Motion of Points around a Triangle [closed]

I came across this problem in the book "Problems in General Physics by IE Irodov"-

Three points are located at the vertices of an equilateral triangle whose side equals s. They all start moving simultaneously with velocity v constant in modulus, with the first point heading continually for the second, the second for the third, and the third for the first. How soon will the points converge?

The problem is famously solved in the reference frame of a particle (say $$A$$). According to particle $$A$$, particle $$B$$ is approaching it with a constant relative speed of $$(v\cos(\pi/3)+v)=3v/2,$$ and since the initial side length of the triangle was "$$s$$", time taken by them to meet will be $$t=2s/3v$$.

But from Ground Frame, it is clear that the particles are accelerating (since particles follow the triangular spiral shown in the figure below).

And, since the whole system should be symmetric around the triangle, their acceleration vectors should be symmetric as well. So, how are we changing reference frames (from ground frame to particle $$A$$'s frame) without taking into account the acceleration vectors? For their acceleration vectors to cancel in Frame of Particle $$A$$, their magnitude and direction should be equal. But their acceleration vectors cannot possibly be unidirectional, since that will not be symmetric. This implies that particle $$B$$ should be accelerating with respect to particle $$A$$'s frame.

What I have got so far:

At any instant of time, the particles are at the vertices of an equilateral triangle, and instantaneously, any particle is in a circular motion about the centroid of the triangle with the radius of the circle decreasing with time. Thus, the acceleration vector of a particle at any instant in time will be directed towards the centroid of the triangle.

From the figure above, we get that the magnitude of acceleration vector at any time is $$a=\frac{v_{tangential}^2}{r}=\frac{(v\sin(\pi/6))^2}{R-v\cos(\pi/6)t}$$ where R is the initial distance of a vertex from the centroid of the triangle. Thus, at any instant in time, according to particle $$A$$, particle $$B$$ should have an acceleration of $$|\overrightarrow{a_B}-\overrightarrow{a_A}|=2a\cos(\pi/6)=\sqrt3a$$.

So, why according to particle $$A$$, particle $$B$$ is in uniform motion? What happens to their acceleration vectors while frames are changed? I might be missing a very obvious fact here, and if that is the case, please let me know.

• @ACuriousMind I'm not active on this stackexchange, and am overall not very invested, but I must say I find this closing ridiculous. The question is tagged kinematics, and it clearly is about kinematics. The situation is certainly physical - you, I, and a friend could implement a setting in a garden that is reasonably modeled by the situation of this question. The OP is asking for clarifications about the analysis of the situation in a certain frame. Notice that none of the answers so far actually provide this (instead they supply alternate methods). Sep 29 '20 at 19:35
• The question is also clearly posed, and shows some effort on the part of the OP. In fact the only thing I see wrong is the Newtonian mechanics tag, but the correct action there would be to remove the tag, not to close the question. Sep 29 '20 at 19:35
• @ACuriousMind Mate, you need to justify your assertions. If not to me, then to your community which upvoted the question, and provided answers for it before they were removed. I came up with a 'more mechanical' development of the above question, which is long, and this is precisely the point of this formulation IMO - it abstracts the salient features of a setting concisely (the construction I had in mind was that of an aperture closing/opening linkage). I would also point out that mathematics in a physical context is on topic for this stackexchange, and passing it off to MSE is wrong. Sep 29 '20 at 20:35
• Also I ill appreciate the strawman - 'of their own volition' is very different from the prescribed set of motion in the question, which three people in a garden can very much approximate. Or three robots, if you like. In any case, I find that this is eating too much time and I don't intend to follow up, but I do hope that you'll rethink both your approach to answering issues that folk raise in general, and in particular to this question. Sep 29 '20 at 20:38
• @ACuriousMind In the question, it's stated: I came across this problem in the book "Problems in General Physics by IE Irodov". Why do you think it's a non-physical question? I think it's a great question! Sep 30 '20 at 16:42