# median bisector?

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?

• 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? – Blue Jul 7 at 10:32
• yes, that is how point M is constructed. – A Z Jul 7 at 10:43
• 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 ... – Blue Jul 7 at 10:46
• geogebra.org/m/ar49cc3g This is an updated sketch. I corrected the error concerning point M supposedly lying on the line TYZ. – A Z Jul 7 at 13:04
• @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.) – 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:

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.

• 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 – A Z Jul 8 at 8:22