This problem was posted here before and remained unsolved.The fixed point is the orthocenter of $\triangle ABC$.

Suppose, $ H $ is the orthocenter of $ ABC $. Note that, $ A_2 $ lies on $ \odot BHC $ and so. $ A'B'C' $ be the anticomplementary triangle of $ ABC $. Then note that, if $ l $ is the line through $ A' $ and parallel to $ AX $, then $ l $ and $ A'A_2 $ are isogonl conjugates wrt $ \angle B'A'C' $. So $ A'A_2,B'B_2,C'C_2 $ concurr at some point $ Q $. So clearly, $ A_2,B_2,C_2 $ lie on the circle with diameter $ HQ $.

Just saying this is called Hagge circle look it up.

$RSM$'s post made me realize there's a really easy proof by inversion. Firstly note that $\frac{A_1B}{A_1C} \cdot \frac{C_1A}{C_1B} \cdot \frac{CB_1}{AC_1} = 1$ is equivalent to trigonometric ceva. From this we have $\frac{A_2B}{A_2C} \cdot \frac{C_2A}{C_2B} \cdot \frac{CB_2}{AC_2} = 1$ (*). As $RSM$ said, $A_2$ is in the circumcircle of $BHC$. Similarly for the others. We perform an inversion centered at $H$. We have, for example, $B', C', A_2'$ are collinear. We need to show that $A_2', B_2', C_2'$ are collinear. Applying menelaus for this points in triangle $A'B'C'$ we are left to prove $\frac{A_2B}{A_2C} \cdot \frac{C_2A}{C_2B} \cdot \frac{CB_2}{AC_2} = 1$ which is what we have in (*) and we are done.

It is easy to guess the fixed point. Take $X=H$,where $H$ is the orthocenter of $ABC$.

(Solution with EulerMacaroni) Bulgaria 2012 wrote: We are given an acute-angled triangle $ABC$ and a random point $X$ in its interior, different from the centre of the circumcircle $k$ of the triangle. The lines $AX,BX$ and $CX$ intersect $k$ for a second time in the points $A_1,B_1$ and $C_1$ respectively. Let $A_2,B_2$ and $C_2$ be the points that are symmetric of $A_1,B_1$ and $C_1$ in respect to $BC,AC$ and $AB$ respectively. Prove that the circumcircle of the triangle $A_2,B_2$ and $C_2$ passes through a constant point that does not depend on the choice of $X$. Answer: The fixed point is the orthocenter $H$ of $\triangle ABC$. (Proof) Fix $A_1 \in \odot (ABC)$ and vary $X$ along $AA_1;$ notice that the maps $X \rightarrow B_1 \rightarrow B_2$ and $X \rightarrow C_1 \rightarrow C_2$ are projective, so $B_2 \rightarrow C_2$ is also projective. Note that $B_2 \in \odot (AHB)$ and $C_2 \in \odot (AHC)$ so for $C' \in \odot(AHC)$ such that $H, A_2, B_2, C'$ are concyclic, inversion at $H$ tells us that $B_2 \rightarrow C'$ is also projective. It is sufficient to show $C_2=C'$ for three choices of $X$. Evidently, $X \in \{A, A_1, AA_1 \cap BC\}$ work, so we are done! $\square$

Let $H$ be the orthocenter of $ABC$. Obviously $A_2$, $B_2$, $C_2$ lie on circles $\odot BCH$, $\odot CAH$, $\odot ABH$ respectively. Now define $Y$ as the isogonal conjugate of $X$ with respect to $ABC$. If $M_a$, $M_b$, $M_c$ are the midpoints of the sides of $ABC$ and lines $AY$, $BY$, $CY$ meet the circumcircle again at $A_1'$, $B_1'$, $C_1'$, then it is not difficult to see that $A_1'$ and $A_2$ are symmetric about $M_a$ and so on. Now we claim the following. Lemma. Let $P$ be a point lying on the circumcircle and $Q$ its reflection through $M_a$. Also, let $F$ be the nine-point center of $ABC$. Then the perpendicular bisector of $HQ$ and line $AP$ are symmetric with respect to $F$. Proof. It suffices to show that $AP$ is the perpendicular bisector of the image of segment $HQ$ through reflection in $F$. Now, as a well-known fact, the image of $H$ is the circumcenter $O$ of $ABC$ and clearly $AO = PO = R$ (where $R$ is the circumradius). Let $Q'$ be the image of $Q$. The homothety centred at $Q$ with ratio $1/2$ maps $Q'$ to $F$, $P$ to $M_a$ and $A$ to $A'$. $M_a$ definitely lies on $ABC$'s nine-point circle; we show that $A'$ does as well. Indeed, consider another homothety with center $A$ and ratio $2$. This one maps $M_b$ to $B$, $M_c$ to $C$ and $Q'$ to $Q$ again. Since $Q \in \odot BCH$ and the midpoint of $AH$ belongs to the nine-point circle, it follows easily what was wanted. So we have $FA' = FM_a = R/2$, implying that $Q'A = Q'P = R$. Thus $AOPQ'$ is a rhombus and $AP$ is the perpendicular bisector of $OQ'$. Back to the main problem, let's consider the circle with center $K$ going through $B_2$, $C_2$ and $H$. Let $K'$ be the reflection of $K$ through $F$ (defined as above). The lemma tells us that $K'$ lies simultaneously on $BB_1'$ and $CC_1'$. But this implies that $K' = Y$, hence $K'$ also lies on $AA_1'$. Applying the lemma again gives us that the perpendicular bisector of $HA_2$ passes through $K$, which means points $H$, $A_2$, $B_2$, $C_2$ are concyclic. Then $H$ is the fixed point we were looking for.

