Points in space

Question

Let $P =\left \{ A_{1}, A_{2} \cdots A_{n} : A \in \mathbb{R}^{3}\right \}$ (where $A_{0}=A_n, A_{1}=A_{n+1})$. By middling of a point $A_{i+1}$ we mean setting it's value to a rectangular projection of that point on the line $A_iA_{i+2}$. We middle every point of $P$ and start the process again ... ad infinitum. Decide whether all points of $P$ will be collinear after a finite amount of steps.

I came with this myself, please be kind.

Answer 1

Convergence looks to be (it's not a mathematical proof!)

• towards a point when points $A_i$ are smoothly spaced (i.e. with distances $A_kA_{k+1}$ of the same order of magnitude (Fig. 1).

• towards a line, indeed, as you have assumed it, when there is large gap between two points. (Fig. 2).

Fig. 1: Viviani's curve, in black (intersection of a sphere and a certain cylinder) shrinks into almost a dot after 300 iterations, with graduated colors, the last ones being more reddish. Only one out of ten of these curves have been plotted.

Fig; 2: The case of a helix with three periods (in black, with connected endpoints). Here, after 500 steps, this still intermediate stage (in red) in the form of a bow with its string displays a trend to a limit which would be approximately the line segment connecting the initial endpoints.

Matlab program:

clear all;close all;hold on;axis equal;
% t=0:pi/20:6*pi;A=[cos(t);sin(t);t]; % helix
t=-2*pi:pi/20:2*pi; A=[1+cos(t);sin(t);2*sin(t/2)]; % Viviani's curve
n=size(A,2);
pf=300; % number of steps
A=[A,A(:,1:2)]; % enables cyclicity
plot3(A(1,:),A(2,:),A(3,:),'k','linewidth',2); % initial  "curve"
for p=1:pf;
    if mod(p,10)==0
        plot3(A(1,:),A(2,:),A(3,:),'color',[p/pf,0,1-p/pf]);
    end;
    for q=1:n 
        v=A(:,q+2)-A(:,q);v=v/norm(v); % unit vector of [A_k, A_k+2] 
        w=A(:,q+1)-A(:,q);
        B(:,q)=A(:,q)+(w'*v)*v; % orth. projection
    end;
    A=[B(:,1:n),B(:,1:2)];
end;