Nice problem, but here goes my solution We prove several claims. $\textbf{Claim 1.}$ The points $A,B_1,A_0,B$ are concyclic. See figure 1. $\emph{Proof.}$ By the $\textbf{Incenter-Excenter lemma}$ in $\triangle AB_1B$, we have that the angle bisector of $\angle BAB_1$ (i.e. $AI$) meets the perpendicular bisector of $BB_1$ at a point (i.e. $A_0$) on $\odot(ABB_1)$, thus proving that the points $A,B_1,A_0,B$ are concyclic. $\square$ Similarly, by Claim 1, it follows that the points $C,B_1,C_0,B$ are concyclic. $\textbf{Claim 2.}$ The circumcircle of $\triangle A_0IC_0$ passes through point $B$. $\emph{Proof.}$ Using Claim 1, we obtain $$\angle A_0BB_1 = \angle A_0AB_1 = \frac{A}2.$$Similarly, we obtain $$\angle C_0BB_1 = \frac{C}2.$$Thus, we have that $$\angle A_0BC_0=\frac{A+C}2=90^{\circ}-\frac{B}2.$$We also have $$\angle A_0IC_0=\angle AIC=90^{\circ}+\frac{B}2,$$which thus proves that points $A_0,B,C_0,I$ are concyclic. $\square$ Now, let $\ell$ be a tangent to $\odot(ABC)$ at $B$. We now observe by $\textbf{Alternate-Segment theorem}$ that $$\angle TBA = \angle BCA.$$Now, observe that \begin{align*}\angle C_0BA & = \angle B_1BA - \angle B_1BC_0 \\&= \frac{B}2 - \frac{C}2 \\&= 90^{\circ} - C - \frac{A}2.\end{align*}Hence, it follows that \begin{align*}\angle TBC_0 &= \angle TBA + C_0BA \\&= C + 90^{\circ}-C-\frac{A}2 \\&=90^{\circ}-\frac{A}2\\&=\angle BIC_0,\end{align*}thus proving that $\ell$ is also tangent to $\odot(A_0IC_0)$, as desired. $\blacksquare$

Nice Problem! $\textbf{Diagram : }$ [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 = -7.3402144305384365, xmax = 4.296608841190585, ymin = -2.66677388193589, ymax = 4.447057703045623; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw((-1.687652905217388,3.097232083385058)--(-3.4260953247324983,-0.7469293231483581), linewidth(1) + wrwrwr); draw((-3.4260953247324983,-0.7469293231483581)--(-6.08272916300559,1.5179428431213295), linewidth(1) + wrwrwr); draw((-6.08272916300559,1.5179428431213295)--(-1.687652905217388,3.097232083385058), linewidth(1) + wrwrwr); draw(circle((-3.6570499162113452,1.672683111887678), 2.430609873901235), linewidth(1) + wrwrwr); draw((-2.68246274608465,0.8974413085095616)--(-6.08272916300559,1.5179428431213295), linewidth(1) + wrwrwr); draw((-4.686984235486233,0.3280220523790201)--(-1.687652905217388,3.097232083385058), linewidth(1) + wrwrwr); draw((-4.149292437820713,2.4861645188592862)--(-3.4260953247324983,-0.7469293231483581), linewidth(1) + wrwrwr); draw((-4.5139738971013985,0.4877583108203598)--(-6.08272916300559,1.5179428431213295), linewidth(1) + wrwrwr); draw((-6.08272916300559,1.5179428431213295)--(-4.149292437820713,2.4861645188592862), linewidth(1) + wrwrwr); draw((-4.149292437820713,2.4861645188592862)--(-4.686984235486233,0.3280220523790201), linewidth(1) + wrwrwr); draw((-4.149292437820713,2.4861645188592862)--(-4.5139738971013985,0.4877583108203598), linewidth(1) + wrwrwr); draw(circle((-4.91101054722856,1.5926897634998154), 1.1741003435245778), linewidth(1) + dotted + wrwrwr); draw((xmin, -15.675811255419713*xmin-93.83377143399176)--(xmax, -15.675811255419713*xmax-93.83377143399176), linewidth(1) + wrwrwr); /* line */ /* dots and labels */ dot((-1.687652905217388,3.097232083385058),dotstyle); label("$A$", (-1.6754966869420451,3.184572159414644), NE * labelscalefactor); dot((-6.08272916300559,1.5179428431213295),dotstyle); label("$B$", (-6.341204130795663,1.406114263169266), NE * labelscalefactor); dot((-3.4260953247324983,-0.7469293231483581),dotstyle); label("$C$", (-3.4539545831874237,-0.9541847966625632), NE * labelscalefactor); dot((-4.686984235486233,0.3280220523790201),linewidth(4pt) + dotstyle); label("$A_1$", (-4.782308937790454,0.1436287195382876), NE * labelscalefactor); dot((-2.68246274608465,0.8974413085095616),linewidth(4pt) + dotstyle); label("$B_1$", (-2.6086381757127683,0.8462293699068321), NE * labelscalefactor); dot((-4.092670631736183,2.233033513131786),linewidth(4pt) + dotstyle); label("$C_1$", (-3.991883206125841,2.1306711838618275), NE * labelscalefactor); dot((-3.8412175056196505,1.1088980058295457),linewidth(4pt) + dotstyle); label("$I$", (-3.871123719343747,0.8791637753928576), NE * labelscalefactor); dot((-4.149292437820713,2.4861645188592862),linewidth(4pt) + dotstyle); label("$C_0$", (-4.332205396148105,2.646643536476227), NE * labelscalefactor); dot((-4.5139738971013985,0.4877583108203598),linewidth(4pt) + dotstyle); label("$A_0$", (-4.485899288416224,0.38514769310247476), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] $$\textbf{Solution : }$$Let $\angle BAC = A , \angle ACB = C$ and $\angle CBA = B. $ Let the perpendicular bisector of $BB_1$ be $\ell $ We know that $\angle BIA_1 = \tfrac{B}{2} + \tfrac{C}{2} = 90 - \tfrac{A}{2}$ and $\angle BIC_1 = \tfrac{C}{2} + \tfrac{B}{2} = 90 - \tfrac{A}{2}$ $\textbf{Claim 1 : }$ $BCB_1C_0$ is a cyclic quadrilateral. $\textbf{(Proof) : }$Let $C_0'$ be the intersection of CI and $\odot(BCB_1)$ $\angle B_1CC_0' = \angle C_0'CB = \angle C_0'BB_1 \Longrightarrow C_0'B_1 = C_0'B $ Thus, $C_0' \in \ell \Longrightarrow C_0'(\neq A_0)$ is the intersection of $\ell$ and $\odot(BCB_1) \Longrightarrow C_0' \equiv C_0 $ $\textbf{Claim 2 : }$ $IC_0AB_0$ is cyclic quadrilateral. $\linebreak$ $\textbf{(Proof) : }$ $\angle BICC_0 = \angle B_1BC_0 = \tfrac C2 \Longrightarrow \angle BC_0A_0 = 90 - \tfrac C2 = \angle BIA_1$ $\textbf{Main Problem : } $ Let $XY$ be the tangent to $\odot(ABC)$ at $B $ By, Alternate Segment Theorem, $\angle CBY = A $ Thus, $\angle C_0BX = 180 - (\tfrac B2 + \tfrac C2 + A) = 90 - \tfrac A2 = \angle BIA_0 $ Since, $XY$ is the common tangent to $\odot(ABC)$ and $\odot(BCB_1C_0)$ , they are tangent to each other at $B$.

Both the solutions' ideas are okay, but i didnt take any chance and used directed angles! There were two possible configurations, remember, whatever you are subtracting angles and stuff, depends on the configuration Like in NJOY's last line, he gives an addition of angles argument, but a different configuration would have made that invalid

Really elaborated proof coming!. Personally felt that this was trivial for a P18. Heres my solution. We break the problem into 2 cases. First when $\angle B > \angle C$ [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 = -18.159382950767856, xmax = 14.3850727306359, ymin = -10.215540365030877, ymax = 11.16044397688852; /* image dimensions */ pen dbwrru = rgb(0.8588235294117647,0.3803921568627451,0.0784313725490196); pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902); pen wvvxds = rgb(0.396078431372549,0.3411764705882353,0.8235294117647058); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen wwffqq = rgb(0.4,1,0); /* draw figures */ draw((-6.445327294818059,6.97811327507128)--(-9.56,-0.65), linewidth(1.2) + dbwrru); draw((-9.56,-0.65)--(3.72,-0.61), linewidth(1.2) + dbwrru); draw((3.72,-0.61)--(-6.445327294818059,6.97811327507128), linewidth(1.2) + dbwrru); draw(circle((-6.83621772903952,0.06466692308734225), 2.81597916727979), linewidth(1.2) + linetype("4 4") + sexdts); draw(circle((-2.9251832695215163,1.0908454811431896), 6.859397643835345), linewidth(1.2) + linetype("4 