The problem is pretty easy with complex bashing. We can choose X such that it lies on the x-axis and thus its conjugate is the number itself. We are allowed to do it because for a given configuration of triangle and point, we can rotate it around O, so that X is on the x-axis and the circumcircle is unchanged. Now a1=(a-x)/(ax-1) and a2=b+c-bca1. Then the "most difficult" part is probably to guess that the constant point is indeed the orthocenter. From then on it's 2-3-minute computation to show that it always lies on the desired circle.

Guessing that the point is the orthocenter is actually very easy, it follows by plugging in $P=H$. I solved this problem without a diagram. We start with complex numbers with $(ABC)$ as the unit circle. I'm sure one can give a version of this preliminary argument without complex numbers with rotations, translations etc, but its a lot easier to encode this way. We claim the fixed point is the orthocenter $H$, with complex coordinates $h=a+b+c$. Note that $a_2=b+c-bc\overline{a}_1=b+c-bc/a_1\equiv b+c-a_1'$ by standard complex number formulas. Thus, $a_2=h-(a+a_1')\equiv h-x$, and similar formulas for $b_2,c_2$. Thus, it suffices to show that $O,X,Y,Z$ are concylic. We have that $x=a+a_1'$, and since both $A,A_1'$ are on the unit circle, we see that $X$ is the reflection of $O$ in $AA_1'$. We see that $AA_1$ and $AA_1'$ are isogonal in $\triangle ABC$ since $a_1'=bc/a_1$, so we have that $AA_1',BB_1',CC_1'$ all concur at the isogonal conjugate of $P$. We now finish by the following lemma. Lemma: If lines $\ell_1,\ell_2,\ell_3$ concur at some point $Q$, and if we let $O_i$ be the reflection of some point $O$ in $\ell_i$, then $O,O_1,O_2,O_3$ are concyclic. Proof of Lemma: This follows by inversion at $O$. It is not hard to see that $O_i$ gets mapped to the center of the circle $\ell_1'$ that is the image of $\ell_1$. The result is then just the statement that coaxial circles have collinear centers. $\blacksquare$ Thus the reflections of $O$ in $AA_1',BB_1',CC_1'$ are concyclic with $O$, as desired. Note: I think I replaced $X$ with $P$

