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); dot((-2.3223836468452834,2.441178544388017),dotstyle); label("$R$", (-2.204152990371713,2.7313052349756726), NE * labelscalefactor); dot((-2.7916013606178858,6.723704791200539),dotstyle); label("$M$", (-2.6850183599721373,7.002521164955902), NE * labelscalefactor); dot((-3.26088654950535,9.277294081659807),dotstyle); label("$N$", (-3.137597531360772,9.548279004016965), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]
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

Comment

J
U VIEW ATTACHMENTS T PREVIEW J CLOSE PREVIEW rREFRESH
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5 Comments

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The "Alastair Behena " strikes again with a high lavil post :D

by Aryan-23, Nov 19, 2019, 11:20 AM

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Hawww I go offine for a month and you do this? :bomb:

by Pluto1708, Dec 15, 2019, 9:29 PM

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lol you have discord? add me @bakekaga#3392

Edit: Yeah I do have, but I rarely use discord. Anyway, I'll do it
This post has been edited 1 time. Last edited by AlastorMoody, Jan 3, 2020, 6:27 AM

by Kagebaka, Jan 3, 2020, 3:54 AM

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Hope you bring this series back with some Bevan point (V) configurations :jump:

1) $\angle VAk = 90^{\circ}$, this has its application in ELMOSL/2013
2) $(AVT_{A}L)$ exists

and many more (probably)

by Dr_Vex, Jun 20, 2020, 10:25 AM

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Quadrilateral $BCYX$ is circumscribed and $D,E,Q,F$ are points of tangency of incircle with sides $BC,CY,YX,XB$ respectively. Then $DQ,FE,CX,BY$ are concurrent.

Proof
Brianchon's theorem in degenerated hexagon $BDCEYQX.\blacksquare$

by WolfusA, Sep 24, 2020, 8:38 PM

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

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  • Kukuku first shout of 2025

    by HoRI_DA_GRe8, Yesterday at 6:41 PM

  • what a goat, u used to be friends with my brother :)

    by bookstuffthanks, Jul 31, 2024, 12:05 PM

  • hello fellow moody!!

    by crazyeyemoody907, Oct 31, 2023, 1:55 AM

  • @below I wish I started earlier / didn't have to do JEE and leave oly way before I could study conics and projective stuff which I really wanted to study :( . Huh, life really sucks when u are forced due to peer pressure to read sh_t u dont want to read

    by kamatadu, Jan 3, 2023, 1:25 PM

  • Lots of good stuffs here.

    by amar_04, Dec 30, 2022, 2:31 PM

  • But even if he went to jee he could continue with this.

    Doing JEE(and completely leaving oly) seems like a insult to the oly math he knows

    by HoRI_DA_GRe8, Feb 11, 2022, 2:11 PM

  • Ohhh did he go for JEE? Good for him, bad for us :sadge:. Hmmm so that is the reason why he is inactive
    Btw @below finally everyone falls to the monopoly of JEE :) Coz IIT's are the best in India.

    by BVKRB-, Feb 1, 2022, 12:57 PM

  • Kukuku first shout of 2022,why did this guy left this and went for trashy JEE

    by Commander_Anta78, Jan 27, 2022, 3:42 PM

  • When are you going to br alive again ,we miss you

    by HoRI_DA_GRe8, Aug 11, 2021, 5:10 PM

  • kukuku first shout o 2021

    by leafwhisker, Mar 6, 2021, 5:10 AM

  • wow I completely forgot this blog

    by Math-wiz, Dec 25, 2020, 6:49 PM

  • buuuuuujmmmmpppp

    by DuoDuoling0, Dec 22, 2020, 10:54 PM

  • This site would work faster if not all diagrams were displayed on the initial page. Anyway I like your problem selection taste.

    by WolfusA, Sep 24, 2020, 7:58 PM

  • nice blog :)

    by Orestis_Lignos, Sep 15, 2020, 9:09 AM

  • Hello everyone, nice blog :)

    by Functional_equation, Sep 12, 2020, 6:22 PM

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