4") + sexdts); draw((-6.844699555525887,2.8806333165610347)--(-9.56,-0.65), linewidth(1.2) + wvvxds); draw((-9.56,-0.65)--(-4.229188432951489,-0.9998221186094326), linewidth(1.2) + wvvxds); draw((-6.844699555525887,2.8806333165610347)--(-2.5531849979306167,4.0727452751119015), linewidth(1.2) + wvvxds); draw((-2.5531849979306167,4.0727452751119015)--(-4.229188432951489,-0.9998221186094326), linewidth(1.2) + wvvxds); draw(circle((-2.913004353452675,-2.9525546537128955), 7.0345084446998865), linewidth(1.2) + linetype("4 4") + sexdts); draw(circle((-6.668574040024547,2.6193279036340686), 4.3644987138846), linewidth(1.2) + linetype("4 4") + sexdts); draw((-9.56,-0.65)--(-2.5531849979306167,4.0727452751119015), linewidth(1.2) + dtsfsf); draw((-6.844699555525887,2.8806333165610347)--(-4.229188432951489,-0.9998221186094326), linewidth(1.2)); draw((-7.9669829896091136,3.2514417740137023)--(3.72,-0.61), linewidth(1.2) + dtsfsf); draw((-6.445327294818059,6.97811327507128)--(-4.229188432951489,-0.9998221186094326), linewidth(1.2) + wwffqq); /* dots and labels */ dot((-6.445327294818059,6.97811327507128),dotstyle); label("$A$", (-6.862783833010253,7.188482850417983), NE * labelscalefactor); dot((-9.56,-0.65),dotstyle); label("$B$", (-9.46804865789953,-0.4351199568399848), NE * labelscalefactor); dot((3.72,-0.61),dotstyle); label("$C$", (3.8145310230933416,-0.3924106974155704), NE * labelscalefactor); dot((-5.15173246753228,2.3212679233349856),linewidth(4pt) + dotstyle); label("$I$", (-5.068994937184849,2.4904643137324007), NE * labelscalefactor); dot((-4.330738556392406,-0.6342492125192541),linewidth(4pt) + dotstyle); label("$A_1$", (-4.364292156682012,-0.3924106974155704), NE * labelscalefactor); dot((-2.5531849979306167,4.0727452751119015),linewidth(4pt) + dotstyle); label("$B_1$", (-2.463730112295572,4.241543950133391), NE * labelscalefactor); dot((-7.9669829896091136,3.2514417740137023),linewidth(4pt) + dotstyle); label("$C_1$", (-7.8878060591961985,3.430068021069517), NE * labelscalefactor); dot((-6.844699555525887,2.8806333165610347),linewidth(4pt) + dotstyle); label("$C_0$", (-6.756010684449217,3.0456846862497877), NE * labelscalefactor); dot((-4.229188432951489,-0.9998221186094326),linewidth(4pt) + dotstyle); label("$A_0$", (-4.193455118984354,-1.3960782938893084), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Proof: Observe that since $C_0$ lies on perpendicular bisector of $BB_1 \implies C_0B=C_0B_1$. Similiarly we have $A_0B=A_0B_1$. Now we claim that $BC_0B_1C$ is concyclic. Let $CI \cap \odot(CB_1B) =C_0'$ Its well known that $C_0'=C_0'B_1$ . But we also have $C_0B=C_0B_1$ and since both these point lie on $CI \implies C_0=C_0'$ and hence the four points are concyclic. By a similiar arguement we can prove that $A_0BB_1A$ is concylic.Now since $\angle A_0BI\equiv \angle A_0BB_1=\angle A_0AB_1$ and $\angle C_0BI\equiv \angle C_0IB_1=\angle C_0CB_1$ and so $\angle A_0C_0=\angle A_0BI+\angle C_0BI=\angle A_0BB_1+\angle C_0IB_1=180-\angle AIC=180-\angle C_0IA_0 \implies C_0IA_0B $ is cyclic. Now let $l$ be the line tangent to $\odot(ABC)$ at $B$ . We claim that $l$ is also tangent to $\odot(BC_0A_0)$.Note that $\angle(l,AB)=\angle ACB=\angle C$ and $\angle C_1BC_0=\angle C_1BI-\angle C_0BI=\frac{\angle B}{2}-\angle C_0BB_1=\frac{\angle B}{2}-\angle C_0CB_1=\frac{\angle B}{2}-\frac{\angle C}{2}$ and so $\angle(l,BC_0)=\angle(l,AB) +\angle C_1BC_0=\frac{\angle B}{2}-\frac{\angle C}{2}+\angle C=\frac{\angle C+\angle B}{2}=\angle C_0IB$ by exterior angle property $\implies l$ is tangent to $\odot(C_0BA_0)$ by alternate segment theorem and so we are done. $\blacksquare$. Now when $\angle B< \angle C$. [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 = -18.159382950767856, xmax = 14.3850727306359, ymin = -10.21554036503088, ymax = 11.160443976888521; /* image dimensions */ pen dbwrru = rgb(0.8588235294117647,0.3803921568627451,0.0784313725490196); pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902); pen wvvxds = rgb(0.396078431372549,0.3411764705882353,0.8235294117647058); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen wwffqq = rgb(0.4,1,0); /* draw figures */ draw((0.31237175029136255,7.20983748013019)--(-9.56,-0.65), linewidth(1.2) + dbwrru); draw((-9.56,-0.65)--(3.72,-0.61), linewidth(1.2) + dbwrru); draw((3.72,-0.61)--(0.31237175029136255,7.20983748013019), linewidth(1.2) + dbwrru); draw(circle((-5.11051768827992,0.5548503068581906), 4.609724189606807), linewidth(1.2) + linetype("4 4") + sexdts); draw(circle((-2.9253510821999438,1.1465592903812543), 6.873586512609023), linewidth(1.2) + linetype("4 4") + sexdts); draw((-5.124402336710762,5.164553585897318)--(-9.56,-0.65), linewidth(1.2) + wvvxds); draw((-9.56,-0.65)--(-2.2393313815836318,-3.0515116129282656), linewidth(1.2) + wvvxds); draw((-5.124402336710762,5.164553585897318)--(1.9727011916890935,3.399707546996864), linewidth(1.2) + wvvxds); draw((1.9727011916890935,3.399707546996864)--(-2.2393313815836318,-3.0515116129282656), linewidth(1.2) + wvvxds); draw(circle((-2.918532250712892,-1.1172927633197254), 6.657886781214573), linewidth(1.2) + linetype("4 4") + sexdts); draw(circle((-4.335407353398249,2.9176645751013086), 6.326499730764072), linewidth(1.2) + linetype("4 4") + sexdts); draw((-9.56,-0.65)--(1.9727011916890935,3.399707546996864), linewidth(1.2) + dtsfsf); draw((-5.124402336710762,5.164553585897318)--(-2.2393313815836318,-3.0515116129282656), linewidth(1.2)); draw((-3.5487658701898472,4.1358128229343905)--(3.72,-0.61), linewidth(1.2) + dtsfsf); draw((0.31237175029136255,7.20983748013019)--(-2.2393313815836318,-3.0515116129282656), linewidth(1.2) + wwffqq); draw((-5.124402336710762,5.164553585897318)--(-3.5487658701898472,4.1358128229343905), linewidth(1.2) + dtsfsf); /* dots and labels */ dot((0.31237175029136255,7.20983748013019),dotstyle); label("$A$", (-0.11472084395278126,7.423383777252262), NE * labelscalefactor); dot((-9.56,-0.65),dotstyle); label("$B$", (-9.46804865789953,-0.4351199568399848), NE * labelscalefactor); dot((3.72,-0.61),dotstyle); label("$C$", (3.8145310230933416,-0.3924106974155704), NE * labelscalefactor); dot((-0.8846004665763552,2.3963661877441926),linewidth(4pt) + dotstyle); label("$I$", (-0.7980689947434113,2.5758828325812293), NE * labelscalefactor); dot((-1.636209342124067,-0.6261331606690483),linewidth(4pt) + dotstyle); label("$A_1$", (-1.673608812943906,-0.3924106974155704), NE * labelscalefactor); dot((1.9727011916890935,3.399707546996864),linewidth(4pt) + dotstyle); label("$B_1$", (2.063451386692352,3.5795504290549673), NE * labelscalefactor); dot((-3.5487658701898472,4.1358128229343905),linewidth(4pt) + dotstyle); label("$C_1$", (-3.46739770876931,4.305607839270012), NE * labelscalefactor); dot((-5.124402336710762,5.164553585897318),linewidth(4pt) + dotstyle); label("$C_0$", (-5.047640307472642,5.330630065455957), NE * labelscalefactor); dot((-2.2393313815836318,-3.0515116129282656),linewidth(4pt) + dotstyle); label("$A_0$", (-2.1861199260368784,-3.4674773759734063), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Proof: The proof of the fact that $BC_0B_1C,A_0BB_1A,C_0A_0IB$ are concyclic is similiar to the previous part. Now IF $l$ be the line tangent to $\odot(ABC$ at $B$ then we have $\angle(l,AB)=\angle ACB=\angle C$ and $\angle C_0BA=\angle C_0BB_1-\angle C_1BI=\frac{\angle C}{2}-\frac{\angle B}{2}$. Now $\angle(l,BC_0)=\angle(l,AB)-\angle C_0BA=\angle C -\frac{\angle C-\angle B}{2}=\frac{\angle C+\angle B}{2}=\angle BIC_0$ by exterior angle property and so by alternate segment theorem we are done again.$\blacksquare$.

