Does this triangle center exist? (centroids of 3 circular segments) Consider a triangle ABC that is inscribed into a circle.
Then consider 3 circular segments corresponding to the sides of the triangle ABC.
For each circular segment the exact coordinate of its center of mass can be calculated.
In this case A1 - is the center mass for the circular segment BCG.
B1 is the center mass for segment AEC.  C1 is the centroid for segment AJB.

Are the lines AA1,BB1,CC1 always concurrent?
What is the exact point of their concurrence? That must be a well known triangle center I guess.
Usually I've been checking all these centers in the Kimberling Encyclopedia with the help of Geogebra and extremely useful search function. However in this case there is no a compass-and-straightedge construction of A1,B1,C1. So this hypothetical point X  apparently can be constructed only with the help of some advanced math software...
 A: Problems like this make me want to look for general principles instead of bogging-down in messy specifics. In this case, we have a triangle $\triangle ABC$, say, with circumcenter $O$, and circular-segment-centroids $A'$, $B'$, $C'$ (OP's $A_1$, $B_1$, $C_1$), with $\overline{OA'}$, $\overline{OB'}$, $\overline{OC'}$ bisecting $\angle BOC$, $\angle COA$, $\angle AOB$, respectively; a little calculus tells us the centroids' distances from the circumcenter. (We'll get to that later.)
Stepping back, we see that we have six points $A$, $B$, $C$, $A'$, $B'$, $C'$ arranged about a common center, $O$, and that we know the distances of these point from $O$ and angles determined by these points and $O$. We can establish a condition on those lengths and angles that guarantees the concurrence of $\overleftrightarrow{AA'}$, $\overleftrightarrow{BB'}$, $\overleftrightarrow{CC'}$. So let's do that.
Define
$$
a := |OA| \quad b := |OB| \quad c := |OC| \quad 
a' := |OA'| \quad b' := |OB'| \quad c' := |OC'| 
$$
(Note that we're generalizing beyond $O$ being the circumcenter, which would required $a=b=c$. We're also generalizing beyond, say, $\overline{OA'}$ bisecting $\angle BOC$, etc; the general rule turns out to be pretty nice without these assumptions.) We'll also use various angles, $\angle XOY$, taken to be oriented "from" $X$ "to" $Y$; this allows us to write $\angle XOY+\angle YOZ=\angle XOZ$ and $\angle XOY=-\angle YOX$.
Now, let's coordinatize. Abusing notation to define $\operatorname{cis}\theta := (\cos\theta, \sin\theta)$ we can take
$$\begin{align}
A &:= a \operatorname{cis}0 
&& A' := a'\operatorname{cis}\angle AOA'\\
B &:= b \operatorname{cis}\angle AOB
&& B' := b'\operatorname{cis}\angle AOB' \\
C &:= c\operatorname{cis}\angle AOC 
&& C' := c'\operatorname{cis}\angle AOC'
\end{align}$$
From here, the process is straightforward, if tedious. (It helps to have a computer algebra system to crunch the symbols.) We determine the equations of the lines $\overleftrightarrow{AA'}$, $\overleftrightarrow{BB'}$, $\overleftrightarrow{CC'}$, find the intersection of any two, and substitute the intersection into the third. When the dust settles (and barring degeneracies), we get a relation that we can express thusly:
$$\begin{align}
0 &= \phantom{+}a a' \sin\angle AOA'\; \left( 
 b c \sin\angle BOC
+c b' \sin\angle COB'
+b' c' \sin\angle B'OC' 
+c' b \sin\angle C'OB 
 \right) \\[4pt]
&\phantom{=} +b b'\sin\angle BOB'\; \left( 
 c a \sin\angle COA 
+a c' \sin\angle AOC' 
+c' a' \sin\angle C'OA' 
+a' c \sin\angle A'OC
\right) \\[4pt]
&\phantom{=}+ c c' \sin\angle COC'\; \left( 
 a b \sin\angle AOB
+b a' \sin\angle BOA'
+a' b' \sin\angle A'OB' 
+b' a \sin\angle B'OA
\right) 
\end{align} \tag{$\star$}$$
This may seem a little daunting at first glance, but, glancing again, we notice that every "$\sin\angle XOY$" is multiplied by the corresponding lengths "$x$" and "$y$"; conveniently, each such product is thus twice the (signed) area $|\triangle XOY|$, so that we can write
$$\begin{align}
0 &= \phantom{+}|\triangle AOA'|\; \left( 
 |\triangle BOC|+|\triangle COB'|+|\triangle B'OC'|+|\triangle C'OB| 
\right) \\[4pt]
&\phantom{=} +|\triangle BOB'|\; \left( 
 |\triangle COA|+|\triangle AOC'|+|\triangle C'OA'|+|\triangle A'OC|
\right) \\[4pt]
&\phantom{=}+ |\triangle COC'|\; \left( 
 |\triangle AOB|+|\triangle BOA'|+|\triangle A'OB'|+|\triangle B'OA|
\right) 
\end{align} \tag{$\star\star$}$$
Even better, each long factor is the sum of the (signed) areas of adjacent triangles that form a quadrilateral; so each factor gives the (signed) area of that quadrilateral. (This interpretation is a bit nuanced in cases where such a quadrilateral is self-intersecting. Be that as it may ...) This gives us this streamlined expression:
$$
 |\triangle AOA'|\;|\square BCB'C'|
+|\triangle BOB'|\;|\square CAC'A'|
+|\triangle COC'|\;|\square ABA'B'|
=0 
\tag{$\star\star\star$}$$
Pretty nifty! $\square$

Now that we have generalized the problem, let's work our way towards the specifics of OP's ostensible concurrence.
