The point $O$ is the incenter of the triangle $ABC$, $N$ is the centroid. Then it is possible to construct the point $M$ that can be loosely defined as the intersection point of the "median bisectors" of the triangle $ABC$. Сause in a sense cevians $AR$, $BP$, $CQ$ lie somewhat in the middle of bisectors and medians drawn from the vertices $A,B,C$. Perhaps it is even possible to express $AR,BP, CQ$ as a good looking function of the sides $a,b,c$ of the original triangle $ABC$.

I wonder what is known about this point $M$? Symmedian is a classic concept in planimetry, so perhaps this "medianbisector" or "bisectormedian" concept might also have some geometric sense?

enter image description here

  • $\begingroup$ To be clear: You're using the medians and bisectors of $\triangle ABC$ to define points $D$, $E$, $F$, $G$, $H$, $I$, which in turn determine sides of $\triangle JKL$. And, apparently, $\overleftrightarrow{AK}$, $\overleftrightarrow{BL}$, $\overleftrightarrow{CJ}$ are concurrent at your point of interest, $M$. Correct? $\endgroup$ – Blue Jul 7 at 10:32
  • $\begingroup$ yes, that is how point M is constructed. $\endgroup$ – A Z Jul 7 at 10:43
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    $\begingroup$ Curiously, the side-lines of $\triangle JKL$ aren't defined consistently (from a which cevians-yield-which-points perspective). Points $E$ and $F$ on $\overline{JL}$ are determined by medians; points $I$ and $G$ on $\overline{KL}$ are determined by bisectors; points $D$ and $H$ on $\overline{JK}$ are determined by one of each. It's actually surprising that we get concurrence out of that. Also surprising: according to my GeoGebra sketch, it doesn't matter that $O$ and $N$ are incenter and centroid; they can be any points of concurrency. Intriguing ... $\endgroup$ – Blue Jul 7 at 10:46
  • $\begingroup$ geogebra.org/m/ar49cc3g This is an updated sketch. I corrected the error concerning point M supposedly lying on the line TYZ. $\endgroup$ – A Z Jul 7 at 13:04
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    $\begingroup$ @AZ: To search the ETC, you essentially calculate coordinates of a target point for a specific triangle (say, the $6$-$9$-$13$) and lookup a corresponding value in this "Tables" page. The page explains exactly what calculations to make, although it's not presented as clearly as I think it could be. The linked "Triangle Converter" page is almost-certainly helpful; I haven't used it. (It's relatively new.) $\endgroup$ – Blue Jul 7 at 18:07

There's a bit of ambiguity in the construction (which points connect to which points?) that I'll address while proving a generalization of the result to arbitrary points of concurrency.

Given $\triangle ABC$, let the cevians through points $P_+$ and $P_-$ meet the opposite sides at appropriate points $D_\pm$, $E_\pm$, $F_\pm$, as shown:

enter image description here

Specifically, if we define barycentric coordinates $$P_\pm = \frac{\alpha_\pm A+\beta_\pm B+\gamma_\pm C}{\alpha_\pm+\beta_\pm+\gamma_\pm} \tag1$$ then $$D_\pm =\frac{0 A+\beta_\pm B+\gamma_\pm C}{0+\beta_\pm+\gamma_\pm} \qquad E_\pm = \frac{\alpha_\pm A+0 B+\gamma_\pm C}{\alpha_\pm+0+\gamma_\pm} \qquad F_\pm = \frac{\alpha_\pm A+\beta_\pm B+0 C}{\alpha_\pm+\beta_\pm+0} \tag2$$ Now we want to specify the lines through the appropriate pairs of the $D$, $E$, $F$ cevian-points. This is where the ambiguity arises: Does $D_+$ connect to $E_+$? or $E_-$? or $F_+$? or $F_-$? And then what about $D_-$? Once these decisions are made, the final line is uniquely determined, but we still need to get a handle on those first two. We'll say that our lines are $$\ell_F:=\overleftrightarrow{D_+E_s} \quad\text{and}\quad \ell_E:=\overleftrightarrow{D_-F_t} \quad\left(\text{and}\quad\ell_D:=\overleftrightarrow{E_{-s}F_{-t}}\right) \tag3$$ where $s$ and $t$ are each "$\pm$"; or, treating them as "$\pm 1$", we can write $$E_s := \tfrac12\left( (E_++E_-)+s(E_+-E_-)\right) \qquad F_t := \tfrac12\left(F_-+F_+)+t(F_--F_+)\right) \tag4$$ (Note that $s=t=+1$ gives the lines $\overleftrightarrow{D_+E_+}$ and $\overleftrightarrow{D_-F_-}$; the "$+1$" indicates the "same sign" in the subscripts.) From here, we (and, by "we", I mean "Mathematica") can determine the equations of the lines and find their points of intersection $$D := \ell_E\cap\ell_F \qquad E:=\ell_F\cap\ell_D\qquad F:=\ell_D\cap\ell_E \tag5$$ the associated cevians $\overleftrightarrow{AD}$, $\overleftrightarrow{BE}$, $\overleftrightarrow{CF}$. As it turns out, all four $s$-$t$ sign choices cause their triplets of cevians to meet at a point of concurrency. The barycentric coordinates are (surprisingly?) uncomplicated:

