Incenter-Related Configurations -Part (III)

by AlastorMoody, Nov 17, 2019, 12:41 PM

Here's the link to Part (II). We'll try to deal with problems with our developed lemmas in the configuration!

Quote:

Let $\Gamma$ be circumcircle of $\odot (ABC)$ . Let the Incircle be tangent to $BC, AC, AB$ at $D, E, F$ . Let $M$ denote the foot of the altitude from $D$ to $EF$ . Let $X$ be the second intersection of $\Gamma$ with the circle with diameter $AM$ . If $AX$ intersects $BC$ at $R$ . Prove, that, $RI \perp AI$ .

Solution: Let $\stackrel{\longrightarrow}{IM}$ $\cap$ $\odot (ABC)$ $=$ $R'$ . Then, from what we proved earlier, $AR'$ $\perp$ $R'I$ $\implies$ $R'$ $\in$ circle with diameter $AM$ $\implies$ $R'$ $\equiv$ $X$ . So, $RI$ is tangent to $\odot (BIC)$ [Proved Earlier] $\implies$ $RI$ $\perp$ $AI$ $\qquad \blacksquare$
$[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 = -18.03392976649076, xmax = 6.863423985529888, ymin = -8.188949997929846, ymax = 5.953563239898243; /* 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 ttffqq = rgb(0.2,1,0); draw((-3.96,4.81)--(-7.18,-4.03)--(4.54,-4.11)--cycle, linewidth(0.4) + rvwvcq); /* draw figures */ draw((-3.96,4.81)--(-7.18,-4.03), linewidth(0.4) + rvwvcq); draw((-7.18,-4.03)--(4.54,-4.11), linewidth(0.4) + rvwvcq); draw((4.54,-4.11)--(-3.96,4.81), linewidth(0.4) + rvwvcq); draw(circle((-1.3001726900206425,-1.1652990880241554), 6.54055659024936), linewidth(0.4)); draw(circle((-2.755367825051787,-0.9551106647405635), 3.1050192667475303), linewidth(0.4) + ubqqys); draw((-5.672865888390546,0.10759799584706942)--(-0.5075050499992787,1.1869111818815956), linewidth(0.4) + dtsfsf); draw((-3.96,4.81)--(-2.755367825051787,-0.9551106647405635), linewidth(0.4) + ttffqq); draw(circle((-3.8460683632385777,2.6615587519321715), 2.1514600192089053), linewidth(0.4)); draw((-2.755367825051787,-0.9551106647405635)--(-5.783619551041061,3.596799688215214), linewidth(0.4) + dtsfsf); draw((-3.7321367128921836,0.5131175045847532)--(-2.776562002065541,-4.060057597255526), linewidth(0.4) + dtsfsf); draw((-2.776562002065541,-4.060057597255526)--(-5.672865888390546,0.10759799584706942), linewidth(0.4) + dtsfsf); draw((-2.776562002065541,-4.060057597255526)--(-0.5075050499992787,1.1869111818815956), linewidth(0.4) + dtsfsf); draw((-17.145577039226463,-3.96197558334999)--(-3.96,4.81), linewidth(0.4)); draw((-17.145577039226463,-3.96197558334999)--(-7.18,-4.03), linewidth(0.4) + rvwvcq); draw((-17.145577039226463,-3.96197558334999)--(-2.755367825051787,-0.9551106647405635), linewidth(0.4) + ttffqq); /* dots and labels */ dot((-3.96,4.81),dotstyle); label("$A$", (-3.8710088616527853,5.014810557445007), NE * labelscalefactor); dot((-7.18,-4.03),dotstyle); label("$B$", (-7.095420249209558,-3.821709257821316), NE * labelscalefactor); dot((4.54,-4.11),dotstyle); label("$C$", (4.618580614446059,-3.903339925860728), NE * labelscalefactor); dot((-2.755367825051787,-0.9551106647405635),dotstyle); label("$I$", (-2.666956508071459,-0.7605592063433749), NE * labelscalefactor); dot((-2.776562002065541,-4.060057597255526),dotstyle); label("$D$", (-2.6873641750813118,-3.862524591841022), NE * labelscalefactor); dot((-0.5075050499992787,1.1869111818815956),dotstyle); label("$E$", (-0.42211313698763,1.382245829691184), NE * labelscalefactor); dot((-5.672865888390546,0.10759799584706942),dotstyle); label("$F$", (-5.585252890480437,0.321047145178831), NE * labelscalefactor); dot((-3.7321367128921836,0.5131175045847532),dotstyle); label("$M$", (-3.6465245245444025,0.7087928183660369), NE * labelscalefactor); dot((-5.783619551041061,3.596799688215214),dotstyle); label("$X$", (-5.707698892539555,3.810758203863684), NE * labelscalefactor); dot((-17.145577039226463,-3.96197558334999),dotstyle); label("$R$", (-17.05436175001782,-3.7604862567917574), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

