1.7217647061562313,6.056134432340744), linewidth(0.4) + ubqqys); draw(circle((-8.97208648572443,-1.6517971930379418), 5.716909402476542), linewidth(0.4) + linetype("4 4")); draw(circle((-9.665565083909069,1.122758166353172), 6.392348795598595), linewidth(0.4) + linetype("4 4")); draw((-4.8127515399252045,5.283566240744204)--(-4.92061560412313,-3.160649070750248), linewidth(0.4)); draw((-3.3551493764184945,5.2649472350059785)--(-3.4630134406164252,-3.1792680764884764), linewidth(0.4)); draw((-4.883396507062627,-0.2469254722998166)--(-3.425794343555926,-0.26554447803803993), linewidth(0.4)); draw(circle((-8.199910707029291,-4.741214012502913), 6.5439899781189395), linewidth(0.4) + linetype("4 4")); draw((-6.218395061190416,0.5666524344894496)--(-1.3531342778462379,1.7826866420666359), linewidth(0.4) + ttffqq); /* dots and labels */ dot((-4.8127515399252045,5.283566240744204),dotstyle); label("$A$", (-4.704406413413728,5.591163455766509), NE * labelscalefactor); dot((-7.32,-3.13),dotstyle); label("$B$", (-7.209756689653024,-2.8304356655680385), NE * labelscalefactor); dot((3.64,-3.27),dotstyle); label("$C$", (3.7473776510079952,-2.9813603810041416), NE * labelscalefactor); dot((-3.425794343555926,-0.26554447803803993),dotstyle); label("$I$", (-3.3158990314015875,0.03713392771791844), NE * labelscalefactor); dot((-0.33709122877537645,-12.623211755897882),dotstyle); label("$I_A$", (-0.20684989341788235,-12.308507794955307), NE * labelscalefactor); dot((-1.88144278616565,-6.444378116967967),dotstyle); label("$M_A$", (-1.7462819908661247,-6.150779405162305), NE * labelscalefactor); dot((-1.7217647061562313,6.056134432340744),dotstyle); label("$M_{BC}$", (-1.5953572754300225,6.345787032947024), NE * labelscalefactor); dot((-4.873891246542604,-5.637753652618531),dotstyle); label("$T$", (-4.764776299588169,-5.335785941807348), NE * labelscalefactor); dot((-6.321988149529278,-11.009962827199013),dotstyle); label("$J$", (-6.21365356777475,-10.708705811332615), NE * labelscalefactor); dot((-14.51837862789293,-3.03804990803786),dotstyle); label("$K$", (-14.393773144411488,-2.739880836306377), NE * labelscalefactor); dot((-3.4630134406164252,-3.1792680764884764),dotstyle); label("$D$", (-3.346083974488808,-2.89080555174248), NE * labelscalefactor); dot((-1.3531342778462379,1.7826866420666359),dotstyle); label("$E$", (-1.2331379583833773,2.089710057648919), NE * labelscalefactor); dot((-6.218395061190416,0.5666524344894496),dotstyle); label("$F$", (-6.092913795425868,0.8823123341600952), NE * labelscalefactor); dot((-6.755455378061811,3.6178896713466084),dotstyle); label("$L$", (-6.636242770995835,3.930991585969376), NE * labelscalefactor); dot((1.209544047603321,-5.67180992537143),dotstyle); label("$A'$", (1.33258220403036,-5.365970884894569), NE * labelscalefactor); dot((-3.3551493764184945,5.2649472350059785),dotstyle); label("$G$", (-3.2253442021399263,5.560978512679289), NE * labelscalefactor); dot((-4.92061560412313,-3.160649070750248),dotstyle); label("$H_A$", (-4.794961242675389,-2.8606206086552595), NE * labelscalefactor); dot((-4.883396507062627,-0.2469254722998166),dotstyle); label("$G'$", (-4.764776299588169,0.06731887080513904), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]
'K, good! How about merging another configuration in this big picture? Also lemme' think about some lemmas I left out in these two configuration. Just to mention: $L$ is the miquel point of $BFEC$
Oops, looks like I am about to run outta post limit, so I'll continue this in Part (II) over here
This post has been edited 6 times. Last edited by AlastorMoody, Nov 17, 2019, 10:42 AM

Comment

J
U VIEW ATTACHMENTS T PREVIEW J CLOSE PREVIEW rREFRESH
J

7 Comments

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Waiting for part 2 :D

Edit ~(AlastorMoody): Sure! Once I get rid of this boring SST!!!
This post has been edited 2 times. Last edited by AlastorMoody, Nov 16, 2019, 10:26 AM

by aops29, Nov 16, 2019, 8:20 AM

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Link to My second blog: https://artofproblemsolving.com/community/c1008894

by AlastorMoody, Nov 16, 2019, 1:45 PM

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So super!, Cool!!

