Coordinates of centres of central triangles with respect to the reference triangle In Kimberling's Encyclopedia of Triangle Centers, a lot of centres are described as the centres of certain central triangles of the reference triangle, whether as a main or alternate definition. For example, $X_{164}$ is defined as the incentre of the excentral triangle (whose vertices are the triangle's excentres) and the Kiepert antipode $X_{114}$ can be defined as the Tarry point $X_{98}$ of the medial triangle (whose vertices are the triangle's medians).
For a programmatic implementation of triangle centres these indirect definitions, when used judiciously,  reduce the workload and error bounds associated with the centre's computation as compared to simply using trilinear/barycentric coordinates (which may be very complicated).
However, I cannot figure out how to combine the pieces together to generate the centre's coordinates in either system with respect to the reference triangle, which is the question I am asking here. For example, I can easily look up the trilinear coordinates of the vertices of the excentral triangle ($-1:1:1$ and permutations) and the trilinear coordinates of the incentre ($1:1:1$), but how do I combine these two parts to obtain trilinear coordinates for $X_{164}$ ($\sin B/2+\sin C/2-\sin A/2$ and cyclic permutations of $A,B,C$)?
 A: This answer is an answer for the very specific question:

How to obtain or verify $X(164)$ knowing it is the $X(1)$-point of the excentral triangle?

Explicitly:
How to obtain the barycentric coordinates of $X(164)$, knowing it is the incenter of the circle through the points with (inhomogeneous) barycentric coordinates
$$\begin{aligned}
S&(-a:b:c)\ ,\\
T&(a:-b:c)\ ,\\
U&(a:-b:c)\ ?
\end{aligned}
$$


As a reference, the following collects a lot of information in a beautiful presentation:
Max Schindler, Evan Chen, Barycentric Coordinates for the Impatient

We fix the frame, we will work usually with baricentric normalized coordinates $(x,y,z)$ of a point $P=xA+yB+zC$, considered w.r.t. the triangle $ABC$ with sides $a=BC$, $b=CA$, $c=AB$. The word normalized refers to the relation $x+y+z=1$. While working with points in barycentric coordinates, often expressions get complicated, so it is natural to show the point $(x,y,z)$ only up to the common denominator, which is omitted. In such a case we write $(x:y:z)$, and this point stays for $(\ x/(x+y+z),\ y/(x+y+z),\ z/(x+y+z)\ )$.
The formula for the squared distance $OP^2$ between two points with barycentric normalized coordinates $P(x,y,z)$ and $O(x_0,y_0,z_0)$ is algebraic, explicitly
$$
PO^2
=
-a^2(y-y_0)(z-z_0) 
-b^2(x-x_0)(z-z_0) 
-c^2(x-x_0)(y-y_0) \ ,
$$
and it factorizes through building the "displacement" (vector) $(x-x_0,y-y_0,z-z_0)$.
In our case, let $S,T,U$ be the ex-centers of $\Delta ABC$. Let us compute first $TU^2$. To have an easy typing, let me please type sage code:
sage: var('a,b,c');
sage: def d2(P, Q):
....:     x1, y1, z1 = P
....:     x2, y2, z2 = Q
....:     x12, y12, z12 = x2-x1, y2-y1, z2-z1
....:     return -a^2*y12*z12 -b^2*x12*z12 -c^2*x12*y12
....: 
sage: S, T, U = vector([-a,b,c]), vector([a,-b,c]), vector([a,b,-c])
sage: S, T, U = S/sum(S), T/sum(T), U/sum(U)
sage: S, T, U
((a/(a - b - c), -b/(a - b - c), -c/(a - b - c)),
 (a/(a - b + c), -b/(a - b + c), c/(a - b + c)),
 (a/(a + b - c), b/(a + b - c), -c/(a + b - c)))
sage: d2(T, U).factor()
4*a^2*b*c/((a + b - c)*(a - b + c))

So we can write:
$$
\begin{aligned}
TU^2 &= a^2\cdot \frac{bc}{(p-b)(p-c)}
&&=a^2\cdot\sin^{-2}\frac A2\ ,\text{ and similarly}
\\
US^2 &= b^2\cdot \frac{ac}{(p-a)(p-c)}
&&=b^2\cdot\sin^{-2}\frac B2\ ,
\\
ST^2 &= c^2\cdot \frac{ab}{(p-a)(p-b)}
&&=c^2\cdot\sin^{-2}\frac C2\ .
\end{aligned}
$$
(Above $p=(a+b+c)/2$, since i need $s$ for an other purpose. This is a rare case where i do not use $p$ for a prime number.)

