19
$\begingroup$

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.

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.

$\endgroup$
5
  • 1
    $\begingroup$ The three known pseudovertices also all lie on the Neuberg cubic, although this isn't sufficient and may be irrelevant because so does e.g. X(1) and that isn't a pseudovertex. $\endgroup$ Nov 8 '21 at 17:35
  • 1
    $\begingroup$ We may try to define a triangle centre $t$ by a relation $F(a,b,c,t)=0$, where $F$ is a symmetric equivariant w.r.t. homotheties etc. function. Say, orthocentre corresponds to a function $|(a-b)(c-t)|+|(a-c)(b-t)|+|(b-c)(a-t)|$. We only need to check that such equation defines unique point, that is a certain topological condition (on degree of the map or something like that). $\endgroup$ Nov 10 '21 at 6:42
  • $\begingroup$ @PeterTaylor are $B,C,D$ allowed to be any (real, I assume) constants? $\endgroup$ Nov 11 '21 at 4:38
  • $\begingroup$ For every non-isosceles triangle ABC and every point P in the plane there is a topological triangle center t such that P=t(A,B,C): write P as affine combination of orthocenter, circumcenter and incenter, and use the coefficients in this affine combination to define t. However, I don't know which of these will be pseudovertices. $\endgroup$
    – Goldstern
    Dec 5 '21 at 10:03
  • $\begingroup$ @Goldstern Additionally, "most" triangle centers can't be expressed in terms of $X(1)$, $X(3)$, and $X(4)$. So it's not clear to me how much this helps. $\endgroup$ Dec 5 '21 at 20:23
5
+200
$\begingroup$

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:

  • $P$ and $Q$ fully general polynomials of total degree respectively $3$ and $2$;
  • $P$ a fully general polynomial of total degree $4$; $Q = (z - \overline{z})Q'$ where $Q'$ is a fully general polynomial of total degree $2$;
  • $P$ a polynomial with monomials of total degree $3$ to $6$; $Q = (\overline{z}-z)(z^2+2z\overline{z}-2z-\overline{z})(\overline{z}^2+2z\overline{z}-2\overline{z}-z)$.

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.

$\endgroup$
3
  • $\begingroup$ +1, even though this didn't work out it's quite neat. I'm now starting to wonder if there might actually be only finitely many. You mentioned earlier the Neuberg cubic - is it at all plausible that being on the Neuberg cubic is a necessary condition? $\endgroup$ Nov 13 '21 at 20:14
  • $\begingroup$ They're certainly harder to find than I expected, but there's no evidence to suggest that they aren't extremely common even in the rational class that I considered at higher degrees. The comment about the Neuberg cubic was just looking for anything they had in common: I don't see why they would necessarily be on it. It's probably possible to calculate the $\mu$ corresponding to an arbitrary point on the cubic and apply a similar analysis to the one described in the answer, but I don't know when I'll have time to do that. $\endgroup$ Nov 13 '21 at 22:59
  • $\begingroup$ "is it at all plausible that being on the Neuberg cubic is a necessary condition?" I've been investigating the Neuberg cubic aspect; conveniently, if $P$ is on the cubic for $\triangle ABC$, then $A$, $B$, $C$ are all on the cubic for $\triangle PBC$ (likewise, $\triangle APC$ and $\triangle ABP$), so this is a not-unreasonable avenue to pursue. It's possible to parameterize the two parts of the cubic as a variable point moves along the Euler line (see the very bottom of Gibert's K001 page), but the algebra is a bit messy. $\endgroup$
    – Blue
    Nov 14 '21 at 5:23
4
$\begingroup$

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 $\triangle ABC$ as the locus of points $P$ such that the Euler lines of $\triangle ABC$, $\triangle PBC$, $\triangle APC$, $\triangle ABP$ concur (at $M$ in the figure). Consequently, it's also true that the Neuberg cubics of all four triangles contain all four points $A$, $B$, $C$, $P$ (and a companion point, $P'$, with the same properties).

enter image description here

This kind of interchangeability is at least reminiscent of the pseudovertex property, so it seems not-unreasonable to search the Neuberg cubic for other pseudovertex candidates. An explicit parameterization of the cubic might aid the search, so I'm providing one here.

