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Now asked at MO.

Let $\mathbb{T}\subseteq(\mathbb{R}^2)^3$ be the set of ordered noncollinear triples of points in the plane. Say that a pseudovertex is a function $\mu:C\rightarrow\mathbb{R}^2$ such that:

  • $C$ is a dense open subset of $\mathbb{T}$.

  • $\mu$ is order-invariant and is continuous with respect to the usual topologies on domain and codomain.

  • $\mu$ respects "distortion-free" maps: if $\alpha$ is a composition of translations, rotations, reflections, and scalings $(x,y)\mapsto(\lambda x,\lambda y)$ for nonzero $\lambda$, then $\alpha(\mu(a,b,c))=\mu(\alpha(a),\alpha(b),\alpha(c))$ in the strong sense (if either side is defined, they're both defined and equal).

  • If $a,b,c$ are non-collinear and $(a,b,c)\in C$ then $(\mu(a,b,c),b,c)\in C$ as well and $\mu(\mu(a,b,c),b,c)=a$ (so given $\{a,b,c,\mu(a,b,c)\}$ we cannot tell which three points were the original vertices).

It is, as far as I can tell, not immediately obvious that pseudovertices exist at all. However, there is at least one very natural example: the orthocenter, modulo appropriate domain-tweaking (e.g. we need to throw out all right triangles). This is in fact the only example I know, but I strongly suspect I'm missing an easy proof that there are infinitely many:

Are there infinitely many pseudovertices?

(I am prepared to be highly embarrassed.) I am more ambivalent however about the prospects for additional natural pseudovertices:

Are there any pseudovertices, besides the (map corresponding to the) orthocenter, which are "natural" - e.g. have been given a name in the existing literature?

(Incidentally, unless I'm missing something neither of the "Shermanesque" fourth-vertex candidates suggested at MO are pseudovertices due to the third requirement. That said, see Blue's comment below.)

EDIT: user runway44 observed below that I'm essentially asking about continuous functions $\mu:\subseteq_{\mbox{dense open}}\mathbb{C}\setminus\mathbb{R}\rightarrow\mathbb{C}\setminus\mathbb{R}$ satisfying the following properties:

  • $\mu$ commutes with conjugation and with the map $z\mapsto 1-z$.

  • $\mu({1\over z})={\mu(z)\over z}$.

  • $\mu\circ\mu=\mathsf{id}.$

(I'm including this observation here since I think it's worth keeping around, and comments are temporary.)

SECOND 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.