EMMO Seniors 2016 wrote:

In $\triangle ABC$ , let point $D$ be the tangency point of incircle $\omega$ with side $BC$ . Analogously define $E$ and $F$ for sides $AC$ and $AB$ respectively. Let $P$ be the feet of the perpendicular from $D$ to $EF$ . Let $N$ be the midpoint of arc $\overarc{ABC}$ of circumcircle $\Gamma$ of $\triangle ABC$ . Prove that the lines $AP$ and $ND$ concur on $\Gamma$ .

Anant Mudgal, IMOTC-2016, Senior Batch

Lol! We proved this while proving the Sharky-Devil Lemma--here

Some other problems that were already discussed in the configuration,

ELMO SL 2019 G3 wrote:

Let $\triangle ABC$ be an acute triangle with incenter $I$ and circumcenter $O$ . The incircle touches sides $BC,CA,$ and $AB$ at $D,E,$ and $F$ respectively, and $A'$ is the reflection of $A$ over $O$ . The circumcircles of $ABC$ and $A'EF$ meet at $G$ , and the circumcircles of $AMG$ and $A'EF$ meet at a point $H\neq G$ , where $M$ is the midpoint of $EF$ . Prove that if $GH$ and $EF$ meet at $T$ , then $DT\perp EF$ .

Proposed by Ankit Bisain

Thailand MO 2019 P8 wrote:

Let $ABC$ be a triangle such that $AB\ne AC$ and $\omega$ be the circumcircle of this triangle.
Let $I$ be the center of the inscribed circle of $ABC$ which touches $BC$ at $D$ .
Let the circle with diameter $AI$ meets $\omega$ again at $K$ .
If the line $AI$ intersects $\omega$ again at $M$ , show that $K, D, M$ are collinear.

Belarus TST 2017 #3 P1 wrote:

Let $I$ be the incenter of a non-isosceles triangle $ABC$ . The line $AI$ intersects the circumcircle of the triangle $ABC$ at $A$ and $D$ . Let $M$ be the middle point of the arc $BAC$ . The line through the point $I$ perpendicular to $AD$ intersects $BC$ at $F$ . The line $MI$ intersects the circle $BIC$ at $N$ .
Prove that the line $FN$ is tangent to the circle $BIC$ .

St. Petersburg 2014 Grade 11 P7 wrote:

$M$ is midpoint of arc $BAC$ from circumcircle of triangle $ABC$ . $AL$ is bisector of angle $A$ and $I$ is incenter of
triangle $ABC$ . $MI$ meets circumcircle of triangle $ABC$ again at $K$ . $BC$ intersects with circumcircle of triangle
$AKL$ again at $P$ .Prove that $\angle AIP=90^\circ$ .

Let's Look into another cool configuration!

ELMO ShortList 2010 G4 wrote:

Let $ABC$ be a triangle with circumcircle $\omega$ , incenter $I$ , and $A$ -excenter $I_A$ . Let the incircle and the $A$ -excircle hit $BC$ at $D$ and $E$ , respectively, and let $M$ be the midpoint of arc $BC$ without $A$ . Consider the circle tangent to $BC$ at $D$ and arc $BAC$ at $T$ . If $TI$ intersects $\omega$ again at $S$ , prove that $SI_A$ and $ME$ meet on $\omega$ .

Amol Aggarwal.

