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Attachments:
Shaarygin_P20.pdf (150kb)

[asy][asy] import olympiad; size(9cm); defaultpen(fontsize(10pt)); pair A = dir(130), B = dir(210), C = dir(330), I = incenter(A,B,C), D = foot(I,B,C), P = 2I-D, X = extension(A,A+dir(C--foot(C,A,I)),P,P+dir(B--C)), Y = dir(90), M = (B+C)/2, T = extension(A,P,Y,M), S = extension(A,P,B,C), M1 = dir(270); dot("$A$", A, dir(110)); dot("$B$", B, dir(210)); dot("$C$", C, dir(330)); dot("$D$", D, dir(270)); dot("$I$", I, dir(225)); dot("$P$", P, dir(90)); dot("$X$", X, dir(180)); dot("$Y$", Y, dir(90)); dot("$M$", M, dir(270)); dot("$T$", T, dir(225)); dot("$S$", S, dir(270)); dot("$M_A$", M1, dir(270)); draw(A--B--C--A, linewidth(1.2)); draw(incircle(A,B,C)^^unitcircle); draw(X--I--Y--X, dashed); draw(Y--M); draw(circumcircle(A,I,Y)^^circumcircle(A,P,I), dotted); draw(X--P--I--A--S^^D--I--T^^M--M1--I, dotted); [/asy][/asy] Define $\omega$, $\Omega$, and $\omega_A$ to be the incircle, circumcircle, and $A$-excircle of $ABC$, respectively. Let $P$ be the point on $\omega$ such that $\overline{XP} \parallel \overline{BC}$ (so $\overline{XP}$ is a tangent), $M$ be the midpoint of $\overline{BC}$, $T = \overline{AP} \cap \overline{YM_A}$, $M_A$ be the midpoint of arc $BC$ not containing $A$, and $D$, $S$ where $\omega$, $\omega_A$ touches $\overline{BC}$, respectively. Without loss of generality, assume $AB < AC$ throughout the proof (the other case is identical). Claim 1: $X$, $A$, $Y$ are collinear with $\overline{IA} \perp \overline{XY}$. Proof. It is obvious that $\angle XAI = 90^\circ$ since $\overline{AX}$ is the external bisector and $\overline{AI}$ is the internal bisector. Furthermore, since $A$, $I$, $M_A$ are collinear and $YM_A$ is a diameter of $\Omega$, $\angle YAI = 90^\circ$ as well. The result follows. $\square$ Claim 2: $AITY$ is cyclic with diameter $IY$. Proof. It suffices to show that $\angle ITY = 90^\circ$. Recall that $S$ is the reflection of $D$ over $\overline{YM_A}$ with $BD = CS = s-b$, In addition, $A$, $P$, $S$ are collinear by a homothety centered at $A$ that sends $\omega$ to $\omega_A$. Therefore, since $I$, $M$ are midpoints of $\overline{PD}$, $\overline{DS}$, respectively, so there exists a homothety (with scale factor $2$) centered at $P$ sending $\overline{IT}$ to $\overline{DS}$. We can conclude that $\overline{IT} \parallel \overline{DS} \perp \overline{YM}$ as desired. $\square$ Note that $XIPA$ is cyclic with diameter $\overline{XI}$. Then $$\angle PIY = \angle IYT = \angle IAT = \angle IXP = 90^\circ - \angle XIP,$$or $\angle XIY = 90^\circ$, and we are done. $\blacksquare$