Noah Schweber
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  • $\mu$ needs to be order-invariant: $\mu(a,b,c)=\mu(b,c,a)=...$. – Noah Schweber Oct 20 '21 at 01:34
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    Consider the Lemma within the discussion of "my" Shermanesque fourth vertex candidate. Ignoring $A_\pm$, etc, and related geometry, what matters here are the cyclic points $A$, $B$, $C$, $D$ at "polar angles" $a$, $b$, $c$, $d$ satisfying $$k \sin s= \sin(s-a)+\sin(s-b)+\sin(s-c)+\sin(s-d)$$ where $s:=(a+b+c+d)/2$, and $k$ (here) is just some constant. As noted in the Lemma's proof, any three pts uniquely determine the fourth, satisfying your third bullet. Order- and "distortion-free"-invariance hold as well. (Barring degeneracies,) Does this count? – Blue Oct 20 '21 at 04:02
  • @NoahSchweber: I'm not saying that "my" fourth-vertex candidate ($W$ in that answer, aka Kimberling's $X(1309)$) satisfies your definition of pseudovertex; in fact, I just checked: it doesn't. Rather, I'm saying that the four-point configuration discussed in the accompanying Lemma effectively serves as an example. (Infinitely-many examples, actually, by varying $k$ in the equation in my earlier comment.) – Blue Oct 20 '21 at 18:57
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    @Blue Aaargh, multitasking kills reading comprehension ... – Noah Schweber Oct 20 '21 at 18:57
  • @NoahSchweber: I rescind my suggestion of the Lemma points (for now). The key sine relation depends upon a preferred direction derived from the "related geometry", which makes the fourth point rotation-dependent and therefore disrespectful of "distortion-free" maps. There may be a way to throw a bunch parameters at the configuration to tease-out that direction, but it's not clear that this can be done in a way that gives the same result for all four derived triangles. I'll need to give it more thought. That said, I'm considering another notion that might generate pseudovertices. – Blue Oct 21 '21 at 11:56
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    @NoahSchweber: Found one! Well, as it turned out, I re-found one: Kimberling's X(74) ("Isogonal Conjugate of Euler Infinity Line") entry notes "In Hyacinthos 8129 (10/4/03), Floor van Lamoen noted that if $X(74)$ is denoted by $J$, then each of the points $A$, $B$, $C$, $J$ is [X(74)] of the other three [...]." – Blue Oct 21 '21 at 13:56
  • @Blue Neat! Can you add that as an answer? It at least partially answers the question. – Noah Schweber Oct 21 '21 at 16:56
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    Defining $\mu(z):=\mu(0,1,z)$ (with domain $\Bbb C\setminus\Bbb R$), the conditions (after some work) are equivalent to $$\begin{array}{cc}\mu(\bar{z})=\overline{\mu(z)} & \quad \mu(1/z)=\mu(z)/z \[5pt] \mu(\mu(z))=z & \quad \mu(1-z)=1-\mu(z) \end{array}$$ – anon Oct 22 '21 at 00:58
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    For what it's worth, using runaway44's description, the centers $X_4$ and $X_{74}$ are $$\frac{(z+\overline z)(z-1)}{z-\overline z}$$ and $$-\frac{z (3(z\overline z+1)(z+\overline z) - 2(z^2+4z\overline z+\overline z^2))}{(z-\overline z) (z^2+2\overline zz-2z-\overline z)},$$ respectively. – Carl Schildkraut Oct 23 '21 at 06:27
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    ($X_{1138}$ available at https://pastebin.com/DAdXPRVh using the notation $w=\overline z$). – Carl Schildkraut Oct 23 '21 at 07:02
  • @runway44 That's nice! Potentially stupid question: is it clear that there are infinitely many such $\mu$s? – Noah Schweber Oct 29 '21 at 18:27
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    Not to me it isn't. – anon Oct 30 '21 at 02:44
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    The orthocenter fails on the noncollinearity requirement of the third condition: if $abc$ is has a right angle at $b$, then μ(a,b,c)=b, and μ(a,b,c),b,c are collinear. – Günter Rote Nov 04 '21 at 08:30
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    The involutions $f(z) = 1 - z, g(z) = \frac{1}{z}$ have a relation of $(f \circ g)^3 = \text{id}$, and both commute with conjugation, so we can determine $\mu$ entirely from its action on a fundamental domain, e.g. the domain bounded by the $x$-axis, the line $x = \frac{1}{2}$, and the unit circle. Geometrically, this is saying we can assume WLOG that $AB \geq AC \geq BC$ (as otherwise, we can rearrange vertex names until it is true). It also may be worth noting that the line $x = \frac{1}{2}$ maps to itself, which corresponds to the fact that if $ABC$ is isosceles, then so is $\mu(A)BC$. – user44191 Nov 05 '21 at 17:10
  • This fact appears in other ways with the unit circle and the unit circle centered at $1$, but the way it appears is more complex (because it involves the division by $z$). – user44191 Nov 05 '21 at 17:12
  • @GünterRote Oops. Good point, fixed! – Noah Schweber Nov 07 '21 at 23:34
  • Also worth noting: for any continuous choice of "center", you will have to leave out some part of the domain - there will always be some isosceles triangle between the equilateral and its "third" that goes to itself, violating noncollinearity. So the orthocenter isn't unusual for that. – user44191 Nov 09 '21 at 07:58

1 Answers1

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I'll take the liberty of shifting OP's question to a search for "centric tetrads" or "$X$-centric tetrads". (I believe the following captures the spirit of OP's intentions. However, as noted by OP in a comment, Kimberling's definition of "triangle center" lacks a continuity requirement and imposes a stricter homogeneity requirement than the question, so there's some deviation.)

