Long, but nice!

This was 2 pages of angle chasing

I did with orthic triangles which seem unnecesary, i may lose marks for this on

Angle chasing FTW! [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(8cm); 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 = -7.812451230515015, xmax = 40.32502306282564, ymin = -21.674846778797146, ymax = 10.114641114922083; /* image dimensions */ pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen dbwrru = rgb(0.8588235294117647,0.3803921568627451,0.0784313725490196); pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902); /* draw figures */ draw((7.363889857435007,2.86)--(3.563889857435007,-4.58), linewidth(0.8) + dtsfsf); draw((3.563889857435007,-4.58)--(15.8,-4.34), linewidth(0.8) + dtsfsf); draw((15.8,-4.34)--(7.363889857435007,2.86), linewidth(0.8) + dtsfsf); draw((-0.15678186814230818,1.3664800850548786)--(11.295241694614747,-16.936459564625576), linewidth(0.8) + rvwvcq); draw((11.295241694614747,-16.936459564625576)--(19.21664759651896,5.2138229530930476), linewidth(0.8) + rvwvcq); draw((19.21664759651896,5.2138229530930476)--(-0.15678186814230818,1.3664800850548786), linewidth(0.8) + rvwvcq); draw(circle((9.234836452766224,-2.208569644298099), 8.047579723129681), linewidth(0.8) + dbwrru); draw((2.2866247921184977,1.8517128728387287)--(1.2875649212449662,-0.9419147566512097), linewidth(0.8) + rvwvcq); draw((6.898563651387755,-9.909567553508774)--(14.336439716041642,-8.432489915659678), linewidth(0.8) + rvwvcq); draw((17.12632609122058,-0.6312520839141611)--(14.104335875233762,4.198575958442354), linewidth(0.8) + rvwvcq); draw((2.2866247921184977,1.8517128728387287)--(14.336439716041642,-8.432489915659678), linewidth(0.8) + sexdts); draw((14.104335875233762,4.198575958442354)--(6.898563651387755,-9.909567553508774), linewidth(0.8) + sexdts); draw((1.2875649212449662,-0.9419147566512097)--(17.12632609122058,-0.6312520839141611), linewidth(0.8) + sexdts); /* dots and labels */ dot((7.363889857435007,2.86),dotstyle); label("$A$", (7.504017862820649,3.17534287263529), NE * labelscalefactor); dot((3.563889857435007,-4.58),dotstyle); label("$B$", (3.690529639581973,-4.264084972699201), NE * labelscalefactor); dot((15.8,-4.34),dotstyle); label("$C$", (15.912446814060024,-4.0140201711753525), NE * labelscalefactor); dot((5.4638898574350065,-0.86),linewidth(4pt) + dotstyle); label("$D$", (5.59727375120131,-0.6068872504129175), NE * labelscalefactor); dot((11.581944928717505,-0.74),linewidth(4pt) + dotstyle); label("$E$", (11.692603288345095,-0.48185484965099323), NE * labelscalefactor); dot((9.681944928717504,-4.46),linewidth(4pt) + dotstyle); label("$F$", (9.817117276916239,-4.201568772318239), NE * labelscalefactor); dot((-0.15678186814230818,1.3664800850548786),linewidth(4pt) + dotstyle); label("$I_C$", (-0.02918428308525973,1.6124378631112373), NE * labelscalefactor); dot((19.21664759651896,5.2138229530930476),linewidth(4pt) + dotstyle); label("$I_B$", (19.35083783501293,5.457184186540407), NE * labelscalefactor); dot((11.295241694614747,-16.936459564625576),linewidth(4pt) + dotstyle); label("$I_A$", (11.411280386630766,-16.673550748320178), NE * labelscalefactor); dot((2.2866247921184977,1.8517128728387287),linewidth(4pt) + dotstyle); label("$J$", (2.408947531772254,2.112567466158934), NE * labelscalefactor); dot((14.104335875233762,4.198575958442354),linewidth(4pt) + dotstyle); label("$K$", (14.224509403774052,4.456924980445013), NE * labelscalefactor); dot((17.12632609122058,-0.6312520839141611),linewidth(4pt) + dotstyle); label("$L$", (17.256545122250703,-0.3880805490795501), NE * labelscalefactor); dot((1.2875649212449662,-0.9419147566512097),linewidth(4pt) + dotstyle); label("$M$", (1.4086883256768639,-0.7006615509843606), NE * labelscalefactor); dot((6.898563651387755,-9.909567553508774),linewidth(4pt) + dotstyle); label("$N$", (7.035146359963434,-9.671736305652422), NE * labelscalefactor); dot((14.336439716041642,-8.432489915659678),linewidth(4pt) + dotstyle); label("$O$", (14.4745742052979,-8.171347496509332), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Proof: Let $D,E,F$ be the midpoints of $AB,AC,BC$ respectively and let $I_A,I_B,I_C$ be the $A$-excentre $B$-excentre and $C$-excentre respectively.. Let $DE \cap I_AI_C=M,DE \cap I_AI_B=L,DF \cap I_CI_B=J,DF \cap I_AI_B=O,EF \cap I_CI_B=K,EF \cap I_CI_A=N$. It suffices to prove $M,L,J,O,K,N$ concyclic. Now observe that $\angle CBD=\angle BDM$ as $DE \parallel BC$ by thales theorem and $\angle MBA \equiv \angle I_CBA=90-\frac{\angle CBD}{2} $ because we know that $I_CB$ is the bisector of exterior angle at point $B$. Now these two condition together imply that $DM=DB$. Similarly we have that since $DF \parallel AC$ by thales $\implies \angle DAC \equiv \angle BAC=\angle JDA$ and also since $I_CA$ is the exterior angle bisector $\implies \angle JAD \equiv \angle I_CAB=90-\frac{\angle CAD}{2} $ . Now these two conditions again imply $DJ=DA$ and since $D$ is midpoint we have $DM=DB=DA=DJ$. Now in similiar fashion we can prove that $FB=FN=FO=FC$ and $EC=EK=EL=EA$.Now observe that $\angle MDJ=180- \angle MDB-\angle JDA=180-\angle ABC-\angle BAC$(as proved earlier)$=\angle ACB=\angle LEC$($DE \parallel BC $ by thales theorem)$\implies \angle DMJ=\angle ELC \implies JM \parallel LO$. In a similiar fashion we can prove that $KL \parallel MN$ and $JK \parallel NO \implies JKLOMN$ is a lemoine hexagon with respect to reference triangle $I_AI_BI_C \implies$ by well known fact that $JKLOMN$ is cyclic. Done. $\blacksquare$.

This one was nice, but too easy! Sharygin 2020 CR P7 wrote: Prove that the medial lines of triangle $ABC$ meets the sides of triangle formed by its excenters at six concyclic points. [asy][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 = -11.28, xmax = 11.64, ymin = -6.77, ymax = 5.45; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw((-3.36,3.63)--(-4.34,-1.19), linewidth(0.4) + wrwrwr); draw((2.5,-1.21)--(-3.36,3.63), linewidth(0.4) + wrwrwr); draw((-4.34,-1.19)--(2.5,-1.21), linewidth(0.4) + wrwrwr); draw((xmin, 0.35365403396675704*xmin + 4.818277554128303)--(xmax, 0.35365403396675704*xmax + 4.818277554128303), linewidth(0.4) + wrwrwr); /* line */ draw((xmin, -2.8276222068882113*xmin-5.870810615144388)--(xmax, -2.8276222068882113*xmax-5.870810615144388), linewidth(0.4) + wrwrwr); /* line */ draw((xmin, -1.2274376565186662*xmin-6.5170794292910115)--(xmax, -1.2274376565186662*xmax-6.5170794292910115), linewidth(0.4) + wrwrwr); /* line */ draw((xmin, 0.8147053291783969*xmin + 