Point charges symmetrically spreading out The Problem
There are $3$ positively charged particles fixed in a frictionless horizontal plane, positioned in the vertices of a triangle.
The $i$-th particle has mass $m_i$ and charge $Q_i$.
When they are free to move, their positions always form a triangle that is similar to the first triangle, such as their corresponding sides are always parallel. [So they can't rotate, just spread out]
Determine the largest angle of the triangle, if the charge/mass ratio of the particles is given by:$$\dfrac{Q_1}{m_1}:\dfrac{Q_2}{m_2}:\dfrac{Q_3}{m_3}=1:2:3$$

My Attempt
I tried to approach it with vectors, centering a cartesian referential in the centroid of the triangle. If the position of each particle $i$ is $\vec{r_i}$, then:
$$\begin{cases}\displaystyle\vec{r_{21}}+\vec{r_{23}}+\vec{r_{31}}=0\\\vec{a_1}=\frac{KQ_1}{M_1}\left(\frac{Q_2}{r^3_{21}}\vec{r}_{21}+\frac{Q_3}{r^3_{31}}\vec{r}_{31}\right)\\\vec{a_2}=\frac{2KQ_1}{M_1}\left(\frac{Q_1}{r^3_{12}}\vec{r}_{12}+\frac{Q_3}{r^3_{32}}\vec{r}_{32}\right)\\\vec{a_3}=\frac{3KQ_1}{M_1}\left(\frac{Q_1}{r^3_{13}}\vec{r}_{13}+\frac{Q_2}{r^3_{23}}\vec{r}_{23}\right)\\\vec{a_{1}}+\vec{a_{2}}+\vec{a_{3}}=0\end{cases}$$
There must be an elegant solution to this problem... What am I missing here? Is it possible to represent symmetry here without directly operating the position vectors?
 A: Building up from @Dan's comment, the problem seems friendly enough that you only seem to need the equations of motion at time $0$ to find a geometry that follows the desired rules. In reality, things turned out to be trickier.
First, we start by setting the centroid as the origin of coordinates. Then we place charge $1$ at $\mathbf{r_1}\left( t = 0\right) = (1, 0)$, charge $2$ at $\mathbf{r_2}\left( t = 0\right) = (\cos(\phi),\sin(\phi))$ and charge $3$ at $\mathbf{r_3}\left( t = 0\right) = - \mathbf{r_1}\left( t = 0\right) - \mathbf{r_2}\left( t = 0\right) $.
For the condition established in the problem to be satisfied, the net force acting on charge $1$, $\mathbf{f_1}$, must remain horizontal at all times. In fact, all net forces must remain parallel to the corresponding position vectors: $\mathbf{f_i} \parallel \mathbf{r_i}$. In particular this is true at $t=0$. The equation $0 = (0, 1) \cdot \mathbf{f_1}\left( t = 0\right)$ happens to be enough to give you a value for $\phi$, which, in this setup, fully determines the geometry of the system.
After replacing this value and letting the system evolve, it turns out that this is not a general enough ansatz to have a triangle that genuinely remains symmetrical at all times. The initial position of charge $2$ should be less constrained that in my previous ansatz, so we go instead for
\begin{eqnarray}
\mathbf{r_1}\left( t = 0\right) &=& (1, 0) \, ,
\\
\mathbf{r_2}\left( t = 0\right) &=& l(\cos(\phi),\sin(\phi)) \, ,
\\
\mathbf{r_3}\left( t = 0\right) &=& - \mathbf{r_1}\left( t = 0\right) - \mathbf{r_2}\left( t = 0\right) \, ,
\end{eqnarray}
with $l$ some positive number. We exhausted the degree of freedom associated with scaling invariance by setting the norm of $\mathbf{r_1}\left( t = 0\right)$ equal to $1$, and we exhausted rotational invariance by fixing its orientation along the $x$ axis. So setting the norm of $\mathbf{r_2}\left( t = 0\right)$ to $1$ was a mistake in that first ansatz.
With the more general ansatz we are forced to use an additional equation, so that we can solve for both $\phi$ and $l$. This can come, for instance, from $\mathbf{f_2}\left( t = 0\right)$ having no component orthogonal to $\mathbf{r_2}\left( t = 0\right)$. These equations are not particularly nice, but Mathematica is able to provide a solution:
\begin{eqnarray}
\phi &\approx& 1.802119183 \, \, \mathrm{rad} \approx 103.2538234°
\\
&\mathrm{and}&
\\
l &\approx& 1.381620315 \, .
\end{eqnarray}
Notice that $\phi$ is not an angle of the triangle, but the solution allows you to find the largest triangle angle which happens to be
\begin{equation}
\arccos \left[ \frac{(\mathbf{r_2} - \mathbf{r_1}) \cdot (\mathbf{r_3} - \mathbf{r_1})}{|\mathbf{r_2} - \mathbf{r_1}| |\mathbf{r_3} - \mathbf{r_1}|} \right] \approx 1.470038873 \, \, \mathrm{rad} \approx 84.22702317° \, .
\end{equation}
A: A simpler approach is to set the magnitudes of the mutual accelerations proportional to the separations. This is a necessary condition and seems to also be sufficient.
i.e. $\space a_{12}/r_{12} = a_{23}/r_{23} = a_{31}/r_{31}$
taking $a_{ij} = \frac{KQ_iQ_j/r_{ij}^2}{M_{ij}}$ where $M_{ij}=\frac{M_iM_j}{(M_i+M_j)}$ is the reduced mass
giving $a_{ij}/r_{ij} = K(Q_i/M_i)(Q_j/M_j)(M_i+M_j)/r_{ij}^3$
Substituting the $1:2:3$ ratios for $Q_i/M_i$ gives
$1\cdot2\cdot(M_1+M_2)/r_{12}^3 = 2\cdot3\cdot(M_2+M_3)/r_{23}^3 = 3\cdot1\cdot(M_3+M_1)/r_{31}^3$
This immediately gives you the ratio of the three sides and you can use the cosine formula to get the angles.  But the answer is not independent of the masses!
Assuming the masses are all equal, we get the sides of the triangle in the ratio $2^{(1/3)}:6^{(1/3)}:3^{(1/3)}$ which gives the same answer of $84.2270^{\circ}$ found by @secavara
