Yay I solved this

Interesting problem :0 a) We proceed via contradiction(!). Note that $\triangle A_1B_1C_1$ is the pedal triangle of the circumcenter $Be$ of $\triangle I_AI_BI_C$ (the excentral triangle) w.r.t. $\triangle ABC$. If $Be$ does not lie on the sides of $\triangle ABC$, then the isogonal conjugate of $Be$ w.r.t. $\triangle ABC$ exists and has a pedal triangle with vertices distinct from $A,B,C$ sharing the same circumcircle as $\triangle A_1B_1C_1$. Then $A$ cannot also lie on this circle, implying that $Be$ lies on the sides of $\triangle ABC$. If $Be = B_1$ or $C_1$, then we arrive at a similar contradiction. Hence $Be = A_1$, implying the result. b) Use Pappus on $BACI_CA_1I_B$, immediately implying the result.

With areals we get $A_1=(0,s-b,s-c)$ and similar for $B_1,C_1$. Hence if $a^2 yz+b^2 xz+c^2 xy-(x+y+z)(ux+vy+wz)$ is the circle passing through $A_1,B_1,C_1$ we get (from $A_1$): $$a^2 (s-b)(s-c)=(s-b+s-c)(v(s-b)+w(s-c)) \Rightarrow a(s-b)(s-c)=v(s-b)+w(s-c)$$Summing the similar results for $B_1,C_1$ gives: $$2 \sum_{cyc} u (s-a)=\sum_{cyc} a(s-b)(s-c)$$And then plugging then subtracting this from the first equation gives: $$u(s-a)=\frac{\sum_{cyc} a(s-b)(s-c)}{2}-a(s-b)(s-c)$$Note that $A$ on $\odot A_1 B_1C_1$ is equivalent to $u=0$ which is in turn equivalent to: $$K=b(s-a)(s-c)+c(s-a)(s-b)-a(s-b)(s-c)=0$$Now if $I_B,I_C$ are the excentres of $\triangle ABC$ opposite $B,C$ then: $$B_1,I_B,A_1 \text{ colinear} \Leftrightarrow 0=\begin{vmatrix} 0 & s-b &s-c \\ s-a & 0 & s-c \\ a & -b & c \end{vmatrix}=a(s-b)(s-c)-c(s-a)(s-b)-b(s-a)(s-b)=-K$$Hence if $A$ is on $\odot A_1B_1C_1$ then $A_1,B_1,I_B$ are colinear so $\angle A_1 B_1 A=90^{\circ}$ and hence this circle has diameter $AA_1$ Also: $$I,B_1,C_1 \text{ colinear} \Leftrightarrow 0=\begin{vmatrix} a & b & c \\ s-a & 0 & s-c \\ s-a & s-b & 0 \end{vmatrix}=b(s-a)(s-c)+c(s-a)(s-b)-a(s-b)(s-c)=K$$Hence as we are assuming $K=0$ we get $I,B_1,C_1$ are colinear. In fact noticing that $I_BB_1,I_AA_1,I_CC_1$ concur at the circumcentre of $I_AI_BI_C$ say $O$ we have in fact proved that: $$O \equiv A_1 \Leftrightarrow A,A_1,B_1,C_1 \text{ concyclic} \Leftrightarrow I,B_1,C_1 \text{ colinear}$$as they are all equivalent to $K=0$