Trivial for #20. [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 = -5.365423794143402, xmax = 21.226418881831403, ymin = -10.752711091452602, ymax = 6.808265532876247; /* image dimensions */ pen dbwrru = rgb(0.8588235294117647,0.3803921568627451,0.0784313725490196); pen qqffff = rgb(0,1,1); pen ffqqff = rgb(1,0,1); pen ttffqq = rgb(0.2,1,0); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen wvvxds = rgb(0.396078431372549,0.3411764705882353,0.8235294117647058); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); /* draw figures */ draw((3.8,3.6)--(2.8,-4.86), linewidth(0.8) + dbwrru); draw((2.8,-4.86)--(16.66,-4.74), linewidth(0.8) + dbwrru); draw((16.66,-4.74)--(3.8,3.6), linewidth(0.8) + dbwrru); draw(circle((9.700446354481475,-1.3865539426101035), 7.7253470734119185), linewidth(0.8) + blue); draw(circle((6.29888308407502,-1.7231229152422067), 3.106467294780013), linewidth(0.8) + qqffff); draw((-1.0573471979071332,1.3197703416591051)--(6.29888308407502,-1.7231229152422067), linewidth(0.8) + ffqqff); draw((6.29888308407502,-1.7231229152422067)--(9.633562739421919,6.338503596768529), linewidth(0.8) + ffqqff); draw((3.8,3.6)--(6.29888308407502,-1.7231229152422067), linewidth(0.8) + ttffqq); draw((3.8,3.6)--(6.2719882025521505,1.3832279533632261), linewidth(2) + ttffqq); draw((9.633562739421919,6.338503596768529)--(16.66,-4.74), linewidth(0.8)); draw(circle((7.96622291174847,2.3076903407631613), 4.362049702345759), linewidth(0.8) + blue); draw((9.633562739421919,6.338503596768529)--(9.729999904502197,-4.800000000826822), linewidth(0.8) + green); draw((6.29888308407502,-1.7231229152422067)--(9.703105075499996,-1.693649179319653), linewidth(0.8) + ttffqq); draw(circle((2.6207679430839437,-0.2016762867915511), 3.980368178147804), linewidth(0.8) + blue); draw((-23.59711422485024,-12.029635218247629)--(-26.20255001421502,-23.524960621827038), linewidth(2) + wrwrwr); draw((-23.59711422485024,-12.029635218247629)--(-9.159506315802915,-22.0398821252849), linewidth(2) + wrwrwr); draw((-9.159506315802915,-22.0398821252849)--(-26.20255001421502,-23.524960621827038), linewidth(2) + wrwrwr); draw((xmin, -0.2656619696139111*xmin-24.473214613832162)--(xmax, -0.2656619696139111*xmax-24.473214613832162), linewidth(2) + wrwrwr); /* line */ draw((xmin, -11.476190476190451*xmin-258.9939700229305)--(xmax, -11.476190476190451*xmax-258.9939700229305), linewidth(2) + wrwrwr); /* line */ draw((-23.59711422485024,-12.029635218247629)--(-14.801173565668526,-22.531479686476512), linewidth(2) + wrwrwr); draw(circle((-17.98050433768564,-19.345575772837975), 9.223301622678013), linewidth(2) + wrwrwr); draw((xmin, -11.476190476190451*xmin-225.69326841008703)--(xmax, -11.476190476190451*xmax-225.69326841008703), linewidth(2) + wrwrwr); /* line */ draw((-20.91968770888181,-18.915649173382693)--(-18.039833109541366,-18.664707486303243), linewidth(2) + wrwrwr); draw(circle((-19.300357936945396,-20.849035273469333), 2.5219458206487837), linewidth(2) + wrwrwr); draw((-20.56088276434941,-23.03336306063543)--(-18.039833109541366,-18.664707486303243), linewidth(2) + wrwrwr); draw(circle((-20.91968770888181,-18.915649173382693), 4.133316906008411), linewidth(2) + wrwrwr); draw((9.703105075499996,-1.693649179319653)--(6.325777965597888,-4.829473783847638), linewidth(0.8) + ttffqq); draw((6.2719882025521505,1.3832279533632261)--(13.134221843406504,-4.7705262178060055), linewidth(0.8) + green); draw((6.2719882025521505,1.3832279533632261)--(6.325777965597888,-4.829473783847638), linewidth(0.8) + green); draw((-1.0573471979071332,1.3197703416591051)--(9.633562739421919,6.338503596768529), linewidth(0.8) + wvvxds); draw((-1.0573471979071332,1.3197703416591051)--(6.2719882025521505,1.3832279533632261), linewidth(0.8) + dtsfsf); /* dots and labels */ dot((3.8,3.6),dotstyle); label("$A$", (3.8726514212114815,3.7691977984791203), NE * labelscalefactor); dot((2.8,-4.86),dotstyle); label("$B$", (2.871140463285158,-4.691843052967425), NE * labelscalefactor); dot((16.66,-4.74),dotstyle); label("$C$", (16.73688700147202,-4.570971040803903), NE * labelscalefactor); dot((6.29888308407502,-1.7231229152422067),linewidth(4pt) + dotstyle); label("$I$", (6.376428816027292,-1.5837055973340002), NE * labelscalefactor); dot((3.21389280684353,-1.358466854103733),linewidth(4pt) + dotstyle); label("$D$", (3.285558790702947,-1.221089560843434), NE * labelscalefactor); dot((6.325777965597888,-4.829473783847638),linewidth(4pt) + dotstyle); label("$E$", (6.393696246336366,-4.691843052967425), NE * labelscalefactor); dot((6.2719882025521505,1.3832279533632261),linewidth(4pt) + dotstyle); label("$F$", (6.341893955409143,1.5244318582994247), NE * labelscalefactor); dot((-1.0573471979071332,1.3197703416591051),linewidth(4pt) + dotstyle); label("$X$", (-0.9967639259475417,1.4553621370631262), NE * labelscalefactor); dot((9.633562739421919,6.338503596768529),linewidth(4pt) + dotstyle); label("$Y$", (9.70904286567868,4.4801843570038296), NE * labelscalefactor); dot((13.134221843406504,-4.7705262178060055),linewidth(4pt) + dotstyle); label("$K$", (13.197063788111738,-4.6400407620402015), NE * labelscalefactor); dot((7.989163620631632,0.8832329241004819),linewidth(4pt) + dotstyle); label("$L$", (8.051369556007524,1.0236763793362618), NE * labelscalefactor); dot((9.729999904502197,-4.800000000826822),linewidth(4pt) + dotstyle); label("$P$", (9.795380017224051,-4.657308192349276), NE * labelscalefactor); dot((9.703105075499996,-1.693649179319653),linewidth(4pt) + dotstyle); label("$Q$", (9.778112586914977,-1.549170736715851), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Proof: Let $E,D,L$ be the tangent points of the incircle with$BC,CA,AB$ respectively and let $F$ be the diametrically opposite point of $E$ w.r.t incircle. Its well known (also lemma 4.9 egmo page 61) that if $AF \cap BC=K$ then $K$ is the touch point of the $A$ excircle with $BC$. Now observe that $\angle XAI=\angle XFI=90^\circ$ since we have that exterior angle bisector is perpendicular to interior angle bisector and also $XF$ is tangent to incircle . Now we claim that $Y-A-X$ are collinear. Its well known that $AI$ intersect $\odot(ABC)$ at midpoint of arc $BC$ not containing $A \implies YA\perp AI$ since $Y$ is midpoint of arc $BAC \implies \angle YAX+\angle XAI=90+90=180$. Done . Now its well known that if $P$ is the midpoint of $BC$ then $P$ is also the midpoint of $EK$ since $E$ and $K$ are the touch points of the incircle and $A$ excircle with $BC$(directly follows from lemma 2.20 egmo page 33 which says that $BK=CE \implies BE=CK \implies BP-BE=CP-CK \implies PE=PK$).Observe that since $Y$ is midpoint of arc $BAC \implies YP \perp BC \implies YP \parallel IE \implies Q$ is midpoint of $FK$ by thales theorem $\implies QI \parallel FE$ since $I$ is midpoint of $FE$ and thales theorem implies $QI \parallel FE \implies \angle QIE=\angle IEP=90^\circ \implies \angle QIF=90^\circ$ and also $\angle IQY=\angle EPY=90^\circ \implies AIQY$ cyclic since $\angle YAI=\angle YQI=90^\circ$ and now since $\angle XFI=\angle FIQ=90^\circ \implies XF \parallel IQ \implies \angle AFX=\angle AQI $. Now as we had proved that $AXFI$ is cyclic we have $\angle AFX=\angle AIX=\angle AQI\implies XI $ tangent to $\odot(AIQY)$. Now observe that since $\angle YAI=90^\circ \implies YI $ is diameter $\implies YI \perp XI$ since $XI$ tangent to $\odot(AIYQ)$ which implies the result. Done. $\blacksquare$