AlastorMoody wrote:

Let $I_A$ , $I$ be the $A-$ excenter and Incenter respectively in an acute angled triangle $\Delta ABC$ with circumcenter $O$ . Let $AI \cap \odot (ABC)$ $=$ $M$ . Let $(I_A)$ denote the $A-$ excircle which touches $BC$ at $D$ . Let $MD$ $\cap$ $\odot (ABC)$ $=$ $K$ . Let $\odot (OKM)$ $\cap$ $AO$ $=$ $N$ . Let $AN$ $\cap$ $I_AK$ $=$ $F$ . Prove, $AK$ , $MF$ and $I_AN$ concur

Greece TST 2015 P3 wrote:

Let $ABC$ be an acute triangle with $\displaystyle{AB<AC<BC}$ inscribed in circle $\displaystyle{c(O,R)}$ .The excircle $\displaystyle{(c_A)}$ has center $\displaystyle{I}$ and touches the sides $\displaystyle{BC,AC,AB}$ of the triangle $ABC$ at $\displaystyle{D,E,Z}$ respectively. $\displaystyle{AI}$ cuts $\displaystyle{(c)}$ at point $M$ and the circumcircle $\displaystyle{(c_1)}$ of triangle $\displaystyle{AZE}$ cuts $\displaystyle{(c)}$ at $K$ .The circumcircle $\displaystyle{(c_2)}$ of the triangle $\displaystyle{OKM}$ cuts $\displaystyle{(c_1)}$ at point $N$ .Prove that the point of intersection of the lines $AN,KI$ lies on $\displaystyle{(c)}$ .

Let's try solving the ELMO problem, From the discussion earlier,
$[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 = -15.635599701583377, xmax = 18.76511711617264, ymin = -15.144724829248375, ymax = 4.3960102155916925; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); draw((-5.795507283662416,3.1551608514553613)--(-8.12,-3.49)--(3.38,-3.53)--cycle, linewidth(0.4) + rvwvcq); /* draw figures */ draw((-5.795507283662416,3.1551608514553613)--(-8.12,-3.49), linewidth(0.4) + rvwvcq); draw((-8.12,-3.49)--(3.38,-3.53), linewidth(0.4) + rvwvcq); draw((3.38,-3.53)--(-5.795507283662416,3.1551608514553613), linewidth(0.4) + rvwvcq); draw(circle((-2.3848542149786027,-7.780586806348393), 7.162473896515057), linewidth(0.4)); draw((-5.795507283662416,3.1551608514553613)--(-0.25232298855185054,-14.618228423644108), linewidth(0.4)); draw((-4.517385441405362,-0.9429451890526773)--(-4.526288240285144,-3.5024998669903122), linewidth(0.4)); draw(circle((-2.3639629156563373,-1.7743382511970436), 6.006284887735937), linewidth(0.4)); draw((-5.795507283662416,3.1551608514553613)--(1.06758145234974,-6.703837353849449), linewidth(0.4)); draw((-0.21371175971486162,-3.517500133009687)--(-0.25232298855185054,-14.618228423644108), linewidth(0.4)); draw((-7.1847009419052865,1.8083901832398288)--(-2.384854214978606,-7.7805868063483965), linewidth(0.4)); draw((-7.1847009419052865,1.8083901832398288)--(1.06758145234974,-6.703837353849449), linewidth(0.4)); draw((-12.633794584042043,-3.4742998449250715)--(-2.384854214978606,-7.7805868063483965), linewidth(0.4)); draw((-12.633794584042043,-3.4742998449250715)--(-4.517385441405362,-0.9429451890526773), linewidth(0.4)); draw((-12.633794584042043,-3.4742998449250715)--(-5.795507283662416,3.1551608514553613), linewidth(0.4)); draw((-12.633794584042043,-3.4742998449250715)--(-8.12,-3.49), linewidth(0.4) + rvwvcq); draw(circle((-8.575590012723701,-2.2086225169888727), 4.250995582345481), linewidth(0.4)); /* dots and labels */ dot((-5.795507283662416,3.1551608514553613),dotstyle); label("$A$", (-5.681949671527742,3.4373017141132474), NE * labelscalefactor); dot((-8.12,-3.49),dotstyle); label("$B$", (-7.994128998622819,-3.2172631785018404), NE * labelscalefactor); dot((3.38,-3.53),dotstyle); label("$C$", (3.482175710251525,-3.2454604873688533), NE * labelscalefactor); dot((-4.517385441405362,-0.9429451890526773),dotstyle); label("$I$", (-4.413070772512151,-0.6513080716036498), NE * labelscalefactor); dot((-4.526288240285144,-3.5024998669903122),dotstyle); label("$D$", (-4.413070772512151,-3.2172631785018404), NE * labelscalefactor); dot((-0.25232298855185054,-14.618228423644108),dotstyle); label("$I_A$", (-0.12707982472615564,-14.327002872104995), NE * labelscalefactor); dot((-0.21371175971486162,-3.517500133009687),dotstyle); label("$E$", (-0.09888251585914253,-3.2454604873688533), NE * labelscalefactor); dot((-2.384854214978606,-7.7805868063483965),dotstyle); label("$M$", (-2.2700752986191532,-7.503254126287829), NE * labelscalefactor); dot((1.06758145234974,-6.703837353849449),dotstyle); label("$S$", (1.1699963831564482,-6.431756389341332), NE * labelscalefactor); dot((-7.1847009419052865,1.8083901832398288),dotstyle); label("$T$", (-7.0636178060113854,2.0838308884966192), NE * labelscalefactor); dot((-3.7182321165428314,-3.505310496985938),dotstyle); label("$L$", (-3.5953488153687703,-3.2172631785018404), NE * labelscalefactor); dot((-12.633794584042043,-3.4742998449250715),dotstyle); label("$F$", (-12.533895726211933,-3.1890658696348275), NE * labelscalefactor); dot((-6.639267823040818,-5.993014121839311),dotstyle); label("$R$", (-6.527868937538136,-5.698626358798991), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
Here $R$ is $A-$ Mixtilinear Point and $S$ is $A-$ antipode. Only thing remaining is: $ME \cap I_AS$ $\in$ $\odot (ABC)$ . To prove this, First note from $\Psi _{\odot (BIC)}$ If, Let $ME \cap \odot (ABC)=T'$ , then $\Psi _{\odot (BIC)} (T')=E$ , $\Psi _{\odot (BIC)} (S)$ $=$ $MS$ $\cap$ $BC$ , Hence Since, $I_AME\Psi(S)$ is cyclic $\implies$ $T'$ $\in$ $I_AS$ $\qquad \square$

