#AOT, ATheory which I believe should be true, posting here for future reference

by AlastorMoody, Sep 2, 2020, 1:13 PM

I read a theory, which I believe explains lots of stuffs. I will post here and let's see whether it turns out correct or not. I will delete this post if this turns out false.
Claim
Proof
if there are some contradictions or some instances which can't be explained using this, pls post below
my theory for beast titan power

Hello Peeps

by aops29, Jul 21, 2020, 7:45 PM

I am a contributor now! Along with SenatorPauline, of course. You can expect real good stuff from us soon! Some really good stuff ;)
This post has been edited 1 time. Last edited by aops29, Jul 21, 2020, 7:49 PM

Incenter-Related Configurations -Part (VI)

by AlastorMoody, Nov 19, 2019, 9:52 AM

Here's the link to Part (V). Now, Let's take a look at the below diagram!
[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); 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 = -22.781863434478517, xmax = 10.798848194270466, ymin = -9.038432588976118, ymax = 10.036512623091987; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen ttffqq = rgb(0.2,1,0); pen qqqqcc = rgb(0,0,0.8); pen ubqqys = rgb(0.29411764705882354,0,0.5098039215686274); draw((-2.3833838000602188,7.634178367214)--(-6.24,-3.97)--(7.32,-3.93)--cycle, linewidth(0.4) + rvwvcq); /* draw figures */ draw((-2.3833838000602188,7.634178367214)--(-6.24,-3.97), linewidth(0.4) + rvwvcq); draw((-6.24,-3.97)--(7.32,-3.93), linewidth(0.4) + rvwvcq); draw((7.32,-3.93)--(-2.3833838000602188,7.634178367214), linewidth(0.4) + rvwvcq); draw(circle((0.5276880979739716,0.22373478682363146), 7.961721776957119), linewidth(0.4)); draw((-4.5538811657489,1.1033655325293221)--(2.0402804374723105,2.362198681202362), linewidth(0.4) + ttffqq); draw((-0.893806415611854,-3.9542295174501834)--(2.0402804374723105,2.362198681202362), linewidth(0.4) + ttffqq); draw((-0.893806415611854,-3.9542295174501834)--(-4.5538811657489,1.1033655325293221), linewidth(0.4) + ttffqq); draw(circle((-0.9051484478953363,-0.10928057334941683), 3.844965672725561), linewidth(0.4) + qqqqcc); draw((-2.310996558686184,3.4694555322923124)--(-0.893806415611854,-3.9542295174501834), linewidth(0.4)); draw((-3.986231755908378,2.811366707739868)--(7.32,-3.93), linewidth(0.4) + red); draw((-6.24,-3.97)--(0.26280803581714984,4.480532689119575), linewidth(0.4) + red); draw(circle((-1.6442661239777778,3.7624488969322916), 3.9416473751650316), linewidth(0.4)); draw((-21.36248018711416,-4.014609086097727)--(-6.24,-3.97), linewidth(0.4) + ubqqys); draw((-21.36248018711416,-4.014609086097727)--(-0.9051484478953369,-0.10928057334941663), linewidth(0.4) + ubqqys); draw((-21.36248018711416,-4.014609086097727)--(0.26280803581714984,4.480532689119575), linewidth(0.4) + ubqqys); draw((-21.36248018711416,-4.014609086097727)--(-2.3833838000602188,7.634178367214), linewidth(0.4) + ubqqys); draw(circle((0.5511739007385885,-7.7379523503817556), 7.766434694557108), linewidth(0.4)); /* dots and labels */ dot((-2.3833838000602188,7.634178367214),dotstyle); label("$A$", (-2.2756092021686842,7.91707426619553), NE * labelscalefactor); dot((-6.24,-3.97),dotstyle); label("$B$", (-6.129133487434961,-3.6985489365356963), NE * labelscalefactor); dot((7.32,-3.93),dotstyle); label("$C$", (7.440777031395571,-3.643498589603321), NE * labelscalefactor); dot((-0.9051484478953369,-0.10928057334941663),dotstyle); label("$I$", (-0.7892498349945487,0.15497534873058744), NE * labelscalefactor); dot((-0.893806415611854,-3.9542295174501834),dotstyle); label("$D$", (-0.7892498349945487,-3.6710237630695084), NE * labelscalefactor); dot((2.0402804374723105,2.362198681202362),dotstyle); label("$E$", (2.1559437258875342,2.632240960687484), NE * labelscalefactor); dot((-4.5538811657489,1.1033655325293221),dotstyle); label("$F$", (-4.450097905997512,1.366082981242848), NE * labelscalefactor); dot((-1.9540534033919768,1.5996756876363232),dotstyle); label("$P$", (-1.835206426709681,1.8615361036342273), NE * labelscalefactor); dot((-2.310996558686184,3.4694555322923124),dotstyle); label("$Q$", (-2.193033681770121,3.7332478993349936), NE * labelscalefactor); dot((-3.986231755908378,2.811366707739868),dotstyle); label("$X$", (-3.87206926320757,3.100168909612676), NE * labelscalefactor); dot((0.26280803581714984,4.480532689119575),dotstyle); label("$Y$", (0.3668074505853344,4.75167931758394), NE * labelscalefactor); dot((-4.761848155218107,6.174349870984253),dotstyle); label("$K$", (-4.6427741202608255,6.45824007248758), NE * labelscalefactor); dot((-21.36248018711416,-4.014609086097727),dotstyle); label("$L$", (-21.240453720372003,-3.726074110001884), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]
Everything looks familiar except, those $X,Y$. Define: $CP,BP$ $\cap$ $AB,AC$ $=X,Y$. $$-1= ( \overline{PY} ~ \cap ~ (I), \overline{PB} ~ \cap ~ (I) ; F , D)$$Projecting this through $P$ onto $(I)$ $\implies$ $QY$ is tangent to $(I)$. Similarly, we can show for $X$ and conclude, $XY$ is tangent to $(I)$ at $Q$ [Credits, to Wizard_32 for this approach]
$$\angle BXY=180^{\circ}-\angle FIQ=180^{\circ}-2\angle FDP=2\angle DFE=\angle DIE=180^{\circ}-\angle YCB$$Hence, $BXYC$ is cyclic. Cool! Redefine: $\odot (AXY)$ $\cap$ $\odot (ABC)$ $=$ $K$ and $XY$ $\cap$ $BC$ $=$ $L$. By Radical Axes, $K$ $\in$ $AL$. Notice $K$ is the Miquel Point of $BXYC$ $\implies$ $K,P,I$ are collinear and $\overline{KPI}$ $\perp$ $AL$. But we already know, $\odot (AFE)$ & $IP$ concur on $\odot (ABC)$. Hence, $KAFE$ is cyclic.
[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 = -23.930559818900786, xmax = 11.986676614850097, ymin = -9.877842359161582, ymax = 10.524325549518245; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen ttffqq = rgb(0.2,1,0); pen qqqqcc = rgb(0,0,0.8); pen ubqqys = rgb(0.29411764705882354,0,0.5098039215686274); draw((-2.3833838000602188,7.634178367214)--(-6.24,-3.97)--(7.32,-3.93)--cycle, linewidth(0.4) + rvwvcq); /* draw figures */ draw((-2.3833838000602188,7.634178367214)--(-6.24,-3.97), linewidth(0.4) + rvwvcq); draw((-6.24,-3.97)--(7.32,-3.93), linewidth(0.4) + rvwvcq); draw((7.32,-3.93)--(-2.3833838000602188,7.634178367214), linewidth(0.4) + rvwvcq); draw(circle((0.5276880979739716,0.22373478682363146), 7.961721776957119), linewidth(0.4)); draw((-4.5538811657489,1.1033655325293221)--(2.0402804374723105,2.362198681202362), linewidth(0.4) + ttffqq); draw((-0.893806415611854,-3.9542295174501834)--(2.0402804374723105,2.362198681202362), linewidth(0.4) + ttffqq); draw((-0.893806415611854,-3.9542295174501834)--(-4.5538811657489,1.1033655325293221), linewidth(0.4) + ttffqq); draw(circle((-0.9051484478953363,-0.10928057334941683), 3.844965672725561), linewidth(0.4) + qqqqcc); draw((-2.310996558686184,3.4694555322923124)--(-0.893806415611854,-3.9542295174501834), linewidth(0.4)); draw((-3.986231755908378,2.811366707739868)--(7.32,-3.93), linewidth(0.4) + red); draw((-6.24,-3.97)--(0.26280803581714984,4.480532689119575), linewidth(0.4) + red); draw(circle((-1.6442661239777778,3.7624488969322916), 3.9416473751650316), linewidth(0.4)); draw((-21.36248018711416,-4.014609086097727)--(-6.24,-3.97), linewidth(0.4) + ubqqys); draw((-21.36248018711416,-4.014609086097727)--(-0.9051484478953369,-0.10928057334941663), linewidth(0.4) + ubqqys); draw((-21.36248018711416,-4.014609086097727)--(0.26280803581714984,4.480532689119575), linewidth(0.4) + ubqqys); draw((-21.36248018711416,-4.014609086097727)--(-2.3833838000602188,7.634178367214), linewidth(0.4) + ubqqys); draw(circle((0.5511739007385885,-7.7379523503817556), 7.766434694557108), linewidth(0.4)); draw(circle((0.5355948372904149,-2.4566498414501257), 6.942543777427655), linewidth(0.4)); draw(circle((-2.3754770607437763,4.9537937389402416), 2.6803962901057847), linewidth(0.4)); draw((-4.761848155218107,6.174349870984253)--(-0.9051484478953369,-0.10928057334941663), linewidth(0.4)); draw(circle((-11.13381431750469,-2.0619448297235485), 10.413380967305114), linewidth(0.4) + linetype("4 4")); draw(circle((-13.819843764765764,2.314339996688904), 9.846164770723794), linewidth(0.4) + linetype("4 4")); /* dots and labels */ dot((-2.3833838000602188,7.634178367214),dotstyle); label("$A$", (-2.262456527719926,7.933574069050966), NE * labelscalefactor); dot((-6.24,-3.97),dotstyle); label("$B$", (-6.119143390688259,-3.6659268775866276), NE * labelscalefactor); dot((7.32,-3.93),dotstyle); label("$C$", (7.42342116629978,-3.636486519854045), NE * labelscalefactor); dot((-0.9051484478953369,-0.10928057334941663),dotstyle); label("$I$", (-0.7904386410907912,0.19075998538170924), NE * labelscalefactor); dot((-0.893806415611854,-3.9542295174501834),dotstyle); label("$D$", (-0.7904386410907912,-3.6659268775866276), NE * labelscalefactor); dot((2.0402804374723105,2.362198681202362),dotstyle); label("$E$", (2.153597132167478,2.663750034918658), NE * labelscalefactor); dot((-4.5538811657489,1.1033655325293221),dotstyle); label("$F$", (-4.441042999931045,1.397814652417601), NE * labelscalefactor); dot((-1.9540534033919768,1.5996756876363232),dotstyle); label("$P$", (-1.8502915194637681,1.8983007338715072), NE * labelscalefactor); dot((-2.310996558686184,3.4694555322923124),dotstyle); label("$Q$", (-2.2035758122547606,3.753043271024219), NE * labelscalefactor); dot((-3.986231755908378,2.811366707739868),dotstyle); label("$X$", (-3.881676203011974,3.1053554009073987), NE * labelscalefactor); dot((0.26280803581714984,4.480532689119575),dotstyle); label("$Y$", (0.3871756682125165,4.783455791664614), NE * labelscalefactor); dot((-4.761848155218107,6.174349870984253),dotstyle); label("$K$", (-4.647125504059124,6.4615561824218295), NE * labelscalefactor); dot((-21.36248018711416,-4.014609086097727),dotstyle); label("$L$", (-21.251487265235763,-3.724807593051793), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]
HMMT November 2016 Team P10 wrote:
Let $ABC$ be a triangle with incenter $I$ whose incircle is tangent to $\overline{BC}$, $\overline{CA}$, $\overline{AB}$ at $D$, $E$, $F$. Point $P$ lies on $\overline{EF}$ such that $\overline{DP} \perp \overline{EF}$. Ray $BP$ meets $\overline{AC}$ at $Y$ and ray $CP$ meets $\overline{AB}$ at $Z$. Point $Q$ is selected on the circumcircle of $\triangle AYZ$ so that $\overline{AQ} \perp \overline{BC}$.

