sahadian wrote: perpendicular from $I$ to $AI$ meet $AB$ and $AC$ at ${B}'$ and ${C}'$ respectively . How can it be? perpendicular from $I$ to $AI$?

A line passing through point $ I $ perpendicularly to $ AI $.

let $K,L$ are circumcenters of $(AB'B'')$ and $(AC'C'')$ and $B_1,C_1$ are lie on bisector of angles $B$ and $C$ respectively, such that are on $(ABC)$. We have the circumcenter of $(ATI)$ lies on the line $B_1C_1$ and $LK$. The problem is equivalent to $LK$, $B_1C_1$, $BC$ are concurrent. It's very easy! Let $O=B_1C_1\cap BC$. We have \[ \frac{CO}{BO}=\frac{\sin \frac{B}{2} \cos \frac{A-B}{2}}{\sin \frac{C}{2} \cos \frac{A-C}{2}} \] and $ \frac{BK}{IK}=\frac{\cos \frac{A-C}{2}}{\sin \frac{B}{2}\cos \frac{A}{2} $, $ \frac{CL}{IL}=\frac{\cos \frac{A-B}{2}}{\sin \frac{C}{2}\cos \frac{A}{2}} $. So \[ \frac{CO}{BO}\cdot \frac{BK}{IK}\cdot \frac{IL}{CL}=1 \] and by Menelaus theorem, we get that $O,L,K$ are collinear, $LK$, $B_1C_1$, $BC$ are concurrent.

nice problem let o1,o2 are circumcenter of AB'B'' and AC'C'' it is easy to see o1 and o2 on BI and CI. let o1,o2 meet BC at S. use menelaus in IBC and line s,o1,o2

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Denote the circumcentres of $\Delta A{B}'{B}'',\Delta A{C}'{C}'',\Delta A{B}'{C}'$ are $X,Y,Z$. $U,V$ are the midpoints of $A{B}',A{C}'$. Then $U,Z,X$ on the perpendicular bisectors of $A{B}'$, and $V,Z,Y$ on the perpendicular bisectors of $A{C}'$. By Desargues theorem $XY,UV,BC$ are concurrent.

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Very nice problem! We will show that the circles $(AIT)$, $(C, AC)$ and $(B, AB)$ all have a second common point $P$. This obviously implies the thesis, since the centers of the latter two circles both lie on $BC$. Let us perform an inversion with center $A$ and ray $\sqrt{AB \cdot AC}$ followed by a simmetry with respect to the internal bisector of angle $A$. Now $C$ and $B$ are fixed with respect to this double tranformation. The image of line $BC$ is the circumcircle of $ABC$, the image of $(C, AC)$ is the axis of $AC$ and similarly with the other circle. The image of the incircle is the mixtilinear incirle of triangle $ABC$ with respect to the vertex $A$. Now, note that this implies that the image of the points of tangency of the incircle with $AB$, $AC$ (let's call them $R$, $S$) are the points of tangency of the mixtiliear incircle (let's call them $T$, $U$) , and since $ARIS$ is cyclic we get that the image of $I$ lies on the midpoint of $UT$, wich is notoriously the incircle of $ABC$. Now, points $B^'$ and $C^'$ go into the intersections of the sides with the circle of diameter $AI$, wich are obviously the points of tangency $R$, $S$ of the incirle in the triangle $ABC$. Finally, for what we said before, the images of point $B^''$ and $C^''$ are the midpoint of the arches $AC$, $AB$ not containing the respective vertexes. Let's call the $B_1$, $C_1$. Now, the thesis becomes the following: If $K= RB_1 \cap SC_1$, then $K$, $I$ and $O$ (the circumcenter) are collinear. Now, the tangent line to the incirle at $R$ and the one at the circumcirle at $B_1$ are parallel; this implies that the one of the two centers of the homoteties that send the incircle into the circumcircle lies on $RB_1$ (just look at the triangles that the line joining the tangency points forms with the line connecting the centers and the two tangent lines). But this point obviously lies also on $SC_1$, meaning it is point $K$. This finishes the proof, since the center of that homotety clearly lies on $OI$.

It suffices to prove that perpendicular bisector of $AT$, perpendicular bisector of $AI$ and $BC$ are concurrent. If $O_1$ and $O_2$ are circumcenters of $AC'C''$ and $AB'B''$, then perpendicular bisector of $AT$ is $O_1O_2$. And if $M$ and $N$ are midpoints of arcs $AC$ and $AB$ then $MN$ is perpendicular bisector of $AI$. So it suffices to prove that $MN$, $O_1O_2$ and $BC$ are concurrent. From $BB''=BA$ and $CC''=CA$ we get $O_1$ and $O_2$ are on angle bisectors of $C$ and $B$. So if we prove $(CI,O_1N)=(BI,O_2N)$ then $MN$, $O_1O_2$ and $BC$ would be concurrent. If $E$ , $F$ and $G$ are feet of perpendiculars from $I$, $O_1$ and $N$ to $AC$ then $(CI,O_1N)=(CE,FG)$ and $CE=p-c$, $CF=\frac{CA+CC'}{2}=\frac{b+(b-\frac{bc}{p})}{2}=b-\frac{bc}{a+b+c}=\frac{b(a+b)}{a+b+c}$ and$CG=\frac{a+b}{2}$ so $(CE,FG)=\frac{\frac{CF}{FE}}{\frac{CG}{GE}}=\frac{\frac{CF}{CF-CE}}{\frac{CG}{CG-CE}}=\frac{\frac{\frac{b(a+b)}{a+b+c}}{\frac{b(a+b)}{a+b+c}-p+c}}{\frac{\frac{a+b}{2}}{\frac{a+b}{2}-\frac{a+b-c}{2}}}=\frac{\frac{b(a+b)}{b(a+b)-(a+b+c)(p-c)}}{\frac{a+b}{c}}=\frac{2bc}{2ab+2b^2-(a+b+c)(a+b-c)}$ By the same approach $(BI,O_2N)=\frac{2cb}{2ac+2c^2-(a+b+c)(a+c-b)}$ so it suffices to prove $2ac+2c^2+(a+b+c)(a+c-b)=2ab+2b^2-(a+b+c)(a+b-c)$ but after expanding it they're both $c^2+b^2-a^2$. QED.