Cool! Let's try the Greece TST, Rename $I$ as $I_A$ and instead let $I$ be incenter. Let $A'$ be the $A-$ antipode in $\odot (ABC)$ .
$[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 = -14.844640701355988, xmax = 20.726680065924224, ymin = -14.337880150385272, ymax = 5.867796318241894; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); draw((-4.763226340596997,3.592620619458683)--(-6.36,-3.09)--(4.14,-3.23)--cycle, linewidth(0.4) + rvwvcq); /* draw figures */ draw((-4.763226340596997,3.592620619458683)--(-6.36,-3.09), linewidth(0.4) + rvwvcq); draw((-6.36,-3.09)--(4.14,-3.23), linewidth(0.4) + rvwvcq); draw((4.14,-3.23)--(-4.763226340596997,3.592620619458683), linewidth(0.4) + rvwvcq); draw(circle((-1.0787977677203862,-0.819832579028965), 5.748456935341297), linewidth(0.4)); draw((-4.763226340596997,3.592620619458683)--(0.9391122126469792,-12.466531270282518), linewidth(0.4)); draw(circle((-1.15543704811387,-6.567778608540223), 6.2595862139733836), linewidth(0.4)); draw(circle((-1.912057063975011,-4.436955325411917), 8.520754432466513), linewidth(0.4)); draw((-8.085229198084875,-10.31021693213839)--(-6.36,-3.09), linewidth(0.4) + rvwvcq); draw((-8.085229198084875,-10.31021693213839)--(0.9391122126469792,-12.466531270282518), linewidth(0.4) + rvwvcq); draw((0.9391122126469792,-12.466531270282518)--(6.5827142704640424,-5.1018734211126935), linewidth(0.4) + rvwvcq); draw((6.5827142704640424,-5.1018734211126935)--(4.14,-3.23), linewidth(0.4) + rvwvcq); draw(circle((3.3903407481024335,-3.7539050358648547), 5.3462117668096525), linewidth(0.4)); draw((4.164789203090494,1.535916307634432)--(0.9391122126469792,-12.466531270282518), linewidth(0.4) + linetype("4 4") + dtsfsf); draw((-4.763226340596997,3.592620619458683)--(5.480119558329307,-8.67475910603405), linewidth(0.4) + linetype("4 4") + dtsfsf); /* dots and labels */ dot((-4.763226340596997,3.592620619458683),dotstyle); label("$A$", (-4.639753595988713,3.885132537770542), NE * labelscalefactor); dot((-6.36,-3.09),dotstyle); label("$B$", (-6.243378712546428,-2.7917793111697486), NE * labelscalefactor); dot((4.14,-3.23),dotstyle); label("$C$", (4.25307659583134,-2.9375634126749954), NE * labelscalefactor); dot((-3.2499863088747176,-0.6690259467979268),dotstyle); label("$I$", (-3.123598940334147,-0.3717632261826566), NE * labelscalefactor); dot((0.9391122126469792,-12.466531270282518),dotstyle); label("$I_A$", (1.045826362715911,-12.180275448107624), NE * labelscalefactor); dot((6.5827142704640424,-5.1018734211126935),dotstyle); label("$E$", (6.7022495011194865,-4.80359991194215), NE * labelscalefactor); dot((-8.085229198084875,-10.31021693213839),dotstyle); label("$Z$", (-7.96363111030834,-10.022670745829975), NE * labelscalefactor); dot((-1.1554370481138698,-6.567778608540222),dotstyle); label("$M$", (-1.024307878658593,-6.290597747295665), NE * labelscalefactor); dot((2.6056308051562267,-5.23228577751661),dotstyle); label("$A'$", (2.707765119875724,-4.949384013447397), NE * labelscalefactor); dot((4.164789203090494,1.535916307634432),dotstyle); label("$K$", (4.28223341613239,1.8149982963960412), NE * labelscalefactor); dot((-1.0787977677203864,-0.8198325790289649),dotstyle); label("$O$", (-0.9659942380564944,-0.5175473276879031), NE * labelscalefactor); dot((5.480119558329307,-8.67475910603405),dotstyle); label("$N$", (5.594290329679611,-8.389888808971214), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
Turns out, Introducing Phantom Points finishes this problem with neat angle chase. Redefine $K,N$ as $I_AA'$ $\cap \odot (ABC)$ and $AA'$ $\cap$ $\odot (AZE)$ , this finishes off the problem. For more details, see here. Anyway, this was a cute configuration!