Here's my solution: [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 = -12.05, xmax = 14.15, ymin = -3.97, ymax = 7.27; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen qqccqq = rgb(0,0.8,0); /* draw figures */ draw((-4.781523029254206,2.683377195388904)--(6.871149068786946,5.218374928896586), linewidth(0.4) + wrwrwr); draw(circle((0.7015597483382036,1.4355753715441475), 2.546602274766395), linewidth(0.4) + blue); draw((-4.781523029254206,2.683377195388904)--(2.0614153440904808,-0.7175562789934126), linewidth(0.4) + wrwrwr); draw((2.0614153440904808,-0.7175562789934126)--(6.871149068786946,5.218374928896586), linewidth(0.4) + wrwrwr); draw((-0.66,3.58)--(-2.76,-0.64), linewidth(0.4) + qqccqq); draw((-2.76,-0.64)--(4.7,-0.76), linewidth(0.4) + qqccqq); draw((4.7,-0.76)--(-0.66,3.58), linewidth(0.4) + qqccqq); draw((-0.66,3.58)--(2.0614153440904808,-0.7175562789934126), linewidth(0.4) + wrwrwr); /* dots and labels */ dot((-0.66,3.58),dotstyle); label("$A$", (-0.89,4.11), NE * labelscalefactor); dot((-2.76,-0.64),dotstyle); label("$B$", (-3.17,-1.33), NE * labelscalefactor); dot((4.7,-0.76),dotstyle); label("$C$", (4.93,-1.19), NE * labelscalefactor); dot((6.871149068786946,5.218374928896586),linewidth(4pt) + dotstyle); label("$I_B$", (7.13,5.61), NE * labelscalefactor); dot((-4.781523029254206,2.683377195388904),linewidth(4pt) + dotstyle); label("$I_C$", (-5.53,3.07), NE * labelscalefactor); dot((-1.8356826708257379,1.2174376805311355),linewidth(4pt) + dotstyle); label("$C_1$", (-2.65,0.99), NE * labelscalefactor); dot((3.0875794770264293,0.545579304049496),linewidth(4pt) + dotstyle); label("$B_1$", (3.59,0.49), NE * labelscalefactor); dot((2.0614153440904808,-0.7175562789934126),linewidth(4pt) + dotstyle); label("$A_1$", (2.17,-1.31), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $V$ be the Bevan Point of $\triangle ABC$. Then $\measuredangle B_1VC_1=\measuredangle B_1AC_1=\measuredangle B_1A_1C_1$, which means that $V$ lies on $\gamma$ such that $V$ is the antipode of $A$. Suppose $V \neq A_1$. Then, $VA_1 \perp AA_1$ along with the fact that $VA_1 \perp BC$ gives that $A$ must lie on $BC$, i.e. a contradiction. Thus, $V=A_1,$ proving Part (a). Part (b) follows by applying Pappus' Theorem on $I_C,A,I_B$ and $B,A_1,C$ (with symbols having their usual meanings). Hence, done. $\blacksquare$

My first USA TST Let $I_A,I_B,I_C$ be the execenters. Let $A_1'$ be the antipode of $A_1$ in $A_1B_1C_1$ .Note that $I$ is the orthocenter of $I_AI_BI_C$ with orthic triangle $ABC$. Let $V$ be the circumcenter of $I_AI_BI_C$.Since $V$ and $I$ are isogonal conjugates in $I_AI_BI_C \implies I_AV,I_BV,I_CV$ are perpendicular to $BC,CA,AB$ which clearly implies $V = I_AA_1 \cap I_BB_1 \cap I_CC_1$.This proves the existense of the Bevan point of $ABC$.Obviously $AB_1C_1V$ etc is concyclic but we know that $AA_1B_1C_1$ is concyclic .Therfore $V \in A_1B_1C_1 \implies V = ${$A_1,A_1'$} The later is clearly impossible therefore The bevan point coincides with $A_1$ this finishes $(a)$.Now the previous result also implies $A_1B_1I_B$ and $A_1C_1I_C$ collinear.Now by Pappus theorm on $I_CAI_BBA_1C$ concludes $(b)$

Nice bary practice CantonMathGuy wrote: Let the excircle of a triangle $ABC$ opposite the vertex $A$ be tangent to the side $BC$ at the point $A_1$. Define points $B_1$ on $\overline{CA}$ and $C_1$ on $\overline{AB}$ analogously, using the excircles opposite $B$ and $C$, respectively. Denote by $\gamma$ the circumcircle of triangle $A_1B_1C_1$ and assume that $\gamma$ passes through vertex $A$. Show that $\overline{AA_1}$ is a diameter of $\gamma$. Show that the incenter of $\triangle ABC$ lies on line $B_1C_1$. Solution: Let $I_b$ be the $B-$ excenter. We know $I_b = (a:-b:c)$, $A_1 = (0:s-b:s-c)$. We have $A,B_1,C_1,A_1$ are conyclic. Let the equation of $\odot(AB_1C_1)$ be \[a^2yz+b^2zx+c^2xy = (vy+wz)(x+y+z).\]We have $B_1$ lies on the circle so, \[c^2xy = vy(s-a+s-b) \implies v = c(s-a).\]Therefore equation of circle is \[a^2yz+b^2zx+c^2xy = (s-a)(cy+bz)(x+y+z).\]As $A_1$ lies on the circle, \[a(s-b)(s-c) = b(s-a)(s-c)+c(s-a)(s-b).\] (a) We need to show $A_1B_1\perp AC$ or $A_1,B_1,I_b$ are collinear. We have, \[\begin{vmatrix} a&-b & c\\s-a&0&s -c \\ 0 &s-b&s-c \end{vmatrix} = -a(s-b)(s-c)+b(s-a)(s-c)+c(s-a)(s-b)=0~~~\square\] (b) We need to show $I,B_1,C_1$ are collinear. We have \[\begin{vmatrix} a&b&c \\ s-a & s-b & 0 \\ s-a & 0 & s-c \end{vmatrix}=-a(s-b)(s-c)+b(s-a)(s-c)+c(s-a)(s-b)=0~~~\square\]

I think this is too bash-favored, whether it's barycentric or side bash. And once a. is proved, b. follows almost immediately.