Why so easy problems this year? Sharygin 2020 CR P18 wrote: Bisectors $AA_1$, $BB_1$, and $CC_1$ of triangle $ABC$ meet at point $I$. The perpendicular bisector to $BB_1$ meets $AA_1,CC_1$ at points $A_0,C_0$ respectively. Prove that the circumcircles of triangles $A_0IC_0$ and $ABC$ touch.

Claim:- $C_0IA_0B$ is a cyclic quadrilateral.

Now draw a tangent to $\odot(ABC)$ at $B$, we will show that this line is also tangent to $\odot(BA_0IC_0)$ at $B$. Let $K$ be any point on this tangent and $K$ lies on the side of $BC$ where $A$ lies. Now, it suffices to show that $\angle C_0IB=\angle C_0BK$. $$\angle C_0B_1B=\angle C_0BB_1=\frac{\angle C}{2}\implies\angle C_0BA=\angle{\frac{C}{2}-\frac{B}{2}}\implies \angle KBC_0=\angle C_0BK=\angle {\frac{B}{2}+\frac{C}{2}}=\angle C_0IB\implies BK \text{ is tangent to } \odot(C_0IA_0B)$$So, $\odot(ABC), \odot(A_0IC_0)$ are internally tangent at $B$. $\blacksquare$

Solution: [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(24.817cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ real xmin = -3.42, xmax = 34.76, ymin = -11.32, ymax = 6.56; /* 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); pen dcrutc = rgb(0.8627450980392157,0.0784313725490196,0.23529411764705882); pen zzttqq = rgb(0.6,0.2,0.); pen ffxfqq = rgb(1.,0.4980392156862745,0.); pen xdxdff = rgb(0.49019607843137253,0.49019607843137253,1.); draw((5.42,3.42)--(4.52,-6.02)--(17.96,-6.)--cycle, linewidth(2.) + green); /* draw figures */ draw((5.42,3.42)--(4.52,-6.02), linewidth(2.) + green); draw((4.52,-6.02)--(17.96,-6.), linewidth(2.) + green); draw((17.96,-6.)--(5.42,3.42), linewidth(2.) + green); draw((xmin, 0.9105555208241746*xmin-10.135710954125267)--(xmax, 0.9105555208241746*xmax-10.135710954125267), linewidth(1.6) + dotted); /* line */ draw((xmin, -2.2651507818922987*xmin + 15.69711723785626)--(xmax, -2.2651507818922987*xmax + 15.69711723785626), linewidth(1.6) + dotted); /* line */ draw((xmin, -0.33293172429312495*xmin-0.020546231695475318)--(xmax, -0.33293172429312495*xmax-0.020546231695475318), linewidth(1.6) + dotted); /* line */ draw((xmin, -1.098230670321856*xmin + 5.058344193945416)--(xmax, -1.098230670321856*xmax + 5.058344193945416), linewidth(2.) + red); /* line */ draw(shift((11.233879749888848,-1.8971919253071998)) * scale(7.878688197709038, 7.878688197709038)*unitcircle, linewidth(2.) + red); draw(shift((6.640181456876914,-4.718055140119127)) * scale(2.4880172483837417, 2.4880172483837417)*unitcircle, linewidth(2.) + yqqqyq); draw((4.52,-6.02)--(6.636479054356646,-2.2300406464976406), linewidth(2.)); draw((4.52,-6.02)--(9.116967767050626,-4.954189428165346), linewidth(2.)); draw((9.116967767050626,-4.954189428165346)--(10.607598163548047,-0.47690388362221803), linewidth(2.)); draw((10.607598163548047,-0.47690388362221803)--(6.636479054356646,-2.2300406464976406), linewidth(2.)); draw(shift((11.2419015727768,-7.287856906017579)) * scale(6.840425563386806, 6.840425563386806)*unitcircle, linewidth(2.) + dcrutc); draw(shift((5.867542350655218,-1.385570774956543)) * scale(4.826365571394927, 4.826365571394927)*unitcircle, linewidth(2.) + zzttqq); draw((xmin, -1.6284725430467961*xmin + 1.3406958945715184)--(xmax, -1.6284725430467961*xmax + 1.3406958945715184), linewidth(2.) + ffxfqq); /* line */ /* dots and labels */ dot((5.42,3.42),linewidth(3.pt) + ududff); label("$A$", (4.92,3.5), NE * labelscalefactor,ududff); dot((4.52,-6.02),linewidth(4.pt) + ududff); label("$B$", (3.7,-6.36), NE * labelscalefactor,ududff); dot((17.96,-6.),linewidth(3.pt) + ududff); label("$C$", (18.2,-6.58), NE * labelscalefactor,ududff); dot((8.134514255894613,-2.7287840891974744),linewidth(3.pt) + uuuuuu); label("$I$", (8.22,-2.6), NE * labelscalefactor,uuuuuu); dot((9.584166074057297,-6.012464038580272),linewidth(3.pt) + uuuuuu); label("$A_1$", (9.32,-6.56), NE * labelscalefactor,uuuuuu); dot((10.607598163548047,-0.47690388362221803),linewidth(3.pt) + uuuuuu); label("$B_1$", (10.68,-0.36), NE * labelscalefactor,uuuuuu); dot((4.935327747072867,-1.6636734084801483),linewidth(3.pt) + uuuuuu); label("$C_1$", (4.6,-1.44), NE * labelscalefactor,uuuuuu); dot((9.116967767050626,-4.954189428165346),linewidth(3.pt) + uuuuuu); label("$A_0$", (8.3,-4.86), NE * labelscalefactor,uuuuuu); dot((6.636479054356646,-2.2300406464976406),linewidth(3.pt) + uuuuuu); label("$C_0$", (6.38,-1.86), NE * labelscalefactor,uuuuuu); dot((0.1114009724408102,1.1592824696829462),linewidth(3.pt) + xdxdff); label("$X$", (0.2,1.28), NE * labelscalefactor,xdxdff); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* re-scale y/x */ currentpicture = yscale(1.5384615384615383) * currentpicture; /* end of picture */ [/asy][/asy] Claim 1: $B,C_0,B_1,C$ are concyclic . Proof: In $\triangle B_1CB$, $CC_0$ which is the angle bisector of $\angle B_1CB$ meets the perpendicular bisector of $BB_1$ at $C_0$. So, by Incentre-Excentre Lemma, $B,C_0,B_1,C$ are concyclic . Similarly , $A,B_1,A_0,B$ are concyclic. Claim 2: $A_0,I,C_0,B$ are concyclic . Proof: As, $BC_0B_1C$ and $AB_1A_0B$ are cyclic quadrilaterals , $\angle C_0BA_0 = \angle C_0BI+\angle A_0BI= \angle C_0BB_1+\angle A_0BB_1=\angle C_0CB_1+\angle A_0AB_1= \frac{A}{2}+\frac{C}{2} $ (Let the angles of $\triangle ABC$ be $A,B,C$ respectively) $\implies \angle C_0BA_0 = \frac{A+C}{2}= 180^{0}-\angle AIC= 180^{0}-\angle C_0IA_0 $. So, $C_0IB_0B$ is cyclic quadrilateral . Claim 3: $\odot (ABC)$ and $\odot (A_0IC_0)$ touch at B . Proof: We have already proved that they intersect at B. Now we will prove that the tangent to $\odot (ABC)$ at $B$ is also tangent to $\odot (A_0IC_0)$ at $B$. Let $X$ be a point on the tangent to $\odot (ABC)$ at $B$ as shown . $\angle XBA= C$ ---(Alternate segment theorem). Now, $\angle XBC_0 = \angle XBA +\angle ABC_0= C + \angle ABC_0 $ $\angle ABC_0=\angle ABI-\angle IBC_0= \frac {B}{2}-\frac {C}{2}$--(as $\angle IBC_0=\angle ICA=\frac{C}{2}$) $\implies \angle XBC_0=C+\angle ABC_0=C+\frac{B}{2}-\frac{C}{2} = \frac{B+C}{2} $ We also note that , $\angle C_0IB=\frac{B+C}{2}$ , $\implies XBC_0=\angle C_0IB$ , $\implies XB$ is also tangent to $\odot (A_0IC_0B) $ $\implies \odot (ABC)$ and $\odot (A_0IC_0)$ touch at B . $\blacksquare$

Minimal angle chasing solution. First, note that $A_0$ lie on the perpendicular bisector of $BB_1$ as well as the angle bisector of $\angle B_1AB$. Hence it must be the midpoint of arc in $\odot(B_1AB)$, which means $B,B_1,A,A_0$ are concyclic. Similarly, $B,B_1,C,C_0$ are concyclic hence $$IB\cdot IB_1 = IA\cdot IA_0 = IC\cdot IC_0$$or $A,A_0,C,C_0$ are concyclic. Hence $\angle IBC_0 = \angle BCI = \angle IA_0C_0$ or $A_0,B,C_0,I$ are concyclic. To finish, let $M = BI\cap\odot(ABC)$. Notice that by Reim's (on $IIA_0C_0$ and $ACA_0C_0$), we get that the tangent at $I$ of $\odot(IA_0C_0)$ is parallel to AC, which is also parallel to the tangent at $M$ of $\odot(BAC)$. Hence these two circles are tangent at $B$.