Another problem,

Korea Winter MOP 2018 Day 2 P1 wrote:

Let $\Delta ABC$ be a triangle with circumcenter $O$ and circumcircle $w$ . Let $S$ be the center of the circle which is tangent with $AB$ , $AC$ , and $w$ (in the inside), and let the circle meet $w$ at point $K$ . Let the circle with diameter $AS$ meet $w$ at $T$ . If $M$ is the midpoint of $BC$ , show that $K,T,M,O$ are concyclic.

Anyway the start is familiar, Let $I$ be the incenter WRT $\Delta ABC$ . Let $F,E$ $\in$ $AB, AC$ such $FE \perp AI$ at $I$ . Let $M$ be the midpoint of $\overline{BC}$ . Let $K$ be the $A-$ mixtilinear incircle touch point. Let $D,G$ be the midpoint of arc $BC$ and arc $BAC$ . Also, $FE$ is tangent to $\odot (BIC)$ . Let $EF$ $\cap$ $BC$ $=$ $L$ . By La Hire's Theorem, $KI$ is polar of $L$ WRT $\odot (BIC)$ $\implies$ $L$ $\in$ $KD$ . Let $LG$ $\cap$ $\odot (ABC)$ $=$ $T'$ and Let $T'D$ $\cap$ $BC$ $=$ $N$ . Since, $N$ is orthocenter WRT $\Delta LDG$ $\implies$ $N$ $\in$ $KG$ . Apply Pascal on $DAT'GKK$ $\implies$ $KK$ $\cap$ $AT$ $\in$ $EF$ $\implies$ By Radical Axes Theorem, $AT'FE$ is cyclic $\implies$ $T' \equiv T$ . Now Pascal on $TTGKKD$ $\implies$ $TBKC$ is harmonic, what follows next is reflection argument and nice angle chase.