Prove that $P$, $I$, $Q$ are collinear.
Aryan-23's quickie approach
To solve the above problem, If $KI$ $\cap$ $\odot (AYZ)$ $=$ $R$, then we only need to show, $AR$ $\perp$ $BC$ and we're done! But this easily follows from the fact that $AR$ is the diameter of $\odot (AFE)$ and if $\odot (BXR)$ $\cap$ $BC$ $=R'$, then $\angle XRR'$ $=$ $180^{\circ}$ $-$ $\angle ABC$ $=$ $180^{\circ}$ $-$ $\angle AYX$ $=$ $180^{\circ}$ $-$ $\angle ARX$ $\qquad \square$
A bit unrelated though
MODS Advanced September 2019 P3 wrote:
$\Delta ABC$ is a triangle with incentre $I$. The feet of the altitudes from $I$ to $BC$, $AC$, $AB$ are $D$, $E$, $F$ respectively, and the line through $D$ parallel to $AI$ intersects $AB$ and $AC$ at $X$ and $Y$ respectively. Prove that the circles with diameters $XF$ and $YE$ have a common point on the circumcircle of $\Delta ABC$
Add 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 = -24.465390983050167, xmax = 10.043770835333204, ymin = -8.470029256893008, ymax = 11.132306103877182; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen ttffqq = rgb(0.2,1,0); pen qqqqcc = rgb(0,0,0.8); pen ubqqys = rgb(0.29411764705882354,0,0.5098039215686274); draw((-2.3391553000189,8.126768970246303)--(-6.24,-3.97)--(7.32,-3.93)--cycle, linewidth(0.4) + rvwvcq); /* draw figures */ draw((-2.3391553000189,8.126768970246303)--(-6.24,-3.97), linewidth(0.4) + rvwvcq); draw((-6.24,-3.97)--(7.32,-3.93), linewidth(0.4) + rvwvcq); draw((7.32,-3.93)--(-2.3391553000189,8.126768970246303), linewidth(0.4) + rvwvcq); draw(circle((0.5267986791578804,0.5252477654785609), 8.123842492151903), linewidth(0.4)); draw((-4.579409621102252,1.179597000833472)--(2.2247310819420094,2.4300261387433983), linewidth(0.4) + ttffqq); draw((-0.8293018073848326,-3.9540392383698673)--(2.2247310819420094,2.4300261387433983), linewidth(0.4) + ttffqq); draw((-0.8293018073848326,-3.9540392383698673)--(-4.579409621102252,1.179597000833472), linewidth(0.4) + ttffqq); draw(circle((-0.8408890511355429,-0.025963606878789763), 3.9280927217839157), linewidth(0.4) + qqqqcc); draw((-2.226653604971686,3.649572956980028)--(-0.8293018073848326,-3.9540392383698673), linewidth(0.4)); draw((-3.9983128158337555,2.9816153694999445)--(7.32,-3.93), linewidth(0.4) + red); draw((-6.24,-3.97)--(0.44172394912175833,4.655614650885135), linewidth(0.4) + red); draw(circle((-1.590022175577222,4.050402681683757), 4.144630569383107), linewidth(0.4)); draw((-22.564170777099925,-4.018153896097674)--(-6.24,-3.97), linewidth(0.4) + ubqqys); draw((-22.564170777099925,-4.018153896097674)--(-0.8408890511355438,-0.025963606878788865), linewidth(0.4) + ubqqys); draw((-22.564170777099925,-4.018153896097674)--(0.44172394912175833,4.655614650885135), linewidth(0.4) + ubqqys); draw((-22.564170777099925,-4.018153896097674)--(-2.3391553000189,8.126768970246303), linewidth(0.4) + ubqqys); draw(circle((0.5507627120400265,-7.598559381568989), 7.6994091589423626), linewidth(0.4)); draw(circle((0.5351845057446993,-2.3175474474505875), 6.973788391205323), linewidth(0.4)); draw(circle((-2.3307694734320914,5.283973757317162), 2.8428075813780302), linewidth(0.4)); draw((-4.8361407088992365,6.6273537918696475)--(-0.8408890511355438,-0.025963606878788865), linewidth(0.4)); draw(circle((-11.7025299141177,-2.0220587514882142), 11.043533766994019), linewidth(0.4) + linetype("4 4")); draw(circle((-14.421258052415205,2.50545610226356), 10.4338159391735), linewidth(0.4) + linetype("4 4")); draw((-3.26088654950535,9.277294081659807)--(-2.226653604971686,3.649572956980028), linewidth(0.4)); draw((-3.26088654950535,9.277294081659807)--(-2.3391553000189,8.126768970246303), linewidth(0.4) + rvwvcq); draw(circle((-3.869012743120453,7.845619827806222), 1.5554769803851212), linewidth(0.4) + ubqqys); draw(circle((-3.685505490860074,3.951650896017007), 2.9126186485474266), linewidth(0.4) + ubqqys); draw((-2.3391553000189,8.126768970246303)--(-0.8408890511355438,-0.025963606878788865), linewidth(0.4) + ttffqq); /* dots and labels */ dot((-2.3391553000189,8.126768970246303),dotstyle); label("$A$", (-2.232439188583503,8.416831075545382), NE * labelscalefactor); dot((-6.24,-3.97),dotstyle); label("$B$", (-6.1359345418104745,-3.6896617591005665), NE * labelscalefactor); dot((7.32,-3.93),dotstyle); label("$C$", (7.441440599848557,-3.633089362676987), NE * labelscalefactor); dot((-0.8408890511355438,-0.025963606878788865),dotstyle); label("$I$", (-0.7332706833586516,0.2704059905499773), NE * labelscalefactor); dot((-0.8293018073848326,-3.9540392383698673),dotstyle); label("$D$", (-0.704984485146862,-3.661375560888777), NE * labelscalefactor); dot((2.2247310819420094,2.4300261387433983),dotstyle); label("$E$", (2.3499249217264198,2.703019036763883), NE * labelscalefactor); dot((-4.579409621102252,1.179597000833472),dotstyle); label("$F$", (-4.467048847314885,1.4584263154451405), NE * labelscalefactor); dot((-1.864427924622838,1.678542035972305),dotstyle); label("$P$", (-1.751573818983079,1.9675778832573532), NE * labelscalefactor); dot((-2.226653604971686,3.649572956980028),dotstyle); label("$Q$", (-2.119294395736344,3.9193255598708356), NE * labelscalefactor); dot((-3.9983128158337555,2.9816153694999445),dotstyle); label("$X$", (-3.8730386848673026,3.2687430009996747), NE * labelscalefactor); dot((0.44172394912175833,4.655614650885135),dotstyle); label("$Y$", (0.5678944343836722,4.937628695495261), NE * labelscalefactor); dot((-4.8361407088992365,6.6273537918696475),dotstyle); label("$K$", (-4.721624631220992,6.917662570320533), NE * labelscalefactor); dot((-22.564170777099925,-4.018153896097674),dotstyle); label("$L$", (-22.4570709100131,-3.746234155524146), NE * labelscalefactor); 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We just let those $X,Y$ in the question be $M,N$. We'll observe some cyclic quadrilateral using which the conclusion would follow.

By Converse of Reim's Theorem on $\odot (AKFI)$ $\implies$ $KMPF$ is cyclic with diameter $FM$. Notice: $\angle NAK$ $=$ $\angle KXY$ $=$ $\angle KFE$ $=$ $\angle KMN$ $\implies$ $ANKM$ is cyclic. From $\angle KNP$ $=$ $\angle KAF$ $=$ $\angle KEP$ $\implies$ $NKPE$ is cyclic with diameter $NE$. This solves the MODS problem $\qquad \square$
Another Similar Problem
SORY P6 wrote:
Let $\Delta ABC$ be a triangle with incenter $I$. Let the incircle be tangent to the sides $BC, CA, AB$ at $D, E, F$ respectively. Let $P$ be the foot of the perpendicular from $D$ onto $EF$. Assume that $BP$, $CP$ intersect the sides $AC$, $AB$ in $Y$, $Z$ respectively. Finally, let the rays $IP$, $YZ$ meet the circumcircle of $\Delta ABC$ in $R$, $X$ respectively. Prove that the tangent from $X$ to the incircle and the line $RD$ meet on the circumcircle of $\Delta ABC$

Proposed by Wizard_32
Though we won't solve it here, 'coz I hate Poncelet's Theorem (which is the only main step left to prove). Anyway here's Aryan-23's proof on the Contest
Aryan-23 wrote:
my solution to SORY p6 avoids poncelet ... but is a bit wordy
The solution is divided into the following claims ...
Notation :- Let $DP$ intersect the incircle again at $S$ . Let the tangency point of the A-mixtilinear incircle with $(ABC)$ be $T$ .Let $M$ be the midoint of arc $BC$ not containing $A$ of $\odot (ABC)$

Claim :- $R= \odot (AI) \cap \odot (ABC)$
Proof

Claim: $YZ$ is tangent to the incircle at $S$
Proof

Claim : $AS$ passes through $T$ .
Proof

Claim :- $R,D,M $ are collinear
Proof

A Lame Lemma

We now show that , if $\odot (A,AI)$ meets the circumcircle of $\Delta{ABC}$ at $X_1 , X_2$ , then $X_1X_2$ is tangent to the incircle at $S$. This combined with the "Lame Lemma " implies the conclusion , as then $X_1X_2$ is just $YZ$ and $MX_2$ , $MX_1$ are tangents to the incircle . As $I$ is the incenter of $\Delta MX_1X_2 $ , therefore
$\Delta MX_1X_2 $ and $\Delta ABC $ would share the same incircle

Endgame
This post has been edited 5 times. Last edited by AlastorMoody, Aug 25, 2020, 2:44 PM

Incenter-Related Configurations -Part (V)

by AlastorMoody, Nov 18, 2019, 2:40 AM

Here's the link to Part (IV). We'll dive deeper as we take down some Monsters in the Incenter Configuration! :diablo: ,
Thanks to Aryan-23 for notifying me this problem
USAJMO 2014 P6 wrote:
Let $ABC$ be a triangle with incenter $I$, incircle $\gamma$ and circumcircle $\Gamma$. Let $M,N,P$ be the midpoints of sides $\overline{BC}$, $\overline{CA}$, $\overline{AB}$ and let $E,F$ be the tangency points of $\gamma$ with $\overline{CA}$ and $\overline{AB}$, respectively. Let $U,V$ be the intersections of line $EF$ with line $MN$ and line $MP$, respectively, and let $X$ be the midpoint of arc $BAC$ of $\Gamma$.

(a) Prove that $I$ lies on ray $CV$.