Again Trivial if you know your lemmas well. Sharygin 2020 CR P20 wrote: The line touching the incircle of triangle $ABC$ and parallel to $BC$ meets the external bisector of angle $A$ at point $X$. Let $Y$ be the midpoint of arc $BAC$ of the circumcircle. Prove that the angle $XIY$ is right.

Notations:- $M$ is the midpoint of $\widehat{BC}$ not containg $A$. $T_A$ is the $A-\text{Mixtillinear Incircle}$ Touch Point with $\odot(ABC)$. $D'$ is the $D-\text{antipode}$ WRT $\odot(I)$ and $I_D'$ is the $A-\text{excircle}$ touchpoint with $BC$. Claim:- $\overline{Y-A-X}$

Proof

Also notice that the line $\|BC$ touching $\odot(I)$ must be tangent to the incircle at $D-\text{antipode}$ It's well known that:-

$\overline{A-D'-I_D'}$ Proof

$AT_A,AI_D'$ are isogonal. Proof

So now we angle chase. $$\angle D'XI=\angle D'AI=\angle IAT_A=\angle T_AYM=\angle T_AID=\angle D'IY\implies \angle XIY=90^\circ.\blacksquare$$

Let $M$ be the other midpoint of arc $BC$ and $K$ be the midpoint of $BC$. Let $P$ be the point on the incircle such that $XP\parallel BC$. We make note of the following three observations. $A,X,I,P$ are concyclic since $\angle XPI = \angle XAI=90^{\circ}$, by the well known lemma, $IK\parallel AP$, and $MK\cdot MY = MB^2 = MI^2$ so $\angle KIM = \angle MYI$. With these three observations, we can angle chase $$\angle IXP = \angle IAP = \angle MAP = \angle MIK = \angle MYI = \angle PIY$$But $\angle XIP = 90^{\circ}-\angle IXP = \angle PIY$ hence the result follows.