hatchguy wrote: $RSM$'s post made me realize there's a really easy proof by inversion. Firstly note that $\frac{A_1B}{A_1C} \cdot \frac{C_1A}{C_1B} \cdot \frac{CB_1}{AC_1} = 1$ is equivalent to trigonometric ceva. From this we have $\frac{A_2B}{A_2C} \cdot \frac{C_2A}{C_2B} \cdot \frac{CB_2}{AC_2} = 1$ (*). As $RSM$ said, $A_2$ is in the circumcircle of $BHC$. Similarly for the others. We perform an inversion centered at $H$. We have, for example, $B', C', A_2'$ are collinear. We need to show that $A_2', B_2', C_2'$ are collinear. Applying menelaus for this points in triangle $A'B'C'$ we are left to prove $\frac{A_2B}{A_2C} \cdot \frac{C_2A}{C_2B} \cdot \frac{CB_2}{AC_2} = 1$ which is what we have in (*) and we are done. Here's another solution which is essentially the above one, but fleshed out more. [asy][asy] unitsize(2.5inches); pair A=dir(115); pair B=dir(220); pair C=dir(-40); pair H=orthocenter(A,B,C); pair P=0.27*A+0.35*B+0.38*C; pair D=2*foot(H,B,C)-H; pair A1=2*foot(0,A,P)-A; pair X=foot(A,B,C); pair Y=foot(B,A,C); pair Z=foot(C,A,B); pair A2=2*foot(A1,B,C)-A1; pair A3=extension(Y,Z,A2,H); pair U=extension(A,H,Y,Z); draw(circumcircle(A,B,C)); draw(A--D); draw(A--B--C--cycle); draw(C--Z); draw(B--Y); draw(A--A1); draw(A1--A2); draw(A2--A3); draw(Y--A3); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$H$",H,dir(125)); dot("$P$",P,dir(0)); dot("$D$",D,dir(D)); dot("$A_1$",A1,dir(A1)); dot("$X$",X,dir(X)); dot("$Y$",Y,dir(Y)); dot("$Z$",Z,dir(140)); dot("$U$",U,dir(145)); dot("$A_2$",A2,dir(45)); dot("$A_3$",A3,dir(A3)); [/asy][/asy] We'll relabel $X$ in the problem to $P$. Let $H$ be the orthocenter of $\triangle ABC$ with orthic triangle $XYZ$. Let $D=AH\cap(ABC)$, and define $E$, $F$ similarly. Also let $U=AH\cap YZ$, and define $V$, $W$ similarly. Finally, let $A_3=HA_2\cap YZ$, and define $B_3$ and $C_3$ similarly. We claim that $H$ is the desired fixed point. To see this, we'll invert about $H$ with (negative) power $HA\cdot HD$. Note that $H$ and $D$ are reflections over $BC$, so $A_2\in(BHC)$. The image of $(BHC)$ is $YZ$, so the image of $A_2$ is $A_3$. Thus, it suffices to show that $A_3$, $B_3$, $C_3$ collinear. By Menalaus's theorem, this is equivalent to showing that $\prod_{\mathrm{cyc}}YA_3/A_3Z=-1$. We already know that $\prod_{\mathrm{cyc}}YU/UZ=1$ by Ceva, so it suffices to show that $\prod_{\mathrm{cyc}}(YZ;UA_3)=-1$. We now compute \[(YZ;UA_3)\stackrel{H}{=}(B,C;X,HA_2\cap BC)=(B,C;X,DA_1\cap BC)\stackrel{D}{=}(BC;A_1A).\]It's easy to see that the concurrence of $AA_1$, $BB_1$, $CC_1$ is equivalent to $\prod_{\mathrm{cyc}}(BC;A_1A)=-1$ (say using trig Ceva), so we're done.

Here is a very easy solution. Consider the homothety with center $G$ and coefficient $-\frac{1}{2}$, i.e. which maps the orthocenter $H$ to the circumcenter $O$. It suffices to show that the images $A_3$, $B_3$, $C_3$ of $A_2$, $B_2$, $C_2$ are concyclic with $O$. Note that $G$ is the centroid of triangle $CC_1C_2$ since if $CG \cap C_1C_2 \cap AB = C_0$, then $C_0C_1 = C_0C_2$ and $CG = 2GC_0$. Hence $C_3$ is the midpoint of $CC_1$ and as $CC_1$ is a chord, we get $\angle XC_3O = \angle CC_3O = 90^{\circ}$. Therefore $C_3$ lies on the circle with diameter $OX$; similarly this holds for $A_3$ and $B_3$ and we are done.

We prove a stronger statement, by deleting the condition that $P$ must lie in the interior of $\triangle ABC.$ We prove the fixed point is $H,$ the orthocenter of $\triangle ABC.$ Fix $\triangle ABC$ and $A_1,$ and animate $P$ on $\overline{AA_1}.$ Observe the maps $P \xmapsto{B} B_1 \mapsto{B_2}$ and $P \xmapsto{C} C_1 \mapsto{C_2}$ are projective, implying the map $B_2 \mapsto C_2$ is projective. Note that $AC_2HB$ is cyclic, say, by angle chasing, and analogously, $AC_2A_2C$ and $CA_2HB$ are also cyclic. Consequently, by second intersection of circles, it follows the map $B_2 \to (HB_2A_2) \cap (AHB) \neq H$ is projective. It suffices to check these maps concur for three positions of $P.$ It turns out this is quite simple, with three special cases of $P=A, P=\overline{AA_1} \cap \overline{BC},$ and $P=A_1$ being easy.

We claim that the fixed point is the orthocenter. Invert negative around $H$ swapping the nine-point circle and circumcircle. Note that since $A_2$ lies on $(BHC)$, $A_2'$ lies on $B'C'$. We wish to show that $A_2',B_2',C_2'$ are collinear, so we will use Menalaus. We have by inversion distance formula $$\prod_{cyc} \frac{A_2'C'}{A_2'B'}=\prod_{cyc} \frac{\frac{A_2C}{HA_2\cdot HC}}{\frac{A_2B}{HA_2\cdot HB}}=\prod_{cyc} \frac{A_2C}{A_2B}\cdot \frac{HB}{HC}=\prod_{cyc} \frac{A_2C}{A_2B}=\prod_{cyc} \frac{A_1C}{A_1B}=\prod_{cyc} \frac{\sin\angle CAA_1}{\sin\angle BAA_1}=1$$where the last equality is by Ceva.

Just see China TST 2006