2.3458211286342427)--(xmax, 0.8147053291783969*xmax + 2.3458211286342427), linewidth(0.4) + wrwrwr); /* line */ draw((xmin, 2.7683446963858023*xmin-8.130861740964505)--(xmax, 2.7683446963858023*xmax-8.130861740964505), linewidth(0.4) + wrwrwr); /* line */ draw((xmin, -0.36122669308686334*xmin-0.30693326728284176)--(xmax, -0.36122669308686334*xmax-0.30693326728284176), linewidth(0.4) + wrwrwr); /* line */ draw((xmin, -0.0029239766081871374*xmin + 1.2087426900584792)--(xmax, -0.0029239766081871374*xmax + 1.2087426900584792), linewidth(0.4) + wrwrwr); /* line */ draw((xmin, -0.8259385665529011*xmin-1.9598634812286688)--(xmax, -0.8259385665529011*xmax-1.9598634812286688), linewidth(0.4) + wrwrwr); /* line */ draw((xmin, 4.918367346938776*xmin + 3.3248979591836734)--(xmax, 4.918367346938776*xmax + 3.3248979591836734), linewidth(0.4) + wrwrwr); /* line */ draw(circle((-1.4720630335387321,0.3610256932361879), 4.91417211346492), linewidth(0.4) + red); /* dots and labels */ dot((-3.36,3.63),dotstyle); label("$A$", (-3.28,3.83), NE * labelscalefactor); dot((-4.34,-1.19),dotstyle); label("$B$", (-4.26,-0.99), NE * labelscalefactor); dot((2.5,-1.21),dotstyle); label("$C$", (2.82,-1.05), NE * labelscalefactor); dot((-3.85,1.22),linewidth(4pt) + dotstyle); label("$M_C$", (-4.22,1.71), NE * labelscalefactor); dot((-0.92,-1.2),linewidth(4pt) + dotstyle); label("$M_A$", (-0.92,-1.93), NE * labelscalefactor); dot((-0.43,1.21),linewidth(4pt) + dotstyle); label("$M_B$", (-0.1,1.63), NE * labelscalefactor); dot((-6.309298332962932,1.2271909307981368),linewidth(4pt) + dotstyle); label("$P$", (-6.74,0.79), NE * labelscalefactor); dot((-5.7461712055254734,2.7861209274305963),linewidth(4pt) + dotstyle); label("$S$", (-5.82,3.17), NE * labelscalefactor); dot((0.32715736839392257,4.933978077202762),linewidth(4pt) + dotstyle); label("$X$", (0.4,5.09), NE * labelscalefactor); dot((3.370154803840357,1.1988884362460794),linewidth(4pt) + dotstyle); label("$Q$", (3.62,1.37), NE * labelscalefactor); dot((1.716892578658477,-3.377911276571167),linewidth(4pt) + dotstyle); label("$R$", (2.22,-3.43), NE * labelscalefactor); dot((-1.6014138722165585,-4.551443738861035),linewidth(4pt) + dotstyle); label("$Y$", (-2.02,-4.87), NE * labelscalefactor); dot((-7.169322975784636,2.282817562931512),linewidth(4pt) + dotstyle); label("$I_C$", (-8.02,2.33), NE * labelscalefactor); dot((0.4038714247037156,-7.012806424364196),linewidth(4pt) + dotstyle); label("$I_A$", (0.48,-6.85), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Let $I_A,I_B,I_C$ be the $A,B,C\text{ excenters}$ of $\triangle ABC$ respectively and let $M_AM_BM_C$ be the medial triangle of $\triangle ABC$ where $M_A\in BC, M_B\in CA$ and $M_C\in AB$. Now consider the following Notations. $$\text{Notations} \begin{cases} M_BM_C\cap I_AI_C=P \\ M_BM_C\cap I_AI_B=Q \\ M_AM_B\cap I_CI_A=Y \\ M_AM_B\cap I_CI_B=X \\ M_AM_C\cap I_AI_B=R \\ M_AM_C\cap I_BI_C=S\end{cases}$$So we need to prove that $P,S,X,Q,R,Y$ are concyclic. For this do some length chasing and apply Power of Point. Notice that $\angle M_CAS=90^\circ-A/2\implies \angle M_CSA=A\implies M_CAS$ is an isosceles triangle. Similarly $\{AM_BX, PM_CB, BM_AY, RM_AC,CM_BQ\}$ are isosceles triangles. So, $SM_C=c/2$ and similarly we get lengths for the below sidelengths. $$\begin{cases}M_CS.M_CR=\frac{c}{2}.