(b) Prove that line $XI$ bisects $\overline{UV}$.
Part (a) trivially follows from Right-Angles on Incircle Chord Lemma. Also, $M$ is the center of $BVUC$. By Reim's Theorem, there exists a Homothety at $I$ taking $UV$ to $M_BM_C$, where $BI$ $\cap$ $\odot (ABC)$ $=$ $M_B$ and $CI$ $\cap$ $\odot (ABC)$ $=$ $M_C$. Let $IX$ $\cap$ $M_BM_C$ $=$ $J$, then, since, $I$ is orthocenter WRT $\Delta M_AM_BM_C$ and since, $XM_CIM_B$ is parallelogram, it follows $J$ is midpoint of $M_BM_C$. Hence, $IX$ $\cap$ $EF$ must be the midpoint of $UV$ which ofcourse lies on perpendicular from $M$ to $EF$ $\qquad \square$
Aryan-23's Approach Using Lengths
This above problem, especially part (b), will heavily reduce our burden in below problems!
AlastorMoody wrote:
Let $AT$ be the isogonal conjugate of the $A-$Nagel cevian, where $T \in \odot (ABC)$. Let $(I)$ be the incircle. Let $(I)$ touch $BC,$ $CA,$ and $AB$ at $D,$ $E,$ $F$. Let $D'$ be the first intersection of ray $TA$ with $(I)$ and Let $AT$ intersect $EF$ at $L$. Let $EF$ intersect $ID$ at $J$, prove, $JLD'D$ cyclic
STEMS 2019 Cat B P6 wrote:
In triangle $ABC$, with circumcircle $\Gamma$, the incircle $\omega$ has center $I$ and touches sides $BC, CA, AB$ at points $D, E, F$ respectively. Point $Q$ lies on $EF$ and point $R$ lies on $\omega$, such that and $DQ \perp EF$ and $D, Q, R$ are collinear. Ray $AR$ meets $\omega$ again at $P$ and $\Gamma$ again at $S$. Ray $AQ$ meets $BC$ at $T$. Let $M$ be the midpoint of $BC$ and $O$ be the circumcenter of triangle $MPD$. Prove that $O, T, I, S$ are collinear.
Mathematical Reflections O451 wrote:
Let $ABC$ be a triangle, $\Gamma$ its circumcircle, $\omega$ its incircle and $I$ the incenter. Let $M$ be the midpoint of $BC$. The incircle $\omega$ is tangent to $AB$ and $AC$ at $F$ and $E$; respectively. Suppose $EF$ meets $\Gamma$ at distinct points $P$ and $Q$. Let $J$ denote the point on $EF$ such that $MJ$ is perpendicular on $EF$. Show that $IJ$ and the radical axis of $(MPQ)$ and $(AJI)$ intersect on $\Gamma$.
Taiwan TST R2 2019 Day 1 P2 wrote:
Let $ABC$ be a scalene triangle and $I$ be its incenter, $\Omega$ be its circumcircle. Let $M$ be the midpoint of $BC$. The incircle $\omega$ touches $CA,AB$ at $E,F$, respectively. Suppose the line $EF$ intersects $\Omega$ at two points $P,Q$, and $R$ be the point on the circumcircle $\Gamma$ of triangle $MPQ$ such that $MR$ is perpendicular to $PQ$. Prove that $AR,\Gamma,\omega$ intersects at a point.
Let's try the first result, which is indeed a crucial step in proving the Taiwan TST problem.
[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 = -9.8030712381575, xmax = 12.410272150317091, ymin = -9.35947015194124, ymax = 3.258437199052897; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen ubqqys = rgb(0.29411764705882354,0,0.5098039215686274); pen qqqqcc = rgb(0,0,0.8); pen wwccqq = rgb(0.4,0.8,0); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); draw((-3.3647375827835124,1.7629912237603584)--(-5.96,-4.53)--(5.1,-4.73)--cycle, linewidth(0.4) + rvwvcq); /* draw figures */ draw((-3.3647375827835124,1.7629912237603584)--(-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)--(-3.3647375827835124,1.7629912237603584), linewidth(0.4) + rvwvcq); draw(circle((-0.4030572637793516,-3.1400666869981144), 5.728134704092408), linewidth(0.4)); draw((-0.2994912894680226,2.587131692418541)--(-0.5066232380906821,-8.86726506641477), linewidth(0.4)); draw((-0.2994912894680226,2.587131692418541)--(-4.466063214608107,-7.177827184075416), linewidth(0.4)); draw((-3.3647375827835124,1.7629912237603584)--(-0.5066232380906821,-8.86726506641477), linewidth(0.4)); draw((-3.3647375827835124,1.7629912237603584)--(-4.466063214608107,-7.177827184075416), linewidth(0.4) + ubqqys); draw(circle((-2.3157953254666253,-2.138365994357423), 2.4571311144143166), linewidth(0.4) + qqqqcc); draw((-4.587338065571309,-1.2015699155330901)--(-0.8203120282457429,-0.18874474366376615), linewidth(0.4) + wwccqq); draw((-0.8203120282457429,-0.18874474366376615)--(-2.3602208131726323,-4.595095464499592), linewidth(0.4) + wwccqq); draw((-2.3602208131726323,-4.595095464499592)--(-4.587338065571309,-1.2015699155330901), linewidth(0.4) + wwccqq); draw((-2.287674213911024,-0.5832685253326907)--(-2.3602208131726323,-4.595095464499592), linewidth(0.4) + wwccqq); draw(circle((-2.506741820946712,-2.5858764920697612), 2.014554363846041), linewidth(0.4) + linetype("4 4") + dtsfsf); /* dots and labels */ dot((-3.3647375827835124,1.7629912237603584),dotstyle); label("$A$", (-3.284729489736267,1.9474857859625971), NE * labelscalefactor); dot((-5.96,-4.53),dotstyle); label("$B$", (-5.888424657401731,-4.3523640602769), NE * labelscalefactor); dot((5.1,-4.73),dotstyle); label("$C$", (5.1818317198052775,-4.5526483039434735), NE * labelscalefactor); dot((-2.3157953254666253,-2.138365994357423),dotstyle); label("$I$", (-2.2468929543731098,-1.9489531362780166), NE * labelscalefactor); dot((-0.5066232380906821,-8.86726506641477),dotstyle); label("$M_A$", (-0.4261271028587992,-8.685786786880946), NE * labelscalefactor); dot((-0.2994912894680226,2.587131692418541),dotstyle); label("$M_{BC}$", (-0.225842859192225,2.7668304191440347), NE * labelscalefactor); dot((-4.466063214608107,-7.177827184075416),dotstyle); label("$T$", (-4.395396659159997,-6.992474544972643), NE * labelscalefactor); dot((-2.3602208131726323,-4.595095464499592),dotstyle); label("$D$", (-2.283308271403396,-4.406987035822329), NE * labelscalefactor); dot((-0.8203120282457429,-0.18874474366376615),dotstyle); label("$E$", (-0.753864956131375,-0.0007336751577098196), NE * labelscalefactor); dot((-4.587338065571309,-1.2015699155330901),dotstyle); label("$F$", (-4.522850268765998,-1.020362552005721), NE * labelscalefactor); dot((-4.058655585491474,-3.87039700662281),dotstyle); label("$D'$", (-3.994828171826848,-3.69688835373175), NE * labelscalefactor); dot((-3.700541138956399,-0.9631403849929725),dotstyle); label("$L$", (-3.630675001523986,-0.7836629913088612), NE * labelscalefactor); dot((-2.287674213911024,-0.5832685253326907),dotstyle); label("$J$", (-2.2104776373428234,-0.40130216249085704), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]
Recall, we already earlier proved that, $D'L \cap (I)$ is reflection of orthocenter WRT $\Delta DEF$ over $EF$ and $DI$ $\cap$ $(I)$ is the $D-$antipode. Hence, by Converse of Reim's theorem, we get $DD'LJ$ is cyclic $\qquad \square$
Let's take a look at the STEMS problem! This problem is tough but beautiful tough! Luckily, since we have already done the basics on the configuration earlier, we won't have any much difficulties,
[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 = -11.515515234203619, xmax = 13.331794051704854, ymin = -8.868184822118293, ymax = 5.2459015181560025; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen qqqqcc = rgb(0,0,0.8); pen ttffqq = rgb(0.2,1,0); pen ubqqys = rgb(0.29411764705882354,0,0.5098039215686274); pen ffqqff = rgb(1,0,1); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); draw((-4.620593695741067,2.849901959146017)--(-5.96,-4.53)--(5.1,-4.73)--cycle, linewidth(0.4) + rvwvcq); draw((-4.562293637057545,-3.9115788828959563)--(-2.8426691930227572,-4.58637126233232)--(-0.43,-4.63)--cycle, linewidth(0.4) + ffqqff); /* draw figures */ draw((-4.620593695741067,2.849901959146017)--(-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)--(-4.620593695741067,2.849901959146017), linewidth(0.4) + rvwvcq); draw(circle((-0.3775890583776725,-1.731674928285277), 6.244512401151261), linewidth(0.4)); draw((-0.2646868582142609,4.511816740751381)--(-0.490491258541085,-7.975166597321936), linewidth(0.4)); draw((-4.620593695741067,2.849901959146017)--(-0.490491258541085,-7.975166597321936), linewidth(0.4)); draw(circle((-2.7947366275701775,-1.9357003928046626), 2.651104220771672), linewidth(0.4) + qqqqcc); draw((-2.8426691930227572,-4.58637126233232)--(-5.403227101238288,-1.4622755960564502), linewidth(0.4) + ttffqq); draw((-5.403227101238288,-1.4622755960564502)--(-1.164512226750342,0.15492678399300508), linewidth(0.4) + ttffqq); draw((-1.164512226750342,0.15492678399300508)--(-2.8426691930227572,-4.58637126233232), linewidth(0.4) + ttffqq); draw((-4.54096052412599,-6.385734786766322)--(-0.2646868582142609,4.511816740751381), linewidth(0.4)); draw((-4.620593695741067,2.849901959146017)--(-4.54096052412599,-6.385734786766322), linewidth(0.4)); draw((-4.596101855188915,0.009405492775725544)--(-2.8426691930227572,-4.58637126233232), linewidth(0.4)); draw((-3.8278869149203847,-4.568555390326938)--(-4.562293637057545,-3.9115788828959563), linewidth(0.4) + rvwvcq); draw((-4.620593695741067,2.849901959146017)--(-3.8278869149203847,-4.568555390326938), linewidth(0.4) + ubqqys); draw((-4.562293637057545,-3.9115788828959563)--(-2.8426691930227572,-4.58637126233232), linewidth(0.4) + ffqqff); draw((-2.8426691930227572,-4.58637126233232)--(-0.43,-4.63), linewidth(0.4) + ffqqff); draw((-0.43,-4.63)--(-4.562293637057545,-3.9115788828959563), linewidth(0.4) + ffqqff); draw((-3.5567082582136305,-7.106350382800299)--(-0.2646868582142609,4.511816740751381), linewidth(0.4) + dtsfsf); draw((-3.8278869149203847,-4.568555390326938)--(-3.5567082582136305,-7.106350382800299), linewidth(0.4) + ubqqys); draw(circle((-3.7076651616556227,0.4571007831706772), 2.561041970779934), linewidth(0.4)); draw((-4.54096052412599,-6.385734786766322)--(-2.8426691930227572,-4.58637126233232), linewidth(0.4)); draw(circle((-3.357415583887273,-5.8016687306679335), 1.3198150565329196), linewidth(0.4)); /* dots and labels */ dot((-4.620593695741067,2.849901959146017),dotstyle); label("$A$", (-4.529755328411319,3.046303646944424), NE * labelscalefactor); dot((-5.96,-4.53),dotstyle); label("$B$", (-5.873954027485056,-4.326422551005496), NE * labelscalefactor); dot((5.1,-4.73),dotstyle); label("$C$", (5.1851352694397805,-4.530089020562124), NE * labelscalefactor); dot((-2.7947366275701775,-1.9357003928046626),dotstyle); label("$I$", (-2.7171237493573406,-1.7398583876363254), NE * labelscalefactor); dot((-0.490491258541085,-7.975166597321936),dotstyle); label("$M_A$", (-0.41569264336745726,-7.768385886512503), NE * labelscalefactor); dot((-0.2646868582142609,4.511816740751381),dotstyle); label("$M_{BC}$", (-0.19165952685516777,4.716368697308771), NE * labelscalefactor); dot((-4.54096052412599,-6.385734786766322),dotstyle); label("$S$", (-4.468655387544331,-6.179787423970808), NE * labelscalefactor); dot((-2.8426691930227572,-4.58637126233232),dotstyle); label("$D$", (-2.757857043268666,-4.387522491872485), NE * labelscalefactor); dot((-1.164512226750342,0.15492678399300508),dotstyle); label("$E$", (-1.0877919929043258,0.3579062487969392), NE * labelscalefactor); dot((-5.403227101238288,-1.4622755960564502),dotstyle); label("$F$", (-5.324054559682164,-1.2510588607004192), NE * labelscalefactor); dot((-4.208518560529974,-1.0064568980053574),dotstyle); label("$Q$", (-4.122422389298066,-0.8029926276758382), NE * labelscalefactor); dot((-4.596101855188915,0.009405492775725544),dotstyle); label("$R$", (-4.509388681455657,0.21533972010729985), NE * labelscalefactor); dot((-4.562293637057545,-3.9115788828959563),dotstyle); label("$P$", (-4.489022034499994,-3.7154231423356134), NE * labelscalefactor); dot((-3.8278869149203847,-4.568555390326938),dotstyle); label("$T$", (-3.7558227440961374,-4.367155844916822), NE * labelscalefactor); dot((-0.43,-4.63),dotstyle); label("$M$", (-0.3545927025004692,-4.4282557857838105), NE * labelscalefactor); dot((-3.5567082582136305,-7.106350382800299),dotstyle); label("$V$", (-3.47068968671686,-6.892620067419005), NE * labelscalefactor); dot((-6.266299237125552,0.3460708595441685),dotstyle); label("$L$", (-6.179453731819996,0.5412060713979041), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]
Highly refer to Part (I) & (II). We already know, $S,I, M_{BC}$ are collinear. Let $AT$ $\cap$ $\odot (ABC)$ $=$ $V$. We want to show firstly, $AV$ & $SM_{BC}$ concur on $BC$. But this is obvious, from Pascal on $AVM_ASM_{BC}L$, where $L$ $= \odot (AEF)$ $\cap$ $\odot (ABC)$. Then, from Shooting Lemma, $STDV$ is cyclic. Also, $\angle TSD$ $=$ $\angle TVD$ $=$ $\angle TSP$ $\implies$ Line $SD$ is reflection of Line $AS$ over $SM_{BC}$. Since, $IP=ID$ and also from Reflection argument, $TP$ is tangent to $(I)$ $\implies$ $IT$ is perpendicular bisector of $PD$ $\implies$ $SP=SD$. Moreover, since, $O$ the circumcenter WRT $\Delta MPD$ must lie on perpendicular bisector of $PD$ $\implies$ $O,T,I,S$ are collinear $\qquad \square$
Let's take a look at the Mathematical Reflections Problem which was proposed by AoPS User: tworigami Apparently, this problem is beautiful, tough and cleverly stated. Anyway,
[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(19cm); 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 = -15.017491981658836, xmax = 11.131477483886478, ymin = -8.250811492972314, ymax = 6.602660506718562; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen qqqqcc = rgb(0,0,0.8); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); draw((-4.618108299402482,3.6910718535547)--(-5.96,-4.53)--(5.1,-4.73)--cycle, linewidth(0.4) + rvwvcq); /* draw figures */ draw((-4.618108299402482,3.6910718535547)--(-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)--(-4.618108299402482,3.6910718535547), linewidth(0.4) + rvwvcq); draw(circle((-0.3683841730197502,-1.2226447679921784), 6.496519544127112), linewidth(0.4)); draw((-0.2509256265684152,5.272812850766571)--(-0.48584271947108304,-7.718102386750928), linewidth(0.4)); draw((-4.618108299402482,3.6910718535547)--(-0.48584271947108304,-7.718102386750928), linewidth(0.4)); draw(circle((-2.6431144226071637,-1.7618809705153757), 2.827636594269825), linewidth(0.4) + qqqqcc); draw((-6.837851891540301,-1.8148893342365993)--(4.979844423017278,2.465338023691826), linewidth(0.4)); draw(circle((-1.44702002458779,1.7554659401255956), 6.465949659788316), linewidth(0.4)); draw((-2.0744613450109326,-0.08964661424579373)--(-0.43,-4.63), linewidth(0.4)); draw(circle((-5.7066256861518205,0.21268804151315282), 3.6447255102583913), linewidth(0.4)); draw((-4.235015582748999,-6.4431742170539055)--(-0.2509256265684152,5.272812850766571), linewidth(0.4)); draw((-4.235015582748999,-6.4431742170539055)--(-0.48584271947108304,-7.718102386750928), linewidth(0.4)); draw((-13.93604048454686,-4.385767803172747)--(-5.96,-4.53), linewidth(0.4) + rvwvcq); draw((-13.93604048454686,-4.385767803172747)--(-6.837851891540301,-1.8148893342365993), linewidth(0.4)); draw((-3.2969221131279314,-7.021647416773472)--(-0.2509256265684152,5.272812850766571), linewidth(0.4)); draw((-13.93604048454686,-4.385767803172747)--(-0.48584271947108304,-7.718102386750928), linewidth(0.4)); draw((-13.93604048454686,-4.385767803172747)--(-0.2509256265684152,5.272812850766571), linewidth(0.4)); draw((-7.740992957065166,3.2368199586366244)--(-4.235015582748999,-6.4431742170539055), linewidth(0.4) + linetype("4 4") + dtsfsf); /* dots and labels */ dot((-4.618108299402482,3.6910718535547),dotstyle); label("$A$", (-4.536470613911576,3.902029234047494), NE * labelscalefactor); dot((-5.96,-4.53),dotstyle); label("$B$", (-5.865352668717978,-4.307032491611389), NE * labelscalefactor); dot((5.1,-4.73),dotstyle); label("$C$", (5.194375400315942,-4.521368306902744), NE * labelscalefactor); dot((-2.6431144226071637,-1.7618809705153757),dotstyle); label("$I$", (-2.5645811132311103,-1.542100474352914), NE * labelscalefactor); dot((-0.48584271947108304,-7.718102386750928),dotstyle); label("$M_A$", (-0.39978937878842435,-7.500636139452573), NE * labelscalefactor); dot((-0.2509256265684152,5.272812850766571),dotstyle); label("$M_{BC}$", (-0.16401998196793383,5.488114267203518), NE * labelscalefactor); dot((-4.235015582748999,-6.4431742170539055),dotstyle); label("$T$", (-4.150666146387137,-6.236054829233581), NE * labelscalefactor); dot((-0.43,-4.63),dotstyle); label("$M$", (-0.3354886342010179,-4.414200399257067), NE * labelscalefactor); dot((-2.694238733401516,-4.589055357443012),dotstyle); label("$D$", (-2.6074482762893814,-4.3713332361987955), NE * labelscalefactor); dot((-0.791370694583703,0.37507337503804045),dotstyle); label("$E$", (-0.6998595201963214,0.5798240970314968), NE * labelscalefactor); dot((-5.433819446773386,-1.3063656827946843),dotstyle); label("$F$", (-5.350946712018725,-1.0919952622410694), NE * labelscalefactor); dot((-6.837851891540301,-1.8148893342365993),dotstyle); label("$P$", (-6.744129511412533,-1.6064012189403205), NE * labelscalefactor); dot((4.979844423017278,2.465338023691826),dotstyle); label("$Q$", (5.065773911141129,2.6803150868867722), NE * labelscalefactor); dot((-2.0744613450109326,-0.08964661424579373),dotstyle); label("$J$", (-1.9858744119444516,0.12971888491965208), NE * labelscalefactor); dot((-5.3325830164503545,-3.412793461180097),dotstyle); label("$X$", (-5.243778804373048,-3.192486252096345), NE * labelscalefactor); dot((-7.740992957065166,3.2368199586366244),dotstyle); label("$Y$", (-7.66577351716536,3.4519240219356493), NE * labelscalefactor); dot((-13.93604048454686,-4.385767803172747),dotstyle); label("$G$", (-13.86007857908552,-4.178431002436576), NE * labelscalefactor); dot((-3.2969221131279314,-7.021647416773472),dotstyle); label("$V$", (-3.2075885591051754,-6.814761530520238), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]
