# Property of triangle centers

$$M$$ is the intersection of 3 cevians in the triangle $$ABC$$.

$$AB_1 = x,\quad CA_1 = y,\quad BC_1= z.$$

It can be easily proven that for both Nagel and Gergonne points the following equation is true: $$S = xyz / r,$$ where $$S$$ is the area of the triangle $$ABC$$ and $$r$$ is the radius of the inscribed circle.

I wonder what other triangle centers might possibly have the same property and what is the geometric place for them?

Also, please note that for the case where point $$M$$ is the centroid the formula looks as follows: $$S = 2xyz/R$$, where $$R$$ is the radius of the circumcircle. Substitution $$x = b/2$$, $$y = a/2$$, $$z = c/2$$ brings it back to the classic $$S = abc/4R$$. Perhaps, some other triangle centers might exist, so that this equation $$S = 2xyz/R$$ holds true for them as well. I wonder in what particular relation these hypothetical points might be to the centroid of $$ABC$$?

• Fixed $xyz$ defines a cubic curve. There are some known triangle-related cubics, possibly the cubics $xyz=Sr$ and $xyz=SR/2$ were also studied. – Fedor Petrov Aug 19 '20 at 7:26
• So it must be a cubic that is passing through Nagel and Gergonne points and some other known triangle centers are probably lying on it as well. – A Z Aug 19 '20 at 7:32
• I checked that the Triangle Center X(883) satisfies the condition đť‘†=đť‘Ąđť‘¦đť‘§/đť‘ź, so that its isotomic conjugate X(885) must also satisfy the same condition and the curve in question is inevitably Tucker-Gergonne-Nagel cubic: bernard-gibert.pagesperso-orange.fr/Exemples/k013.html – A Z Aug 19 '20 at 9:58
• "It is the locus of point M such that the cevian triangles of X(7) and M have the same area." This interpretation is a bit different from mine though. I wonder whether it is trivial or not that both geometric interpretations of Tucker-Gergonne-Nagel cubic are the same. – A Z Aug 19 '20 at 10:23
• Please use TeX like $S = x y z/r$, not Markdown fakery like S = xyz/r, which reads badly (e.g., compare $a$ $a$ vs. a *a*). I have edited accordingly. – LSpice Aug 22 '20 at 13:07

This is just a coda to the above comments but too long for a comment. If $$M$$ has barycentric coordinates $$(\lambda,\mu,\nu)$$ (not necessarily positive and normalised so that $$\lambda+\mu+\nu=1$$), then both conditions reduce to a cubic equation of the form $$\frac{\lambda\mu\nu}{(\mu+\nu)(\nu+\lambda)(\lambda+\mu)}$$ is a constant which depends on the (shape of the) triangle and can easily be computed explicitly.
In order to verify if a given centre (with centre function $$f$$ from the Encyclopedia of Triangle Centers, normalised to be homogeneous with $$f(a,b,c)+f(b,a,c)+f(c,a,b)=1$$), it should be easy to write a small programme, say in Mathematica, to check this on the spot.