Simply angle chasing,

Problem 18. Bisectors $AA_1, BB_1,$ and $CC_1$ of triangle $ABC$ meet at point $I$. The perpendicular bisector to $BB_1$ meets $AA_1, CC_1$ at points $A_0, C_0$ respectively. Prove that the circumcircles of triangles $A_0IC_0$ and $ABC$ touch. [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 = -4.672515923566868, xmax = 25.489904458598705, ymin = -14.702286624203845, ymax = 4.404082802547774; /* image dimensions */ pen qqccqq = rgb(0,0.8,0); draw((11.276942675159237,2.0840764331210195)--(6.056942675159237,-8.19592356687898)--(21.736942675159238,-8.57592356687898)--cycle, linewidth(0.8) + blue); /* draw figures */ draw((11.276942675159237,2.0840764331210195)--(6.056942675159237,-8.19592356687898), linewidth(0.8) + blue); draw((6.056942675159237,-8.19592356687898)--(21.736942675159238,-8.57592356687898), linewidth(0.8) + blue); draw((21.736942675159238,-8.57592356687898)--(11.276942675159237,2.0840764331210195), linewidth(0.8) + blue); draw(circle((16.88892478122831,-6.810413198344613), 5.159486831195207), linewidth(0.8) + linetype("4 4") + qqccqq); draw(circle((13.960965644504844,-5.744133673881322), 8.275557550965881), linewidth(0.8)); draw((17.0139266624592,-1.6524408359753173)--(21.736942675159238,-8.57592356687898), linewidth(0.8)); draw((21.736942675159238,-8.57592356687898)--(13.206233402313177,-10.424010930375796), linewidth(0.8)); draw((21.736942675159238,-8.57592356687898)--(6.582193443118025,-1.9973252214261483), linewidth(0.8) + red); draw((6.056942675159237,-8.19592356687898)--(17.0139266624592,-1.6524408359753173), linewidth(0.8) + red); draw((13.206233402313177,-10.424010930375796)--(11.276942675159237,2.0840764331210195), linewidth(0.8) + red); draw((17.0139266624592,-1.6524408359753173)--(13.206233402313177,-10.424010930375796), linewidth(0.8)); /* dots and labels */ dot((11.276942675159237,2.0840764331210195),dotstyle); label("$A$", (11.36506369426751,2.3021719745222926), NE * labelscalefactor); dot((6.056942675159237,-8.19592356687898),dotstyle); label("$C$", (5.542770700636942,-8.270439490445873), NE * labelscalefactor); dot((21.736942675159238,-8.57592356687898),dotstyle); label("$B$", (22.000732484076415,-8.438592356687912), NE * labelscalefactor); dot((12.28854858071465,-4.474424505087679),linewidth(4pt) + dotstyle); label("$I$", (11.554235668789802,-4.592095541401284), NE * labelscalefactor); dot((12.888100651743736,-8.361474589270697),linewidth(4pt) + dotstyle); label("$A_1$", (12.96251592356687,-8.186363057324856), NE * labelscalefactor); dot((8.730859083030401,-2.9300498747495616),linewidth(4pt) + dotstyle); label("$B_1$", (8.506464968152864,-2.5742611464968226), NE * labelscalefactor); dot((15.70839110360469,-2.432103322828171),linewidth(4pt) + dotstyle); label("$C_1$", (15.589904458598715,-2.9105668789808994), NE * labelscalefactor); dot((13.206233402313177,-10.424010930375796),linewidth(4pt) + dotstyle); label("$A_0$", (12.500095541401266,-10.792732484076451), NE * labelscalefactor); dot((17.0139266624592,-1.6524408359753173),linewidth(4pt) + dotstyle); label("$C_0$", (16.682898089171964,-1.2710764331210244), NE * labelscalefactor); dot((15.23390087909482,-5.752986720814271),linewidth(4pt) + dotstyle); label("$X$", (15.253598726114638,-6.420757961783451), NE * labelscalefactor); dot((6.582193443118025,-1.9973252214261483),linewidth(4pt) + dotstyle); label("$M$", (6.3204777070063685,-1.6073821656051012), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Solution. Notations and some trivial facts. Let $M$ be the mid-point