Above it is important that we could write the expressions $TU^2$, $US^2$, $ST^2$ also as squares of "elements" directly connected to the original triangle $\Delta ABC$. (Taking the sine of half an angle brings us slightly out of the world of algebraic expressions in $a,b,c$.) Else i could not continue. The continuation is now clear. Let $s,t,u$ be the sides of $\Delta STU$, so
$$
\begin{aligned}
s & = TU = a\cdot\sin^{-1}\frac A2\ ,
\\
t & = US = b\cdot\sin^{-1}\frac B2\ ,
\\
u & = ST = c\cdot\sin^{-1}\frac C2\ .
\\[3mm]
&\qquad\text{ Then we use the formula for $X(1)$ in $\Delta STU$:}
\\[3mm]
X(1)_{\Delta STU}
&=
\frac 1{s+t+u}(sS+tT+uU)\ ,\text{ and here we plug in }
\\
S &=
\frac{-a}{-a+b+c}A +
\frac{b}{-a+b+c}B +
\frac{c}{-a+b+c}C \ ,\\
\\
T &=
\frac{a}{a-b+c}A +
\frac{-b}{a-b+c}B +
\frac{c}{a-b+c}C \ ,\\
\\
U &=
\frac{a}{a+b-c}A +
\frac{b}{a+b-c}B +
\frac{-c}{a+b-c}C \ ,
\\[3mm]
&\qquad\text{ and we compute of the $A$-coefficient in $X(1)_{\Delta STU}$,}
\\[3mm]
\text{($A$-coefficient)}
&=\frac a{s+t+u}
\left(
\frac{-s}{2(p-a)} +
\frac{t}{2(p-b)} +
\frac{u}{2(p-c)} 
\right)
\\
&\sim
a
\left(
-
\frac{a}{(p-a)\sin\frac A2} +
\frac{b}{(p-b)\sin\frac B2} +
\frac{c}{(p-c)\sin\frac B2} 
\right)
\\
&=
a
\left(
-
\frac{a}{r\cot\frac A2\sin\frac A2} +
\frac{b}{r\cot\frac B2\sin\frac B2} +
\frac{c}{r\cot\frac C2\sin\frac C2} 
\right)
\\
&\sim
a
\left(
-
\frac{\sin A}{\cos\frac A2} +
\frac{\sin B}{\cos\frac B2} +
\frac{\sin C}{\cos\frac C2} 
\right)
\\
&\sim
a
\left(
-
\sin\frac A2 +
\sin\frac B2 +
\sin\frac C2 
\right)\ .
\end{aligned}
$$
The corresponding coefficients of $A,B,C$ are the needed barycentric coefficients, and they match the one in the ETC.
Above, the symbol $\sim$ denotes equality up to a factor, which is a symmetric polynomial in $a,b,c$.

Conclusion:
The OP reopens implicitly the following door. Given the triangle $ABC$, we associate other triangles $STU$, where $S\cong S(a,b,c)$ is an asymmetric expression, a linear combination in $A,B,C$ with polynomial (or slightly more general) weights in $a,b,c$, maybe symmetric w.r.t. $b\leftrightarrow c$, and $T,U$ are obtained correspondingly using cyclic permutations of $(a,b,c)$ (and implicitly $A,B,C$), so $T\cong S(b,c,a)$, $U\cong S(c,a,b)$.
We compute the squared sides of $STU$ as above, and if the explicit expressions for $TU^2$, $US^2$, $ST^2$ admit a radical, then we can proceed as above to write all centers $X(k)_{STU}$ in terms of $S,T,U$, then using the expressions for $S,T,U$ in terms of $A,B,C$ we obtain weights w.r.t. the initial triangle, and we may want to match them with existing centers.

Application:
Let $STU$ be constructed based on $S=(0,\frac 12,\frac 12)=\frac 12(B+C)$, so $S$ is the mid point of $BC$, and we chose $T,U$ similarly. The sides of $STU$ are $s=a/2$, $t=b/2$, $u=c/2$, so
$$
\begin{aligned}
X(98)_{\Delta STU}
&=
\left(\ 
\frac 1{t^4+u^4-s^2(t^2+u^2)}\ :\ 
\dots
\ \right)
\\
&=
\left(\ 
\frac 1{b^4+c^4-a^2(b^2+c^2)}\ :\ 
\dots
\ \right)
\\
&\sim \frac 1{b^4+c^4-a^2(b^2+c^2)}S+\dots
\\
&\sim \frac 1{b^4+c^4-a^2(b^2+c^2)}(B+C)+\dots
\\
&\sim 
\left(
\frac 1{a^4+b^4-c^2(a^2+b^2)}+
\frac 1{a^4+c^4-b^2(a^2+c^2)}
\right)A+\dots
\\
&\sim
(b^4+c^4-a^2(b^2+c^2))
\cdot\Big(\ 
(a^4+b^4-c^2(a^2+b^2))+
(a^4+c^4-b^2(a^2+c^2))
\ \Big)A+\dots
\\
&\qquad\text{ matching again the ETC, and giving}
\\
&=X(114)_{\Delta ABC}\ .
\end{aligned}
$$
