Barycenter of Cevian Triangle

Problem

Barycenter of Cevian Triangle

Proof 1

Assume point $P\,$ has barycentric coordinates $P=x:y:z.\,$ This means that the points $D\,$ on $BC,\,$ $E\,$ on $AC,\,$ and $F\,$ on $AB\,$ are defined as $D=0:y:z,\,$ $E=x:0,z,\,$ $F=x:y:0.\,$ If normalized, they appear as

$\displaystyle D=\left(0,\frac{y}{y+z},\frac{z}{y+z}\right),\,E=\left(\frac{x}{x+z},0,\frac{z}{x+z}\right),\,F=\left(\frac{x}{x+y},\frac{y}{x+y},0\right).$

The barycenters of triangles $ABC\,$ and $DEF\,$ are, respectively,

$\displaystyle\begin{align} G &= \left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right)\\ &= \left(\frac{1}{3}\left(\frac{x}{x+z}+\frac{x}{x+y}\right),\frac{1}{3}\left(\frac{y}{x+y}+\frac{y}{y+z}\right),\frac{1}{3}\left(\frac{z}{x+z}+\frac{z}{y+z}\right)\right), \end{align}$

yielding three equations:

$\displaystyle\frac{x}{x+z}+\frac{x}{x+y}=1,\,\frac{y}{x+y}+\frac{y}{y+z}=1,\,\frac{z}{x+z}+\frac{z}{y+z}=1.$

The first one reduces to $x^2=yz,\,$ the other two to $y^2=xz\,$ and $z^2=xy,\,$ respectively. From these, $x=y=z,\,$ meaning that $P\,$ is the barycenter of $\Delta ABC.$

Proof 2

Taking any point $O\,$ as the origin, we have $OD=\lambda OB+(1-\lambda)OC,\,$ $OE=\mu OC+(1-\mu)OA,\,$ $OF=\nu OA+(1-\nu)OB,\,$ for some $\lambda,\,$, $\mu,\,$ $\nu,$ between $0\,$ and $1.\;$ If in addition the two triangles $ABC\,$ and $DEF\,$ have the same centroid, we must have $OA+OB+OC = OD+OE+OF,\;$ whence $(\nu-\mu)OA+(\lambda-\nu)OB+(\mu-\lambda)OC = 0.\;$ This is true even if $O\,$ is off the plane of $ABC,\,$ in which case $OA,\,$ $OB\,$ and $OC\,$ are linearly independent; so $\lambda=\mu=\nu.\,$ But by Ceva's theorem, $\displaystyle\frac{\lambda}{1-\lambda}\cdot\frac{\mu}{1-\mu}\cdot\frac{\nu}{1-\nu} = 1.\,$ These are only possible if $\displaystyle\lambda=\mu=\nu=\frac{1}{2}.\,$ i.e. if $D,\,$ $E,\,$ $F\,$ are the midpoints of the sides and the three Cevians are the medians of the triangle.

Proof 3

Here is a picture as locus. We calculate the cross product of $G\,$ and the centroid of cevian triangle of $P,\,$ then we get two loci whose intersections are $G\,$ and the vertices of the anticomplementary triangle.

loci of the cross-product of two barycenters

Note that only $G\,$ has a cevian triangle with ordinary (not infinite) points.

Corollaries

For all cevian triangles (except for the medial one), if their centroid coincides with that of $\Delta ABC,\,$ the latter is necessarily equilateral.

  • Intouch triangle

    The intouch triangle is the cevian triangle of Gergonne's point, $Ge=(s-b)(s-c):(s-c)(s-a):(s-a)(s-b).\,$ It coincides with $G(1,1,1)\,$ only if, say, $(s-b)(s-c)=(s-c)(s-a),\,$ or $a=b,\,$ etc., making $\Delta ABC\,$ equilateral.

    This supplies a third proof for Marian Dinca's criterion of equilaterality of $\Delta ABC.$

  • Incentral triangle

    The incentral triangle is the cevian triangle of the incenter $I=a:b:c.\,$ It is immediate that it may coincide with the barycenter $G=1:1:1\,$ only if $\Delta ABC\,$ is equilateral.

  • Extouch triangle

    The extouch triangle is the cevian triangle of the Nagel point $I=s-a:s-b:s-c.\,$ It is immediate that it may coincide with the barycenter $G=1:1:1\,$ only if $\Delta ABC\,$ is equilateral.

  • Orthic triangle

    The orthic triangle is the cevian triangle of the orthocenter $H=(a^2+b^2-c^2)(c^2+a^2-b^2):(b^2+c^2-a^2)(a^2+b^2-c^2): \ldots.\;$ It is immediate that it may coincide with the barycenter $G=1:1:1\,$ only if $\Delta ABC\,$ is equilateral.

  • Cevian triangle of the symmedian point

    The symmedian point $K\,$ has the barycentric coordinates $K=a^2:b^2:c^2,\,$ with the obvious conclusion that it coincides with the barycenter $G=1:1:1\,$ only if $\Delta ABC\,$ is equilateral.

  • Cevian triangle of the circumcenter

    The circumcenter $O\,$ has the barycentric coordinates $K=a^2(b^2+c^2-a^2): b^2(c^2+a^2-b^2):c^2(a^2+b^2-c^2).\;$ It's a little more involved than for other points but is not too difficult to prove that it coincides with the barycenter $G=1:1:1\,$ only if $\Delta ABC\,$ is equilateral.

Acknowledgment

The statement has been inspired by Marian Dinca's criterion of equilaterality of $\Delta ABC.$ Proof 2 is by Angelos Tsirimokos; Proof 3 is by Francisco Javier García Capitán.

 

Barycenter and Barycentric Coordinates

  1. 3D Quadrilateral - a Coffin Problem
  2. Barycentric Coordinates
  3. Barycentric Coordinates: a Tool
  4. Barycentric Coordinates and Geometric Probability
  5. Ceva's Theorem
  6. Determinants, Area, and Barycentric Coordinates
  7. Maxwell Theorem via the Center of Gravity
  8. Bimedians in a Quadrilateral
  9. Simultaneous Generalization of the Theorems of Ceva and Menelaus
  10. Three glasses puzzle
  11. Van Obel Theorem and Barycentric Coordinates
  12. 1961 IMO, Problem 4. An exercise in barycentric coordinates
  13. Centroids in Polygon
  14. Center of Gravity and Motion of Material Points
  15. Isotomic Reciprocity
  16. An Affine Property of Barycenter
  17. Problem in Direct Similarity
  18. Circles in Barycentric Coordinates
  19. Barycenter of Cevian Triangle
  20. Concurrent Chords in a Circle, Equally Inclined

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