Iterativ centroid-triangle sequence Three points that are not on a straight line - $A_1$, $A_2$ and $A_3$ - are given; for 
$n = 4, 5, 6, ...$
$A_n$ is  the centroid of the triangle $A_{n−3},A_{n−2},A_{n−1}$.
Q:Is there a point that all triangles have in common (it certainly looks so!) and if yes then is there a way to define its location?
Here is a Mathematica application for clarification if needed:

Dreieck[{{x1_, y1_}, {x2_, y2_}, {x3_, y3_}}] := {{x2, y2}, {x3, 
      y3}, {(x1 + x2 + x3)/3, (y1 + y2 + y3)/3}};
  Zeichner[{{x1_, y1_}, {x2_, y2_}, {x3_, y3_}}] := 
    Graphics[{Hue[x1/x2 + x3, 1, 0.5], 
      Polygon[{{x1, y1}, {x2, y2}, {x3, y3}}]}];
  Manipulate[
   Show[Zeichner /@ 
     NestList[Dreieck, {pt1, pt2, pt3}, n]], {{pt1, {0, 0}}, {0, 
     0}, {10, 10}}, {{pt2, {4, 3}}, {0, 0}, {10, 
     10}}, {{pt3, {7, 1}}, {0, 0}, {10, 10}}, {n, 0, 20, 1}]

 A: The answer is yes and yes, the common point can be calculated explicitly.  See also this related question on MO.
The operation you are iterating is a linear map, so we can work out everything rather easily.  Let $z_n=x_n+iy_n$ be the coordinates of the $n$th point in the sequence, written as a complex number.  Then let $v_n$ be the 3-dimensional complex vector whose coordinates are $(z_n,z_{n+1},z_{n+2})$. (Alternatively, you could carry this all out with real, 6-dimensional vectors involving $x_n,y_n,x_{n+1},y_{n+1},x_{n+2},y_{n+2}$).
Note that $v_n$ gets mapped to $v_{n+1}=(z_{n+1},z_{n+2},(z_n+z_{n+1}+z_{n+2})/3)$, or in matrix form:
$$v_{n+1}=\begin{pmatrix}
0 & 1 & 0 \\
0 & 0 & 1 \\
\frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\
\end{pmatrix}v_n.$$
Let $M$ be the matrix there.  Then in the limit, the behavior will be controlled by the (right) eigenvectors of $M$, call them $e_1,e_2,e_3$ and let their corresponding eigenvalues be $\lambda_1,\lambda_2,\lambda_3$.  This is because if we decompose $v_1$ into the eigenvectors of $M$:
$$ v_1=a_1e_1+a_2e_2+a_3e_3$$
Then 
$$v_{2}=a_1 M e_1 + a_2 M e_2 + a_3 M e_3 $$
$$v_{2}=a_1\lambda_1 e_1 + a_2\lambda_2 e_2 + a_3\lambda_3 e_3$$
Continuing, to the $n$th iteration:
$$v_{n}=a_1\lambda_1^{n-1}e_1 + a_2\lambda_2^{n-1}e_2+a_3\lambda_3^{n-1}e_3.$$
The eigenvalues of $M$ are $1$, $\frac{-1\pm i\sqrt{2}}{3}$ (the absolute values of these two are both equal to $1/\sqrt{3}$) and their corresponding (non-normalized) eigenvectors are $(1,1,1)$ and $(-1\pm 2i\sqrt{2},-1\mp 2i\sqrt{2},1)$.  
Note that the absolute values of $\lambda_2,\lambda_3$ are smaller than 1, so under iteration, the coefficients $\lambda_2^n$ and $\lambda_3^n$ will both approach zero.  As $\lambda_1=1$, $v_n$ thus approaches $a_1 e_1$.  And since $e_1$ is $(1,1,1)$, this means that all three points are converging to the same point, whose coordinates are given by the complex coefficient $a_1$.  This coefficient $a_1$ can be calculated via a change of basis matrix as follows:
If $E$ is the matrix whose columns are the right eigenvectors of $M$, then $E^{-1}\cdot v_1$ is a vector whose components are $a_1,a_2,a_3$.  Here's $E^{-1}$:
$$E^{-1}=\begin{pmatrix}
\frac{1}{6} & \frac{1}{3} & \frac{1}{2} \\
\frac{-1-i\sqrt{2}}{12} & \frac{-2+i\sqrt{2}}{12}  & \frac{1}{4} \\
\frac{-1+i\sqrt{2}}{12} & \frac{-2-i\sqrt{2}}{12}  & \frac{1}{4} \\
\end{pmatrix}.$$
As we just want $a_1$, this means that we can just use the first row of $E^{-1}$: $a_1=\frac{1}{6}z_1+\frac{1}{3}z_2+\frac{1}{2}z_3$.  
Or, returning to real coordinates, the limit point is $\left(\frac{x_1}{6}+\frac{x_2}{3}+\frac{x_3}{2},\frac{y_1}{6}+\frac{y_2}{3}+\frac{y_3}{2}\right)$!
A: Any nested sequence of closed subsets of a compact topological space has nonempty intersection. Therefore, a common point exists. 