$$\begin{align} (s,t)=(+,+)\qquad K_A &= \left(\frac12:\frac{\beta_+\beta_-}{\alpha_+\beta_- + \alpha_-\beta_+}: \frac{\gamma_+\gamma_-}{\alpha_+\gamma_- +\alpha_-\gamma_+} \right) \\[4pt] &=\left(\frac12:\frac{1}{\dfrac{\alpha_+}{\beta_+}+\dfrac{\alpha_-}{\beta_-}}:\frac{1}{\dfrac{\alpha_+}{\gamma_+}+\dfrac{\alpha_-}{\gamma_-}}\right) \tag6\\[4pt] (s,t)=(+,-)\qquad K_B &= \left(\frac{1}{\dfrac{\beta_+}{\alpha_+}+\dfrac{\beta_-}{\alpha_-}}:\frac12:\frac{1}{\dfrac{\beta_+}{\gamma_+}+\dfrac{\beta_-}{\gamma_-}}\right) \tag7 \\[4pt] (s,t)=(-,+)\qquad K_C &= \left(\frac{1}{\dfrac{\gamma_+}{\alpha_+}+\dfrac{\gamma_-}{\alpha_-}}:\frac{1}{\dfrac{\gamma_+}{\beta_+}+\dfrac{\gamma_-}{\beta_-}}:\frac12\right) \tag8 \\[4pt] (s,t)=(-,-)\qquad K\phantom{_X} &= \left(\frac1{\beta_+\gamma_- + \beta_-\gamma_+}: \frac1{\gamma_+\alpha_-+\gamma_-\alpha_+}:\frac1{\alpha_+\beta_-+\alpha_- \beta_+}\right) \tag9 \end{align}$$

One should notice that $K_A$, $K_B$, and $K_C$ are cyclic transformations of each other. Our derivation was driven by points $D_\pm$, prioritizing vertex $A$, and yielding the "naturally-associated" point $K_A$; the point $K_B$ is likewise naturally-associated with vertex $B$, and $K_C$ with $C$. And, yes, the corresponding cevians concur; namely, at the point with barycentric coordinates

$$K_\star:= \left(\frac1{\beta_+\gamma_-} + \frac1{\beta_-\gamma_+} : \frac1{\gamma_+\alpha_-} + \frac1{\gamma_-\alpha_+} : \frac1{\alpha_+\beta_-} + \frac1{\alpha_-\beta_+}\right) \tag{10}$$

On the other hand, $K$ is fully symmetric, being naturally associated to each of $A$, $B$, $C$, so a distinguishing subscript is unnecessary.

The reader may observe that, in choosing my lines $\ell_D$, $\ell_E$, $\ell_F$ in $(3)$, I ignored the possibility that $D_+$ could be connected to an $F$-point instead of an $E$-point (and vice-versa for $D_-$). This is addressed simply by flipping "$+$" and "$-$" in the discussion and the results. But our results are symmetric in their "$+$"s and "$-$"s (except for the specific $(s,t)$ associations), so we find that the same four points arise.

Also, the above guarantees that the order of the defining points of concurrency is not important. For instance, the $K_A$-point for $(P_+,P_-):=(\text{orthocenter},\text{incenter})$ is same as the $K_A$-point for $(P_+,P_-):=(\text{incenter},\text{orthocenter})$.

It's important to keep in mind that these $K$-points are determined "abstractly", mixing and matching cevian-points $D\pm$, $E_\pm$, $F_\pm$ based on their subscripts (that is to say, their originating concurrency points $P+$ and $P_-$). Properties such as the orderings of the points along the edges of $\triangle ABC$ are not a consideration, so we want to be careful not to say that (using OP's original image for reference) the side-lines of $\triangle JKL$ were chosen to be closest to the vertices of $\triangle ABC$. (After all, such a basis for construction would not yield consistent results under changes to the shape of the triangle.)

In a comment to the question, OP links to a GeoGebra sketch that appears to demonstrate the collinearity of the circumcenter, the $K_A$-point for the centroid and circumcenter, and the $K_A$-point for the centroid and orthocenter. (Or, maybe different $K$ points are in play; it's hard to tell. :) We can verify this by noting that these barycentric coordinates

$$\begin{align} \text{circumcenter} &= (\sin 2A:\sin2B:\sin2C) \\ \text{centroid} &= (1:1:1) \\ \text{orthocenter} &= (\tan A:\tan B:\tan C) \end{align}$$ lead to $K_A$-points with these coordinates. $$\begin{align} K_A(\text{centroid},\text{circumcenter}) &= \left(1:\frac{\sin2B}{\sin C\cos(A-B)}:\frac{\sin2C}{\sin B\cos(A-C)}\right) \\ K_A(\text{centroid},\text{orthocenter}) &= \left(\frac1{2\cos A} : \frac{\sin B}{\sin C}:\frac{\sin C}{\sin B}\right) \end{align}$$ One can show that the determinant whose entries are the coordinates for the circumcenter and the two $K_A$-points vanishes, indicating collinearity. $\square$

Verifying other collinearities and such is left as an exercise to the reader.

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  • $\begingroup$ I just found another surprising conjecture from the same construction: Point Z that is constructed for centroid and orthocenter, Lemoine point and one of the vertices of a triangle ABC are always collinear: geogebra.org/m/zcmzfjfm $\endgroup$ – A Z Jul 8 at 8:22

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