Solution : [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(28.635cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ real xmin = -4.3, xmax = 33.88, ymin = -11.58, ymax = 6.3; /* image dimensions */ pen ududff = rgb(0.30196078431372547,0.30196078431372547,1.); pen uuuuuu = rgb(0.26666666666666666,0.26666666666666666,0.26666666666666666); pen yqqqyq = rgb(0.5019607843137255,0.,0.5019607843137255); draw((9.14,2.18)--(5.72,-5.96)--(17.92,-6.26)--cycle, linewidth(2.)); draw(shift((10.220030357929938,-3.0487003177606953)) * scale(3.021042928161183, 3.021042928161183)*unitcircle, linewidth(2.) + red); draw((-0.291981327211736,0.23174793930823478)--(10.220030357929938,-3.0487003177606953)--(10.29429584889354,-0.028570351907468514)--(9.14,2.18)--cycle, linewidth(2.) + green); draw((-0.291981327211736,0.23174793930823478)--(9.14,2.18)--(11.155900439337517,-0.04975735003314001)--(8.291888984785917,-9.227320755202093)--cycle, linewidth(2.) + yqqqyq); /* draw figures */ draw((9.14,2.18)--(5.72,-5.96), linewidth(2.)); draw((5.72,-5.96)--(17.92,-6.26), linewidth(2.)); draw((17.92,-6.26)--(9.14,2.18), linewidth(2.)); draw((xmin, 0.20655809135997363*xmin + 0.2920590449698429)--(xmax, 0.20655809135997363*xmax + 0.2920590449698429), linewidth(1.6)); /* line */ draw((xmin, -4.841253099387313*xmin + 46.429053328400045)--(xmax, -4.841253099387313*xmax + 46.429053328400045), linewidth(1.2)); /* line */ draw((xmin, -0.024590163934426208*xmin + 0.22456807060630687)--(xmax, -0.024590163934426208*xmax + 0.22456807060630687), linewidth(2.)); /* line */ draw(shift((11.877817888253235,-3.7587392110351434)) * scale(6.5394395941797425, 6.5394395941797425)*unitcircle, linewidth(2.)); draw((-0.291981327211736,0.23174793930823478)--(10.220030357929938,-3.0487003177606953), linewidth(2.) + green); draw((10.220030357929938,-3.0487003177606953)--(10.29429584889354,-0.028570351907468514), linewidth(2.) + green); draw((10.29429584889354,-0.028570351907468514)--(9.14,2.18), linewidth(2.) + green); draw((9.14,2.18)--(-0.291981327211736,0.23174793930823478), linewidth(2.) + green); draw(shift((4.9640245153591,-1.4084761892262305)) * scale(5.505990611049515, 5.505990611049515)*unitcircle, linewidth(2.) + green); draw((xmin, 3.2044436800640645*xmin-35.79821200829216)--(xmax, 3.2044436800640645*xmax-35.79821200829216), linewidth(2.)); /* line */ draw(shift((5.349229704438432,-3.273352004756682)) * scale(6.64145974327148, 6.64145974327148)*unitcircle, linewidth(2.) + yqqqyq); draw((5.72,-5.96)--(8.291888984785917,-9.227320755202093), linewidth(2.)); draw((11.155900439337517,-0.04975735003314001)--(9.14,2.18), linewidth(2.)); draw((-0.291981327211736,0.23174793930823478)--(9.14,2.18), linewidth(2.) + yqqqyq); draw((9.14,2.18)--(11.155900439337517,-0.04975735003314001), linewidth(2.) + yqqqyq); draw((11.155900439337517,-0.04975735003314001)--(8.291888984785917,-9.227320755202093), linewidth(2.) + yqqqyq); draw((8.291888984785917,-9.227320755202093)--(-0.291981327211736,0.23174793930823478), linewidth(2.) + yqqqyq); /* dots and labels */ dot((9.14,2.18),ududff); label("$A$", (9.22,2.38), NE * labelscalefactor,ududff); dot((5.72,-5.96),ududff); label("$B$", (5.3,-6.28), NE * labelscalefactor,ududff); dot((17.92,-6.26),ududff); label("$C$", (18.02,-6.7), NE * labelscalefactor,ududff); dot((10.220030357929938,-3.0487003177606953),linewidth(4.pt) + uuuuuu); label("$I$", (10.46,-3.2), NE * labelscalefactor,uuuuuu); dot((10.145764866966335,-6.068830283613925),linewidth(3.pt) + uuuuuu); label("$D$", (10.22,-5.94), NE * labelscalefactor,uuuuuu); dot((10.29429584889354,-0.028570351907468514),linewidth(4.pt) + uuuuuu); label("$E$", (10.38,0.14), NE * labelscalefactor,uuuuuu); dot((-0.291981327211736,0.23174793930823478),linewidth(4.pt) + uuuuuu); label("$X$", (-0.62,0.4), NE * labelscalefactor,uuuuuu); dot((12.038575184206048,2.778724157712987),linewidth(4.pt) + uuuuuu); label("$Y$", (12.08,2.2), NE * labelscalefactor,uuuuuu); dot((8.291888984785917,-9.227320755202093),linewidth(4.pt) + uuuuuu); label("$T$", (8.36,-9.92), NE * labelscalefactor,uuuuuu); dot((9.592634128019872,-0.011316375164673423),linewidth(3.pt) + uuuuuu); label("$J$", (9.68,0.1), NE * labelscalefactor,uuuuuu); dot((11.467554035012409,-0.05742096304153899),linewidth(3.pt) + uuuuuu); label("$K$", (11.66,0.12), NE * labelscalefactor,uuuuuu); dot((11.155900439337517,-0.04975735003314001),linewidth(3.pt) + uuuuuu); label("$G$", (11.08,-0.74), NE * labelscalefactor,uuuuuu); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* re-scale y/x */ currentpicture = yscale(1.3333333333333333) * currentpicture; /* end of picture */ [/asy][/asy] Claim 1: $Y,A,X$ are collinear. Proof : Let $Z$ be the midpoint of arc $BC$ not containing $A$. As $X$ and $Y$ are midpoints of arc $BC$ , they both lie on perpendicular bisector of $BC$. So , $\angle ZAY = 90^{0}$ , but as internal and external angle bisectors are perpendicular, Y lies on the external bisector . $\implies Y,A,X $ are collinear. Let $E$ be the touchpoint of line parallel to $BC$ with the incircle and $D$ be the touchpoint of incicle with $BC$ . So, $\angle XEI = \angle XAI = 90^{0} \implies XAEI $ is cyclic quadrilateral . So, $\angle XIE =\angle YAE$ , so to prove that $\angle XIY=90^{0}$, we need to prove that $\angle EAI = \angle EIY$ ---(as $\angle YAI=90^{0}$) ----(1) Let $YI \cap (ABC)$ at $T$ Thus by a well known property of mixtillinear incircles , we know that T is the touch point of $A$-mixtillinear incircle and $(ABC)$ . Proof: (Euclidean Geometry In Mathematical Olympiads----Lemma 4.40c) Let $AE \cap BC = F$ . By Lemma 4.9 of Euclidean Geometry In Mathematical Olympiads by Evan Chen , $F$ is the touchpoint of $A$-excircle and $BC$. Now by Lemma 4.40d of Euclidean Geometry In Mathematical Olympiads by Evan Chen, $AT$ and $AF$ are isogonal with respect to $\angle BAC \implies AT$ and $AE$ are isogonal with respect to $\angle BAC$ . So, $\angle EAI=\angle IAT$ ----(2)(as $AI$ is the internal angle bisector of $\angle BAC$) Let $YT\cap XE=G$. Claim 2 : $AGTX$ is a cyclic quadrilateral . Proof: In $AYTB,\angle ATY=\angle ATG = \angle ABY =\angle ABC-\angle YBC= B-(90^{0}-\frac{\angle BAC}{2})$ (as $YB=YC$ and $\angle BYC=\angle BAC$) $ \implies \angle ATY=\angle ATG=B-90^{0}+ \frac{A}{2} $ ---(Let the angles of $\triangle ABC$ be $A,B,C$ respectively . Let $XG$ intersect $AI$ and $AC$ at $J$ and $K$ respectively . In $\triangle AXJ$ , $\angle AXJ+\angle XAJ = \angle AXG +90^{0} = \angle AJK = 180^{0}-\angle JAK-\angle JKA$ $\angle AXJ = \angle AXG = 180^{0}-\frac{A}{2}-C-90^{0}= A+B+C-\frac{A}{2}-C-90^{0} = B+\frac{A}{2}-90^{0} =\angle ATG $. $\implies AGTX$ is a cyclic quadrilateral . So, $\angle XAI-\angle TAI=90^{0}-\angle TAI=\angle XAT=\angle XGT=\angle EGI $. In $\triangle IEG,\angle EIG= 90^{0}-\angle EGI=90^{0}-(90^{0}-\angle TAI) = \angle TAI $ $\implies \angle EIG=\angle TAI \implies \angle EIY=\angle TAI =\angle EAI $----(from 2) So, from (1), we get that $XI\perp YI$. $\blacksquare$