Let $G$ be orthocenter WRT $\Delta M_ADM_{BC}$, $$GP \times GQ=GV \times GM_A = GD \times GM$$Hence, $PDMQ$ is cyclic. Let $AT$ $\cap$ $\odot (I)$ $=$ $D'$. So from the above STEMS example, we know $TD=TD'$. Let $\Delta M_AM_BM_C$ be circum-Incentral Triangle WRT $\Delta ABC$. Since, $I$ is Orthocenter WRT $\Delta M_AM_BM_C$ $\implies$ $IM_CM_{BC}M_B$ is parallelogram & $PM_CM_BQ$ is Isosceles Trapezium. Suppose if $Z$ is midpoint of $M_BM_C$ $\implies$ $ZD=ZD'$. Hence, $Z$ is the center of $\odot (PD'DMQ)$. Let $DI \cap EF=I'$.
[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 = -14.525882318406245, xmax = 9.402893439360872, ymin = -7.792735443295693, ymax = 5.689086995836438; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen qqqqcc = rgb(0,0,0.8); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen ttffqq = rgb(0.2,1,0); draw((-4.618108299402482,3.6910718535547)--(-5.96,-4.53)--(5.1,-4.73)--cycle, linewidth(0.4) + rvwvcq); draw((-6.780053115248595,-0.1760944526782966)--(-2.6431144226071637,-1.7618809705153757)--(3.8860130660730166,3.6870263329294897)--(-0.2509256265684152,5.272812850766571)--cycle, linewidth(0.4) + ttffqq); /* draw figures */ draw((-4.618108299402482,3.6910718535547)--(-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)--(-4.618108299402482,3.6910718535547), linewidth(0.4) + rvwvcq); draw(circle((-0.3683841730197502,-1.2226447679921784), 6.496519544127112), linewidth(0.4)); draw((-0.2509256265684152,5.272812850766571)--(-0.48584271947108304,-7.718102386750928), linewidth(0.4)); draw((-4.618108299402482,3.6910718535547)--(-0.48584271947108304,-7.718102386750928), linewidth(0.4)); draw(circle((-2.6431144226071637,-1.7618809705153757), 2.827636594269825), linewidth(0.4) + qqqqcc); draw((-6.837851891540301,-1.8148893342365993)--(4.979844423017278,2.465338023691826), linewidth(0.4)); draw(circle((-1.44702002458779,1.7554659401255956), 6.465949659788316), linewidth(0.4)); draw((-2.0744613450109326,-0.08964661424579373)--(-0.43,-4.63), linewidth(0.4)); draw(circle((-5.7066256861518205,0.21268804151315282), 3.6447255102583913), linewidth(0.4)); draw((-4.235015582748999,-6.4431742170539055)--(-0.2509256265684152,5.272812850766571), linewidth(0.4)); draw((-4.235015582748999,-6.4431742170539055)--(-0.48584271947108304,-7.718102386750928), linewidth(0.4)); draw((-13.93604048454686,-4.385767803172747)--(-5.96,-4.53), linewidth(0.4) + rvwvcq); draw((-13.93604048454686,-4.385767803172747)--(-6.837851891540301,-1.8148893342365993), linewidth(0.4)); draw((-3.2969221131279314,-7.021647416773472)--(-0.2509256265684152,5.272812850766571), linewidth(0.4)); draw((-13.93604048454686,-4.385767803172747)--(-0.48584271947108304,-7.718102386750928), linewidth(0.4)); draw((-13.93604048454686,-4.385767803172747)--(-0.2509256265684152,5.272812850766571), linewidth(0.4)); draw((-7.740992957065166,3.2368199586366244)--(-4.235015582748999,-6.4431742170539055), linewidth(0.4) + linetype("4 4") + dtsfsf); draw((-4.618108299402482,3.6910718535547)--(-4.235015582748999,-6.4431742170539055), linewidth(0.4)); draw((-4.618108299402482,3.6910718535547)--(-3.2969221131279314,-7.021647416773472), linewidth(0.4)); draw((-6.780053115248595,-0.1760944526782966)--(3.8860130660730166,3.6870263329294897), linewidth(0.4) + dtsfsf); draw((-6.780053115248595,-0.1760944526782966)--(-2.6431144226071637,-1.7618809705153757), linewidth(0.4) + ttffqq); draw((-2.6431144226071637,-1.7618809705153757)--(3.8860130660730166,3.6870263329294897), linewidth(0.4) + ttffqq); draw((3.8860130660730166,3.6870263329294897)--(-0.2509256265684152,5.272812850766571), linewidth(0.4) + ttffqq); draw((-0.2509256265684152,5.272812850766571)--(-6.780053115248595,-0.1760944526782966), linewidth(0.4) + ttffqq); draw((-4.235015582748999,-6.4431742170539055)--(-2.694238733401516,-4.589055357443012), linewidth(0.4)); draw(circle((-1.5232123491588767,-2.4579696086554956), 2.431630987892209), linewidth(0.4) + linetype("4 4")); draw((-2.616424698317758,-0.2859392173109916)--(-2.694238733401516,-4.589055357443012), linewidth(0.4)); draw(circle((-2.8715322243189374,-2.4335876940956407), 2.1627468251410904), linewidth(0.4) + linetype("4 4")); /* dots and labels */ dot((-4.618108299402482,3.6910718535547),dotstyle); label("$A$", (-4.545831941386301,3.879838097078445), NE * labelscalefactor); dot((-5.96,-4.53),dotstyle); label("$B$", (-5.888177898529334,-4.329871959651366), NE * labelscalefactor); dot((5.1,-4.73),dotstyle); label("$C$", (5.181312675592202,-4.543869141224892), NE * labelscalefactor); dot((-2.6431144226071637,-1.7618809705153757),dotstyle); label("$I$", (-2.5614944395226864,-1.5673628884294863), NE * labelscalefactor); dot((-0.48584271947108304,-7.718102386750928),dotstyle); label("$M_A$", (-0.40206833455345875,-7.520375394020297), NE * labelscalefactor); dot((-0.2509256265684152,5.272812850766571),dotstyle); label("$M_{BC}$", (-0.16861686374597465,5.241638343455429), NE * labelscalefactor); dot((-4.235015582748999,-6.4431742170539055),dotstyle); label("$T$", (-4.156746156707161,-6.255846593813099), NE * labelscalefactor); dot((-0.43,-4.63),dotstyle); label("$M$", (-0.3437054668515877,-4.42714340582115), NE * labelscalefactor); dot((-2.694238733401516,-4.589055357443012),dotstyle); label("$D$", (-2.6198573072245575,-4.388234827353236), NE * labelscalefactor); dot((-0.791370694583703,0.37507337503804045),dotstyle); label("$E$", (-0.7133369622967708,0.5726089273057726), NE * labelscalefactor); dot((-5.433819446773386,-1.3063656827946843),dotstyle); label("$F$", (-5.362912089212496,-1.1199142360484775), NE * labelscalefactor); dot((-6.837851891540301,-1.8148893342365993),dotstyle); label("$P$", (-6.7636209140574,-1.625725756131357), NE * labelscalefactor); dot((4.979844423017278,2.465338023691826),dotstyle); label("$Q$", (5.06458694018846,2.6542178753391608), NE * labelscalefactor); dot((-2.0744613450109326,-0.08964661424579373),dotstyle); label("$J$", (-1.9973200517379333,0.10570598569080704), NE * labelscalefactor); dot((-5.3325830164503545,-3.412793461180097),dotstyle); label("$X$", (-5.246186353808754,-3.2209774733158225), NE * labelscalefactor); dot((-7.740992957065166,3.2368199586366244),dotstyle); label("$Y$", (-7.658518218819422,3.432389444697437), NE * labelscalefactor); dot((-13.93604048454686,-4.385767803172747),dotstyle); label("$G$", (-13.864436484451707,-4.193691935013668), NE * labelscalefactor); dot((-3.2969221131279314,-7.021647416773472),dotstyle); label("$V$", (-3.222940273477225,-6.820020981597849), NE * labelscalefactor); dot((-4.326081493888807,-4.034137678335941),dotstyle); label("$D'$", (-4.254017602876946,-3.843514728802443), NE * labelscalefactor); dot((-6.780053115248595,-0.1760944526782966),dotstyle); label("$M_C$", (-6.705258046355529,0.027888828754979448), NE * labelscalefactor); dot((3.8860130660730166,3.6870263329294897),dotstyle); label("$M_B$", (3.955692453852911,3.879838097078445), NE * labelscalefactor); dot((-1.4470200245877902,1.7554659401255965),dotstyle); label("$Z$", (-1.374782796251309,1.9538634629167124), NE * labelscalefactor); dot((-2.616424698317758,-0.2859392173109916),dotstyle); label("$I'$", (-2.5420401502887295,-0.08883690664876194), NE * labelscalefactor); dot((-4.442765056660414,-0.9474177118156701),dotstyle); label("$L$", (-4.370743338280688,-0.7502827406032965), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]
Our next step is to show, $J$ which is the foot from $M$ to $EF$ lies on $TI$. Redefine: $TI$ $\cap$ $EF$ $=$ $J$. Taking Homothety at $I$ which sends $J$ to $Z$ and using Right Angle on Incircle Chords Lemma, we have $MJ$ $\perp$ $EF$. Let $D'A$ $\cap$ $\odot (PQMD)$ $=$ $N$
$$\angle JMD=\angle LI'D=180^{\circ}-\angle LD'D=\angle NMD$$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(19cm); 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 = -16.703984590817605, xmax = 14.263615087265507, ymin = -8.740605115142586, ymax = 8.706993727874869; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen qqqqcc = rgb(0,0,0.8); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen ttffqq = rgb(0.2,1,0); draw((-4.618108299402482,3.6910718535547)--(-5.96,-4.53)--(5.1,-4.73)--cycle, linewidth(0.4) + rvwvcq); draw((-6.780053115248595,-0.1760944526782966)--(-2.6431144226071637,-1.7618809705153757)--(3.8860130660730166,3.6870263329294897)--(-0.2509256265684152,5.272812850766571)--cycle, linewidth(0.4) + ttffqq); /* draw figures */ draw((-4.618108299402482,3.6910718535547)--(-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)--(-4.618108299402482,3.6910718535547), linewidth(0.4) + rvwvcq); draw(circle((-0.3683841730197502,-1.2226447679921784), 6.496519544127112), linewidth(0.4)); draw((-0.2509256265684152,5.272812850766571)--(-0.48584271947108304,-7.718102386750928), linewidth(0.4)); draw((-4.618108299402482,3.6910718535547)--(-0.48584271947108304,-7.718102386750928), linewidth(0.4)); draw(circle((-2.6431144226071637,-1.7618809705153757), 2.827636594269825), linewidth(0.4) + qqqqcc); draw((-6.837851891540301,-1.8148893342365993)--(4.979844423017278,2.465338023691826), linewidth(0.4)); draw(circle((-1.44702002458779,1.7554659401255956), 6.465949659788316), linewidth(0.4)); draw((-2.0744613450109326,-0.08964661424579373)--(-0.43,-4.63), linewidth(0.4)); draw(circle((-5.7066256861518205,0.21268804151315282), 3.6447255102583913), linewidth(0.4)); draw((-4.235015582748999,-6.4431742170539055)--(-0.2509256265684152,5.272812850766571), linewidth(0.4)); draw((-4.235015582748999,-6.4431742170539055)--(-0.48584271947108304,-7.718102386750928), linewidth(0.4)); draw((-13.93604048454686,-4.385767803172747)--(-5.96,-4.53), linewidth(0.4) + rvwvcq); draw((-13.93604048454686,-4.385767803172747)--(-6.837851891540301,-1.8148893342365993), linewidth(0.4)); draw((-3.2969221131279314,-7.021647416773472)--(-0.2509256265684152,5.272812850766571), linewidth(0.4)); draw((-13.93604048454686,-4.385767803172747)--(-0.48584271947108304,-7.718102386750928), linewidth(0.4)); draw((-13.93604048454686,-4.385767803172747)--(-0.2509256265684152,5.272812850766571), linewidth(0.4)); draw((-7.740992957065166,3.2368199586366244)--(-4.235015582748999,-6.4431742170539055), linewidth(0.4) + linetype("4 4") + dtsfsf); draw((-4.618108299402482,3.6910718535547)--(-4.235015582748999,-6.4431742170539055), linewidth(0.4)); draw((-4.618108299402482,3.6910718535547)--(-3.2969221131279314,-7.021647416773472), linewidth(0.4)); draw((-6.780053115248595,-0.1760944526782966)--(3.8860130660730166,3.6870263329294897), linewidth(0.4) + dtsfsf); draw((-6.780053115248595,-0.1760944526782966)--(-2.6431144226071637,-1.7618809705153757), linewidth(0.4) + ttffqq); draw((-2.6431144226071637,-1.7618809705153757)--(3.8860130660730166,3.6870263329294897), linewidth(0.4) + ttffqq); draw((3.8860130660730166,3.6870263329294897)--(-0.2509256265684152,5.272812850766571), linewidth(0.4) + ttffqq); draw((-0.2509256265684152,5.272812850766571)--(-6.780053115248595,-0.1760944526782966), linewidth(0.4) + ttffqq); draw((-4.235015582748999,-6.4431742170539055)--(-2.694238733401516,-4.589055357443012), linewidth(0.4)); draw(circle((-1.5232123491588767,-2.4579696086554956), 2.431630987892209), linewidth(0.4) + linetype("4 4")); draw((-2.616424698317758,-0.2859392173109916)--(-2.694238733401516,-4.589055357443012), linewidth(0.4)); draw(circle((-2.8715322243189374,-2.4335876940956407), 2.1627468251410904), linewidth(0.4) + linetype("4 4")); draw((-4.754955270674418,7.3111899623043834)--(-4.618108299402482,3.6910718535547), linewidth(0.4)); draw((-4.754955270674418,7.3111899623043834)--(-2.0744613450109326,-0.08964661424579373), linewidth(0.4)); /* dots and labels */ dot((-4.618108299402482,3.6910718535547),dotstyle); label("$A$", (-4.51836000204344,3.9485576797792), NE * labelscalefactor); dot((-5.96,-4.53),dotstyle); label("$B$", (-5.852736248334826,-4.284291990735529), NE * labelscalefactor); dot((5.1,-4.73),dotstyle); label("$C$", (5.199927376607032,-4.485707273194605), NE * labelscalefactor); dot((-2.6431144226071637,-1.7618809705153757),dotstyle); label("$I$", (-2.5545609980674375,-1.5148318569232349), NE * labelscalefactor); dot((-0.48584271947108304,-7.718102386750928),dotstyle); label("$M_A$", (-0.38934671163235834,-7.456582689465975), NE * labelscalefactor); dot((-0.2509256265684152,5.272812850766571),dotstyle); label("$M_{BC}$", (-0.16275451886589656,5.534703029144423), NE * labelscalefactor); dot((-4.235015582748999,-6.4431742170539055),dotstyle); label("$T$", (-4.14070634743267,-6.19773717409675), NE * labelscalefactor); dot((-0.43,-4.63),dotstyle); label("$M$", (-0.33899289101758906,-4.3849996319650675), NE * labelscalefactor); dot((-2.694238733401516,-4.589055357443012),dotstyle); label("$D$", (-2.604914818682207,-4.334645811350298), NE * labelscalefactor); dot((-0.791370694583703,0.37507337503804045),dotstyle); label("$E$", (-0.691469635320974,0.625205519204447), NE * labelscalefactor); dot((-5.433819446773386,-1.3063656827946843),dotstyle); label("$F$", (-5.324021131879748,-1.0616474713903141), NE * labelscalefactor); dot((-6.837851891540301,-1.8148893342365993),dotstyle); label("$P$", (-6.733928109093288,-1.565185677538004), NE * labelscalefactor); dot((4.979844423017278,2.465338023691826),dotstyle); label("$Q$", (5.074042825070109,2.7148890747173597), NE * labelscalefactor); dot((-2.0744613450109326,-0.08964661424579373),dotstyle); label("$J$", (-1.9754920609975908,0.1720211336715261), NE * labelscalefactor); dot((-5.3325830164503545,-3.412793461180097),dotstyle); label("$X$", (-5.2233134906502094,-3.151331026903227), NE * labelscalefactor); dot((-7.740992957065166,3.2368199586366244),dotstyle); label("$Y$", (-7.640296880159135,3.495373294246279), NE * labelscalefactor); dot((-13.93604048454686,-4.385767803172747),dotstyle); label("$G$", (-13.833816815775757,-4.1332305288912226), NE * labelscalefactor); dot((-3.2969221131279314,-7.021647416773472),dotstyle); label("$V$", (-3.183983755752054,-6.776806111166594), NE * labelscalefactor); dot((-4.326081493888807,-4.034137678335941),dotstyle); label("$D'$", (-4.216237078354824,-3.7807537845878394), NE * labelscalefactor); dot((-6.780053115248595,-0.1760944526782966),dotstyle); label("$M_C$", (-6.6835742884785185,0.07131349244198812), NE * labelscalefactor); dot((3.8860130660730166,3.6870263329294897),dotstyle); label("$M_B$", (3.9914356818525696,3.9485576797792), NE * labelscalefactor); dot((-1.4470200245877902,1.7554659401255965),dotstyle); label("$Z$", (-1.3460693033129747,2.0099355861105943), NE * labelscalefactor); dot((-2.616424698317758,-0.2859392173109916),dotstyle); label("$I'$", (-2.504207177452668,-0.029394148787549854), NE * labelscalefactor); dot((-4.442765056660414,-0.9474177118156701),dotstyle); label("$L$", (-4.342121629891747,-0.6839938167795466), NE * labelscalefactor); dot((-4.754955270674418,7.3111899623043834),dotstyle); label("$N$", (-4.6442445535803625,7.574032764042567), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]
Hence, $N,J,M$ are collinear. Moreover, $\angle D'IT=\angle D'DB=\angle D'NJ$ $\implies$ $D'NJI$ is cyclic. Hence, by Radical Axes Theorem, we're done! $\qquad \square$