\frac{a+b}{2}=M_CP.M_CQ\implies S,P,R,Q\text{ are concyclic} \\ M_AY.M_AX=\frac{a}{2}.\frac{b+c}{2}=M_AR.M_AS\implies X,S,Y,R\text{ are concyclic} \\ M_BX.M_BY=\frac{b}{2}.\frac{a+c}{2}=M_BQ.M_BP\implies X,Q,Y,P\text{ are concyclic}\end{cases}$$ Now $\angle PSX=\angle PSR+\angle RSX=\angle PQR+RSX=90^\circ-\angle C/2+90^\circ-\angle A/2=90^\circ+\angle B/2=180^\circ-\angle PYX\implies \{S,X,P,Y\}$ are concyclic. Similarly we get that $\{X,Q,R,S\},\{Q,Y,R,P\}$ are concyclic. So, as Three Unique Points determine a circle we get that $\{P,S,X,Q,R,Y\} \text{ are concyclic.}$

[asy][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 = -12.18752694271128, xmax = 17.648615995462283, ymin = -8.625492909812813, ymax = 9.445304594441282; /* image dimensions */ /* draw figures */ draw((-10.83024789548552,-7.373480672750021)--(9.487129307843334,-7.435985044022045), linewidth(1)); draw((-10.83024789548552,-7.373480672750021)--(-2.1656586469707997,7.0133918454982735), linewidth(1)); draw((-2.1656586469707997,7.0133918454982735)--(9.487129307843334,-7.435985044022045), linewidth(1)); draw(circle((-2.0107608258807215,-1.4528634932016813), 6.875135138901312), linewidth(1) + linetype("4 4")); draw((-0.7951312195201095,5.313947265072489)--(-5.477533615786717,-7.389947760863962), linewidth(1)); draw((-5.477533615786717,-7.389947760863962)--(-8.547518676349371,-3.5831874081244757), linewidth(1)); draw((-8.547518676349371,-3.5831874081244757)--(-2.21,-7.4), linewidth(1)); draw((-5.45,1.56)--(-5.477533615786717,-7.389947760863962), linewidth(1)); draw((1.4194170022348027,-7.4111655370443925)--(1.45,2.53), linewidth(1)); draw((4.859989700360052,-1.6983637667609792)--(1.4194170022348027,-7.4111655370443925), linewidth(1)); draw((-0.7951312195201095,5.313947265072489)--(-3.1847329885559694,5.321298635018977), linewidth(1)); draw((-5.45,1.56)--(-0.7951312195201095,5.313947265072489), linewidth(1)); draw((-2.21,-7.4)--(4.859989700360052,-1.6983637667609792), linewidth(1)); draw((1.45,2.53)--(-3.1847329885559694,5.321298635018977), linewidth(1)); draw((-5.45,1.56)--(-2.21,-7.4), linewidth(1)); draw((-2.21,-7.4)--(1.45,2.53), linewidth(1)); draw((1.45,2.53)--(-5.45,1.56), linewidth(1)); draw((-3.1847329885559694,5.321298635018977)--(1.4194170022348027,-7.4111655370443925), linewidth(1)); draw((-8.547518676349371,-3.5831874081244757)--(4.859989700360052,-1.6983637667609792), linewidth(1)); draw((-2.1656586469707997,7.0133918454982735)--(-2.21,-7.4), linewidth(1)); draw((9.487129307843334,-7.435985044022045)--(-5.45,1.56), linewidth(1)); draw((-10.83024789548552,-7.373480672750021)--(1.45,2.53), linewidth(1)); draw(circle((-3.83,-2.92), 4.7639059604488425), linewidth(1)); draw(circle((-2,2.045), 3.4839237936556535), linewidth(1)); draw(circle((-0.38,-2.435), 5.29151443350578), linewidth(1)); /* dots and labels */ dot((-2.21,-7.4),dotstyle); label("$A$", (-2.16161656267725,-8.166475326148605), NE * labelscalefactor); dot((-5.45,1.56),dotstyle); label("$B$", (-6.1719807146908625,1.7627997731140057), NE * labelscalefactor); dot((1.45,2.53),dotstyle); label("$C$", (1.534682926829272,2.7774702212138345), NE * labelscalefactor); dot((-2.1656586469707997,7.0133918454982735),linewidth(4pt) + dotstyle); label("$I_{A}$", (-2.0649812819058377,7.198534316505946), NE * labelscalefactor); dot((9.487129307843334,-7.435985044022045),linewidth(4pt) + dotstyle); label("$I_{B}$", (9.57957005104935,-7.2484401588201886), NE * labelscalefactor); dot((-10.83024789548552,-7.373480672750021),linewidth(4pt) + dotstyle); label("$I_{C}$", (-11.462762336925687,-7.465869540555866), NE * labelscalefactor); dot((-2,2.045),linewidth(4pt) + dotstyle); label("$M$", (-1.4610107770845104,1.5695292115711812), NE * labelscalefactor); dot((-0.38,-2.435),linewidth(4pt) + dotstyle); label("$N$", (0.08515371525808697,-2.9240113442994895), NE * labelscalefactor); dot((-3.83,-2.92),linewidth(4pt) + dotstyle); label("$L$", (-3.4420340328984635,-3.4796642087351097), NE * labelscalefactor); dot((-3.1847329885559694,5.321298635018977),linewidth(4pt) + dotstyle); label("$Z_{1}$", (-4.04600453771979,5.869799205899027), NE * labelscalefactor); dot((-0.7951312195201095,5.313947265072489),linewidth(4pt) + dotstyle); label("$Y_{2}$", (-0.6879285309132117,5.507416903006232), NE * labelscalefactor); dot((4.859989700360052,-1.6983637667609792),linewidth(4pt) + dotstyle); label("$Y_{1}$", (4.96523539421441,-1.4986409529211584), NE * labelscalefactor); dot((1.4194170022348027,-7.4111655370443925),linewidth(4pt) + dotstyle); label("$X_{2}$", (1.534682926829272,-8.04568122518434), NE * labelscalefactor); dot((-5.477533615786717,-7.389947760863962),linewidth(4pt) + dotstyle); label("$X_{1}$", (-5.785439591605213,-8.04568122518434), NE * labelscalefactor); dot((-8.547518676349371,-3.5831874081244757),linewidth(4pt) + dotstyle); label("$Z_{2}$", (-9.360944980147469,-3.527981849120816), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $\measuredangle$ denote a directed angle modulo 180 degrees, $I_A,I_B,I_C$ be the $A,B,C$ excenters and $M,N,L$ be the midpoints of $BC,CA,AB.$ Also, let $X_1,Y_2$ be the intersections of $ML$ with $I_BI_C, I_BI_A; Y_1,Z_2$ be the intersections of $LN$ with $I_AI_B,I_AI_C;Z_1,X_2$ be the intersections of $NM$ with $I_CI_A,I_CI_B.$ Observe that since $MN||BA$ and $B$ is famously the foot of the altitude from $I_B$ to $I_AI_C,$ we get that $\measuredangle BZ_1M=\measuredangle Z_2BA=\measuredangle MBZ_1;$ this means that $BM=Z_1M=CM,$ so $\measuredangle BZ_1C=90^\circ,$ and the same argument replicated for $Y_2$ gives us $Z_1,Y_2$ lie on a circle with diameter $CB$. By symmetry, we find that $X_1Z_2BA$ and $Y_1X_2AC$ are concyclic as well, so this leads to our main claim: Claim: $X_1X_2Y_1Z_2,Y_1Y_2Z_1X_2,Z_1Z_2X_1Y_2$ are cyclic. Proof: We just angle chase: \begin{align*} \measuredangle AX_1Z_2 &= \measuredangle ABZ_2\\ &= 90^\circ-\measuredangle ABI_B \\ &= 90^\circ-\measuredangle ACI_C + 90^\circ-\measuredangle I_AAC\\ &= \measuredangle X_2CA +\measuredangle NAY_1 \\ &= \measuredangle X_2Y_1A+\measuredangle AY_1Z_2 \\ &= \measuredangle X_2Y_1Z_2, \end{align*}so $X_1X_2Y_1Z_2$ is cyclic, and so are the other two quadrilaterals by symmetry. $\Box$ Now, we can finally complete the proof: simply observe that again using $MN||BA$, we get that $\measuredangle Z_2Y_1X_2=\measuredangle Z_2X_1X_2=\measuredangle Z_2BA=\measuredangle Z_2Z_1X_2,$ so $X_1,X_2,Y_1,Y_2,Z_1,Z_2$ are concyclic as desired. $\blacksquare$

Here is my solution ....

