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

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

M is the intersection of 3 Cevians in the triangle ABC

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 $ABC$?

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    $\begingroup$ 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. $\endgroup$
    – Fedor Petrov
    Aug 19, 2020 at 7:26
  • $\begingroup$ 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. $\endgroup$
    – A Z
    Aug 19, 2020 at 7:32
  • $\begingroup$ I checked that the Triangle Center X(883) satisfies the condition 𝑆=𝑥𝑦𝑧/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 $\endgroup$
    – A Z
    Aug 19, 2020 at 9:58
  • $\begingroup$ "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. $\endgroup$
    – A Z
    Aug 19, 2020 at 10:23
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    $\begingroup$ 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. $\endgroup$
    – LSpice
    Aug 22, 2020 at 13:07

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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.

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GeoGebra found X(7) X(8) X(506) X(507) and some more if you let outlying intersections of cevians.

PS: a bug was found in GeoGebra.
I hope it is fixed soon. [Edit: now fixed]

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    $\begingroup$ Geogebra seems to show that X(7) is the same point as X(506) and X(507)... it must be a bug! It shows X(883), X(885) correctly though. $\endgroup$
    – A Z
    Aug 21, 2020 at 3:14
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    $\begingroup$ You can use the Contact Us form to have your accounts merged, so you can freely edit your post. $\endgroup$
    – Glorfindel
    Aug 22, 2020 at 11:48

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