Following whatever we have discussed, anyone can figure out that all these configurations do blend with mixtilinear configuration easily. Look at the below diagram, you'll know why, if you haven't figured out yet.
$[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 = -16.659301143352742, xmax = 18.482449503760346, ymin = -14.828963858869182, ymax = 5.132702697236808; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen ubqqys = rgb(0.29411764705882354,0,0.5098039215686274); draw((-3.16,3.47)--(-5.96,-4.53)--(5.1,-4.73)--cycle, linewidth(0.4) + rvwvcq); /* draw figures */ draw((-3.16,3.47)--(-5.96,-4.53), linewidth(0.4) + rvwvcq); draw((-5.96,-4.53)--(5.1,-4.73), linewidth(0.4) + rvwvcq); draw((5.1,-4.73)--(-3.16,3.47), linewidth(0.4) + rvwvcq); draw(circle((-0.38230008984725977,-1.9921949685534592), 6.1279026318434004), linewidth(0.4)); draw(circle((-0.49309395894018165,-8.119095929392039), 6.539776085781137), linewidth(0.4)); draw((-3.16,3.47)--(0.9735202751624851,-14.492298730220067), linewidth(0.4)); draw((-0.27150622075433767,4.134705992285119)--(-0.49309395894018127,-8.119095929392037), linewidth(0.4)); draw((-0.27150622075433767,4.134705992285119)--(-5.116996181789503,-12.743833704194408), linewidth(0.4) + dtsfsf); draw(circle((-2.3209451552846554,-0.17613036069431207), 3.741427486884803), linewidth(0.4)); draw((-13.468109192426345,-4.394229490191206)--(-5.116996181789503,-12.743833704194408), linewidth(0.4) + dtsfsf); draw((-13.468109192426345,-4.394229490191206)--(-0.27150622075433767,4.134705992285119), linewidth(0.4) + dtsfsf); draw((-13.468109192426345,-4.394229490191206)--(1.3312052758570734,-0.9885815087200975), linewidth(0.4) + dtsfsf); draw((-13.468109192426345,-4.394229490191206)--(-5.96,-4.53), linewidth(0.4) + rvwvcq); draw((-13.468109192426345,-4.394229490191206)--(-0.49309395894018127,-8.119095929392037), linewidth(0.4) + ubqqys); /* dots and labels */ dot((-3.16,3.47),dotstyle); label("$A$", (-3.034671589185127,3.7500764422684276), NE * labelscalefactor); dot((-5.96,-4.53),dotstyle); label("$B$", (-5.857533526412244,-4.228829237444932), NE * labelscalefactor); dot((5.1,-4.73),dotstyle); label("$C$", (5.203476513334826,-4.430462232961155), NE * labelscalefactor); dot((-1.959708193042848,-1.7458931285640056),dotstyle); label("$I$", (-1.8536783297329655,-1.4635767275081724), NE * labelscalefactor); dot((0.9735202751624851,-14.492298730220067),dotstyle); label("$I_A$", (1.08440246207485,-14.195260158675339), NE * labelscalefactor); dot((-0.49309395894018127,-8.119095929392037),dotstyle); label("$M_A$", (-0.38463793382905775,-7.829418443091756), NE * labelscalefactor); dot((-0.27150622075433767,4.134705992285119),dotstyle); label("$M_{BC}$", (-0.15420022466766048,4.412584856107443), NE * labelscalefactor); dot((-3.5383521874161747,-7.2448634163792045),dotstyle); label("$T$", (-3.4091328665723974,-6.965277033736518), NE * labelscalefactor); dot((-5.116996181789503,-12.743833704194408),dotstyle); label("$J$", (-4.993392117057004,-12.466977339964865), NE * labelscalefactor); dot((-5.250621661942771,-2.5032047484079136),dotstyle); label("$E$", (-5.137415685282877,-2.2124992822827116), NE * labelscalefactor); dot((1.3312052758570734,-0.9885815087200975),dotstyle); label("$F$", (1.4588637394621207,-0.7146541727336331), NE * labelscalefactor); dot((-6.0140385669906244,0.4233199642126803),dotstyle); label("$D$", (-5.886338240057419,0.7255815095250963), NE * labelscalefactor); dot((-13.468109192426345,-4.394229490191206),dotstyle); label("$K$", (-13.346759074157656,-4.113610382864234), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