of the arc $AC$ and so, obviously, $MB_1B$ are collinear as $BB_1$ is the angle bisector of $\angle ABC$. Also, let $CC_0\cap BB_1$ be $X$ and so, $\angle C_0XB=90$. We denote the circumcircle of $\triangle PQR$ by $(PQR)$. We also denote $\angle ABC=B, \angle BCA= C,\angle CAB=A$. Proof. Let $(CB_1B)\cap A_0C_0=J$. Thus, $J$ lies on perpendicular bisector of $BB_1\implies J$ is the mid-point of arc $B_1B$ not containing $C$ of $(CBB_1)$. Thus, $J$ lies on the angle bisector of angle $\angle BCB_1\implies J=CI\cap A_0C_0\implies J\equiv C_0\implies C_0\in (BCB_1)\implies BCB_1C_0$ is cyclic. Hence, $C/2=\angle B_1CI=\angle C_0BI$. Now note that $\angle AIB_1=\angle IAB+\angle IBA=(A+B)/2=90-C/2$. Thus, $\angle A_0IB=90-C/2\implies \angle IAC_0=C/2$ because $\angle IXA_0=90$. Hence, $\angle IA_0C_0=\angle IBC_0=C/2\implies IA_0BC_0$ is cyclic. Next, we observe that, $\angle MAB=\angle MAC+\angle CAB=\angle MBC+\angle CAB=B/2+A$. Also, $\angle CIB_1=\angle BCI+\angle IBC=(B+C)/2=90-A/2\implies \angle C_0IB=90-A/2$. But, $\angle C_0BI=C/2\implies \angle IC_0B=180-(90-A/2+C/2)=90+(A-C)/2=B/2+A$ because $A+B+c=180$. Thus, $\angle IC_0B=\angle MAB$. Now, construct a line through $B$ tangent to $(IA_0C_0)$ at $B$ and let $Y$ be a point on this line such that $Y$ and $C_0$ are on opposite sides of line $BM$. Thus, $\angle YBM=\angle YBI=\angle IC_0B=\angle MAB\implies \angle MBY=\angle MAB\implies YB$ is also tangent to $(ABC)$ at $B$ and thus, $(IA_0C_0)$ and $(ABC)$ are internally tangent at $B$ and this completes the proof. Feedback: This was a simple angle chasing problem after noting that the tangancy point is $B$ for both the circles.

WLOG, suppose $BA \ge BC$. By Phantom Points, it's easy to see $ABA_0B_1$ and $CBC_0B_1$ are cyclic. Now, observe $$\angle BA_0C_0 = \frac{\angle BA_0B_1}{2} = \frac{180^{\circ} - \angle BAB_1}{2} = \frac{\angle B}{2} + \frac{\angle C}{2} = \angle BIC_0$$so $B$ lies on $A_0IC_0$. Let $X$ be a point on the tangent to $(ABC)$ at $B$ such that $A, X$ are on opposite sides of $BC$. Then, it's easy to see $$\angle XBA_0 = \angle XBI - \angle IBA_0 = \angle XBC + \angle CBI - \angle B_1BA_0$$$$= \angle A + \frac{\angle B}{2} - \angle B_1AA_0 = \frac{\angle A}{2} + \frac{\angle B}{2}.$$But $$\angle BIA_0 = \angle IBA + \angle IAB = \frac{\angle A}{2} + \frac{\angle B}{2}$$so $XB$ is also tangent to $(A_0IC_0B)$, as desired. $\blacksquare$ Remark: To any American students who might be reading this: please try AIME I 2020/13.

Notice $A_0$ is the intersection of the angle bisector of $\angle BAB_1$ and perpendicular bisector of $\overline{BB_1}$ so it is the midpoint of arc $BB_1$ (not containing $A$) of $(ABB_1).$ Similarly, $C_0$ lies on $(CBB_1).$ Notice \begin{align*}\measuredangle A_0B_1C&=\measuredangle A_0B_1A\\&=\measuredangle A_0B_1B+\measuredangle BB_1A\\&=\measuredangle A_0AB+\measuredangle B_1BA+\measuredangle BAC\\&=\measuredangle IBA+\measuredangle BAI\\&=90-\measuredangle B_1CI\end{align*}so $\overline{CI}\perp\overline{A_0B_1}.$ Hence, $$\measuredangle A_0BI=\measuredangle IB_1A_0=90-\measuredangle CIB=90-\measuredangle C_0IB=\measuredangle A_0C_0I$$and $B$ lies on $(A_0C_0I).$ Let $\ell$ be the line tangent to $(ABC)$ at $B.$ We claim $\ell$ is also tangent to $(A_0C_0I)$ at $B.$ Indeed, \begin{align*}\measuredangle ACB+\measuredangle CBB_1=\measuredangle AB_1B&\implies \measuredangle (\overline{AB},\ell)+\measuredangle B_1BA=\measuredangle IA_0B\\&\implies\measuredangle (\overline{IB},\ell)=\measuredangle IA_0B.\end{align*}$\square$