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}]


2 Answers 2


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 $M$:

$$ v_1=a_1e_1+a_2e_2+a_3e_3$$


$$v_{2}=a_1 M e_1 + a_2 M e_2 + a_3 M e_3 $$ $$v_{2}=a_1\lambda_1 e_1 + a_2\lambda_2 e_2 + a_3\lambda_3 e_3$$

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 $\left(\frac{x_1}{6}+\frac{x_2}{3}+\frac{x_3}{2},\frac{y_1}{6}+\frac{y_2}{3}+\frac{y_3}{2}\right)$!

  • $\begingroup$ If I knew of a simpler way, I would've written that up instead... maybe someone else will come up with one though! $\endgroup$
    – j.c.
    Jul 15, 2014 at 22:07

Any nested sequence of closed subsets of a compact topological space has nonempty intersection. Therefore, a common point exists.


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