Every point $M$ on the Euler line of $\triangle ABC$ corresponds to two points on the cubic (see animation and description below), so that the latter can be parameterized by the former.

enter image description here

Following a construction from Gibert, define circumcenter $O:=X(3)$ and orthocenter $D:=X(4)$, as well as $E:=X(74)$ and $F:=X(1138)$. Let $E'$ be the dilation of $E$ in $O$ by factor $-2$. For $M$ on the Euler line (through $O$ and $D$), let $O'$ be the reflection of $O$ in $M$. Then, we find $P$ and $P'$ on the cubic via the intersection of $\overleftrightarrow{O'E'}$ and the (necessarily rectangular) circumhyperbola through $A$, $B$, $C$, $D$, $M$. In particular, if $M = O + m(D-O)$, we can parameterize the barycentric $A$-coordinate of $P$ and $P'$ (with $B$- and $C$-coordinates following cyclically) as (deep breath) ... $$(\sin A)\left(\begin{array}{c} 2\left(m(m-1)^2\sigma^2+m\sigma_a\sigma_b\sigma_c+\sigma\right)\left( 2m(2\cos A-\cos(B-C)) - \cos A \right) \\ +\left(1-m^2\pm n\right)\left(m\sigma(2\cos A-\cos(B-C))+\sigma_b\sigma_c\cos A+2\sigma\cos B\cos C\right) \end{array}\right)$$ where $$\begin{align} \sigma_a &:= \frac{(-a^2+b^2+c^2)^2-b^2c^2}{b^2c^2}=1+2\cos 2A, \quad\sigma_b := \cdots, \quad\sigma_c := \cdots \\[0.5em] \sigma\phantom{_x} &:=\sigma_a+\sigma_b+\sigma_c \\[0.5em] n^2 &:= \left(1-m^2\sigma\right)^2+4m(2m-1)(m(m-1)\sigma^2+m\sigma_a\sigma_b\sigma_c+\sigma) \end{align}$$ (As a sanity check, the animation (created in GeoGebra) uses the parameterization to determine $P$ and $P'$, not GeoGebra's geometric construction features.)

The known pseudovertices $X(4)$, $X(74)$, $X(1138)$ correspond to $m = 0, \frac12, \infty$. (The "other" hyperbola intersection points for those $m$ are $X(3)$, $X(1263)$, $X(30)$.) Of course, it's not the case that every (Kimberling-esque) triangle center on the Neuberg cubic corresponds to a "scalar", triangle-independent $m$. (For instance $X(399)$ and $X(8487)$ arise from $m=6\sigma/(3\sigma^2-4\sigma_a\sigma_b\sigma_c)$.) However, it might be sensible to make such a restriction in a first pass at searching for (or ruling out) other pseudovertices on the cubic.

So, in that first pass, "all we have to do" is find an appropriate $m$ (and $\pm$ sign) to yield a $P$ from $\triangle ABC$ that in turn yields $A$, $B$, $C$ from $\triangle PBC$, $\triangle APC$, $\triangle ABP$.

Clearly, this is easier typed than done, as the second-level application of the parameterization explodes in complexity. Perhaps at least some of the thorny algebra can be avoided by mining the geometric lore of Neuberg cubics; Gibert's page about the curve, and associated papers, themselves provide a good deal of lore to consider. (Perhaps the pseudovertex question itself has already been addressed.)


Addendum. [Moved to this non-answer.]

