For an arbitrary triangle A₁B₁C₁ - what is the nearest equilateral triangle A₂B₂C₂ which has the same centroid O₁ and known side length l. Where "nearest" means minimal |A₁A₂| + |B₁B₂| + |C₁C₂| (sum of orange line lengths)
Original formulation: Suppose we have a triangle consisting of 3 idealized springs - they can stretch infinitely long and always pull/push with linear force; and also there is no inertia - as soon as the forces cancel each other, any motion stops. At initial state the triangle's nodes are at A₀, B₀ and C₀ points. All springs are at neutral length, and so the the triangle is equilateral. Now, imagine we move each node to 3 arbitrary points (A₁, B₁ and C₁ with centroid O₁), then release them and let the springs do their job. The question is - at what points A₂, B₂ and C₂ will the triangle nodes settle?
What I've figured so far is that the triangle will come back to equilateral shape and will probably have the same centroid as in distorted state (O₁). So the question really boils down to finding the angle between lines O₁-A₁ and O₁-A₂.
Bonus question: What if I want to do the same trick with 6 springs forming a tetrahedron in 3D space?