*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.*

## 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:

$G$ is a

*dense open*subset of $\mathbb{T}$ and $t$ is*continuous*;$G$ 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 $t(a,b,c)=t(a,c,b)=t(b,a,c)$;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 $t(\alpha(a),\alpha(b),\alpha(c))=\alpha(t(a,b,c))$.and $t$ is

*iterable*: if $(a,b,c)\in G$ then $(t(a,b,c),b,c)\in G$;- EDIT: I should have also required that $G$ be
*connected*: if we allow disconnected $G$, the symmetry requirement becomes vacuous since the set of scalene triangles is dense open. The iterability requirement then needs to be*weakened*to only hold on a dense open subset of $G$ (otherwise the orthocenter doesn't count). That said, I may be having a silly moment but it's not entirely obvious to me that the question is trivial without these modifications, even if omitting them was obviously a mistake. For now: I prefer answers addressing the modified case, but I'm also interested in the looser question.

- EDIT: I should have also required that $G$ be

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):

The orthocenter, $X(4)$.

The isogonal conjugate of the Euler infinity point, $X(74)$.

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.