Consider what has happened after a small interval of time, $\delta t$. The first particle will have travelled a distance of $v \delta t$ towards the second particle whereas the second particle has travelled
a distance of $v \delta t$ perpendicular to the line joining them.
The distance between them is the hypotenuse of a right-angled triangle with the other sides having lengths $a-v \delta t$ and $v \delta t$. Ignoring distances as small as $ (\delta t)^2$, the distance between the two particles is just $a-v \delta t$.
Therefore the distance between two particles will have decreased by $v \delta t$. This will be the same throughout the motion and so the particles will meet when $a=vt$, i.e. after time $\frac {a}{v}$.
In your solution, the relative velocity should just be $v$ along the line joining the particles since the second particle is moving at right angles to the first at every instant.
The issue in this interesting problem is that the acceleration vector does not have a single component, but two components. In particular, as shown below, as a result of this double component the acceleration vector is perpendicular to the velocity vector. It is known that the perpendicular directions of the acceleration and velocity vectors represent the typical situation occurring whenever velocity is a constant. This explains the apparent contradiction described in the OP.
As correctly noted, the scenario of this problem resembles that of a uniform circular motion, i.e. a type of motion where an object moves among a circular path with a constant speed. Differently from one-dimensional problems, where objects with constant velocity have zero acceleration, in 2D or 3D problems an object can have acceleration if it motion follows a curved trajectory. This is the case of uniform circular motion, in which a particle with constant velocity $v$ moving on a circular trajectory with radius $R$ is subjected to a centripetal acceleration with magnitude $v^2/R$, directed along the radial director towards the center of the circle. The role of centripetal acceleration is to change the direction of the velocity vector, so that the motion remains tangential to the path.
However, in the problem described by the OP, we are not dealing with a true uniform circular motion. The similarity comes from the fact that, at any point in the path of the particle, we rotate and scale the equilateral triangle to reproduce the initial one. Actually the three points do not travel a circular path, but rather a spiral. Therefore, another acceleration component must necessarily exist that transforms the circular path into a spiral.
To better illustrate this, it is easier to use the polar coordinate system. The fundamental components of this system are the unit radial vector $\hat {\textbf{r}}$ and the unit tangential vector $\hat{\boldsymbol{\theta}}$. In our case of spiral path, we have that the velocity vector results from a radial component (forming an angle of $5\pi/6$ with $\textbf{v}$) and a tangential component (forming an angle of $\pi/3$ with $\textbf{v}$). Here is a picture:
Assuming that, for the radial vector, positive values are directed externally, the velocity vector $\textbf{v}=v$ is given by
$$\textbf{v}=v \cdot \cos\left(\frac{5\pi}{6}\right) \hat {\textbf{r}}+v \cdot \cos\left(\frac{\pi}{3}\right) \hat{\boldsymbol{\theta}}\\
= - \frac{v\,\sqrt{3}}{2} \hat {\textbf{r}}+ \frac{v}{2} \hat{\boldsymbol{\theta}} $$
Note that, using the standard dot notation for time derivatives, the coefficients of the last equation satisfy the relations $-v\sqrt{3}/2=\dot{r}$ and $v/2=r\,\dot{\theta}$. This last relation can also be written as $\dot{\theta}=v/(2r)$, and will be used in the next steps.
The acceleration vector is obtained by differentiating the velocity equation:
$$\textbf{a}= - \frac{v\, \sqrt{3}}{2}\, \dot {\hat{\textbf{r}}}+ \frac{v}{2} \dot{\hat{\boldsymbol{\theta}}} $$
Since it is known that the derivative of the radial and tangential vector can be expressed as $\dot {\hat{\textbf{r}}}=\dot{\theta} \hat{\boldsymbol{\theta}}$ and as $\dot {\hat{\boldsymbol{\theta}}}=-\dot{\theta} \hat{\boldsymbol{r}}$, respectively, we get
$$\textbf{a}= - \frac{v}{2} \dot{\theta} \hat{\boldsymbol{r}} - \frac{v\, \sqrt{3}}{2}\, \dot{\theta} \hat{\boldsymbol{\theta}} $$
and substituting $\dot{\theta}=v/(2r)$ we finally obtain
$$\textbf{a}= - \frac{v^2}{4r} \hat{\boldsymbol{r}} - \frac{v^2\, \sqrt{3}}{4r}\, \hat{\boldsymbol{\theta}} $$
The figure below shows the acceleration vector, whose magnitude is $v^2/(2r)$:
It is also clear that, if we consider two of the three particles and try to describe the motion of one of them from the others' point of view, the resulting vectors are still perpendicular.
In conclusion, the scenario of the OP is characterized by moving points whose acceleration vectors are perpendicular to the corresponding velocity vectors. As stated at the beginning of this answer, this is a classical situation where velocity is constant, and this explains why, according to any of the three particles and from its frame, the other two are in uniform motion. Just to provide a very intuitive and simplified scenario that illustrates well the situation of a constant speed with velocity vector perpendicular to the acceleration vector, we can think of an observer stationary at the center of an Archimedean spiral and a second observer who travels the spiral at a constant speed, so that the distance between them decreases linearly. From the point of view of the stationary observer, the running observer has a constant speed and the distance between them decreases uniformly, although the running observer has not zero acceleration.
Best Answer
Because the particles are situated on an n-sided regular polygon at $t=0$ they admit a radial symmetry at an angle of $\alpha = \frac{2\pi}{n}$. The velocity vectors preserve this symmetry, so the particles will stay on an n-sided regular polygon.
Therefore it suffices to consider one point $(x, y)$, or in the complex plane $z=x+yi$, if we take $0$ as the center of the polygon, the next point is at $e^{-\alpha i} z$. So the differential equation to figure out the path of $z$ is $$\frac{dz}{dt} = z - e^{-\alpha i} z = (1 - e^{-\alpha i}) z.$$
We won't care about the velocity of the point for now, because this differential equation is easy to solve. The solutions are $$z(t) = C e^{(1-e^{-\alpha i})t}$$ for some $C$ in $\mathbb{C}$ which is the equation for a spiral. We can rewrite this by noticing $e^{-\alpha i} = \cos(\alpha) - i\sin(\alpha)$ so that $$z(t) = C e^{(1-\cos(\alpha))t}(e^{i\sin(\alpha)t}).$$ In polar coordinates $$r(\theta) = R e^{\frac{1-\cos(\alpha)}{\sin(\alpha)}\theta}.$$ Let's introduce the variable $\beta = \frac{1-\cos(\alpha)}{\sin(\alpha)} = \tan(\frac{\alpha}{2})$, then we see that the trajectory of the point is a classic logarithmic spiral $r(\theta) = R e^{\beta \theta}$.
The arc length parametrization of this function is known (check Wikipedia for a full explanation). This also explains the way to calculate the arc length to the origin, it requires the following calculation.
We know that $\beta = \tan(\frac{\alpha}{2}) = \tan(\frac{\pi}{n})$ so that the formula for the arc length between a point on the curve $(r, \theta)$ and the origin is equal to $\frac{r}{\sin(\frac{\pi}{n})}$. Meaning that the time it takes to reach the origin from a point at distance $r$ is equal to $\frac{r}{v\sin(\frac{\pi}{n})}$
Some values of $\frac{1}{\sin(\frac{\pi}{n})}$ for specific values of $n$.