As I previously explained, the vertices converge to a point (the centroid, i.e. (a+b+c)/3), but they do *not* converge to an equilateral triangle. The shape of the triangle is preserved by each step (i.e., the triangles are all similar to each other). Each step scales the triangle by 1/2 and rotates it by 180 degrees about the centroid. It is *not* reflected. Reversing a step is identical, except it is scaled by 2 rather than 1/2. Performing two consecutive steps scales by 1/4 with no rotation. The mappings of old vertices to corresponding new vertices are: a -> (b+c)/2 b -> (a+c)/2 c -> (a+b)/2 Consider the difference between the new a and the new b: (b+c)/2 - (a+c)/2 = (b-a)/2 In other words, the new difference is -1/2 times the old difference (a-b). The length is halved, and the vector is negated (i.e., it is rotated by 180 degrees). This also shows that the ratios of the edge lengths are preserved. For example, consider the ratio of a pair of edge lengths: |a-b| / |a-c| After one iteration, the new ratio is: |(b-a)/2| / |(c-a)/2| = |a-b| / |a-c| Since the edge length ratios never change, clearly the triangles retain their exact shape, and do not converge to equilateral (unless they were equilateral to begin with). You can easily see this by drawing a right triangle, and inside that triangle drawing the triangle formed by the midpoints of each edge (i.e., one iteration). The new triangle is clearly a right triangle, with the exact same shape as the original (but scaled by 1/2 and rotated by 180 degrees). So all triangles will be right triangles, and hence will not converge to an equilateral triangle. Tom Brad Klee writes:
The triangle inequality a+b>c transforms to 2b>0, and similar for cyclic permutation. If (a,b,c) are positive lengths, even if they do not satisfy the triangle inequality, they will after one iteration. Thereafter the lengths can be folded into a triangle (or it’s mirror image). Similarly, if (a,b,c) are positive angles, choose angular units where a+b+c=pi. All subsequent iterations preserve the sum, so each triple determines a unique triangle (again, up to reflection). By the limiting property, both sequences converge to equilateral.
Another possibility is:
a = (a+b+c)/3 b = [(a*b)^(1/2)+(b*c)^(1/2)+(c*d)^(1/2)]/3 c = (a*b*c)^(1/3)
Is this the best generalization of the agm to three (or more) variables? Does it even converge?
—Brad
On Sep 18, 2020, at 12:03 AM, Henry Baker <hbaker1@pipeline.com> wrote:
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?