Briefly, I'd like to know whether there are infinitely many "generalized triangle centers" which – like the orthocenter – are indistinguishable from a vertex of the original triangle. This is basically a refinement of this MSE question of mine; note that the type of "generalized triangle center" I'm interested in is not the standard one (see 1,2), although I would also be interested in the situation for that definition.
Separately, since MO bounties have time limits, I'll "informally bounty" this question: I'll reward the first complete answer to the question with a 1000 point bounty, whenever – if ever – that should happen.
Definitions
Let $\mathbb{T}$ be the set of noncollinear ordered triples of points in $\mathbb{R}^2$. Say that a topological triangle center representative (ttcr) is a function $t:G\rightarrow \mathbb{R}^2$ such that:
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$G$ is a connected dense open subset of $\mathbb{T}$ and $t$ is continuous;
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$G$ and $t$ are each symmetric: if $(a,b,c)\in G$, then $(a,c,b)$ and $(b,a,c)$ are in $G$ as well and we have $t(a,b,c)=t(a,c,b)=t(b,a,c)$;
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both $G$ and $t$ are homothety-etc.-invariant: if $\alpha:\mathbb{R}^2\rightarrow\mathbb{R}^2$ is a composition of rotations, reflections, translations, and homotheties, and $(a,b,c)\in G$, then $(\alpha(a),\alpha(b),\alpha(c))\in G$ and $t(\alpha(a),\alpha(b),\alpha(c))=\alpha(t(a,b,c))$.
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and $t$ is (usually) iterable: for a dense open $H\subseteq G$, if $(a,b,c)\in H$ then $(t(a,b,c),b,c)\in H$.
Each classical triangle center that I'm familiar with corresponds to a ttcr, possibly after tweaking the domain. For instance, in the case of the orthocenter we need to throw out right triangles to satisfy the iterability requirement.
A topological triangle center is then an equivalence class of ttcrs with respect to the relation $t\sim s\iff t_{\upharpoonright \operatorname{dom}(t) \cap \operatorname{dom}(s)}=s_{\upharpoonright \operatorname{dom}(t)\cap \operatorname{dom}(s)}$. Finally, a pseudovertex is a topological triangle center with a representative $t$ satisfying $$t(t(a,b,c),b,c)=a$$ for every $(a,b,c)\in \operatorname{dom}(t)$.
Question
My question is simply, how many pseudovertices are there? Specifically:
Are there infinitely many pseudovertices?
I strongly suspect that the answer is yes (indeed that there should be continuum-many due to the existence of at least one not-too-interesting continuously-parameterized family), and I suspect that in fact there is an easy proof of this fact, but I can't see it at the moment.
So far I know of three distinct pseudovertices (modulo appropriate definitional abuse):
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The orthocenter, $X(4)$.
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The isogonal conjugate of the Euler infinity point, $X(74)$.
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The isogonal conjugate of Parry's reflection point, $X(1138)$.
As a curiosity, note that these three centers are nontrivially related to each other: it turns out that $X(74)$ is the crosspoint of $X(4)$ and $X(1138)$. This fact, as well as the examples of $X(74)$ and $X(1138)$, was found by MSE user Blue at the above-linked question.
Best Answer
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 scaling, $z^{-1} t(z,0,1) = t(1,0,z^{-1}) = t(z^{-1},0,1)$ so $\mu(z^{-1}) = z^{-1} \mu(z)$. Similarly, $z \to 1 - z$ is a half-rotation around $\tfrac12 + 0i$, so $\mu(1-z) = 1 - \mu(z)$.
The three known pseudovertices have the following $\mu$: $$\mu_4 = \frac{(z + \overline{z})(z-1)}{z - \overline{z}} \\ \mu_{74} = \frac{-z(3z\overline{z}^2 + 3z^2\overline{z} - 2\overline{z}^2 - 8z\overline{z} - 2z^2 + 3\overline{z} + 3z)}{(z-\overline{z})(z^2+2z\overline{z}-2z-\overline{z})} \\ {\mu_{1138} = \frac{(9z^2\overline{z}^3+9z^3\overline{z}^2+\overline{z}^4-7z\overline{z}^3-24z^2\overline{z}^2-7z^3\overline{z}+z^4+9z\overline{z}^2+9z^2\overline{z})(3z\overline{z}^2-\overline{z}^2-4z\overline{z}-z^2+3z)}{3(\overline{z}-z)(z^2+2z\overline{z}-2z-\overline{z})(\overline{z}^2+2z\overline{z}-2\overline{z}-z)^2}}$$
I therefore considered candidates of the form $\mu(z) = \frac{P(z, \overline{z})}{Q(z, \overline{z})}$ where $P$ is a polynomial of total degree $k$ and $Q$ is a polynomial of total degree $k-1$, both having real coefficients.
The general approach was to use Sage to expand $(z\overline{z})^k z\mu(z^{-1}) - (z\overline{z})^k \mu(z)$ and $\mu(1-z) - 1 + \mu(z)$ for $z = a+bi$ with the coefficients of $P$ and $Q$ as variables; then since the results should be identically zero, I consider both values as polynomials in $a$, $b$; separately take the real and imaginary part of each coefficient; and form the ideal given by all of these subcoefficients. Finally I ask Sage for the minimal associated prime ideals.
The approaches I then took to filter down the prime ideals were rather more ad hoc: for the simpler ones I just expanded $\mu(\mu(z)) - z$ in the non-eliminated variables to get a new ideal and look for its primes; for more complicated ones I took a small number of non-real values of $z$, calculated $\mu(\mu(z)) - z$ for those values, and obtained an ideal that way; and for the most recent cases treated it occurred to me that $f(b) = \mathfrak{Im}(\mu(\tfrac12 + bi)) : \mathbb{R} \to \mathbb{R}$ should be an involution and I obtained ideals from a combination of $f(f(b)) - b$ and the obvious fact that the equilateral triangle gives $\mu(\tfrac12 + \tfrac{\sqrt 3}2i) = \tfrac12 + \tfrac{\sqrt 3}6i$ or a pole; there is a slim possibility that the cases treated by the second approach contained a solution which was missed due to having a pole at one of the sample points. But I didn't find any solutions other than $X(4)$ and $X(74)$ in the following cases:
The almost-fully-general quartic case took several hours of calculations just to yield the initial prime ideals, before taking into account the involutive requirement, so I don't think I can push this approach any further.