Let $P$ be a point inside the triangle $\Delta ABC$; $P_a, P_b, P_c$ are reflection of $P$ around $BC, CA, AB$ respectively. What conditions are on $P$ such that $AP_a, BP_b, CP_c$ are concurrent?
I inspired this problem from this problem: Let $O$ be the circumcenter of triangle $ABC$; $O_a, O_b, O_c$ are reflections of $O$ around $BC, CA, AB$ respectively. Prove that $AO_a, BO_b, CO_c$ are concurrent.
This problem can be solved through considering the intersection of the parallelograms in the diagram. But I thought of generalizing it for any point $P$.
We shall use trilinear coordinates. See https://en.wikipedia.org/wiki/Trilinear_coordinates and the reference therein.
Let $(p,q,r)$ be a point (in trilinear coordinates, relative to a reference triangle.) The reflection of this point in the Line $BC$ (of the reference triangle) is given by \begin{eqnarray*} (-p,q+2p \cos(C),r+2p \cos(B)). \end{eqnarray*} The equation of the line through this point and $A=(1,0,0)$ is \begin{eqnarray*} (q+2p \cos(C))z=(r+2p \cos(B))y. \end{eqnarray*} The other $2$ lines can be found similarly and the condition fo their concurrency is \begin{eqnarray*} \begin{vmatrix} 0 & r+2p \cos(B) & q+2p \cos(C) \\ r+2q \cos(A) & 0 & p+2q \cos(C) \\ q+2r \cos(A) & p+2r \cos(B) & 0 \\ \end{vmatrix}=0. \end{eqnarray*} This is a cubic in $p,q$ and $r$ and, strictly speaking, this answers your question.
Similar cubics to this are known see for example https://faculty.evansville.edu/ck6/encyclopedia/ETCpart3.html#X3610.
Note that if $(p,q,r)=(1,1,1)$ (the incenter) then this gives the central point \begin{eqnarray*} \left(\frac{1}{1+2 \cos(A)},\frac{1}{1+2 \cos(A)},\frac{1}{1+2 \cos(A)} \right). \end{eqnarray*} This is illustrated in the diagram below
I don't know which of CK's central point it is but I have it coded up as $X14$ ... which it is not !
For the Circumcenter $(p,q,r)=(\cos(A),\cos(B),\cos(C))$ this construction gives \begin{eqnarray*} ( \cos(B-C),\cos(C-A),\cos(A-B)). \end{eqnarray*} This is illustrated in the diagram below (again I don't know which KFC point this is, although other brands of fried chicken are available.)
The orthocenter is easy ... it gives the orthocenter.
These are the only central points, for which, this construction gives concurrence. I did try a few others.