Given triangle vertices $A$, $B$, $C$, we seek a triangle center $D$ —either found in, or destined for, Clark Kimberling's Encyclopedia of Triangle Centers (ETC)— such that $A$, $B$, $C$ are the centers of the same type for respective triangles $\triangle DBC$, $\triangle ADC$, $\triangle ABD$.

We identify the type of center (thus also the tetrad) by Kimberling's $X$ number (if available).

I'm marking this answer as Community Wiki so that it might serve as a master list of tetrads that may appear in other answers.

List of $X$-Centric Tetrads

(It's pretty short right now, but at least it shows that the tetrad phenomenon is not unique!)

$X(4)$, $X(74)$, $X(1138)$.

In more detail (included is the (vertex) centroid of $ABCD$ —ie, the average of the four points— which is necessarily a triangle center shared by $\triangle ABC$, $\triangle DBC$, $\triangle ADC$, $\triangle ABD$):

  • $X(4)$ = orthocenter.
    As OP mentions this as a "natural" example; typically called an "orthocentric system" (I like tetrad better in this context), this foursome is well known in the literature. $$\text{barycentric coords of $D$} = \frac{1}{-a^2+b^2+c^2}:\frac{1}{-b^2+c^2+a^2}:\frac{1}{-c^2+a^2+b^2}$$ Centroid of $ABCD$: $X(5)$ = the nine-point center. In fact, the four triangles share a common nine-point circle.

  • $X(74)$ = isogonal conjugate of Euler Infinity point.
    The ETC notes that Floor van Lamoen recognized this tetrad in 2003. $$\text{barycentric coords of $D$} = \frac{a^2}{a^2(2a^2-b^2-c^2) - (b^2 - c^2)^2}:\cdots:\cdots$$ Centroid of $ABCD$: $X(6699)$ = centroid of $(A,B,C,X(74))$.

    The tetrad's four triangles also share a common circumcircle, and thus also a common circumcenter, $X(3)$. Defining $P_3:=P-X(3)$ for the complex coordinate of $P$ relative to the shared $X(3)$, it happens that $$A_3 B_3+A_3 C_3+A_3D_3+B_3C_3+B_3D_3+C_3D_3=0$$

  • $X(1138)$ = isogonal conjugate of $X(399)$.
    The ETC doesn't (yet?) mention the tetrad-ic property of $X(1138)$, although it was brought to the attention of Kimberling on 30 October, 2021. (I don't doubt that someone has observed it before me.) However, the very few properties it does mention include two notes relating to $X(4)$ and $X(74)$ (whose tetrad-ic nature is documented):

