An Affine Property of Barycenter

The following problem has been posted by Emmanuel José García at the CutTheKnotMath facebook page:

Consider any tetrahedron $ABCD.$ Take an arbitrary point $P$ in the space. Now, reflect $P$ around the four centroids of the four triangular faces of the tetrahedron.

Then, the line segments joining the vertices with the symmetric image of $P$ corresponding to the opposite faces of the vertices are concurrent.

The problem appears to be of affine character and admits a slight generalization:

Given $n$ points $\{A_{1},\ldots,A_{n}\}$ in an affine space. To simplify notations, let's agree to denote a sum $\displaystyle\sum_{k=1}^{n}$ simply as $\sum$ and that with omitted $i-th$ term $\sum^{i}.$ Introduce the barycenter $G_i$ of $n-1$ points $A_{1},\ldots,A_{i-1},A_{i+1},\ldots,A_n:$ $G_{i}=\frac{1}{n-1}\sum^{i}A_{k}.$

Let $P_{i}$ stand for the reflection of $P$ in $G_i.$

Then the $n$ lines $A_{i}P_{i},$ $i=1,\ldots,n,$ are concurrent.

Proof

As a reflection, $P_{i}=2G_{i}-P.$ The line $A_{i}P_{i}$ is defined by the equation, say,

$\begin{align} l_{i}(t)&=tA_{i}+(1-t)P_{i}\\ &=tA_{i}+(1-t)(2G_{i}-P)\\ &=[tA_{i}+(1-t)2G_{i}]-(1-t)P. \end{align}$

Introduce $G=\sum A_{k},$ the barycenter of the whole collection of points. We know that the line joining $A_{i}$ to $G_{i}$ passes through $G.$ With this in mind, let's see if we can find $t$ such that $tA_{i}+(1-t)2G_{i}$ is independent of $i.$ Since $G=\frac{1}{n}\sum A_k$ and $G_{i}=\frac{1}{n-1}\sum^{i} A_k,$ we have that $nG=(n-1)G_{i}+A_{i}.$ By reverse engineering, let's set $\displaystyle t=\frac{2}{n+1},$ then

$\begin{align}\displaystyle l_{i}(\frac{2}{n+1})&=\frac{2}{n+1}A_{i}+\bigg(1-\frac{2}{n+1}\bigg)2G_{i}-(1-\frac{2}{n+1})P\\ &=\frac{2}{n+1}A_{i}+\bigg(\frac{n-1}{n+1}\bigg)2G_{i}-\frac{n-1}{n+1}P\\ &=\frac{2}{n+1}\bigg( A_{i}+ (n-1)G_{i}\bigg)-\frac{n-1}{n+1}P\\ &=\frac{2}{n+1}\bigg( A_{i}+ \sum\,^{i}A_{k}\bigg)-\frac{n-1}{n+1}P\\ &=\frac{2}{n+1}\sum A_{k}-\frac{n-1}{n+1}P\\ &=\frac{2n}{n+1}G-\frac{n-1}{n+1}P. \end{align}$

It follows that, for $\displaystyle t=\frac{2}{n+1},$ line $A_{i}P_{i}$ passes through $\displaystyle P'=\frac{2n}{n+1}G-\frac{n-1}{n+1}P,$ for all $i=1,2,\ldots,n.$

Extra

If, say, $Q_{i}=2P-A_{i},$ ($i=1,\ldots,n$) is the reflection of $A_i$ in $P$ then the lines $m_{i}(t)=tP_{i}+(1-t)Q_{i},$ joining $P_i$ and $Q_i,$ are also concurrent. Indeed,

$\begin{align} m_{i}(t) &= t(2G_{i}-P)+(1-t)(2P-A_{i})\\ &= [2tG_{i}-(1-t)A_{i}]+[(1-t)2-t]P \end{align}$

and the task is to find $t$ (if possible) such that $2t(n-1)=-(1-t).$ For this $t,$ point $m_{i}(t)$ will be common to all lines $m_{i}.$ This happens for $t=\displaystyle\frac{1}{3-2n}.$

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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