I guess this is also a valid solution Part (a): Let $Be$ be the Bevan Point of $\Delta ABC$, then, $\Delta A_1B_1C_1$ is the Pedal Triangle of Bevan Point WRT $\Delta ABC$, also, from All-Russian 2005, if $\odot (AB_1C_1) \cap I_BI_C =M$, then, $M$ is the midpoint of $I_BI_C$, And, (well-known) that, $Be \in \odot (AB_1C_1)$ $\implies $ $MBe \perp AM$, $\implies $ $ABe$ is the diameter of $\odot (AB_1C_1)$, Now, according to the question, $A_1 \in \odot (AB_1BeC_1M)$ $\implies$ $\angle AA_1Be =90^{\circ}$ also recall, that, $Be$ lies on $I_AA_1$ and $BeA_1 \perp BC$ and also $Be \in \odot (AB_1A_1C_1)$, forces $\boxed{A_1=Be}$ $\implies$ $ABe=AA_1$ is the diameter of $\odot (AB_AA_1C_1)$ As for Part (b), Pappus' Theorem works the most elegant, I couldn't find much better proof for part(b)

Notice that the lines $I_aA,I_bB,I_cC$ and the lines $AA_1,BB_1,CC_1$ are concurrent. By Cevian Nest Theorem, lines $I_aA_1,I_bB_1,I_cC_1$ are concurrent as well. Let them concur at point $X.$ Then, $$\angle BXC=\angle BAC+\angle ABX+\angle ACX=\angle BAC+\angle C_1A_1B_1=180^{\circ.}$$Therefore, $X$ lies on the $BC$ which means that $X=A_1.$ Since, $I_c,C_1,A_1$ is collinear and $I_cA_1 \perp AB,$ we have $\angle AC_1A_1=90^{circ},$ which implies that $AA_1$ is diameter of $\gamma.$ For b, applying the pappus theorem yields our desired result.

Storage. USTAST EGMO 2019 P5 wrote: Let the excircle of a triangle $ABC$ opposite the vertex $A$ be tangent to the side $BC$ at the point $A_1$. Define points $B_1$ on $\overline{CA}$ and $C_1$ on $\overline{AB}$ analogously, using the excircles opposite $B$ and $C$, respectively. Denote by $\gamma$ the circumcircle of triangle $A_1B_1C_1$ and assume that $\gamma$ passes through vertex $A$. Show that $\overline{AA_1}$ is a diameter of $\gamma$. Show that the incenter of $\triangle ABC$ lies on line $B_1C_1$. $a)$ Let $Be$ be the Bevan point of $\triangle ABC$. So, we need to prove that $A_1\equiv Be$. Assume to the contrary. Notice that $\overline{I_A-A_1-Be}$ where $I_A$ is the $A-\text{Excenter}$ of $\triangle ABC$. Now $\angle B_1BeC_1=\angle B_1A_1C_1=180^\circ-\angle BAC\implies Be\equiv A_1$. $\blacksquare$ $b)$ By Pappu's Theorem on $\overline{I_CAI_B}$ and $\overline{BA_1C}$ we get that $I\in B_1C_1$. $\blacksquare$