Attached my solution

Attachments:
P20.pdf (78kb)

Let the $A$-external bisector intersect $BC$ at $K$ and $I_a$ be the $A$- excentre. A $\sqrt{bc}$ inversion followed by a flip across the angle bisector swaps $\{Y,K\}$ and $\{I,I_a\}$. So,we have $$AY \cdot AK = AI \cdot AI_a$$Also, $\frac{AX}{AK}=\frac{AI}{AI_a}$. Multiplying these two relations, $$AY \cdot AX=AI^2 \implies \frac{AI}{AY}=\frac{AX}{AI}$$Since $\angle XAI=\angle YAI=90^{\circ}$, we deduce $\triangle AXI \sim \triangle AIY$ $$\implies \angle XIY=\angle XIA+\angle AIY=\angle AYI+\angle AIY=90^{\circ}$$

Problem 20. The line touching the incircle of triangle $ABC$ and parallel to $BC$ meets the external bisector of angle $A$ at point $X$. Let $Y$ be the midpoint of arc $BAC$ of the circumcircle. Prove that the angle $XIY$ is right. [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 = -18.617353501451386, xmax = 13.419810840361194, ymin = -12.21528660840042, ymax = 8.078638399758269; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0); pen qqzzqq = rgb(0,0.6,0); pen yqqqyq = rgb(0.5019607843137255,0,0.5019607843137255); draw((-1.57,2.09)--(-5.11,-7.21)--(7.09,-7.21)--cycle, linewidth(0.8) + zzttqq); /* draw figures */ draw((-1.57,2.09)--(-5.11,-7.21), linewidth(0.8) + zzttqq); draw((-5.11,-7.21)--(7.09,-7.21), linewidth(0.8) + zzttqq); draw((7.09,-7.21)--(-1.57,2.09), linewidth(0.8) + zzttqq); draw(circle((-0.38836935068976236,-3.955141498937952), 3.254858501062048), linewidth(0.8) + qqzzqq); draw(circle((0.99,-4.208193548387097), 6.798591175599901), linewidth(0.8)); draw((-1.57,2.09)--(0.99,-11.006784723986996), linewidth(0.8) + dotted + red); draw((-1.57,2.09)--(2.3683693506897634,-7.21), linewidth(0.8) + linetype("4 4") + blue); draw((-15.844896774289754,-0.700282997875904)--(-0.38836935068976236,-3.955141498937952), linewidth(0.8) + red); draw((-0.3883693506897624,-0.700282997875904)--(-0.38836935068976236,-7.21), linewidth(0.8) + yqqqyq); draw((-0.38836935068976325,-4.189559774609752)--(-0.5390668136216343,-4.0723506367738524), linewidth(0.8) + yqqqyq); draw((-0.38836935068976325,-4.189559774609752)--(-0.23767188775789222,-4.0723506367738524), linewidth(0.8) + yqqqyq); draw((-0.38836935068976325,-3.955141498937952)--(-0.5390668136216343,-3.837932361102052), linewidth(0.8) + yqqqyq); draw((-0.38836935068976325,-3.955141498937952)--(-0.23767188775789222,-3.837932361102052), linewidth(0.8) + yqqqyq); draw((-1.57,2.09)--(-1.7517338669818847,-10.429426334225398), linewidth(0.8) + linetype("4 4") + blue); draw((0.99,-11.006784723986996)--(0.99,2.5903976272128033), linewidth(0.8) + yqqqyq); draw((0.99,-3.973775272715297)--(1.1406974629318658,-4.090984410551197), linewidth(0.8) + yqqqyq); draw((0.99,-3.973775272715297)--(0.8393025370681237,-4.090984410551197), linewidth(0.8) + yqqqyq); draw((0.99,-4.208193548387097)--(1.1406974629318658,-4.325402686222997), linewidth(0.8) + yqqqyq); draw((0.99,-4.208193548387097)--(0.8393025370681237,-4.325402686222997), linewidth(0.8) + yqqqyq); draw((-15.844896774289754,-0.700282997875904)--(8.640848244765575,-0.700282997875904), linewidth(0.8) + red); draw((0.99,2.5903976272128033)--(-15.844896774289754,-0.700282997875904), linewidth(0.8) + red); draw((0.99,2.5903976272128033)--(-1.7517338669818847,-10.429426334225398), linewidth(0.8) + dotted + red); /* dots and labels */ dot((-5.11,-7.21),dotstyle); label("$B$", (-5.021093512487022,-6.991107893428875), NE * labelscalefactor); dot((7.09,-7.21),dotstyle); label("$C$", (7.168656822446546,-6.991107893428875), NE * labelscalefactor); dot((-0.38836935068976236,-3.955141498937952),linewidth(4pt) + dotstyle); label("$I$", (-0.28807689892307253,-3.7762286842156176), NE * labelscalefactor); dot((0.99,-11.006784723986996),linewidth(4pt) + dotstyle); label("$M$", (0.9844794547238383,-11.389241256033123), NE * labelscalefactor); dot((-0.38836935068976236,-7.21),linewidth(4pt) + dotstyle); label("$D$", (-0.28807689892307253,-7.035758993556837), NE * labelscalefactor); dot((-15.844896774289754,-0.700282997875904),linewidth(4pt) + dotstyle); label("$X$", (-15.759683093261831,-0.5166983748743983), NE * labelscalefactor); dot((0.99,2.5903976272128033),linewidth(4pt) + dotstyle); label("$Y$", (1.0737816549797618,2.765157484530802), NE * labelscalefactor); dot((0.99,-7.21),linewidth(4pt) + dotstyle); label("$N$", (1.0737816549797618,-7.035758993556837), NE * labelscalefactor); dot((-1.7517338669818847,-10.429426334225398),linewidth(4pt) + dotstyle); label("$T$", (-1.9178420535936777,-10.898079154625541), NE * labelscalefactor); dot((2.3683693506897634,-7.21),linewidth(4pt) + dotstyle); label("$E$", (2.4579657589465773,-7.035758993556837), NE * labelscalefactor); dot((-1.57,2.09),linewidth(4pt) + dotstyle); label("$A$", (-1.47133105231406,2.2739953831232213), NE * labelscalefactor); dot((-0.3883693506897623,-0.7002829978759041),linewidth(4pt) + dotstyle); label("$F$", (-0.28807689892307253,-0.5166983748743983), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Solution. Notations. Let the parallel line with $BC$ touch the incircle at $F$ and let the incircle touch $BC$ at $D$. Let the $A$-excircle touch $BC$ at $E$. Let $M$ be the mid point of arc $BC$ not containing $A$ and let $N$ be the midpoint of $BC$. Let $T$ denote the touch point of $A$-mixtilinear incircle with circumcircle of $\triangle ABC$. Also, we denote the circumcircle of $\triangle XYZ$ by $(XYZ)$. Some well known results. Following are a few well known results which will be useful in proving $\angle AIY=90^{\circ}$: Angle made by the external angle bisector and internal angle bisector of $\angle A$ in triangle $ABC$ is $90^{\circ}$ (Mixtilinear Incircles lemma) If $T$ deontes the touch point of $A$-mixtilinear incircle with circumcircle of $\triangle ABC$ and $E$ is the touch point of $A$-excircle with $BC$, then $AT$ and $AE$ are isogonal conjugates. (Mixtilinear Incircles lemma) If $T$ deontes the touch point of $A$-mixtilinear incircle with circumcircle of $\triangle ABC$ with incenter $I$, then $TI$ bisects arc $BAC$. (Diameter of incircle lemma) If $F$ is the diametrically opposite to the touch point of incircle with $BC$ in $\triangle ABC$ with respect to the incircle, then $A,F,E$ are collinear where $E$ is the touch point of $A$-excircle with $BC$. (Fact 5 lemma) In $\triangle ABC$, the external angle bisector of $\angle A$ bisects arc $BAC$ and internal angle bisector bisects arc $BC$ not containing $A$. Proof. As $AX$ is the external angle bisector of $\angle A$, it must bisect arc $BAC$ from $(5)$. Thus, $X,A,Y$ are collinear. Also, as $Y$ is the mid point of arc $BAC$, from $(3)$, we have $T,I,Y$ collinear. Also, as $AI$ is the innternal angle bisector of $\angle A$, we know by $(5)$ that, $A,I,M$ are collinear. Also, as $Y$ is mid point of arc $BC$ containg $A$ and $M$ is the midpoint of arc $BC$ not containing $A$, $YM$ must be the diameter of $(ABC)$ and thus, $Y,N,M$ are collinear too. Now, $(1)$ gives us $\angle XAI=90=\angle XFI\implies XAFI$ is cyclic with diameter $XI$. Therefore, to show that $\angle XIY=90$, it is enough to prove $YI$ is tangent to $(XAI)$ which is equivalent of showing $\angle YIF=\angle IXF$ but $\angle IXF=\angle IAF$ because of concyclicity. Now, by $(2)$, we know that $AT,AE$ are isogonal lines and so, $\angle BAT=\angle CAE\implies \angle TAI=\angle EAI$ because $AI$ bisects $\angle A$. Thus, $\angle IAF=\angle IAT=\angle MAT=\angle MYT$ and $\angle YIF=\angle TID$ and so showing $\angle YIF=\angle IXF$ is equivalent to prove $\angle TID=\angle MYT$ but this following because $FD\|YM$ as $YM$ is the diameter of $(ABC)$ and so $\angle YNC=90$ and $\angle FDC=90$. This completes the proof. Feedback. This problem uses the configuration of mixtilinear incircles very beautifully. P.S: Proofs of the well known results can be found in any good book/handout for euclideian geometry. Say, in book \textit{Euclideian Geometry in Mathematical Olympiad} by Evan Chen, $(2) , (3)$ are in \textbf{lemma 4.40} and $(4)$ is \textbf{lemma 4.9}. $(1),(5)$ are trivial properties.