We consider $\triangle ABC$ with interior angles $\alpha := \angle A$, $\beta := \angle B$, $\gamma := \angle C$. Taking $O$ to be the circumcenter and $r$ the circumradius, we have
$$a=b=c=r \qquad \angle BOC = 2\alpha \quad \angle COA = 2\beta \quad \angle AOB = 2\gamma $$
With $A'$, $B'$, $C'$ along the bisectors of $\angle BOC$, $\angle COA$, $\angle AOB$, we have
$$\angle BOA' = \angle A'OC=\alpha \qquad \angle COB'=\angle B'OA=\beta \qquad
\angle AOC'=\angle C'OB=\gamma$$
$$\angle AOA' = 2\gamma+\alpha=\pi-(\beta-\gamma) \qquad
\angle BOB' = \pi-(\gamma-\alpha) \qquad 
\angle COC' = \pi-(\alpha-\beta)$$
Falling back to version $(\star)$ of our concurrence condition, we have
$$\begin{align}
0 &= \phantom{+}r a' \sin(\beta-\gamma)\; \left( 
 r^2 \sin2\alpha
+r b' \sin\beta
+b' c' \sin(\beta+\gamma) 
+c' r \sin\gamma 
 \right) \\[4pt]
&\phantom{=} +r b'\sin(\gamma-\alpha)\; \left( 
 r^2 \sin2\beta 
+r c' \sin\gamma 
+c' a' \sin(\gamma+\alpha)
+a' r \sin\alpha
\right) \\[4pt]
&\phantom{=}+ r c' \sin(\alpha-\beta)\; \left( 
 r^2 \sin2\gamma
+r a' \sin\alpha
+a' b' \sin(\alpha+\beta) 
+b' r \sin\beta
\right) 
\end{align} \tag{1}$$
Since $\alpha+\beta+\gamma=\pi$ and $r\neq 0$, this simplifies to
$$\begin{align}
0 &= \phantom{+}(r a' - b' c') \sin2\alpha \sin(\beta  - \gamma) \\
&\phantom{=}+(r b' - c' a') \sin2\beta  \sin(\gamma - \alpha) \\
&\phantom{=}+(r c' - a' b') \sin2\gamma \sin(\alpha - \beta)
\end{align} \tag2$$
Note that $(2)$ holds for $A'$, $B'$, $C'$ anywhere along the perpendicular bisectors, so it's still a bit of a generalized result. For OP's circular-segment-centroids, we consult Wikipedia's "List of Centroids" to remind ourselves that
$$a' = \frac{4r\sin^3\alpha}{3(2\alpha-\sin2\alpha)} \qquad
b' = \frac{4r\sin^3\beta}{3(2\beta-\sin2\beta)} \qquad
c' = \frac{4r\sin^3\gamma}{3(2\gamma-\sin2\gamma)} \tag{3}$$
Perhaps-unsurprisingly, upon substituting the values from $(3)$ into $(2)$, the mix of "raw and trigged" angles doesn't simply vanish. For the sake of completeness, here's a version of the resulting concurrence condition
$$\begin{align}
&\phantom{=+\,}
 3 \sin^3\alpha \sin(\beta-\gamma)  (\alpha \sin2\beta\sin2\gamma + 2\beta\gamma \sin2\alpha) \\
&\phantom{=}
+3 \sin^3\beta  \sin(\gamma-\alpha) (\beta \sin2\gamma\sin2\alpha + 2\gamma\alpha \sin2\beta) \\
&\phantom{=}
+3 \sin^3\gamma \sin(\alpha-\beta) (\gamma \sin2\alpha\sin2\beta  + 2\alpha\beta \sin2\gamma) \\[6pt]
&= 8 \sin\alpha \sin\beta \sin\gamma \left(\begin{array}{l}
   \phantom{+}
  \alpha \cos\alpha \sin^2\beta  \sin^2\gamma \sin(\beta-\gamma) \\
+ \beta  \cos\beta  \sin^2\gamma \sin^2\alpha \sin(\gamma-\alpha) \\
+ \gamma \cos\gamma \sin^2\alpha \sin^2\beta  \sin(\alpha-\beta) \\
+ \sin\alpha \sin\beta \sin\gamma \sin(\beta-\gamma) \sin(\gamma-\alpha) \sin(\alpha-\beta)\end{array}\right)
\end{align}\tag{4}$$
OP's alternative construction, taking $A'$, $B'$, $C'$ to be the centroids of the "other" circular segments, requires the substitutions $\alpha\to\pi-\alpha=\beta+\gamma$, $\beta\to\gamma+\alpha$, $\gamma\to\alpha+\beta$ in $(3)$, along with changing the sign of each of $a'$, $b'$, $c'$ because each centroid lies on the "other side" of $O$. These adjustments cause some minor sign changes in $(4)$, but also introduce more-complicated "raw" angle expressions. The result doesn't appreciably simplify, so I won't bother TeX-ing it up.
