If you think of the three points as the vertices of a triangle, then the next iteration replaces the points with the center points of the edges. The average is the centroid of the triangle, which is what the sequence converges to. The triangles remain similar to the original, so they do not approach equilateral triangles, but instead retain their original shape. They just keep getting smaller. Each iteration shrinks the triangle and rotates it 180 degrees about the centroid. Tom Henry Baker writes:
Suppose we have a set of 3 distinct numbers, {a,b,c}.
Then averaging them in pairs preserves distinctness: {(a+b)/2,(b+c)/2,(c+a)/2}.
Proof: assume not, e.g., (a+b)/2=(b+c)/2 => a=c. Contradiction. QED.
So we can go backwards:
{a,b,c} => {a+b-c,b+c-a,c+a-b}
So, {(a+b)/2,(b+c)/2,(c+a)/2} => {(a+b+b+c-c-a)/2=b,(b+c+c+a-a-b)/2=c,(c+a+a+b-b-c)/2=a}
Note also that 'averaging' preserves the sum of the triple (hence its average): (a+b)/2+(b+c)/2+(c+a)/2=a+b+c. Thus, 'going forward' and 'going backwards' preserves this sum.
OK, so now let's iterate this averaging process N times.
I contend that this process makes the numbers more and more 'average'; i.e., all three converge towards the same number: (a+b+c)/3.
Yet, we can still recover our original 3 numbers by running the process backwards N times.
Even better, we can go backwards *further* than our original 3 numbers.
Curiously, if we start with {-1,0,1}, we get {-2^k,0,2^k}, for all k.
Note now that a,b,c need not be 'numbers'; they could be vectors of numbers, so we can play this game in higher dimensions. E.g., we can choose triples of (x,y) pairs to represent triangles, so does iterating them converge towards equilateral triangles?