ISL 2016 G2 wrote:

Let $ABC$ be a triangle with circumcircle $\Gamma$ and incenter $I$ and let $M$ be the midpoint of $\overline{BC}$ . The points $D$ , $E$ , $F$ are selected on sides $\overline{BC}$ , $\overline{CA}$ , $\overline{AB}$ such that $\overline{ID} \perp \overline{BC}$ , $\overline{IE}\perp \overline{AI}$ , and $\overline{IF}\perp \overline{AI}$ . Suppose that the circumcircle of $\triangle AEF$ intersects $\Gamma$ at a point $X$ other than $A$ . Prove that lines $XD$ and $AM$ meet on $\Gamma$ .

Proposed by Evan Chen, Taiwan

FBH TST 2016 Day 2 P2 wrote:

Let $k$ be a circumcircle of triangle $ABC$ $(AC<BC)$ . Also, let $CL$ be an angle bisector of angle $ACB$ $(L \in AB)$ , $M$ be a midpoint of arc $AB$ of circle $k$ containing the point $C$ , and let $I$ be an incenter of a triangle $ABC$ . Circle $k$ cuts line $MI$ at point $K$ and circle with diameter $CI$ at $H$ . If the circumcircle of triangle $CLK$ intersects $AB$ again at $T$ , prove that $T$ , $H$ and $C$ are collinear.
.

Also, This article is amazing!!

Iran TST 2012 Day 1 P2 wrote:

Consider $\omega$ is circumcircle of an acute triangle $ABC$ . $D$ is midpoint of arc $BAC$ and $I$ is incenter of triangle $ABC$ . Let $DI$ intersect $BC$ in $E$ and $\omega$ for second time in $F$ . Let $P$ be a point on line $AF$ such that $PE$ is parallel to $AI$ . Prove that $PE$ is bisector of angle $BPC$ .

Solution: We'll First Introduce some Points. Let $K,L$ be the points where the $A-$ Mixtilinear Incircle touch $AB,AC$ . Let $\odot (AKL)$ $\cap$ $\odot (ABC)$ $=$ $M$ . We'll First Focus on Proving, $MBFC$ is Harmonic Quadrilateral.

It's Well-Known, that $F$ is the $A-$ Mixtlinear Incircle Touch Point. Let $M_A$ be midpoint of arc $BC$ not containing $A$ . $KL$ is tangent to $\odot (BIC)$ . Applying Radical Axes on $\odot (IFM_A), \odot (BIC), \odot (ABC)$ $\implies$ $KL,$ $BC,$ $M_AF$ concur sat at $G$ . $DG$ $\cap$ $\odot (ABC)$ $=$ $T'$ . Let $T'M_A$ $\cap$ $BC$ $=$ $N$ . Since, $N$ is the orthocenter WRT $\Delta GM_AD$ $\implies$ $N$ $\in$ $FD$ $\implies$ $N \equiv E$ . Apply Pascal on $M_AAT'DFF$ $\implies$ $FF$ $\cap$ $AT'$ $\in$ $KL$ . Hence, By Power Of Point, $AT'KL$ is cyclic $\implies$ $T'$ $\equiv$ $M$ . Now, Pascal on $MMDFFM_A$ to get $MBFC$ is Harmonic.