We use bary wrt $ABC$. The Nagel Point is $(s-a:s-b:s-c)$, so $A_1=(0:s-b:s-c), B_1=(s-a:0:s-c), C_1=(s-a:s-b:0)$. Since $A\in (A_1B_1C_1)$, the equation for $(A_1B_1C_1)$ is $-a^2yz-b^2xz-c^2xy+(vy+wz)(x+y+z)=0$ for some $v,w$. Since $B_1\in (A_1B_1C_1)$, we have $-b^2(s-a)(s-c) + w(s-c)b=0$, so $w=b(s-a)$. Similarly, $v=c(s-a)$. So the equation of $(AA_1B_1C_1)$ is \[-a^2yz-b^2xz-c^2xy + (c(s-a)y + b(s-a)z)(x+y+z)=0.\]Since $A_1=(0:s-b:s-c)$ is also on this circle, we get \begin{align*} & -a^2(s-b)(s-c) + [c(s-a)(s-b) + b(s-a)(s-c)]a = 0 \\ \implies & a(s-b)(s-c) = c(s-a)(s-b) + b(s-a)(s-c). \end{align*}In order to show that $AA_1$ is a diameter of $(AA_1B_1C_1)$, it suffices to show $\angle AB_1A_1 = 90^\circ$. But $\angle I_BB_1A=90^\circ$ trivially, so it suffices to show $I_B,B_1,A_1$ collinear. The coordinates of $I_B$ are $I_B = (a:-b:c)$. We compute \[ \begin{vmatrix} a & -b & c \\ s-a & 0 & s-c \\ 0 & s-b & s-c \end{vmatrix}=-a(s-b)(s-c) + b(s-a)(s-c) + c(s-a)(s-b) = 0, \]as desired. We also want to show that $I=(a:b:c)$ is collinear with $B_1,C_1$. We compute \[ \begin{vmatrix} a & b & c \\ s-a & 0 & s-c \\ s-a & s-b & 0\end{vmatrix} = -a(s-b)(s-c) + b(s-a)(s-c) +c(s-a)(s-b)=0, \]as desired.

different solution for part a: draw three perpendicular from $A_1 ,B1 , C1$ we know that they are concurrent(according Carnot's theorem) then if they meet at$T$ we get $T=$$A_1$(according simple lemma i say after this post) qed. part b : papus theorem!

we use this simple lemma :if $P$lies inside of triangle $ABC$ then angle$\angle BAC$ less than $\angle BPC$.

Part a) Let $I_A,I_B,I_C$ be the $A,B,C$ excenters respectivily and let $O$ be the center of $(I_AI_BI_C)$, note that since $\angle BI_AC=90-\frac{\angle BAC}{2}$ we get $\angle I_BOI_C=180-\angle BAC$ hence $O$ lies on $(AB_1A_1C_1)$ but using orthic duality we have that $O-A_1-I_A$ hence $O \equiv A_1$ and since by more orthic duality we have $I_B-B_1-O$ and $I_C-C_1-O$ we get $\angle AB_1A_1=90=\angle AC_1A_1$ hence $AA_1$ is diameter of $(AB_1A_1C_1)$ as desired. Part b) We use Pascal on the degenrate conic $\mathcal C$ passing through $A,B.C,I_B,I_C,A_1$ to get the desired result.

[asy][asy] /* Coordinates of I_A,I_B, and I_C from Geogebra rounded to nearest tenth. */ size(7cm); import geometry; pair I_A,I_B,I_C,A,B,C,A_1,B_1,C_1; I_A=(1.2,-14.3); I_B=(8.1,8.1); I_C=(-10.1,2.6); A=foot(I_A,I_B,I_C); B=foot(I_B,I_A,I_C); C=foot(I_C,I_A,I_B); A_1=foot(I_A,B,C); B_1=foot(I_B,A,C); C_1=foot(I_C,A,B); draw(A--B--C--cycle); draw(I_A--I_B--I_C--cycle); draw(A_1--I_A--A); draw(I_B--B); draw(I_C--C); draw(I_C--A_1,dotted); draw(I_B--A_1,dotted); draw(circle(A,B_1,C_1)); dot(I_A^^I_B^^I_C^^A^^B^^C^^A_1^^B_1^^C_1); label("$I_A$",I_A,SE); label("$I_B$",I_B,SE); label("$I_C$",I_C,SW); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,SE); label("$A_1$",A_1,SE); label("$B_1$",B_1,E); label("$C_1$",C_1,W); [/asy][/asy] Let $I_A,I_B,$ and $I_C$ be the excenters of $\triangle ABC.$ By Cevian Nest, $I_AA_1,$ $I_BB_1,$ and $I_CC_1$ concur at a point $A_1',$ which we claim is $A_1.$ Indeed, \begin{align*}\measuredangle I_BA_1'I_C&=\measuredangle AI_BB_1+\measuredangle C_1I_CA\\&=90-\measuredangle I_BAB_1+90-\measuredangle I_CAC_1\\&=\measuredangle B_1AI_B+\measuredangle I_CAC_1\\&=\measuredangle CAB.\end{align*}Thus, $I_B,B_1,$ and $A_1$ are collinear and $\measuredangle AB_1A_I=90.$ Pappus on $A_1I_BBACI_C$ shows $\overline{B_IB}\cap\overline{C_IC},B_1,$ and $C_1$ are collinear. $\square$

View in the problem with the excentral triangle as reference. Part (a) is then extremely similar to the IMO 2013 P3(or P6? idk) and (b) follows from Pappus on $ I_B-A-I_C$ and $B-A_1-C$.