Problem 7. Prove that the medial lines of triangle $ABC$ meet the sides of triangle formed by its excenters at six concyclic points. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); 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 = -21.615631720578712, xmax = 13.226626084303125, ymin = -14.420230824169898, ymax = 11.972475959040283; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0); pen qqzzqq = rgb(0,0.6,0); draw((-2.43,3.85)--(-5.53,-2.55)--(4.23,-3.05)--cycle, linewidth(0.8) + zzttqq); /* draw figures */ draw((-2.43,3.85)--(-5.53,-2.55), linewidth(0.8) + zzttqq); draw((-5.53,-2.55)--(4.23,-3.05), linewidth(0.8) + zzttqq); draw((4.23,-3.05)--(-2.43,3.85), linewidth(0.8) + zzttqq); draw(circle((-0.9830141890045835,-0.7141203844324278), 6.72800005412083), linewidth(0.8) + linetype("4 4") + red); draw((-8.722302598535961,2.8454233067028105)--(8.138630619085153,5.537299654417491), linewidth(0.8) + qqzzqq); draw((8.138630619085153,5.537299654417491)--(0.11496949890156802,-12.09076221156731), linewidth(0.8) + qqzzqq); draw((0.11496949890156802,-12.09076221156731)--(-8.722302598535961,2.8454233067028105), linewidth(0.8) + qqzzqq); draw((-7.530972695450385,0.8319145848079092)--(5.688655122095117,0.1546795531713569), linewidth(0.8) + blue); draw((2.990248525075188,4.715351793703615)--(-2.7801205750444735,-7.197668283962785), linewidth(0.8) + blue); draw((2.743520641112734,-6.315809673224904)--(-6.4493235938603695,3.2083082278733563), linewidth(0.8) + blue); /* dots and labels */ dot((-2.43,3.85),dotstyle); label("$A$", (-2.3370584125116886,4.0813722750078), NE * labelscalefactor); dot((-5.53,-2.55),dotstyle); label("$B$", (-5.930545628624937,-2.91135960499637), NE * labelscalefactor); dot((4.23,-3.05),dotstyle); label("$C$", (4.4614309152701335,-3.178443114302085), NE * labelscalefactor); dot((-8.722302598535961,2.8454233067028105),linewidth(4pt) + dotstyle); label("$D$", (-8.625661040709873,3.0373185568127327), NE * labelscalefactor); dot((8.138630619085153,5.537299654417491),linewidth(4pt) + dotstyle); label("$E$", (8.224880364577927,5.7324339688976735), NE * labelscalefactor); dot((-3.98,0.65),linewidth(4pt) + dotstyle); label("$G$", (-4.1338020205683135,1.1191733536171446), NE * labelscalefactor); dot((-0.65,-2.8),linewidth(4pt) + dotstyle); label("$H$", (-0.7102770376496099,-3.3969659855522147), NE * labelscalefactor); dot((0.9,0.4),linewidth(4pt) + dotstyle); label("$I$", (1.3049894416571443,0.6335669730612996), NE * labelscalefactor); dot((-6.4493235938603695,3.2083082278733563),linewidth(4pt) + dotstyle); label("$J$", (-6.683235518486496,3.6443265325075393), NE * labelscalefactor); dot((-7.530972695450385,0.8319145848079092),linewidth(4pt) + dotstyle); label("$K$", (-8.164334979181822,0.6821276111168841), NE * labelscalefactor); dot((-2.7801205750444735,-7.197668283962785),linewidth(4pt) + dotstyle); label("$L$", (-3.1625892594566243,-7.694582453471444), NE * labelscalefactor); dot((2.743520641112734,-6.315809673224904),linewidth(4pt) + dotstyle); label("$M$", (3.0531724116581844,-6.650528735276377), NE * labelscalefactor); dot((5.688655122095117,0.1546795531713569),linewidth(4pt) + dotstyle); label("$N$", (5.893969737909874,-0.11912291680026032), NE * labelscalefactor); dot((2.990248525075188,4.715351793703615),linewidth(4pt) + dotstyle); label("$O$", (3.0774527306859767,4.906903121952737), NE * labelscalefactor); dot((0.11496949890156802,-12.09076221156731),linewidth(4pt) + dotstyle); label("$P$", (0.21237508540649444,-11.895077645279505), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Solution. Notations. Let the excircles of $\triangle ABC$ be $D,E,P$ as shown in the figure above. Let $G,I,H$ be the midpoints of sides $AB,AC$ and $BC$ respectively. Let medial lines $GI,IH,GH$ meet lines $DE,EP,PD$ at $J,O,N,M,L,K$ as shown. Also, for sake of simplicity, we assume $\angle BAC=2a, \angle ABC = 2b$ and $\angle BCA =2c$. Proof. We begin by noting that $\angle DAB=\angle CAE= 90-a, \angle ACE=\angle BCP=90-c, \angle CBP= \angle ABD=90-b$ because $D,E,P$ are excenters of $\triangle ABC$ and thus bisects the exterior angles. Also, as $\triangle GIH$ is the medial triangle of $ABC$, we must have $GI||BC, IH||AB$ and $GH||AC$. So, we have, $\angle GHB=\angle ACB=2c\implies \angle CHM=2c$ but, $\angle BCP=90-c\implies \angle HMC=90-c\implies HC=HM$. Similarly, we get, $HB=HL, GB=GK,GJ=GA,IA=IO,IN=IC$. But we also know that $G,I,H$ are the midpoints and so, $HG=HB,GB=GA,IA=IC\implies HB=HL=HM=HC, GB=GK=GJ=GA, IA=IO=IN=IC$. Now, as $IA=IO$ and $\angle EAC=90-a$, we have $\angle JOL=90-a$. Also note that, $\angle IHC=\angle 2b,\angle GHB=2c\implies \angle GHI=180-2b-2c=2a\implies \angle LHM=2a$. Now, as we have $HL=HM\implies \angle HLM=\angle HML=90-a=\angle JML$. Thus, we get $\angle JML=90-a=\angle JOL\implies JOML$ is cyclic quadrialteral. Similarly, we get $ONLK, JKMN$ are cyclic quadrialterals. For sake of simplicity, we denote the circlumcircles of $JOML,ONLK, JKMN$ by $C_1, C_2, C_3$ respectively. If possible, we assume $C_1, C_2, C_3$ to be pairwise distinct. Thus, they can intersect at maximum two points which indeed is the case here. Clearly, $C_1\cap C_2=O,L$ and $C_1\cap C_3=J,M$ and $C_2\cap C_3=N,K$. Since these circles are pairwise different, there are no more intersections possible. Thus, radical axis of $(C_1,C_2), (C_1,C_3), (C_2, C_3)$ will be $OL, JM, NK$ respectively. Thus, all these radical axises must concur at the radical center of $C_1,C_2,C_3$ but this is clearly absurd which contradicts our assumption. Thus, at least two amoung $C_1,C_2,C_3$ must be same. Now it is easy to see that if $C_1$ and $C_2$ are same then, $J,O,N,M,L,K$ will be cyclic. If $C_1$ and $C_3$ are same, then again $J,O,N,M,L,K$ will be cyclic and finally if $C_2$ and $C_3$ are same, then again $J,O,N,M,L,K$ will be cyclic. Thus, we conclude that all the six points $J,O,N,M,L,K$ are concyclic which completes the proof. Feedback: This was my first geomtery problem which was solved using the method of contradiction! .

Dunno how much I liked this one...

Jupiter_is_BIG wrote: Feedback: This was my first geomtery problem which was solved using the method of contradiction! . Check out USATST for EGMO 2019 P5