Now, $PE||AM_A$ . Hence, By Converse of Reim's Theorem, $MPEF$ is cyclic. Since, $GMEF$ is cyclic $\implies$ $GP$ $\perp$ $PE$ $\implies$ $GP$ $\perp$ $AI$ . Hence, $P$ must lie on $KL$ .
$-1=(M,F;B,C) \overset{D}{=} (G,E;B,C)$ Hence, $PE$ bisects $\angle BPC$ $\qquad \blacksquare$

I guess I'm on the verge of exceeding post-wordlimit, so let's continue this in Part-(IV) over here.

This post has been edited 38 times. Last edited by AlastorMoody, Nov 19, 2019, 9:59 AM

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Wow, all this is really cool! Really helpful for olympiads too!

by L567, Dec 15, 2020, 3:34 PM

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Kukuku first shout of 2025
by HoRI_DA_GRe8, Yesterday at 6:41 PM
what a goat, u used to be friends with my brother
by bookstuffthanks, Jul 31, 2024, 12:05 PM
hello fellow moody!!
by crazyeyemoody907, Oct 31, 2023, 1:55 AM
@below I wish I started earlier / didn't have to do JEE and leave oly way before I could study conics and projective stuff which I really wanted to study . Huh, life really sucks when u are forced due to peer pressure to read sh_t u dont want to read
by kamatadu, Jan 3, 2023, 1:25 PM
Lots of good stuffs here.
by amar_04, Dec 30, 2022, 2:31 PM
But even if he went to jee he could continue with this.

Doing JEE(and completely leaving oly) seems like a insult to the oly math he knows
by HoRI_DA_GRe8, Feb 11, 2022, 2:11 PM
Ohhh did he go for JEE? Good for him, bad for us :sadge:. Hmmm so that is the reason why he is inactive
Btw @below finally everyone falls to the monopoly of JEE Coz IIT's are the best in India.
by BVKRB-, Feb 1, 2022, 12:57 PM
Kukuku first shout of 2022,why did this guy left this and went for trashy JEE
by Commander_Anta78, Jan 27, 2022, 3:42 PM
When are you going to br alive again ,we miss you
by HoRI_DA_GRe8, Aug 11, 2021, 5:10 PM
kukuku first shout o 2021
by leafwhisker, Mar 6, 2021, 5:10 AM
wow I completely forgot this blog
by Math-wiz, Dec 25, 2020, 6:49 PM
buuuuuujmmmmpppp
by DuoDuoling0, Dec 22, 2020, 10:54 PM
This site would work faster if not all diagrams were displayed on the initial page. Anyway I like your problem selection taste.
by WolfusA, Sep 24, 2020, 7:58 PM
nice blog
by Orestis_Lignos, Sep 15, 2020, 9:09 AM
Hello everyone, nice blog
by Functional_equation, Sep 12, 2020, 6:22 PM
Nice !
by Synthetic_Potato, Mar 10, 2019, 8:30 AM
Thanks dude!
by AlastorMoody, Mar 1, 2019, 1:39 PM
Amazing work bro!!
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@below Thanks a lot
by AlastorMoody, Feb 26, 2019, 2:48 PM
Nice Blog, loved it.
by Delta0001, Feb 26, 2019, 6:07 AM
@below Thanks a lot!
by AlastorMoody, Feb 16, 2019, 3:11 PM
Nice work.
by TheDarkPrince, Feb 16, 2019, 2:39 PM
@below Thanks buddy!
by AlastorMoody, Jan 29, 2019, 9:31 AM
Nice blog! 6th shout!
by PhysicsMonster_01, Jan 29, 2019, 9:21 AM
@below I am noob right now I hope to develop myself within a year or so
by AlastorMoody, Jan 28, 2019, 7:24 PM
Nice blog. But the properties are a bit 'basic'.
by ayan.nmath, Jan 28, 2019, 7:14 PM
Nice!!!!!!!!
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can I have contrib?
by Kagebaka, Dec 23, 2018, 8:24 PM
Math is fun!
by karolina.newgard, Nov 21, 2018, 6:37 PM

121 shouts

Magic of Hogwarts

Incenter-Related Configurations -Part (III)

by AlastorMoody, Nov 17, 2019, 12:41 PM

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