$\endgroup$
4
  • 1
    $\begingroup$ Sorry, very belated comment: is the deep-breath barycentric coordinate to be read as "$(\sin A) (\mathit{stuff})$" or "$\sin (A\cdot \mathit{stuff})$"? I assumed the former but the formatting on further thought suggests the latter. (Also I just read your addendum, that's fascinating! I'm if nothing else quite happy to have a good story for the set $\{4,74,1138\}$.) $\endgroup$ Nov 29 '21 at 7:38
  • 1
    $\begingroup$ @NoahSchweber: I disambiguated the product. :) ... The addendum about how cubic $K279$ "characterizes" Xs 4/74/1138 was a nice surprise to me. There may still be more going on. I'm still looking. :) $\endgroup$
    – Blue
    Nov 29 '21 at 8:20
  • $\begingroup$ I know it's probably a huge hassle, but if you get a chance can you say a bit more about your addendal Mathematica calculations? (If "addendal" isn't a word, it should be - although I'd also be open to "addendual," "addendial," or "addendary.") $\endgroup$ Nov 29 '21 at 15:49
  • $\begingroup$ @NoahSchweber: The addendisestablishmentarianistic calculations are simply brute-force coordinate bashing, from which a final condition for the $K279$ of $\triangle PBC$ to pass through $A$ factors neatly (after considerable processor churn) into an $X(4)$-condition, an $X(74)$-condition, and an $X(1138)$-condition, and a(n extraneous?) factor with tens of thousands of terms. There must be a way to exploit, say, Montesdeoca's condition to tease the result directly out of the geometry. I'm currently checking for other commonalities in those points. $\endgroup$
    – Blue
    Nov 30 '21 at 0:03
3
$\begingroup$

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 triangle of a point $P$, let $A^\star$, $B^\star$, $C^\star$ be the circumcenters of $\triangle AB'C'$, $\triangle A'BC'$, $\triangle A'B'C$, and let $Q$ be the circumcenter of $\triangle A^\star B^\star C^\star$. Then $K279$ is the locus of $P$ such that $Q$ lies on the Euler line.

Unlike with Neuberg, given a point $P$ on this curve, the corresponding cubic for $\triangle PBC$ does not necessarily pass through $A$. Indeed (barring triangle degeneracies), Mathematica-assisted symbol crunching (see below) shows that this occurs for (Kimberling-esque) triangle center $P$ when and only when $P$ is one of $X(4)$, $X(74)$, $X(1138)$. Effectively, then, the property

$P$ lies on the $K279$ of $\triangle ABC$, and $\triangle ABC$ is inscribed in the $K279$s of $\triangle PBC$, $\triangle APC$, $\triangle ABP$.

characterizes the known pseudovertices. This could be useful in the search for (or the ruling-out of?) other candidates.


For the symbol-crunching, we start with the barycentric $u:v:w$ equation for $K279$ relative to $\triangle ABC$:

$$\sum_{cyc} b^2 c^2 ((b^2 - c^2)^2 + a^2 (b^2 + c^2 - 2 a^2))\; u (v^2 - w^2) = 0 \tag{1}$$ Then, considering $P$ with barycentric coordinates $u:v:w$, the $K279$ with respect to $\triangle PBC$ has barycentric $u':v':w'$ equation that derives from $(1)$ by substituting $u\to u'$, $v\to v'$, $w\to w'$, and $$\begin{align} b^2 &\to |PC|^2 = \frac{a^2 v^2 + b^2 u^2 + (a^2+b^2-c^2) u v}{(u + v + w)^2} \\[0.5em] c^2 &\to |PB|^2 = \frac{a^2w^2 + c^2 u^2 + (a^2-b^2+c^2) u w }{(u+v+w)^2} \end{align}$$ Since $A$ has barycentric coordinates $u':v':w' = u+v+w:-v:-w$ relative to $\triangle PBC$, we substitute those, as well. This yields a degree-7 $uvw$ polynomial, whose $184$ terms I won't transcribe here. Mathematica confirms that the polynomial vanishes if $P$ is one of $X(4)$, $X(74)$, $X(1138)$. (It does not vanish when $P$ is $X(2)$, the centroid of $\triangle ABC$, unless $b=c$.)