Edit ~(AlastorMoody): Thanks Ahir! :D
This post has been edited 1 time. Last edited by AlastorMoody, Nov 16, 2019, 5:11 PM

by GAUSSIANGAUSS, Nov 16, 2019, 4:19 PM

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Really nice demo to show that how a problem is proposed and also you merged some of the configurations which I just came accross a week before, TBH this was really helpful for pronoobs like me. :). But the $A-\text{antipode}$ point was not useful here (I'm a pronoob hence I maybe wrong so rectify me if I'm wrong), also notice that the point where $IL$ intersects $EF$ is actually the feet of perpendicular from $D$ onto $EF$ also you can relate some Midpoint Configuration this might be useful ( I'm a noob, hence don't take this suggestion if you think they are not appropriate here). Thanks, really liked this. :)

Edit~ (AlastorMoody): WOWOW!! Are you some psychologist?!! :P You just peaked in my mind. The last line I mentioned about merging other configurations is the configuration you're talking about!! yeah I'll be writing some lemmas on that too :) :D
This post has been edited 2 times. Last edited by AlastorMoody, Nov 16, 2019, 7:06 PM

by amar_04, Nov 16, 2019, 6:30 PM

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Great work man . I'll just read this and its amazing . I'll just leave somethings I think you missed

If $KI$ intersects $AB , AC$ at $X$,$Y$ then $X$ , $Y$ are the tangency points of the A-mixtillinear incircle with $AB , AC$ .This is due to radical axis on $\odot (IM_A), \odot (BIC), \odot (ABC)$

Also quadrilaterals $AFIE$ and $M_{BC}BM_AC$ are similar .. trivially true .. yet useful to know :P

Also I'll add a short proof of mine to the fact that $L,D,M_A$ are collinear ..
Let $P$ be on $EF$ such that $DP \perp EF $ . Let $DP$ intersect the incircle at $S$ . Firstly , note that $\Delta{SFE}$ and $\Delta{IBC}$ are similar . This is proved by angle-chasing .
$\angle{SFE} = \angle{SDE} = \angle{PDE} = 90 - \angle{FED} = \angle{B}/2 = \angle{IBC}$ .
Similarly , $\angle{SEF} = \angle{ICB}$ .
Let $f$ be the spiral similarity at $L$ that sends $F$$\mapsto$$B$ and $E$$\mapsto$$C$ . Due to the above two lines , $f(S)=I$ . Also $f(P)=D$(As $P$ and $D$ are the orthogonal projections of $S$ ,$I$ onto $FE$ and $BC$ respectively ).
Now note that $I$ is the midpoint of arc $FE$ in $\odot (AI)$ . So $f(I) = M$. Now note that $PD$ and $IM$ are parallel . So the spiral center($L$) carrying $PD \mapsto {IM}$ is also $MD \cap {IP}$ . This implies the conclusion . Also note $f(A) = M_{BC}$

by Aryan-23, Nov 18, 2019, 8:48 AM

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Ye, @aryan-23. I actually indeed forgot that $KI$ intersection part, so I have added that thing in Part (III). Also yeah, Thanks for the proof! Also If anybody else has any view or other proofs for any of the fact above or any of the fact that I have missed, they are welcomed to post this in here

Also, pls shoot me a PM if I made any typos or fakesolves :blush: Humans do make mistakes lol :P

Also, let me know some problems that are based on these configurations and I have missed out!
This post has been edited 1 time. Last edited by AlastorMoody, Nov 18, 2019, 9:36 AM

by AlastorMoody, Nov 18, 2019, 9:35 AM

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@above How can you miss SORY Mock P6. :P

by amar_04, Nov 19, 2019, 6:08 AM

I'll talk about all possible non-sense :D

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