Taking a look at the Taiwan TST problem, we realise that the steps required to prove it are already mentioned in the Mathematical Reflections problem, hence are work is reduced (lol)!
Lemme continue this in Part (VI).
This post has been edited 33 times. Last edited by AlastorMoody, Nov 19, 2019, 11:09 AM

Incenter-Related Configurations -Part (IV)

by AlastorMoody, Nov 17, 2019, 10:01 PM

Here's the link to Part (III). We'll perhaps have a look at a bit different configuration in this post. Who would have ever thought $AD \cap \odot (ABC)$ will have any nice properties. I mean it always seemed an ugly and boring point to me. Anyway, our main motivation for stuffs in this post will be based on Ukraine TST 2016
Ukraine TST 2016, Romania TST 2019 Day 5 P2 wrote:
Let $ABC$ be an acute triangle with $AB<BC$. Let $I$ be the incenter of $ABC$, and let $\omega$ be the circumcircle of $ABC$. The incircle of $ABC$ is tangent to the side $BC$ at $K$. The line $AK$ meets $\omega$ again at $T$. Let $M$ be the midpoint of the side $BC$, and let $N$ be the midpoint of the arc $BAC$ of $\omega$. The segment $NT$ intersects the circumcircle of $BIC$ at $P$. Prove that $PM\parallel AK$.
which implies
math90 wrote:
Let $ABC$ be a triangle. $D$ is the touch point of the incircle on $BC$, $M$ is the midpoint of $BC$ and $I_a$ is the $A$-excenter. Prove that $AD\parallel MI_a$.
'K Let's start, We'll start with our own notations (lol). Seeing the diagram, it isn't tough to understand,
[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); 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 = -13.680498113463484, xmax = 22.071098120249008, ymin = -14.66377802009733, ymax = 5.644300824134521; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen ubqqys = rgb(0.29411764705882354,0,0.5098039215686274); pen qqccqq = rgb(0,0.8,0); draw((-5.119473560303962,3.074872234436409)--(-6.8,-3.83)--(3.72,-3.91)--cycle, linewidth(0.4) + rvwvcq); /* draw figures */ draw((-5.119473560303962,3.074872234436409)--(-6.8,-3.83), linewidth(0.4) + rvwvcq); draw((-6.8,-3.83)--(3.72,-3.91), linewidth(0.4) + rvwvcq); draw((3.72,-3.91)--(-5.119473560303962,3.074872234436409), linewidth(0.4) + rvwvcq); draw(circle((-3.600612428890119,-1.3355109844518491), 2.5187461313510067), linewidth(0.4) + ubqqys); draw(circle((-1.5656614953554076,-7.244486639236112), 6.249561471873599), linewidth(0.4)); draw((-5.119473560303962,3.074872234436409)--(0.4692894381793031,-13.153462294020388), linewidth(0.4)); draw((0.4692894381793031,-13.153462294020388)--(0.5397658380593432,-3.8858157097951276), linewidth(0.4)); draw((-3.600612428890119,-1.3355109844518491)--(-3.619765838059343,-3.8541842902048717), linewidth(0.4)); draw(circle((-1.5216556353578594,-1.457716049558527), 5.786937909920206), linewidth(0.4)); draw((-1.47764977536031,4.3290545401190625)--(-1.5656614953554087,-7.244486639236116), linewidth(0.4)); draw((-5.119473560303962,3.074872234436409)--(-2.923231753837296,-7.072360718402652), linewidth(0.4) + qqccqq); draw((-3.690242237939384,-13.121830874430128)--(-1.47764977536031,4.3290545401190625), linewidth(0.4) + qqccqq); /* dots and labels */ dot((-5.119473560303962,3.074872234436409),dotstyle); label("$A$", (-5.006340338726682,3.3585430321430567), NE * labelscalefactor); dot((-6.8,-3.83),dotstyle); label("$B$", (-6.676701802105052,-3.5280349309081482), NE * labelscalefactor); dot((3.72,-3.91),dotstyle); label("$C$", (3.843644958471,-3.6159486921385895), NE * labelscalefactor); dot((-3.600612428890119,-1.3355109844518491),dotstyle); label("$I$", (-3.482501810732379,-1.037145029378989), NE * labelscalefactor); dot((-3.619765838059343,-3.8541842902048717),dotstyle); label("$D$", (-3.5118063978091927,-3.557339517984962), NE * labelscalefactor); dot((0.4692894381793031,-13.153462294020388),dotstyle); label("$I_A$", (0.5908357929446998,-12.846893621334887), NE * labelscalefactor); dot((0.5397658380593432,-3.8858157097951276),dotstyle); label("$D'$", (0.6494449670983269,-3.586644105061776), NE * labelscalefactor); dot((-1.5656614953554087,-7.244486639236116),dotstyle); label("$M_A$", (-1.4604853024322464,-6.956671618895344), NE * labelscalefactor); dot((-1.47764977536031,4.3290545401190625),dotstyle); label("$M_{BC}$", (-1.3725715412018058,4.618640276446043), NE * labelscalefactor); dot((-2.923231753837296,-7.072360718402652),dotstyle); label("$E$", (-2.8084963079656684,-6.780844096434462), NE * labelscalefactor); dot((-2.156221269735208,-1.0228905623751765),dotstyle); label("$F$", (-2.046577043968517,-0.7440991586108527), NE * labelscalefactor); dot((-3.690242237939384,-13.121830874430128),dotstyle); label("$F'$", (-3.5704155719628194,-12.817589034258074), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]
Clearly, $E$ is midpoint of $FF'$ because $\angle M_{BC}EM_A=90^{\circ}$. Also, $BFCF'$ is Harmonic quadrilateral $\implies$ $F'F, F'M$ are Isogonal WRT $\angle BF'C$. Also, $$\angle DEF=\angle IM_AM_{BC}=\angle DIM_A$$Hence, $AIEF'$ is cyclic $\implies$ By Radical Axes, $F' \in ID$. Clearly, $F'$ is Reflection of $I_A$ over $M_AM_{BC}$. Projecting, $BFCF'$ harmonic bundle onto $BC$ from $I_A$ $\implies$ $F,M,I_A$ collinear! By Converse of Reim's Theorem, we have, $IDMF$ is cyclic.
[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 = -21.027961100978466, xmax = 17.875896098022363, ymin = -15.913394148326498, ymax = 6.185272277007704; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen ubqqys = rgb(0.29411764705882354,0,0.5098039215686274); pen qqccqq = rgb(0,0.8,0); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen qqqqcc = rgb(0,0,0.8); draw((-5.119473560303962,3.074872234436409)--(-6.8,-3.83)--(3.72,-3.91)--cycle, linewidth(0.4) + rvwvcq); /* draw figures */ draw((-5.119473560303962,3.074872234436409)--(-6.8,-3.83), linewidth(0.4) + rvwvcq); draw((-6.8,-3.83)--(3.72,-3.91), linewidth(0.4) + rvwvcq); draw((3.72,-3.91)--(-5.119473560303962,3.074872234436409), linewidth(0.4) + rvwvcq); draw(circle((-3.600612428890119,-1.3355109844518491), 2.5187461313510067), linewidth(0.4) + ubqqys); draw(circle((-1.5656614953554076,-7.244486639236112), 6.249561471873599), linewidth(0.4)); draw((-5.119473560303962,3.074872234436409)--(0.4692894381793031,-13.153462294020388), linewidth(0.4)); draw((0.4692894381793031,-13.153462294020388)--(0.5397658380593432,-3.8858157097951276), linewidth(0.4)); draw((-3.600612428890119,-1.3355109844518491)--(-3.619765838059343,-3.8541842902048717), linewidth(0.4)); draw(circle((-1.5216556353578594,-1.457716049558527), 5.786937909920206), linewidth(0.4)); draw((-1.47764977536031,4.3290545401190625)--(-1.5656614953554087,-7.244486639236116), linewidth(0.4)); draw((-5.119473560303962,3.074872234436409)--(-2.923231753837296,-7.072360718402652), linewidth(0.4) + qqccqq); draw((-3.690242237939384,-13.121830874430128)--(-1.47764977536031,4.3290545401190625), linewidth(0.4) + qqccqq); draw((-6.8,-3.83)--(-1.47764977536031,4.3290545401190625), linewidth(0.4) + dtsfsf); draw((-1.47764977536031,4.3290545401190625)--(3.72,-3.91), linewidth(0.4) + dtsfsf); draw(circle((-4.360042994597043,0.8696806249922793), 2.3322960400628223), linewidth(0.4)); draw((-6.67128524058068,1.182353135573616)--(-1.5656614953554087,-7.244486639236116), linewidth(0.4) + dtsfsf); draw((-2.156221269735208,-1.0228905623751765)--(0.4692894381793031,-13.153462294020388), linewidth(0.4) + linetype("4 4") + qqqqcc); draw((-3.619765838059343,-3.8541842902048717)--(-3.690242237939384,-13.121830874430128), linewidth(0.4)); draw((-14.843228418165088,-3.7688347648808738)--(-1.47764977536031,4.3290545401190625), linewidth(0.4) + linetype("4 4") + dtsfsf); draw((-14.843228418165088,-3.7688347648808738)--(-2.156221269735208,-1.0228905623751765), linewidth(0.4) + linetype("4 4") + dtsfsf); draw((-14.843228418165088,-3.7688347648808738)--(-6.8,-3.83), linewidth(0.4) + linetype("4 4") + rvwvcq); draw((-6.8,-3.83)--(-3.690242237939384,-13.121830874430128), linewidth(0.4) + ubqqys); draw((-3.690242237939384,-13.121830874430128)--(3.72,-3.91), linewidth(0.4) + ubqqys); draw((-1.54,-3.87)--(-3.690242237939384,-13.121830874430128), linewidth(0.4) + ubqqys); draw(circle((-2.570306214445059,-2.602755492225924), 1.6332297872643846), linewidth(0.4) + linetype("4 4")); draw(circle((-3.5386771839746043,1.8713810913006024), 3.2074901028820966), linewidth(0.4) + linetype("4 4")); draw(circle((-2.5487078067098263,0.23743512495709704), 4.229481616587153), linewidth(0.4) + linetype("4 4")); draw(circle((-27.35209557735922,-7.048392083479429), 24.42887558205736), linewidth(0.4) + linetype("4 4")); draw(circle((-10.954881994049979,-7.173085722896223), 9.389491981143628), linewidth(0.4) + linetype("4 4")); draw(circle((-8.20444495676025,-5.506660702058498), 6.862469295767354), linewidth(0.4) + linetype("4 4")); /* dots and labels */ dot((-5.119473560303962,3.074872234436409),dotstyle); label("$A$", (-4.988092108275666,3.3790924134732023), NE * labelscalefactor); dot((-6.8,-3.83),dotstyle); label("$B$", (-6.67817770790439,-3.5088036152023934), NE * labelscalefactor); dot((3.72,-3.91),dotstyle); label("$C$", (3.844996780349933,-3.6044688378228877), NE * labelscalefactor); dot((-3.600612428890119,-1.3355109844518491),dotstyle); label("$I$", (-3.4574485463477638,-1.0215078270695392), NE * labelscalefactor); dot((-3.619765838059343,-3.8541842902048717),dotstyle); label("$D$", (-3.4893369538879284,-3.540692022742558), NE * labelscalefactor); dot((0.4692894381793031,-13.153462294020388),dotstyle); label("$I_A$", (0.5923792112531423,-12.820218616930513), NE * labelscalefactor); dot((0.5397658380593432,-3.8858157097951276),dotstyle); label("$D'$", (0.6561560263334715,-3.572580430282723), NE * labelscalefactor); dot((-1.5656614953554087,-7.244486639236116),dotstyle); label("$M_A$", (-1.448478871317393,-6.920863222000026), NE * labelscalefactor); dot((-1.47764977536031,4.3290545401190625),dotstyle); label("$M_{BC}$", (-1.3528136486968994,4.654628715079794), NE * labelscalefactor); dot((-2.923231753837296,-7.072360718402652),dotstyle); label("$E$", (-2.7877919880043067,-6.761421184299203), NE * labelscalefactor); dot((-2.156221269735208,-1.0228905623751765),dotstyle); label("$F$", (-2.022470207040356,-0.7026237516678914), NE * labelscalefactor); dot((-3.690242237939384,-13.121830874430128),dotstyle); label("$F'$", (-3.5531137689682577,-12.788330209390349), NE * labelscalefactor); dot((-6.67128524058068,1.182353135573616),dotstyle); label("$X$", (-6.5506240777437315,1.4976763686034795), NE * labelscalefactor); dot((-1.54,-3.87),dotstyle); label("$M$", (-1.4165904637772284,-3.540692022742558), NE * labelscalefactor); dot((-14.843228418165088,-3.7688347648808738),dotstyle); label("$Z$", (-14.714056408025874,-3.4450268001220636), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]
Since, $\angle DEF=\angle DII_A=\angle EFI_A $ $\implies AD||FI_A$ $\qquad \square$ This was an unexpected, but beautiful configuration!
Some more problems in the Incircle Configuration:
Iran Round 3 2012 G4 wrote:
The incircle of triangle $ABC$ for which $AB\neq AC$, is tangent to sides $BC,CA$ and $AB$ in points $D,E$ and $F$ respectively. Perpendicular from $D$ to $EF$ intersects side $AB$ at $X$, and the second intersection point of circumcircles of triangles $AEF$ and $ABC$ is $T$. Prove that $TX\perp TF$.

