"Pseudovertices" of a triangle

Solution 1:

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.