    1. $X(4)$ and $X(1138)$ are the only two points whose pedal triangle and cevian triangle are similar. (This is actually enough to prove the tetrad-ic nature of those centers: For any point $P$, the pedal triangles of $P$ with respect to $\triangle ABC$ and of, say, $A$ with respect to $\triangle PBC$ are necessarily similar, while the corresponding cevian triangles are the same triangle. Consequently, pedal-cevian similarity is equivalent to each of $P$ and $A$ being one of these centers; a little bookkeeping guarantees that they're the same type, ie, both $X(4)$s or else both $X(1138)$s. (This straightforward result is not noted in Ehrmann's paper about the pedal-cevian similarity property). Maybe the tetrad-ic nature of $X(1138)$ hasn't been observed before.)
    2. $X(1138)$ is the $X(74)$-cross conjugate of $X(4)$. (The $X(74)$ entry identifies that point as the crosspoint of $X(4)$ and $X(1138)$, which amounts to the same thing.) $$\text{barycentric coords of $D$} =\frac{1}{\left(\begin{array}{c} a^8 - 4 a^6 (b^2 + c^2) + a^4 (6 b^4 + b^2 c^2 + 6 c^4) \\ - a^2 (4 b^6 - b^4 c^2 - b^2 c^4 + 4 c^6) \\+ (b^2 - c^2)^2 (b^4 + 4 b^2 c^2 + c^4)\end{array}\right)}:\cdots:\cdots$$ Centroid of $ABCD$: $X(45694)$ (added 1 November, 2021 ... and citing this post!). The first barycentric coordinate (relative to $\triangle ABC$) is $$\begin{align} &(a^2- b^2-c^2 + b c) (a^2 - b^2 - c^2 - b c) \\ &\cdot \left(\,2 a^4 - a^2 b^2 - b^4 - a^2 c^2 + 2 b^2 c^2 - c^4\,\right)\\ &\cdot\left(\begin{array}{c} a^8 - 2 b^8 - 2 c^8 - a^6 b^2 - a^6 c^2 + 5 a^2 b^6 + 5 a^2 c^6 + 8 b^6 c^2 + 8 b^2 c^6\\ - 3 a^4 b^4 - 3 a^4 c^4 - 12 b^4 c^4 + 7 a^4 b^2 c^2 - 5 a^2 b^4 c^2 - 5 a^2 b^2 c^4 \end{array}\right) \end{align}$$

22 October, 2021. I've done a kind of search with GeoGebra's TriangleCenter command: I animated a slider controlling the Kimberling index parameter, checking when calculated points were sufficiently-close to the vertices of the triangle. Up to the documented maximum of $X(3054)$, the only hits I saw were $X(4)$, $X(74)$, and $X(1138)$. This doesn't rule-out other possibilities: there were hundreds of "undefined" instances, which could be due to computational errors, degeneracies associated my particular test triangle, or lack of implementation in GeoGebra. Even in "defined" cases, there's the possibility of numerical error. So, the search is hardly conclusive ... and nowhere-near comprehensive: the ETC currently documents over $44,000$ centers. (Of course, even those don't scratch the surface of possible triangle centers.)

In any case, this may suggest that it could be time to switch to a theoretical approach. I imagine this topic must be discussed in the literature.

Blue
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    This is great! I've made a couple minor format edits, I hope you don't mind. More substantively, I have a "foundational" question: I'm surprised to see that Kimberling's definition of triangle center doesn't require continuity (or am I missing something?); meanwhile, the homogeneity requirement seems a bit stronger than what I'm looking for. Is there a good source explaining why Kimberling's is the right definition of triangle center, as opposed to something more topology-flavored? (I'm not really satisfied with the explanation in Kimberling $1994$.) – Noah Schweber Nov 03 '21 at 16:17
  • @NoahSchweber: I'm not qualified to defend Kimberling's definition in a broader topological context. Geometrically, of course, the homogeneity requirement makes dimensional calculations work. I haven't given much thought to continuity. ... You could take your questions to Kimberling directly; see "Contact the Keeper" at the bottom of the ETC's Links page. Kimberling (or a minion signing his name) responded personally to my submission of what became X(45694). (No mention of why my finding about X(1138) was ignored, though. shrug) – Blue Nov 03 '21 at 21:02
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    @NoahSchweber: Topology would be tough since (1) it isn't quite clear what the domain is (are degenerate triangles allowed? if so, what about equal vertices?) and (2) it isn't quite clear what the target is (projective plane or Möbius plane? some centers need one, some need another, and others are completely singular e.g. at equilateral triangles). If I'd require anything, it would be rationality (in an appropriate sense, maybe using trigonometric functions of $A/2, B/2, C/2$ in order to allow $X_1$). – darij grinberg Nov 04 '21 at 19:48
  • @darijgrinberg I think the topology issue can be fixed (and should be anyways, per orthocenters of right triangles) by saying that the domain should be some dense open subset of (or even just comeager) the space of triples of $\mathbb{R}^2$-points, and the codomain should be $\mathbb{R}^2$. – Noah Schweber Nov 04 '21 at 23:48