Proposed By Pedram Safaei
Lol! We just need to angle chase, since the basics is already proved earlier,
[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 = -13.334821729868441, xmax = 15.25810741404111, ymin = -7.86794781714506, ymax = 8.373773409682249; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen ubqqys = rgb(0.29411764705882354,0,0.5098039215686274); pen ttffqq = rgb(0.2,1,0); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); draw((-5.147826118775256,5.927030441431838)--(-6.12583426335671,-3.509488424318937)--(5.16,-3.51)--cycle, linewidth(0.4) + rvwvcq); /* draw figures */ draw((-5.147826118775256,5.927030441431838)--(-6.12583426335671,-3.509488424318937), linewidth(0.4) + rvwvcq); draw((-6.12583426335671,-3.509488424318937)--(5.16,-3.51), linewidth(0.4) + rvwvcq); draw((5.16,-3.51)--(-5.147826118775256,5.927030441431838), linewidth(0.4) + rvwvcq); draw(circle((-2.726892722308635,-0.44474953922341326), 3.0648929525979303), linewidth(0.4) + ubqqys); draw((-5.7754564699875175,-0.1287940245974699)--(-0.6572769031227592,1.8158386965816027), linewidth(0.4) + ttffqq); draw((-0.6572769031227592,1.8158386965816027)--(-2.7270316508682266,-3.5096424886725965), linewidth(0.4) + ttffqq); draw((-2.7270316508682266,-3.5096424886725965)--(-5.7754564699875175,-0.1287940245974699), linewidth(0.4) + ttffqq); draw(circle((-0.4827274596930421,0.6745957982903701), 7.025042005528744), linewidth(0.4)); draw(circle((-3.9373594205419455,2.741140451104213), 3.408096926145383), linewidth(0.4)); draw((-5.397426520654763,3.5187080626412763)--(-2.7270316508682266,-3.5096424886725965), linewidth(0.4)); draw((-5.397426520654763,3.5187080626412763)--(-7.3169130374474385,2.300979107432811), linewidth(0.4)); draw((-7.3169130374474385,2.300979107432811)--(-5.7754564699875175,-0.1287940245974699), linewidth(0.4)); draw((-5.147826118775256,5.927030441431838)--(-0.48304589788386654,-6.35044620002113), linewidth(0.4)); draw((-7.3169130374474385,2.300979107432811)--(4.182371199389172,-4.577838844851099), linewidth(0.4) + dtsfsf); draw((-7.3169130374474385,2.300979107432811)--(-5.147826118775256,5.927030441431838), linewidth(0.4) + dtsfsf); draw((-10.79258094921728,-3.5092768853171714)--(-7.3169130374474385,2.300979107432811), linewidth(0.4) + dtsfsf); draw((-10.79258094921728,-3.5092768853171714)--(-6.12583426335671,-3.509488424318937), linewidth(0.4) + rvwvcq); draw((-10.79258094921728,-3.5092768853171714)--(-2.726892722308635,-0.44474953922341326), linewidth(0.4) + dtsfsf); /* dots and labels */ dot((-5.147826118775256,5.927030441431838),dotstyle); label("$A$", (-5.061621739868382,6.170711655971185), NE * labelscalefactor); dot((-6.12583426335671,-3.509488424318937),dotstyle); label("$B$", (-6.022531653721081,-3.2743296923858214), NE * labelscalefactor); dot((5.16,-3.51),dotstyle); label("$C$", (5.250582213672767,-3.2743296923858214), NE * labelscalefactor); dot((-2.726892722308635,-0.44474953922341326),dotstyle); label("$I$", (-2.624191714485929,-0.20410533349061583), NE * labelscalefactor); dot((-2.7270316508682266,-3.5096424886725965),dotstyle); label("$D$", (-2.624191714485929,-3.2743296923858214), NE * labelscalefactor); dot((-0.6572769031227592,1.8158386965816027),dotstyle); label("$E$", (-0.5617509237776989,2.0458300745547255), NE * labelscalefactor); dot((-5.7754564699875175,-0.1287940245974699),dotstyle); label("$F$", (-5.670979246213996,0.10057341968219083), NE * labelscalefactor); dot((-7.3169130374474385,2.300979107432811),dotstyle); label("$T$", (-7.217809839245168,2.538003445064644), NE * labelscalefactor); dot((-5.397426520654763,3.5187080626412763),dotstyle); label("$X$", (-5.2959900115397724,3.7567184577558708), NE * labelscalefactor); dot((-4.234083304556044,0.45684480535892547),dotstyle); label("$K$", (-4.147585480349962,0.6864940988606651), NE * labelscalefactor); dot((-0.48304589788386654,-6.35044620002113),dotstyle); label("$M_A$", (-0.397693133607726,-6.110185779609637), NE * labelscalefactor); dot((4.182371199389172,-4.577838844851099),dotstyle); label("$A'$", (4.26623547265293,-4.3524237420742145), NE * labelscalefactor); dot((-10.79258094921728,-3.5092768853171714),dotstyle); label("$N$", (-10.709897087148876,-3.2743296923858214), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]
$$\angle FTK=\angle FAI=\angle FXK$$Hence, $\angle XKF=\angle XTF=90^{\circ}$ $\qquad \square$
Peru IMO Pre-TST 2015 P3 wrote:
Let $M$ be the midpoint of the arc $BAC$ of the circumcircle of the triangle $ABC,$ $I$ the incenter of the triangle $ABC$ and $L$ a point on the side $BC$ such that $AL$ is bisector. The line $MI$ cuts the circumcircle again at $K.$ The circumcircle of the triangle $AKL$ cuts the line $BC$ again at $P.$ Prove that $\angle AIP = 90^{\circ}.$
Greece TST 2019 wrote:
Let a triangle $ABC$ inscribed in a circle $\Gamma$ with center $O$. Let $I$ the incenter of triangle $ABC$ and $D, E, F$ the contact points of the incircle with sides $BC, AC, AB$ of triangle $ABC$ respectively . Let also $S$ the foot of the perpendicular line from $D$ to the line $EF$.Prove that line $SI$ passes from the antidiametric point $N$ of $A$ in the circle $\Gamma$.( $AN$ is a diametre of the circle $\Gamma$).
ltf0501 || https://artofproblemsolving.com/community/q5h1522709p9104820 wrote:
In a triangle $ABC$, an incircle touches $BC,CA,AB$ at $D,E,F$. The foot from $D$ to $EF$ is $K$. $M$ is the midpoint of arc $BAC$. Prove that $AK$ and $MD$ meet on the circumcircle of $ABC$.
I guess, I stop here. Lemme continue this in Part (V), over here.
This post has been edited 26 times. Last edited by AlastorMoody, Nov 18, 2019, 2:41 AM

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! :D
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

Incenter-Related Configurations -Part (II) [Feat. Sharky Devil Lemma]

by AlastorMoody, Nov 17, 2019, 10:40 AM

Ok, Let's review! Here's the link to Part (I).
[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]
Let's dive in to another configuration! From aops29's blogpost, we can have a good intro in this configuration.
aops29 wrote:
\[\textbf{The Sharky-Devil Lemma}\]In triangle \(ABC\), let \(DEF\) be the contact triangle, and let \(M\) be the midpoint of the arc \(BC\) not containing \(A\) in \((ABC)\). Suppose ray \(MD\) meets \((ABC)\) again at \(T\). If \(I\) is the incenter of \(ABC\) and ray \(TI\) intersects \((ABC)\) again at \(S\), then \(S\) is the antipode of \(A\). If \(P=TS\cap EF\), then \(DP\perp EF\).

[asy] unitsize(100); import olympiad; import cse5; //the config pair A,B,C,I; B=origin; C=2*right; A=1.2*dir(65); I=incenter(A,B,C); dot(A); dot(B); dot(C); dot(I); label("$I$",I,S); label("$A$",A,N); label("$B$",B,W); label("$C$",C,E); draw(A--B--C--cycle); draw(circumcircle(A,B,C)); pair D,E,F; D=foot(I,B,C); E=foot(I,C,A); F=foot(I,A,B); dot(D); dot(E); dot(F); label("$D$",D,S); label("$E$",E,NE); label("$F$",F,NW); draw(D--E--F--cycle); pair P=foot(D,E,F); dot(P); label("$P$",P,N); path w = circumcircle(A,B,C); pair T = OP(w,circumcircle(A,E,F)); dot(T); label("$T$",T,NW); pair M = OP(Line(T,D,10),w); dot(M); label("$M$",M,S); draw(D--P); draw(rightanglemark(D,P,E,2),green); pair S = OP(Line(T,I,10),w); dot(S); label("$S$",S,SE); draw(T--S); draw(A--S,dashed+red); draw(A--T); draw(rightanglemark(A,T,S,2),red); draw(M--T,blue); [/asy]
This is well-known, but since I like the name I mentioned this here :D Now Let's prove this!! Referring to aops29's diagram, we just need to show that If $TS \cap EF=S$, then $DP \perp EF$ coz rest all we have proved earlier :P
We'l work with the original diagram and erase some points to clear mess,
[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 = -25.06371542459404, xmax = 14.590283440692431, ymin = -11.695057807796225, ymax = 7.388429146123024; /* 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 qqqqcc = rgb(0,0,0.8); 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((-3.425794343555926,-0.26554447803803993)--(-7.32,-3.13), linewidth(0.4) + ttffqq); draw((3.64,-3.27)--(-3.425794343555926,-0.26554447803803993), linewidth(0.4) + ttffqq); 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((-14.51837862789293,-3.03804990803786)--(-3.425794343555926,-0.26554447803803993), linewidth(0.4) + ttffqq); draw((-14.51837862789293,-3.03804990803786)--(-7.32,-3.13), linewidth(0.4) + rvwvcq); 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)); draw((-14.51837862789293,-3.03804990803786)--(-4.8127515399252045,5.283566240744204), linewidth(0.4) + ttffqq); 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((-3.3551493764184945,5.2649472350059785)--(-3.4630134406164252,-3.1792680764884764), linewidth(0.4)); draw((-6.218395061190416,0.5666524344894496)--(-1.3531342778462379,1.7826866420666359), linewidth(0.4) + ttffqq); draw((-4.8127515399252045,5.283566240744204)--(-1.88144278616565,-6.444378116967967), linewidth(0.4)); draw(circle((-11.902939656253105,-3.0714588000113654), 8.440614750448322), linewidth(0.4) + linetype("4 4")); draw(circle((-8.44211877371571,-10.216230347254088), 13.93656019609169), linewidth(0.4) + linetype("4 4")); draw((-20.342865871889785,-2.9636495235342566)--(-14.51837862789293,-3.03804990803786), linewidth(0.4) + rvwvcq); draw((-20.342865871889785,-2.9636495235342566)--(-6.218395061190416,0.5666524344894496), linewidth(0.4) + ttffqq); draw((-20.342865871889785,-2.9636495235342566)--(-6.755455378061811,3.6178896713466084), linewidth(0.4) + ubqqys); draw((-4.506237629689092,0.994592908428854)--(-3.4630134406164252,-3.1792680764884764), linewidth(0.4) + dtsfsf); draw((-20.342865871889785,-2.9636495235342566)--(-1.88144278616565,-6.444378116967967), linewidth(0.4) + ubqqys); draw((-1.7217647061562313,6.056134432340744)--(-4.003196202301187,-6.044341361975363), linewidth(0.4)); draw((-4.8127515399252045,5.283566240744204)--(-4.003196202301187,-6.044341361975363), linewidth(0.4)); /* dots and labels */ dot((-4.8127515399252045,5.283566240744204),dotstyle); label("$A$", (-4.713503506922719,5.570954198130715), NE * labelscalefactor); dot((-7.32,-3.13),dotstyle); label("$B$", (-7.219415935215128,-2.855520560742719), NE * labelscalefactor); dot((3.64,-3.27),dotstyle); label("$C$", (3.7405087511626602,-2.9932080568027426), NE * labelscalefactor); dot((-3.425794343555926,-0.26554447803803993),dotstyle); label("$I$", (-3.30909104711049,0.008379357305768217), NE * labelscalefactor); dot((-1.88144278616565,-6.444378116967967),dotstyle); label("$M_A$", (-1.7669910912382387,-6.160020466183282), NE * labelscalefactor); dot((-1.7217647061562313,6.056134432340744),dotstyle); label("$M_{BC}$", (-1.6017660959662117,6.342004176066847), NE * labelscalefactor); dot((-14.51837862789293,-3.03804990803786),dotstyle); label("$K$", (-14.4067032295483,-2.7729080631067053), NE * labelscalefactor); dot((-3.4630134406164252,-3.1792680764884764),dotstyle); label("$D$", (-3.3641660455344993,-2.9105955591667287), NE * labelscalefactor); dot((-1.3531342778462379,1.7826866420666359),dotstyle); label("$E$", (-1.2437786062101532,2.0461542989941153), NE * labelscalefactor); dot((-6.218395061190416,0.5666524344894496),dotstyle); label("$F$", (-6.117915966734949,0.8345043336659088), NE * labelscalefactor); dot((-6.755455378061811,3.6178896713466084),dotstyle); label("$L$", (-6.641128451763034,3.8911667461984294), NE * labelscalefactor); dot((1.209544047603321,-5.67180992537143),dotstyle); label("$A'$", (1.3172088205062646,-5.388970488247151), NE * labelscalefactor); dot((-3.3551493764184945,5.2649472350059785),dotstyle); label("$G$", (-3.254016048686481,5.54341669891871), NE * labelscalefactor); dot((-4.506237629689092,0.994592908428854),dotstyle); label("$P$", (-4.3830535163786655,1.2751043210579838), NE * labelscalefactor); dot((-20.342865871889785,-2.9636495235342566),dotstyle); label("$X$", (-20.244653062493253,-2.690295565470691), NE * labelscalefactor); dot((-4.003196202301187,-6.044341361975363),dotstyle); label("$V$", (-3.8873785305625845,-5.774495477215217), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]

Let $EF$ $\cap$ $BC$ $=X$. By Radical Axes Theorem, $X$ $\in$ $LG$. Let $P$ be foot from $D$ to $EF$. Hence, $XLPD$ is cyclic. Let this intersect $\odot (ABC)$ $=V$. Now observe, $\Psi _{\odot (BIC)}$ $( \odot (XLPDV))$ $=$ $\odot (XLPDV)$ $\implies$ $V$ $\in$ $XM_A$ and since, $\angle DVX$ $=$ $90^{\circ}$ $\implies$ $V$ $\in$ $DM_{BC}$. Also, By Converse of Reim's Theorem, $V$ $\in$ $AP$. Applying Reim's Theorem again $\implies$ $P$ $\in$ $LA'$, Hence the Sharky-Devil Lemma is proved!

