# 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 $t=\displaystyle\frac{1}{3-2n}.$

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