For notation purposes, let $B'$ be denoted by $B_0$ and $B''$ be denoted by $B_1$, analogously for $C$. Apply $\sqrt{bc}$ inversion, and let $X'$ denote the image of $X$ under this map. Since $ABB_1$ is a $B$-isosceles triangle with $B_1$ lying on half-line $BC$, it follows that $B_1'$ is the midpoint of the arc $AC'B'$ of the circle $(AC'B')$. Similarly, $C_1'$ is the midpoint of the arc $AB'C'$ of this circle. Note that $I'$ is the $A$ excenter of triangle $AC'B'$, hence, $B_0'$ is the touching point of the $A$ excircle of triangle $AC'B'$ with $AB'$. Similarly, $C_0'$ is the touching point of the excircle with $AC'$. Note also that $T'=B_0'B_1'\cap C_0'C_1'$ and the circumcenter $O_a$ of triangle $AIT$ is mapped to the reflection of $A$ in $I'T'$. Thus, it suffices to show that $I',T',O$ are collinear where $O$ is the circumcenter of triangle $AC'B'$. Note that $OC_1' \perp AC'$ and $I'C_0' \perp AC'$ yielding that $I'C_0' \parallel OC_1'$. Similarly, $I'B_1' \parallel OB_1'$. Finally, note that $B_1'C_1' \parallel B_0'C_0'$ yielding that triangles $OC_1'B_1'$ and $I'C_0'B_0'$ are homothetic. It follows that $I',T',O'$ are collinear and the conclusion is established.

Let $B_1$ and $C_1$ be the reflections of $B'$ and $C'$ over $\overline{BI}$ and $\overline{CI}$ respectively. Since $BB'\cdot BA = BB_1\cdot BB''$ it follows that $B_1$ lies on $(AB'B'')$. Similarly $C_1$ lies on $(AC'C'')$. Now let the incircle touch $\overline{BC}$ at $D$. Since $IB_1=IB'=IC'=IC_1$ and $IB''=IA=IC''$, $\{B_1,C_1\}$ and $\{B'', C''\}$ are symmetric over $D$. Therefore, $DB_1\cdot DB''=DC_1\cdot DC''$, and consequently, it follows that $D$ lies on $\overline{AT}$. Now let $I'$ be the reflection of $I$ over $\overline{BC}$. Notice that $\angle C''IC_1=\angle AIC'=90^\circ$, so \[DA\cdot DT = DC''\cdot DC_1=DI^2=DI\cdot DI'\]so $I'$ lies on $(AIT)$ and consequently, the circumcenter of $\triangle AIT$ lies on $\overline{BC}$ as desired. [asy][asy] defaultpen(fontsize(10pt)); size(12cm); pen mydash = linetype(new real[] {5,5}); pair A = dir(140); pair B = dir(210); pair C = dir(330); pair I = incenter(A,B,C); pair D = foot(I,B,C); pair B1 = extension(rotate(90,I)*A,I,A,B); pair C1 = extension(rotate(90,I)*A,I,A,C); pair B2 = 2*foot(A,B,I)-A; pair C2 = 2*foot(A,C,I)-A; pair B11 = 2*foot(B1,B,I)-B1; pair C11 = 2*foot(C1,C,I)-C1; pair I1 = 2*D-I; pair T = 2*foot(circumcenter(A,I,I1),A,D)-A; draw(A--B--C--cycle, black+1); draw(B1--C1); draw(circumcircle(A,B1,B2)); draw(circumcircle(A,C1,C2)); draw(I--I1); draw(A--T); draw(B--C2); draw(I--A, mydash); draw(I--B2, mydash); draw(I--C2, mydash); draw(I--C11, mydash); draw(I--B11, mydash); dot("$A$", A, dir(A)); dot("$B$", B, dir(225)); dot("$C$", C, dir(C)); dot("$D$", D, dir(225)); dot("$I$", I, dir(60)); dot("$B'$", B1, dir(180)); dot("$C'$", C1, dir(30)); dot("$B''$", B2, dir(315)); dot("$C''$", C2, dir(225)); dot("$B_1$", B11, dir(225)); dot("$C_1$", C11, dir(315)); dot("$T$", T, dir(270)); dot("$I'$", I1, dir(270)); [/asy][/asy]