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

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5 Comments

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There is one property that I expected to find and wildly searched for , but has been missed most probably...

If $M$ is the midpoint of $BC$ , and $IM_{BC}$ intersects $EF$ at say $Z$ then $MZ$ is perpendicular to $EF$ ... I don't know how to latex ratios, otherwise I would have posted complete solution .... I'll leave a hint .. $AFIE$ and $M_{BC}BM_AC$ are similar...

by Aryan-23, Nov 18, 2019, 2:02 PM

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And the property mentioned instasolves JMO 2014 P6 :P

by Aryan-23, Nov 18, 2019, 2:03 PM

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Oh yeah, this is a pretty nice property! Once again, it skipped my mind while writing this blog post, so I included it in Part (V)

Also, I didn't a JMO prob. could be solved by this... I might wanna add that too in that case

by AlastorMoody, Nov 18, 2019, 4:13 PM

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Would you please explain a bit more detail why $AIEF'$ is cyclic? I'm struggling a bit in this part. Thanks!

by lolm2k, Sep 4, 2020, 12:55 AM

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This is really late, but math90's problem actually has a much easier solution. Note that $\overline{AD}$ is the $D$-symmedian of the intouch triangle and $\overline{MI_A}$ is the $I_A$-symmedian of the excentral triangle. However these two triangles are homothetic, as desired.

by i3435, Dec 30, 2020, 8:24 PM

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

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  • 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

120 shouts
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AlastorMoody • aops29 • Pluto1708 • SenatorPauline
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About Owner
  • Posts: 2125
  • Joined: Oct 11, 2017
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  • Blog created: Nov 19, 2018
  • Total entries: 25
  • Total visits: 12261
  • Total comments: 60
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