Applying Mathematica's Resultant to eliminate $u$ from this and $(1)$ gives this equation (ignoring powers of factors):

$$vw(v + w)\cdot(v-w)\cdot f_4 \cdot f_{74} \cdot f_{1138} \cdot f = 0$$

Here, $$f_4 := v(-b^2+c^2+a^2) -w(-c^2+a^2+b^2)$$ so that $$f_4 = 0 \quad\implies\quad v = \frac{k}{-b^2+c^2+a^2} \quad w = \frac{k}{-c^2+a^2+b^2} \quad\left(u = \frac{k}{-a^2+b^2+c^2}\right)$$ (with the $u$-coordinate symmetrically completing the "center"). Thus, this factor can vanish only at center $X(4)$. Likewise, $f_{74}$ and $f_{1138}$ are linear factors in $v$ and $w$ that can vanish only at centers $X(74)$ and $X(1138)$.

Further, our interest in centers allows us to ignore factors $v$, $w$, $v+w$. Factor $v-w$ vanishes when $v=w(=u)$, which corresponds to centroid $X(2)$, a candidate we have already ruled-out.

Factor $f$ is a degree-$6$ polynomial in $v$ and $w$, with $711$ terms. I haven't done a proper analysis of it; however, when, say, $(a,b,c)=(6,9,13)$, the solutions are all non-real. I'm pretty sure we can consider this extraneous in general. (There's probably a neat way to avoid this factor altogether by first reducing the system or something, but I haven't found it.)

Of course, it would be best to devise a more-direct, less-computational argument, perhaps one based on Montesdeoca's geometric description of $K279$. (One may note that the cevian triangle of $P$ with respect to $\triangle ABC$, and that of $A$ with respect to $\triangle PBC$, are always the same triangle, so there's a start! :) That's what I'm investigating now.

$\endgroup$
7
  • 1
    $\begingroup$ @NoahSchweber: I'm not sure, but I (and my humble computing device) lean toward "very" on the intractable side. :) At least for attacking things with brute force. For a while there, I was trying to look at the broader pencil of cubics through $A$, $B$, $C$, $X(4/74/1138)$, but even the basic equation is huge, and that's without trying to generate the corresponding second-level cubic, for $\triangle PBC$. (BTW: No cubic in Gibert's catalog, besides Neuberg and $K279$, is documented as including all three known pseudovertices.) As I suggested before, the geometric lore of cubics may be the key. $\endgroup$
    – Blue
    Dec 5 '21 at 7:59
  • 1
    $\begingroup$ @NoahSchweber: It would be nice if someone with access to the Kimberling database would "just run a check" for pseudovertices beyond $X(1138)$. :) $\endgroup$
    – Blue
    Dec 5 '21 at 8:09
  • $\begingroup$ Following one of the triangle centre questions back in August I did some work with the Kimberling ETC to find lines which contain a lot of points, and I concluded that there isn't actually a database. There are far too many inconsistencies in the rendering for that to be plausible. I'm pretty sure the HTML pages are all there is. $\endgroup$ Dec 5 '21 at 14:48
  • $\begingroup$ @PeterTaylor: "I'm pretty sure the HTML pages are all there is." Hmm ... With the Math.SE version of this question, I submitted what became $X(45694)$ to the ETC, providing bary-coords and description. The entry appeared w/a list of relations to other centers; these couldn't have been derived by hand. (My email asked about accessing some form of the database, but the reply (which may have been automated) made no mention of the request (nor my "tetrad" observation about $X(1138)$). Maybe there's a secret handshake. :) $\endgroup$
    – Blue
    Dec 6 '21 at 0:05
  • $\begingroup$ Coming back to this a bit later, I have a question that might be easier. What if instead of "self-inverting" triangle centers we look for triangle centers which can be inverted by a possibly different triangle center - along the lines of "For 'most' triangles $\triangle ABC$, the $Y$-center of $\triangle ABU$ is $C$, where $U$ is the $X$-center of $\triangle ABC$." There's a clear injectivity requirement here (fix two vertices) which e.g. rules out the circumcenter but doesn't rule out the incenter. Maybe this sort of phenomenon happens more frequently among natural triangle centers? $\endgroup$ Jan 9 at 21:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.