I actually found another (probably cooler) solution to this while doing another problem, but was too lazy to finish it

Attachments:
P20_diagram.pdf (2kb)
P20.pdf (68kb)

I found this somewhat similar to USAMO 2017 P3

Too easy to bash! Let $A,B,C$ be $x^2, y^2, z^2$ and $I$ be $-(xy+yz+zx)$, $Y$ be $yz$. Let $Q'$ be the image of $Q$ when $I$ is translated to the origin. Then, we have: \[ D\colon \frac 12\left(-xy-yz-zx+y^2+z^2 + y^2z^2\cdot \frac{x+y+z}{xyz}\right) = \frac 1{2x} (-x^2(y+z)+y^2(x+z)+z^2(x+y)). \]So, \[ D' \colon \frac 1{2x} (x+y)(y+z)(z+x). \]But now $P'$ is just $-\frac 1{2x}(x+y)(y+z)(z+x)$, and $A'$ is $(x+y)(x+z)$. So, if $O$ is the circumcenter of $(API')$, then $O$ is the midpoint of $XI$. However, the circumcircle of $(A'P'I')$ is \begin{align*} -\frac{(x+y)(x+z)}{2x}\cdot\frac{\begin{vmatrix}y+z & \frac{(y+z)^2}{yz} \\ -2x & 4\end{vmatrix}}{\begin{vmatrix} y+z & \frac{y+z}{yz} \\ -2x & -\frac 2x \end{vmatrix}} &= -\frac{(x+y)(x+z)}{2x} \cdot \frac{\begin{vmatrix} yz & y+z \\ x & -2 \end{vmatrix}}{\begin{vmatrix} yz & 1 \\ x & \frac 1x\end{vmatrix}} \\ &= -\frac{(x+y)(x+z)}{2x} \cdot \frac{-2yz-xy-xz}{\frac{yz}{x}-x} \\ &= - \frac{(x+y)(x+z)(xy+xz+2yz)}{2(yz-x^2)}, \end{align*}so $X'/Y'$ is just $\frac{(x+y)(x+z)}{yz-x^2}$, and \[ \overline{\frac{(x+y)(x+z)}{yz-x^2}} = \frac{(x+y)(x+z)}{x^2-yz} = - \frac{(x+y)(x+z)}{yz-x^2}, \]so $\angle X'I'Y' = \angle XIY = 90^{\circ}$. $\Box$