Let $T_A$ be the foot from $I_A$ to $BC$. Let $D'$ be $D-$antipode in $(I)$. It's Well-Known, $D'$ $\in$ $AT_A$. Also $\sqrt{bc}$ Inversion implies, $AT,AT_A$ are Isogonal. Hence, by Reflection Argument, $AT$ intersects $(I)$ at the reflection of Orthocenter of $\Delta DEF$ over $EF$. Let this point be $Q$. The fact that $Q$ $\in$ $\odot (LIDT)$ follows from Simple POP.
[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(25cm); 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 = -15.008872990343747, xmax = 12.804220510314202, ymin = -8.92128006748609, ymax = 6.761774282356115; /* 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); pen ffdxqq = rgb(1,0.8431372549019608,0); draw((-5.942203631855723,4.138673889328595)--(-7.045966190852679,-2.451453892808012)--(3.5400471222335197,-2.7410498081490866)--cycle, linewidth(2) + rvwvcq); /* draw figures */ draw((-5.942203631855723,4.138673889328595)--(-7.045966190852679,-2.451453892808012), linewidth(0.4) + rvwvcq); draw((-7.045966190852679,-2.451453892808012)--(3.5400471222335197,-2.7410498081490866), linewidth(0.4) + rvwvcq); draw((3.5400471222335197,-2.7410498081490866)--(-5.942203631855723,4.138673889328595), linewidth(0.4) + rvwvcq); draw(circle((-1.680910362053616,0.037464498041687326), 5.914265702725966), linewidth(0.4)); draw(circle((-1.8426432646069566,-5.874589398082089), 6.228356618102593), linewidth(0.4) + linetype("4 4")); draw((-4.202494036745758,-0.11060375484023069)--(-7.045966190852679,-2.451453892808012), linewidth(0.4) + ttffqq); draw((3.5400471222335197,-2.7410498081490866)--(-4.202494036745758,-0.11060375484023069), linewidth(0.4) + ttffqq); draw((-7.045966190852679,-2.451453892808012)--(-1.519177459500276,5.949518394165464), linewidth(0.4) + dtsfsf); draw((-1.519177459500276,5.949518394165464)--(3.5400471222335197,-2.7410498081490866), linewidth(0.4) + dtsfsf); draw((-1.519177459500276,5.949518394165464)--(-1.842643264606957,-5.874589398082089), linewidth(0.4)); draw((-1.519177459500276,5.949518394165464)--(-7.696524173231269,-8.001677659040286), linewidth(0.4) + ubqqys); draw((-7.045966190852679,-2.451453892808012)--(-7.696524173231269,-8.001677659040286), linewidth(0.4) + ubqqys); draw((-7.696524173231269,-8.001677659040286)--(3.5400471222335197,-2.7410498081490866), linewidth(0.4) + ubqqys); draw((-9.740054607429487,-2.3777531541131265)--(-4.202494036745758,-0.11060375484023069), linewidth(0.4) + ttffqq); draw((-9.740054607429487,-2.3777531541131265)--(-7.045966190852679,-2.451453892808012), linewidth(0.4) + rvwvcq); draw((-9.740054607429487,-2.3777531541131265)--(-1.842643264606957,-5.874589398082089), linewidth(0.4) + linetype("4 4")); draw(circle((-4.202494036745758,-0.11060375484023069), 2.4177329700687884), linewidth(0.4) + qqqqcc); draw(circle((-5.072348834300742,2.0140350672441834), 2.295808679558832), linewidth(0.4)); draw((-7.349716905373479,1.72363506380737)--(-1.842643264606957,-5.874589398082089), linewidth(0.4)); draw((-9.740054607429487,-2.3777531541131265)--(-5.942203631855723,4.138673889328595), linewidth(0.4) + ttffqq); draw(circle((-4.178123331091742,0.7802534735805249), 3.308923484338092), linewidth(0.4) + ubqqys); draw((-7.349716905373479,1.72363506380737)--(-1.519177459500276,5.949518394165464), linewidth(0.4) + ubqqys); draw(circle((-6.971274322087621,-1.2441784544766825), 2.991844893732568), linewidth(0.4) + linetype("4 4")); draw(circle((-7.841129119642604,0.8804603676077339), 3.7711899131827487), linewidth(0.4) + linetype("4 4")); draw((-4.087636730328725,4.087939492580288)--(-4.2686099318547495,-2.527432545419236), linewidth(0.4)); draw(circle((-5.79134893601822,-4.126171276097608), 4.318476862126188), linewidth(0.4) + linetype("4 4")); draw((-6.587013075226309,0.2887728464791664)--(-2.7826744728952577,1.846318604545808), linewidth(0.4) + ttffqq); draw((-5.942203631855723,4.138673889328595)--(-1.842643264606957,-5.874589398082089), linewidth(0.4)); draw(circle((-8.57911002453525,-2.409512495349541), 4.312112728953992), linewidth(0.4) + linetype("4 4")); draw(circle((-4.8139720411984035,-7.604351967915097), 9.666506301667258), linewidth(0.4) + linetype("4 4")); draw((-12.889610117215755,-2.2915924452798433)--(-9.740054607429487,-2.3777531541131265), linewidth(0.4) + rvwvcq); draw((-12.889610117215755,-2.2915924452798433)--(-6.587013075226309,0.2887728464791664), linewidth(0.4) + ttffqq); draw((-12.889610117215755,-2.2915924452798433)--(-7.349716905373479,1.72363506380737), linewidth(0.4) + ubqqys); draw((-5.588903275433209,0.6974119943720299)--(-4.2686099318547495,-2.527432545419236), linewidth(0.4) + dtsfsf); draw((-12.889610117215755,-2.2915924452798433)--(-1.842643264606957,-5.874589398082089), linewidth(0.4) + ubqqys); draw((-1.519177459500276,5.949518394165464)--(-5.019984567584007,-4.844043696400092), linewidth(0.4)); draw((-5.942203631855723,4.138673889328595)--(-5.019984567584007,-4.844043696400092), linewidth(0.4)); draw((-7.349716905373479,1.72363506380737)--(2.5803829077484957,-4.0637448932452145), linewidth(0.4) + dtsfsf); draw((-6.587013075226309,0.2887728464791664)--(-4.2686099318547495,-2.527432545419236), linewidth(0.4) + ttffqq); draw((-4.2686099318547495,-2.527432545419236)--(-2.7826744728952577,1.846318604545808), linewidth(0.4) + ttffqq); draw((-1.842643264606957,-5.874589398082089)--(0.5172075075318441,-11.638575041323948), linewidth(0.4)); draw((0.7626908632355884,-2.665071155537863)--(0.5172075075318441,-11.638575041323948), linewidth(0.4) + dtsfsf); draw((-5.942203631855723,4.138673889328595)--(0.7626908632355884,-2.665071155537863), linewidth(0.8) + ffdxqq); draw((-5.942203631855723,4.138673889328595)--(-5.949509104988513,-4.0561407069402575), linewidth(0.8) + ffdxqq); draw((-5.944497144381616,1.5659575734578608)--(-5.588903275433209,0.6974119943720299), linewidth(0.4) + linetype("4 4") + dtsfsf); draw((xmin, -0.02735646619517119*xmin + 3.9761161965489102)--(xmax, -0.02735646619517119*xmax + 3.9761161965489102), linewidth(0.4) + linetype("2 2") + rvwvcq); /* line */ draw((-5.949509104988513,-4.0561407069402575)--(2.7982309662885934,3.899566485713355), linewidth(0.4)); /* dots and labels */ dot((-5.942203631855723,4.138673889328595),dotstyle); label("$A$", (-5.843451617059394,4.362923256261925), NE * labelscalefactor); dot((-7.045966190852679,-2.451453892808012),dotstyle); label("$B$", (-6.952354449876513,-2.2226017304683516), NE * labelscalefactor); dot((3.5400471222335197,-2.7410498081490866),dotstyle); label("$C$", (3.63879913702985,-2.5168004412157523), NE * labelscalefactor); dot((-4.202494036745758,-0.11060375484023069),dotstyle); label("$I$", (-4.100890022632492,0.10835728545336126), NE * labelscalefactor); dot((-1.842643264606957,-5.874589398082089),dotstyle); label("$M_A$", (-1.7473003366533006,-5.639832909149698), NE * labelscalefactor); dot((-1.519177459500276,5.949518394165464),dotstyle); label("$M_{BC}$", (-1.4304709558484094,6.173376860861314), NE * labelscalefactor); dot((-5.949509104988513,-4.0561407069402575),dotstyle); label("$T$", (-5.866082287116886,-3.8293793045503093), NE * labelscalefactor); dot((-7.696524173231269,-8.001677659040286),dotstyle); label("$J$", (-7.608643881543788,-7.76711589455398), NE * labelscalefactor); dot((-9.740054607429487,-2.3777531541131265),dotstyle); label("$K$", (-9.645404186718089,-2.154709720295875), NE * labelscalefactor); dot((-4.2686099318547495,-2.527432545419236),dotstyle); label("$D$", (-4.168782032804969,-2.290493740640829), NE * labelscalefactor); dot((-2.7826744728952577,1.846318604545808),dotstyle); label("$E$", (-2.6977884790679743,2.0772255804551967), NE * labelscalefactor); dot((-6.587013075226309,0.2887728464791664),dotstyle); label("$F$", (-6.499741048726668,0.5157093464882238), NE * labelscalefactor); dot((-7.349716905373479,1.72363506380737),dotstyle); label("$L$", (-7.2691838306814045,1.9414415601102424), NE * labelscalefactor); dot((2.5803829077484957,-4.0637448932452145),dotstyle); label("$A'$", (2.665680324557684,-3.8293793045503093), NE * labelscalefactor); dot((-4.087636730328725,4.087939492580288),dotstyle); label("$G$", (-3.987736672345031,4.31766191614694), NE * labelscalefactor); dot((-5.588903275433209,0.6974119943720299),dotstyle); label("$P$", (-5.50399156619701,0.9230614075230862), NE * labelscalefactor); dot((-12.889610117215755,-2.2915924452798433),dotstyle); label("$X$", (-12.791067324709507,-2.0641870400659053), NE * labelscalefactor); dot((-5.019984567584007,-4.844043696400092),dotstyle); label("$V$", (-4.938224814759705,-4.621452756562542), NE * labelscalefactor); dot((0.7626908632355884,-2.665071155537863),dotstyle); label("$T_A$", (0.8552267199583057,-2.4489084310432756), NE * labelscalefactor); dot((-4.136378141636767,2.306225035738775),dotstyle); label("$D'$", (-4.055628682517508,2.5298389816050437), NE * labelscalefactor); dot((-5.944497144381616,1.5659575734578608),dotstyle); label("$Q$", (-5.843451617059394,1.783026869707796), NE * labelscalefactor); dot((2.7982309662885934,3.899566485713355),dotstyle); label("$A_1$", (2.891987025132606,4.136616555687001), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]

Now We will focus on the Involvement of the $A-$antipode in this diagram. Clear some unnecessary lines & points. Define: $R$ $=$ $AL$ $\cap$ $EF$. Let $M'$ be midpoint of $EF$ $\implies$ $RLM'I$ is cyclic. Let $RA'$ $\cap \odot (ABC)$ $=$ $Y$ $\implies$ $RYM'A$ is cyclic. By POP, $AFEA'$ is also cyclic. We also know, $P$ is orthocenter WRT $\Delta ARI$. Let $R'$ be foot from $A$ to $RI$, then a series of Power of Point follows, $$RP \times RM'=AP \times PR'=LP \times PI = FP \times PE$$Thus, $P$ lies on Radical Axes of $\odot (A'EF)$ and $\odot (ARY)$. Pretty Cool, isn't it? :) (You said no? :mad: Well, this trivialises ELMO SL 2019 G3, so well, yeah this is pretty interesting!!)
[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 = -15.635835279685946, xmax = 8.193920498944122, ymin = -6.769346326188248, ymax = 6.667611163307858; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen ttffqq = rgb(0.2,1,0); pen qqqqcc = rgb(0,0,0.8); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); draw((-4.446753459113518,4.611610926456197)--(-7.045966190852679,-2.451453892808012)--(3.5400471222335197,-2.7410498081490866)--cycle, linewidth(2) + rvwvcq); /* draw figures */ draw((-4.446753459113518,4.611610926456197)--(-7.045966190852679,-2.451453892808012), linewidth(0.4) + rvwvcq); draw((-7.045966190852679,-2.451453892808012)--(3.5400471222335197,-2.7410498081490866), linewidth(0.4) + rvwvcq); draw((3.5400471222335197,-2.7410498081490866)--(-4.446753459113518,4.611610926456197), linewidth(0.4) + rvwvcq); draw(circle((-1.6931921700442931,-0.4114899795963873), 5.728319342097372), linewidth(0.4)); draw((-3.345933085153599,0.055038275929575345)--(-7.045966190852679,-2.451453892808012), linewidth(0.4) + ttffqq); draw((3.5400471222335197,-2.7410498081490866)--(-3.345933085153599,0.055038275929575345), linewidth(0.4) + ttffqq); draw((-1.536544200453946,5.314687095789082)--(-1.8498401396346391,-6.137667054981856), linewidth(0.4)); draw((-13.041975271686374,-2.2874242730822445)--(-3.345933085153599,0.055038275929575345), linewidth(0.4) + ttffqq); draw((-13.041975271686374,-2.2874242730822445)--(-7.045966190852679,-2.451453892808012), linewidth(0.4) + rvwvcq); draw(circle((-3.345933085153599,0.055038275929575345), 2.6067367718043304), linewidth(0.4) + qqqqcc); draw(circle((-3.8963432721335582,2.333324601192886), 2.343830188775013), linewidth(0.4)); draw((-6.001587925321318,3.36360849606723)--(-1.8498401396346391,-6.137667054981856), linewidth(0.4)); draw((-13.041975271686374,-2.2874242730822445)--(-4.446753459113518,4.611610926456197), linewidth(0.4) + ttffqq); draw((-5.792280011836994,0.9552956345685196)--(-1.5804002441062543,1.9728418072278557), linewidth(0.4) + ttffqq); draw((-4.446753459113518,4.611610926456197)--(-1.8498401396346391,-6.137667054981856), linewidth(0.4)); draw((-18.58683965454231,-2.1357363780358547)--(-13.041975271686374,-2.2874242730822445), linewidth(0.4) + rvwvcq); draw((-18.58683965454231,-2.1357363780358547)--(-5.792280011836994,0.9552956345685196), linewidth(0.4) + ttffqq); draw((-4.348502014831068,1.3040973231349209)--(-3.417217522730037,-2.550723633078282), linewidth(0.4) + dtsfsf); draw((-6.001587925321318,3.36360849606723)--(1.0603691190249185,-5.43459088564898), linewidth(0.4) + dtsfsf); draw((-5.792280011836994,0.9552956345685196)--(-3.417217522730037,-2.550723633078282), linewidth(0.4) + ttffqq); draw((-3.417217522730037,-2.550723633078282)--(-1.5804002441062543,1.9728418072278557), linewidth(0.4) + ttffqq); draw((-4.598972421296053,2.3408573894108984)--(-4.348502014831068,1.3040973231349209), linewidth(0.4) + dtsfsf); draw((-10.384072030799011,-0.15403337105498183)--(1.0603691190249185,-5.43459088564898), linewidth(0.4)); draw(circle((-6.865002557976306,-0.04949754756269626), 3.520621776468737), linewidth(0.4) + linetype("2 2")); draw(circle((-7.415412744956264,2.2287887777006015), 3.8066756294725907), linewidth(0.4) + linetype("2 2")); draw(circle((-2.7039828514710456,-2.6021553471088694), 4.710948613364595), linewidth(0.4) + linetype("2 2")); draw((-10.384072030799011,-0.15403337105498183)--(-3.345933085153599,0.055038275929575345), linewidth(0.4) + ttffqq); /* dots and labels */ dot((-4.446753459113518,4.611610926456197),dotstyle); label("$A$", (-4.370507283441791,4.806214454806233), NE * labelscalefactor); dot((-7.045966190852679,-2.451453892808012),dotstyle); label("$B$", (-6.968706855725297,-2.2515813982624286), NE * labelscalefactor); dot((3.5400471222335197,-2.7410498081490866),dotstyle); label("$C$", (3.617986923877646,-2.5424246339658074), NE * labelscalefactor); dot((-3.345933085153599,0.055038275929575345),dotstyle); label("$I$", (-3.2653029877689557,0.2496704287866301), NE * labelscalefactor); dot((-1.8498401396346391,-6.137667054981856),dotstyle); label("$M_A$", (-1.7723077111582843,-5.9355957171718945), NE * labelscalefactor); dot((-1.536544200453946,5.314687095789082),dotstyle); label("$M_{BC}$", (-1.4620749264080148,5.504238220494342), NE * labelscalefactor); dot((-13.041975271686374,-2.2874242730822445),dotstyle); label("$K$", (-12.960077511214873,-2.096465005887293), NE * labelscalefactor); dot((-3.417217522730037,-2.550723633078282),dotstyle); label("$D$", (-3.342861183956523,-2.3485291434968882), NE * labelscalefactor); dot((-1.5804002441062543,1.9728418072278557),dotstyle); label("$E$", (-1.5008540245017985,2.1692357844289307), NE * labelscalefactor); dot((-5.792280011836994,0.9552956345685196),dotstyle); label("$F$", (-5.708386167677327,1.1415896849436586), NE * labelscalefactor); dot((-6.001587925321318,3.36360849606723),dotstyle); label("$L$", (-5.921671207193137,3.5652833158051496), NE * labelscalefactor); dot((1.0603691190249185,-5.43459088564898),dotstyle); label("$A'$", (1.1361246458754912,-5.237571951483785), NE * labelscalefactor); dot((-4.348502014831068,1.3040973231349209),dotstyle); label("$P$", (-4.273559538207331,1.4906015677877134), NE * labelscalefactor); dot((-4.598972421296053,2.3408573894108984),dotstyle); label("$Q$", (-4.5256236758169255,2.5376372163198773), NE * labelscalefactor); dot((-10.384072030799011,-0.15403337105498183),dotstyle); label("$R$", (-10.303709291790693,0.03638538927081889), NE * labelscalefactor); dot((-3.686340127971625,1.4640687208981875),dotstyle); label("$M'$", (-3.6143148706130086,1.6651075092097407), NE * labelscalefactor); dot((-7.301856004362389,-1.5761927222131948),dotstyle); label("$Y$", (-7.220770993334891,-1.3790516911522919), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]
So what did we basically do in this post? We opened a way for dealing with foots on incircle chord and tried involving $A-$antipode. Also, this is what ELMO SL 2019 G3 is based upon. Also, I guess I will soon run outta post-wordlimit, so I'll have to continue this in Part (III) over here
This post has been edited 23 times. Last edited by AlastorMoody, Nov 17, 2019, 12:42 PM

