Spring triangle problem

Question

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?

Not sure about your "zero inertia". The $m$ in $F=ma$ refers to inertial mass, so if that's $0$ and $F$ is non-zero, then guess what $a$'s gonna be. So kind of hard to figure out what happens. And if $m$ is non-zero, then the displaced-from-equilibrium $A_1,B_1,C_1$ is characterized by some potential energy, which is conserved, whereby the configuration oscillates but never comes to rest. — John Forkosh, Nov 05 '20 at 08:38
@JohnForkosh this is not physics but geometry problem - that is why there is no inertia. Imagine that the nodes move only as long as strings are pushing/pulling them. — Vilius Normantas, Nov 05 '20 at 08:44
But >>how<< do they move??? That's what you can't figure with $a\sim\infty$. When you say linear, that's typically $F=-k(x-x_0)$ for the spring, and I can only guess you want their "motion" linear in $x-x_0$. But is that "motion" now supposed to be acceleration or velocity? — John Forkosh, Nov 05 '20 at 08:50
Sorry, @JohnForkosh I confused you with that part about springs. Please see my updated formulation of the problem. — Vilius Normantas, Nov 05 '20 at 08:52
Okay, that sounds well-posed now, and sorry if I was being overly pedantic: my background is more physics, and I read/interpreted the question in that context. (P.S. also sorry that I don't have the answer:) — John Forkosh, Nov 05 '20 at 08:55
Well, maybe not quite well-posed until you define "nearest" in "nearest equilateral triangle" -- what function of $A_1,A_2; B_1,B_2; C_1,C_2$ do you want minimized? (I'd guess you were thinking that the physical springs in the original formulation implicitly answered that.) — John Forkosh, Nov 05 '20 at 09:06
Nearest means "with minimal distance between A1-A2, B1-B2 and C1-C2" — Vilius Normantas, Nov 05 '20 at 09:09
You mean minimal $(A_2-A_1)^2+(B_2-B_1)^2+(C_2-C_1)^2$ for all possible triangles with the same centroid? I'm not sure how to set up the "same centroid" constraint so that the minimization problem remains easily tractable. — John Forkosh, Nov 05 '20 at 09:15
Ok, fair enough. I'll try to make some drawing to illustrate what I'm trying to do here. — Vilius Normantas, Nov 05 '20 at 09:17
I don't think drawing will help (not me, at least). Assuming I guessed "minimal distance" right, then I can easily >>visualize<< the situation, but can't see how to >>algebraically<< express "same centroid" in the context of the resulting minimization problem. Once you have that, I'd imagine you can easily solve it yourself. — John Forkosh, Nov 05 '20 at 09:21
You should really define what does "nearest" mean. At best write down an expression which should be minimized or maximized. — user, Nov 05 '20 at 09:43
I updated the question with a drawing. But in the process of making it I started to suspect that there may be more that one "nearest" triangle.. So maybe my question is not that well defined after all. And even if there is one triangle - it may not be possible to find analytical solution for it.
Oh well.. And it seemed like such a nice little problem :) — Vilius Normantas, Nov 05 '20 at 10:57
Maybe some ideas how to find a quick and dirty way to pick a set of 3 points on the green circle which form a equilateral triangle and are reasonably near A1, B1, C1? — Vilius Normantas, Nov 05 '20 at 11:18
Okay, actually that drawing did help me. Once you pick one equilateral triangle vertex on the circle, the other two vertices are determined --- $120^o$ apart. So you only have a single degree of freedom. You could permute the $A_2,B_2,C_2$ labels any which way, but it's the same triangle regardless. And then your $\left|A_1A_2\right|$, etc, are well-defined with respect to that one degree of freedom, let's call it $\theta_0$. So you can now easily write an expression for $\mbox{distance}(\theta_0)$ in terms of that single variable, whereby minimizing it is trivial. — John Forkosh, Nov 05 '20 at 11:54
@JohnForkosh "Trivial" huh.. I think you are underestimating how rusty my math is :D — Vilius Normantas, Nov 05 '20 at 12:08
Can you figure out the $x,y$-coords of your $O_1$, given the three pairs of $x,y$-coords of your $A_1,B_1,C_1$? And now, given your triangle side $l$, can you figure out the radius $r$ of the circle necessary for the inscribed triangle to have sides of length $l$? Okay. Now, given a $\theta_0$, let's say measured counterclockwise from the $x$-axis (which is basically along the $O_1\to B_1$ line), can you figure out the three pairs of $x,y$-coords for $A_2,B_2,C_2$? Then your distance is just $\sum_{A,B,C}\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$, with the $x_1,y_1$'s given, and.... — John Forkosh, Nov 05 '20 at 12:33
...the $x_2,y_2$'s all functions of $\theta_0$. So differentiate with respect to $\theta_0$, set that expression equal to $0$, and finally solve that for $\theta_0$, giving you the corresponding $x_2,y_2$'s for your "nearest" triangle. — John Forkosh, Nov 05 '20 at 12:35

Spring triangle problem

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