Let $(I)$ the incircle of $\triangle ABC, (I)$ touches $AC$ at $F$ and touches the line touching the incircle of $ABC, $ $\parallel BC $ at $K.$ $AI$ intersects $\odot(ABC)$ at $D.$ $\widehat {IAF}=\widehat {DYC},\widehat{AFI}=\widehat {YCD}\sim\triangle AIF\sim\triangle YDC.$ Notice that $\frac {AI}{YD}=\frac {IF}{DC}. $ $\widehat {DIC}=\widehat {IAC}+\widehat{ACI}=\widehat {DAB}+\widehat {ACI}=\widehat {DCB}+\widehat {ICB}=\widehat {DCI}\implies\triangle DIC $ is isosceles at $D.$ $DC=DI\implies\frac {IF}{CD}=\frac {IK}{ID}.(\clubsuit)$ Let $(I)$ touches $BC$ at $G.$ $IG\perp BC, IK \perp BC. $ Because $IK\perp XK, XK\parallel BC\implies\overline {I, G, K} $ are collinear. Thus, $IK\parallel DY\implies\widehat {AIK}=\widehat {YDI}. $ From $(\clubsuit) $ $\triangle AIK\sim\triangle YDI\implies\widehat {KAI}=\widehat {IYD}=\widehat {KIY} $ cause $IK\parallel YD.$ $\widehat {KIA}+\widehat {KAI}=\widehat {KIY}+\widehat{KIA}\Leftrightarrow 180^{\circ}-\widehat{AKI}=\widehat {AIY}. $ Because $AX$ is the external bisector of $\angle A$ of $\triangle ABC, AX\perp AD.$ Also $AY\perp AD,\implies\overline{X, A, Y} $ are collinear. $\widehat {XAI}=\widehat{XKI}=90^{\circ}\implies AKIX $ is cyclic. $180^{\circ}-\widehat {AKI}=\widehat {YXI}=\widehat {YIA}\implies\triangle YIA\sim\triangle YXI.$ So $\widehat {YAI}=\widehat{YIX}=90^{\circ}. $ $\blacksquare$

Attachments:

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Dear Mathlinkers, according the first figur, the location of T is the kern of the problem...which has been done in several proofs... PX orthogal to YT leads thet the circles with diameter IX and IY are orthogonal and we are done... Sincerely Jean-Louis

Dear Mathlinkers, for the location of T, this problem help https://artofproblemsolving.com/community/c6h358063 Sincerely Jean-Louis

Let $D'$ be the reflection of touchpoint of $(I)$ and $\overline{BC}$ over $I$. Let $T=\overline{YI}\cap (ABC)$. It is well-known that $T$ is the touchpoint of $A$-mixtilinear incircle of $\triangle ABC$ and $(ABC)$. Hence, we conclude that $\overline{AT}$ and $\overline{AD'}$ are isogonal wrt $\angle BAC$. Observe that $XAD'I$ is cyclic as $\measuredangle XAI=\measuredangle XD'I=90^\circ$. Hence, $$\measuredangle YIX= \measuredangle YIA+\measuredangle AIX=90^\circ+ \measuredangle TYA+\measuredangle AD'X=90^\circ+ \measuredangle (AD',BC)+\measuredangle (BC,AD')=90^\circ.$$$\blacksquare$

We will be using the configuration of Sharygin 2022 Correspondence Round Problem 15 to solve this problem.

Let the line through $X$ parallel to $\overline{BC}$ meet $\odot(ABC)$ at points $D,E$, respectively

