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3

Transferring the Addendum to my previous non-answer to a non-answer of its own, as it's beginning to sprawl (and because it actually seems more significant) ... It happens that $X(4)$, $X(74)$, $X(1138)$ also lie on the circumcubic $K279$, for which there is this geometric description attributed to Angel Montesdeoca: Let $\triangle A' B' C'$ be the cevian ...

0

Bringing together complex numbers and laziness. Following the answer by @FedorPetrov, we see that $N=(A+B+C)/2$ and so $A'=B+C$. Take inversion with respect to the unit circle (which is just a map $z\mapsto 1/\bar{z}$), then the circles are mapped to the lines connecting $A$ and $BC/(B+C)$ (and two analogous) and we want to prove they intersect at one point. ...

4

Not an answer. I'm just expanding a comment about @PeterTaylor's observation that the known pseudovertices $X(4)$, $X(74)$, $X(1138)$ lie on the Neuberg cubic ... Bernard Gibert's "Pairs and Triads of points on the Neuberg Cubic connected with Euler Lines and Brocard Axes Isometric Parallel Chords" Proposition 1 characterizes the Neuberg cubic of $\... 5 This is a report on an unsuccessful computational approach which is rather too long for a comment. I work with complex numbers to represent the points in the obvious way. It suffices to consider$\mu(z) = t(z,0,1)$because this can be extended under the invariants to the full$t(z,z',z'')\$. Since multiplication by a complex number is just rotation and ...

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