Is there a conceptual reason why so many triplets of lines in a triangle are concurrent?

Question

One of the striking phenomena one can't help but notice in elementary Euclidean geometry is how easy it appears to be to define triples of lines in a triangle which meet in a point. Now for each given triple of line an elementary proof that they meet is generally easy to give, but the argument is always ad hoc and doesn't explain the wealth of such concurrences.

So, is there some deeper conceptual reason why it seems like “just about any symmetric construction” between $A$ and $\{B,C\}$, when similarly applied between $B$ and $\{C,A\}$ and between $C$ and $\{A,B\}$ gives three concurrent lines? (Of course, stated in such generality it is obviously false: the question is why it seems true.)

For example, is there a general theorem or reasoning that implies the existence of many, if not most, of the points of concurrence of a Euclidean triangle (bisectors, medians, mediators, altitudes, symmedians, etc.) without requiring an ad hoc argument for each? (It could, for example, be something like “any construction of algebraically sufficiently low degree will necessarily give concurrent lines, because the so-many first terms will cancel for reasons of symmetry”.)

Maybe one type of argument which seems to come up is that the line $L_A$ is the locus of $M$ such that $\phi_B(M) = \phi_C(M)$ for some function $\phi$ that can be defined from the vertices or sides of the triangle, and then if $\phi_B(M) = \phi_C(M)$ and $\phi_C(M) = \phi_A(M)$ then clearly $\phi_A(M) = \phi_B(M)$: this works to prove that the mediators are concurrent (with $\phi_X(M) = \operatorname{dist}(M,X)$) and also for the bisectors (with $\phi_X(M)$ being the distance from $M$ to the edge opposite $X$), but I don't know if any more points can be handled that way (and I'm not even sure how to correctly state the underlying principle).

Answer 1

One can view this intriguing question as belonging to the subject of functional equations. As hinted by F. Poloni, when three "central lines' concur (e.g., incenter, centroid, circumcenter, orthocenter), the point of concurrence shows some striking properties that have led to a definition of "triangle center" as a point having barycentric coordinates of the form

\begin{equation}\label{symmetry} f(a,b,c) : f(b,c,a) : f(c,a,b), \hspace{8 mm} (*) \end{equation}

where $f(a,c,b) = f(a,b,c)$.

The four ancient Greek points all have barycentrics that are, as Gro-Tsen mentioned, of low algebraic degree in terms of the sidelengths $a,b,c$ of a triangle $ABC$:

incenter = $a:b:c$

centroid = $1:1:1$

circumcenter = $a^2(a^2 - b^2 - c^2) : b^2(b^2 - c^2 - a^2) : c^2(c^2 - a^2 - b^2)$

orthocenter = $1/(a^2 - b^2 - c^2 ) : : $

Now, one can discover or invent new triangle centers $X$ via $(*)$ and then, using certain determinants, contrive triples of lines that concur in $X$. Indeed, starting with a triangle center $X$, one can "construct" (algebraically, and sometimes geometrically) pairs of triangles $A_1B_1C_1$ and $A_2B_2C_2$ such that the lines $A_1A_2, B_2B_2, C_2C_2$ concur. John Conway coined the word "perspector" for $X$, as a replacement for "center of perpective".

Returning to Gro-Tsen's notion of "symmetric construction", this translates, somewhat, into the algebraic property $f(a,c,b) = f(a,b,c)$. Probably the best known pair of points that have barycentrics $(*)$ such that $f(a,c,b) \neq f(a,b,c)$ are the Brocard points:

$$ac/b : ba/c : cb/a \text{ and } ab/c : cb/a : ac/b.$$

T. Chow mentioned D. Hofstadter's charming discussion. In addition to that, I'd like to mention the a Hofstadter Zero Point, with barycentrics

$$ A : B : C,$$

(not of low algebraic degree in $a,b,c$ !). For more about this transcendental triangle center, see $X(360)$ in the Encyclopedia of Triangle Centers: https://faculty.evansville.edu/ck6/encyclopedia/etc.html .

The intriguing question can be recast like this: how do triangle centers (and bicentric pairs, like the Brocard points) arise ubiquitously as points of concurrence? Well, start with any $f(a,b,c)$ that is homogeneous in $a,b,c$. Let

$$A' = 0 : f(b,c,a) : f(c,a,b), \hspace{2 mm} B' = f(a,b,c) : 0 : f(c,a,b), \hspace{2 mm}C' = f(a,b,c) : f(b,c,a) : 0.$$

The triangles $ABC$ and $A'B'C'$ are perspective, and their perspector is $(*)$.

In response to T. Chow's question, one might conjecture that the pedal triangle of the centroid (call it $A'B'C'$) is perspective to $ABC$; i.e., that the nice lines $AA',BB',CC'$ concur, but they don't. Then one could ask for the locus of $X$ whose pedal triangle (of which the $A$-vertex is the orthogonal projection of $X$ on line $BC$) is perspective to $ABC$. This locus is the Darboux cubic, elegantly portrayed by Bernard Gibert: https://bernard-gibert.pagesperso-orange.fr/Exemples/k004.html .

So, as an example of numerous unsuccessful triples, start with any triangle center $X$ not on the Darboux cubic.

Here's a parting thought: if a triple of lines, $L,M,N$ fail to concur, then their intersections form a triangle, and in some cases, that triangle is interesting.