. Clearly points $A,X,Y$ are collinear and $\angle IAY =90^\circ$. So $\angle XIY =90^\circ$ is equivalent to $XI^2 = XA \cdot XY$. But $XA \cdot XY = XD \cdot XE$, so our problem is equivalent to $XD \cdot XE = XI^2 \iff$ $\overline{XI}$ is tangent to $\odot(DIE)$. [asy][asy] size(200); pair A=dir(115),B=dir(-150),C=dir(-30),Y=dir(90),I=incenter(A,B,C),F=IP(unitcircle,I--2*I-Y),P=foot(I,B,C),Q=2*I-P,X=IP(circumcircle(A,Q,I),2*A-Y--10*A-9*Y),D=IP(unitcircle,X--Q),E=IP(unitcircle,Q--4*Q-3*D); fill(D--E--F--D--cycle,royalblue+white+white); draw(unitcircle,orange); draw(circle(I,abs(I-foot(I,B,C))),orange); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$Y$",Y,dir(Y)); dot("$F$",F,dir(F)); dot("$I$",I,dir(-90)); dot("$X$",X,dir(X)); dot("$D$",D,dir(D)); dot("$E$",E,dir(E)); draw(A--B--C--A,purple); draw(X--E--F--D,royalblue); draw(X--Y--I--X,brown); draw(I--A--D--I--E--A,red+linewidth(0.8)); [/asy][/asy] Claim: $A$ is the $I$-Dumpy point of $\triangle IDE$, or equivalently $\triangle ADI \sim \triangle AIE$. Proof: By Pocelet Porism, we know $\exists$ a point $F \in \odot(ABC)$ such that $\odot(I)$ is the incircle of $\triangle DEF$. Then $A$ is the $F$-mixtilinear touch point of $\triangle DEF$, which implies $\triangle ADI \sim \triangle AIE$. $\square$ Now we simply ignore all other points and only focus on points $A,D,E,I,X$ to solve the problem. Look at $\triangle IDE$ as the reference triangle. $A$ is the $I$-Dumpy point. Redifine $X$ as intersection of $\overline{DE}$ and tangent to $\odot(IDE)$ at $I$. We want to show $\overline{XA}$ is the external angle bisector of $\angle DAE$. We can restate the problem as follows: Restated Problem wrote: Let $ABC$ be triangle and $Q_A$ be the $A$-Dumpy point, $T=\overline{AA} \cap \overline{BC}$. Prove that $\overline{TQ_A}$ is the external angle bisector of $\angle BQ_AC$. [asy][asy] size(200); pair A=dir(125),B=dir(-150),C=dir(-30),K=IP(unitcircle,(B+C)/2--B+C-dir(55)),T=2*A*K/(A+K),QA=(A+K)/2,O=(0,0); draw(unitcircle); fill(A--B--C--A--cycle,red+white+white); draw(circumcircle(B,O,C),brown); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$K$",K,dir(K)); dot("$T$",T,dir(T)); dot("$Q_A$",QA,dir(140)); dot("$O$",O,dir(dir(90))); draw(T--C--A--B,red); draw(A--T--K,royalblue); draw(T--O^^A--K,purple); draw(B--QA--C,green); [/asy][/asy] Let $O$ be center of $\odot(ABC)$ and $K = \overline{AQ_A} \cap \odot(ABC) \ne A$. It is well known that $(A,K ; B,C) = -1$, i.e. $\overline{TK}$ is tangent to $\odot(ABC)$. $Q_A$ is the midpoint of segment $AK$, i.e. points $O,Q_A,T$ are collinear. $Q_A \in \odot(BOC)$. Now $\overline{TQ_A} \equiv \overline{OQ_A}$ is the external angle bisector of $\angle BQ_AC$ since $O$ is the midpoint of arc $\widehat{BQ_AC}$ of $\odot(BQ_AC)$. $\blacksquare$

No, Angle XIY is wrong Let $T$ be the mixtilinear intouch point and let $D'$ be the $A$-excircle touchpoint on $BC$, $I_A$ be the centre of the $A$-excircle and $D$ be the intersection of the given parallel line with the incircle of $\triangle ABC$. Well known Facts: $T$-$I$-$Y$ and $X$-$A$-$Y$ are collinear, The Homothety at $A$ taking $I$ to $I_A$ takes $D$ to $D'$ which means $A$-$D$-$D'$, $AT$ and $AD'$ are isogonal ($\sqrt{bc}$ inversion) We also know because of the perpendiculars at $A$ and $D$ from $I$, $ADIX$ must be cyclic. Now it is just a matter of trivial angle chasing, $$180^{\circ} - \angle XIY = \angle AYI + \angle AXI = 180^{\circ} - \angle ADI + \angle AXT = 180^{\circ} - (\angle ADI-\angle ACT) = 90^{\circ}$$Details left to the reader xD $\blacksquare$

NJOY wrote: The line touching the incircle of triangle $ABC$ and parallel to $BC$ meets the external bisector of angle $A$ at point $X$. Let $Y$ be the midpoint of arc $BAC$ of the circumcircle. Prove that the angle $XIY$ is right. Let $D'$ the symmetric point of $D$ with respect to $I$,$M$ be the midpoint of $BC$ and $G$ be the symmetric point of $D$ with respect to $M$ It is well know that: 1)$G$ is the tangent poin of the A-excircle. 2)$A,D',G$ collinear 3)$\angle IMD=\angle AYI$ Now from midpoints we get that $IM//AG$ and so: $\angle XIY=180-\angle IXY-\angle XYI=180-\angle ID'G-\angle IMD=180-\angle DD'G-\angle D'GD=90$ We use that $X,A,D',I$ are cyclic becayse $\angle XAI=90=\angle XD'I$

Attachments:

Let $\ell$ denote the line parallel to $BC$ and tangent to incircle, let $\ell \cap AB,AC = D,E$, it's well known in incircle config that $ADI \sim AIC$ and $AEI \sim AIB$ and it follows by angle chase that $ADX \sim AYC$ therefore $AX \cdot AY=AE \cdot AB = AI^2$ so $\measuredangle XIY = 90$

nice let $D$ be the incircle tangency with $BC$, $D'$ reflection of $D$ over $I$. Clearly by two rights, $AXID'$ is cyclic. Let $M$ be the midpoint of $BC$, well known $\measuredangle IMB=\measuredangle AYI$, the former angle is $\measuredangle D'ED$, if $E$ is extouch with $BC$ by well known homothety. Clearly, we are done as $\measuredangle IXA=\measuredangle ID'F=90^\circ-\measuredangle D'ED$.

Solved with MaxSze. Let $D$ be the $A$-intouch point, $D'$ be the reflection of $D$ about $I$. Let $M$ be the midpoint of arc $BC$ not containing $A$, and let $T$ be the $A$-mixtilinear intouch point. Note that $\measuredangle XD'I = 90^\circ = \measuredangle XAI$, so $(AD'IX)$ concyclic, and $XI$ is a diameter of the circle. Now since $MY // DD'$, $A,I,M$ collinear, $T,I,Y$ collinear, and lines $AT,AD'$ are isogonal wrt $\angle BAC$, $$\measuredangle YID' = \measuredangle IYM = \measuredangle TAM = \measuredangle IAD',$$so $YI$ is tangent to $(AD'IX)$. It follows that $XI \perp YI$. $\square$