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

The Well-Known GCD Trick

by AlastorMoody, Oct 14, 2019, 5:57 AM

Sometimes, with questions involving divisibility, It's often a good idea to invoke the GCD! I don't know if this works always but here are some problems that work!
Switzerland MO Finals 2014 wrote:
Let $a,b\in\mathbb{N}$ such that :
\[ ab(a-b)\mid a^3+b^3+ab \]Then show that $\operatorname{lcm}(a,b)$ is a perfect square.
Solution: Let $\gcd(a,b)=d$ $\implies$ $\begin{cases} a=xd \\ b=yd \\ \gcd(x,y)=1 \end{cases}$, Hence, now we have,
$$ab(a-b)\mid a^3+b^3+ab \implies \frac{x^3d^3 + y^3d^3+xyd^2}{xyd^2(x-y)d} \in \mathbb{Z}$$$$\frac{x^3d+y^3d+xy}{xyd(x-y)} \in \mathbb{Z}$$Now, since, $\gcd(x,y)=1$ and $x \mid xyd(x-y)$ $\implies$ $x \mid x^3d+y^3d+xy \implies x \mid d$. Similarly, $y \mid d$. Using, $\gcd(x,y)=1$ $\implies$ $xy \mid d$. But, $d$ $\mid$ $xyd(x-y)$ $\implies$ $d $ $\mid$ $x^3d+y^3d+xy$ $\implies$ $d $ $\mid$ $xy$ $\implies$ $\boxed{xy=d}$ $\implies$ $\text{lcm} ~ (a,b)$ $=$ $xyd$ $=$ $(xy)^2$
It Turns out that when we're given some divisibility and they ask us to prove a perfect square, we must always (?) invoke GCD (:P)
BWM 2013 P5 wrote:
Let $m,n \in \mathbb{Z}^+$, such,
$$mn | m^2+n^2+m$$Prove, $m$ is a perfect square
Solution: Let $\gcd (m,n)=d$ $\implies$ $\begin{cases} m=ad \\\ n=bd \\\ \gcd (a,b)=1 \end{cases}$
$$\frac{m^2+n^2+m}{mn} =\frac{a^2d^2+b^2d^2+ad}{abd^2} =\frac{a^2d+b^2d+a}{abd} \in \mathbb{Z}$$Since, $a$ divides $abd$ $\implies$ $a ~ | ~ a^2d+b^2d+a $ but since, $\gcd (a,b)=1$ $\implies$ $a ~ | ~ d$ And, Since, $d$ divides $abd$ $\implies$ $d ~ | ~ a^2d$ $+b^2d+$ $a$ $\implies$ $d ~ | ~ a$
$$\implies a=d \implies m=ad=a^2=d^2$$Hence done! $\qquad \blacksquare$
Bosnia-Herzegovina Regional Olympiad 2016 Grade 10 P2 wrote:
Let $a$ and $b$ be two positive integers such that $2ab$ divides $a^2+b^2-a$. Prove that $a$ is perfect square
Solution: Let $\gcd (a,b)=d$ $\implies$ $\begin{cases} a=md \\\ b=nd \\\ \gcd (m,n)=1 \end{cases}$
$$\frac{a^2+b^2-a}{2ab} =\frac{m^2d^2+n^2d^2-md}{2mnd^2} =\frac{m^2d+n^2d-m}{2mnd} \in \mathbb{Z}$$Since, $m$ divides $2mnd$ $\implies$ $m ~ | ~ m^2d+n^2d-m $ but since, $\gcd (m,n)=1$ $\implies$ $m ~ | ~ d$ And, Since, $d$ divides $2mnd$ $\implies$ $d ~ | ~ m^2d$ $+n^2d-$ $m$ $\implies$ $d ~ | ~ m$
$$\implies m=d \implies a=md=m^2=d^2$$Hence done! $\qquad \blacksquare$
Now, This Trick doesnot only work for perfect squares, but also for some higher powers :o Below's an example (excuse the stupid solution, b'coz I am n00b)
Indonesia MO 2010 P4 wrote:
Given that $m$ and $n$ are positive integers with property:
\[(mn)\mid(m^{2010}+n^{2010}+n)\]Show that there exists a positive integer $k$ such that $n=k^{2010}$
Solution: Let $\gcd (m,n)=d$ $\implies$ $\begin{cases} m=ad \\\ n=bd \\\ \gcd (a,b)=1 \end{cases}$, We somehow want to show exactly $b=d^{2009}$
$$\frac{m^{2010}+n^{2010}+n}{mn}=\frac{a^{2010}d^{2009}+b^{2010}d^{2009}+b}{abd} \in \mathbb{Z}$$Since, $b ~ | ~ abd$ $\implies$ $b ~ | ~ a^{2010}d^{2009}+b^{2010}d^{2009}+b$ $\implies$ $b ~ | ~ d^{2009}$ And since, $d ~ | ~ abd$ $\implies$ $d ~ | ~ a^{2010}d^{2009}+b^{2010}d^{2009}+b$ $\implies$ $d ~ | ~ b$, $\implies$ $d^2 ~ | ~ abd$ $\implies$ $d^2 ~ | ~b$ $\implies$ $d^3 ~ | ~ abd$ $\implies$ $d^3 ~ | ~ b$ $\implies$ $d^4 ~ | ~ abd$ $\implies$ $d^4 ~ | ~ b$....similarly, we can show $d^{2008} ~ | ~ abd$ $\implies$ $d^{2009} ~ | ~ abd$ $\implies$ $d^{2009} ~ | ~b$, and since, $b ~ | ~ d^{2009}$ $\implies$ $b=d^{2009}$ $\implies$ $n=bd=d^{2010}$ $\qquad \blacksquare$
Pls Lemme know if there are some other problems involving this trick too :omighty:

The Dumpty Parabola (So, sad to know :( That it's already discovered :( )

by AlastorMoody, Sep 11, 2019, 4:15 PM

This Parabola is already discovered :( and is also known as Artzt Parabola, But anyway, I'll like to call this "Dumpty Parabola" :P ...
Let's try to motivate this parabola!
uraharakisuke_hsgs wrote:
Let $d_1 , d_2$ be two abitrary lines that are not perpendicular . Let $A,B$ be abitrary points lie on $d_1,d_2$ . How to construct a parabola that is tangent to $d_1,d_2$ at $A,B$
Well, *coincidence* I was also working on the same construction, 'coz It is almost impossible to construct a parabola tangent to two lines at given point on GEOGEBRA... :P While searching for some Lemmas, in SL Loney's Coordinate Geometry book, I found the following Lemma:
Similarity on Parabola Lemma wrote:
Let $\mathcal{P}$ be a Parabola with $X,Y \in \mathcal{P}$. Let the tangents at $X,Y$ meet at $A$. Then, $\Delta TAY \sim \Delta TXA$
Now, this Lemma clearly motivates the construction of a parabola with focus at Dumpty Point and so we have the following Problem
Problem: Let $\Delta ABC$ be a triangle. Let $D$ be the midpoint of $A-$symmedian Chord. Let $M$ be midpoint of $BC$. Let $AM$ intersect Nine-Point Circle at $M'$. Let $\ell$ be perpendicular to $AM$ at $M'$. Prove that, There Exists a Parabola, $\mathcal{P}$ with Focus at $D$ tangent to $AB,AC$ at $B,C$ and Directrix as $\ell$

Proof: (By Supercali)
Supercali wrote:
Let $B'$ and $C'$ be midpoints of $AC$ and $AB$ respectively. Let $\ell$ meet $AC,AB$ at $E,F$ respectively. Let $AM$ meet $BD$ at $P$ and $CD$ at $Q$. By angle chasing (using property of symmedians, midpoint theorem and the fact that $M',B',C',M$ are concyclic), we can easily see that $E,M',D,Q,B'$ are concyclic. Hence $\angle EDC= \angle EM'Q = 90$. Similarly $\angle FDB = 90$. Hence it will be sufficient to show that $B,C$ lie on the parabola with focus $D$ and directrix $\ell$.

Let $G$ be foot from $C$ on $\ell$. $\angle EDC=\angle EGC = 90$. By angle chasing, we can easily see that $\angle ECD=\angle ECG$, hence $EGCD$ is a kite, which implies $CG=CD$, hence $C$ lies on the parabola. Similarly, $B$ also lies on the parabola, and we are done.
Proof: (By MathPassionForever)
MathPassionForever wrote:
Consider an inparabola of $AB'C'$ with directrix as perpendicular bisector of $AM'$. Note that since the foci of the parabola are isogonal conjugates, if the axis is parallel to the $A$ median, the other focus is on the $A$ symmedian. Also, since the isogonal conjugate of a point at infinity is on the circumcircle, the focus lies on $\odot(AB'C')$, which is just a homothety of ratio $\frac{1}{2}$ of $\odot(ABC)$ from $A$. So clearly $D$ lies on the former. So $D$ is the focus of this inparabola!
This discussion could have ended here, but AoPS guys are just too amazing and they found many more properties which I couldn't have managed to find out nor could I have proved them!! >_< - https://artofproblemsolving.com/community/q1h1912112p13117117
Here's a Property that can be killed using the SL Loney Similarity Lemma
Aryan-23 wrote:
Prove that the circumcircle of the triangle formed by tangents at any 3 points on the parabola passes through its focus??
These are complete Supercali's efforts:
Supercali wrote:
Another interesting property: $B'C'$ is tangent to the parabola and point of intersection lies on $AM$.

This gives another interesting problem:
In $\Delta ABC$, Let $X$ be a point such that $ABXC$ is harmonic. Let $\ell$ be the line passing through the orthocentre of the triangle which is perpendicular to the $A$-median. Prove that the parabola with focus $X$ and directrix $\ell$ is tangent to the sides of $\Delta ABC$
Proof: Let $Q$ be the midpoint of $AM$. Using angle chasing, we can get $QF=QD$ (you will have to use concyclicities used in my original solution). Let $\ell$ meet $B'C'$ at $K$. Then, by angle chasing, $FQDK$ is a kite. Hence $\angle QDK=90$ and we are done.
Supercali wrote:
The niceness continues:
1) Let $BB'$ meet the parabola at $X$ Then $CX$ passes through the centroid of $\Delta AB'C'$
2) Let $X_a$ be the $A$-Humpty point and let $H$ be the orthocentre. Then $Q,X_a$, foot from $A$ onto $BC$, $K$ defined above and midpoint of $AH$ are concyclic.
Supercali wrote:
Another thing:
Let $AX$ meet the parabola again at $X'$. Then $BX'$ is parallel to $AC$.
Also, if $CX'$ meets $AM$ at $N$, then $A$ is the midpoint of $MN$.
Supercali wrote:
Let us call it the Artzt parabola from now on. Here are some properties of its tangents and tangency points:
1) (Well known) 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.
2) If $S$ is the point of tangency, and $CS$ meets $LN$ at $N'$, then $N$ is the midpoint of $LN'$.
3) (Unsolved, but my favourite) Apparently $B'$ (midpoint of $AC$), $Q$ (midpoint of $B'C'$), $S$, $B$, $C$, $N$, $Q_B$ (intersection of $LN$ and $AQ$) lie on a conic.
4) (Unsolved) Apparently the locus of the centre of the above conic as $L$ varies on $BC$ is a line passing through the midpoint of $BC$ (not shown in figure).
These are math_pi_rate's efforts at providing such amazing proofs to these properties:
math_pi_rate wrote:
Here are (brief) proofs of some of the properties (Assuming that the properties in post #1 and post #3 are solved; Also, I'll use the diagram in post #10):
Supercali wrote:
Let $BB'$ meet the parabola at $T$ Then $CT$ passes through the centroid of $\triangle AB'C'$
Let $\mathcal{P}$ denote the parabola from now on. Let $AQ \cap \mathcal{P}=Z$ and $BQ \cap A \infty_{BC}=U$. It's easy to see that $AUMB$ is a parallelogram, which directly gives that $AUCM$ is also a parallelogram (i.e. $CU \parallel AM$). By Pascal on $QZCBBQ$, we get that $BQ \cap CZ$ lies on $AU$, which gives $CZ \parallel AM$, or that $Z=\infty_{AM}$.

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 (I am seriously not in the mood of writing that up :D).
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 is from MathPassionForever's Blog,
Quote:
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(12cm,12cm); 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 = 17.707373326322468, 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(1,0,0.2); 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 * labelscalefactor*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.

My partial proof : Consider the inparabola of $\Delta AB'C'$ with its axis parallel to the $A$-median. Then its focus must lie on the $A$-symmedian as well as on $\odot(AB'C')$ and hence is $D$. Since its directrix is perpendicular to $AM$ and passes through $H_a$ (Orthocenter always lies on directrix of inparabola), the directrix must be $l$. From here, I have only Supercali's proof for tangency to $B,C$. For tangency at $Q$, note that $B'C$ is the perpendicular bisector of $DM'$ and that $QM \perp l$

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:
Such a Clean report, I guess I'll ever be able to present this so neatly :P
This post has been edited 9 times. Last edited by AlastorMoody, Oct 14, 2019, 6:50 AM

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

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  • 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

  • pro blogo

    by Aritra12, Sep 8, 2020, 11:17 AM

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