Geometry of 3 points getting closer to each other This is originally a physics problem but I am not interested in solving it, and my question has more to do with maths hence me being here and not on physics.stackexchange.

Three points are located at the vertices of an equilateral triangle whose side equals $a$. 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?

I made a diagram for the movement after a period of time $dt$

$(1)=(2)=(3)$ since the velocites are the same.
One method of solving this problem consists of saying that $a$ will shrink after $dt$ by a value of $vdt+v\cos{(60)}dt$ where $v\cos{(60)}$ is the velocity of one point towards the point that it's not heading towards.
How can we deduce this mathematically? It seems intuitive that if two bodies are heading towards each other with a given constant velocity then the decrement would be as stated, but, and especially trying to use the diagram, I can't seem to deduce it "with rigour".
Thank you for your time!
 A: The complicated version is to write coordinates for each point, so point $1$ would be $(x_1,y_1)$.  You can write differential equations for each point, so 
$$x_1'=v\frac {x_2-x_1}{\sqrt{ (x_2-x_1)^2+(y_2-y_1)^2}}$$
and similarly for the others, then solve the coupled differential equations.  This throws away the knowledge that the triangle stays equilateral, which I think is the "non-rigorous" part.  Of course it will come out to be that in the final solution.
A: Let $A(t)$, $B(t)$, $C(t)$ be your points (as functions of time $t$.
For convenience, take the origin to be the centroid of the triangle at $t=0$, so $A(0)+B(0)+C(0)=0$.  Then if $R$ is a rotation by $\pi/3$ in the appropriate direction,
$B(0) = R A(0)$, $C(0) = R B(0)$ and $A(0) = R C(0)$.
For some $v > 0$ the motion of your points is governed by the system of differential equations $$ \eqalign{\dfrac{dA}{dt}  &= \frac{v(B-A)}{\|B - A\|}\cr \frac{dB}{dt} &= \frac{v(C-B)}{\|C-B\|}\cr \dfrac{dC}{dt} &= \frac{v(A-C)}{\|A-C\|}\cr}$$
The right sides of these equations are locally Lipschitz functions of $(A,B,C)$ as long as 
$B-A$, $C-B$ and $A-C$ are nonzero.
Now note that $(RC, RA, RB)$ satisfy the same system as $(A,B,C$), and with the same initial conditions.  By the uniqueness theorem for systems of differential equations, we conclude that $A(t)=RC(t)$, $B(t) = RA(t)$ and $C(t)=RB(t)$ for all $t$ (until the three points collide). 
A: Modifying a bit your sketch
 
and splitting the speed into a
 - "height compressing" component ( red in the sketch), and
 - a "tangential speed" (blue), corresponding to a rotation   
then it is easy to express the process, specially using complex numbers in a Argand plane.   
And it comes out also clearly that, apart rotation,  the height is compressed at constant speed, which determines the time needed to collapse.
A: As the three points keep forming a centered equilateral triangle and the speed is in direction of the sides, the radial speed is 
$$v_\rho=\frac{\sqrt3}2v$$ i.e. a constant.
Hence the points meet after $$\frac{2\rho_0}{\sqrt3v}=\frac{2a}{3v}$$ seconds, as simple as that !

Trajectory:
The derivative of $\rho$ on $\theta$ is the ratio of the radial and angular speeds, $\dfrac{\dot\rho}{\dot\theta}$, or 
$$\frac{d\rho}{d\theta}=\frac{v_\rho}{\dfrac{v_\theta}\rho}=\sqrt3\rho$$
hence
$$\rho=\rho_0e^{\sqrt3(\theta-\theta_0)}=\frac{a}{\sqrt 3}e^{\sqrt3(\theta-\theta_0)},$$
a logarithmic spiral.
Note that there is a weird singularity, as the angular speed becomes infinite at the origin.
A: Due to polar symmetry each of the figures joining them is an equilateral triangle.
That is, they are shrinking at a constant rate, rotating at a constant rate and the ratio of these rates ( velocity components in circumferential and radial directions) is constant.
$$ \frac{v_{\theta}}{v_{r}}$$
is a constant. If both velocity components are not constant, then we have the instantaneous triangle distorted, not being equilateral. I.e.,
$$ \frac{r\cdot d\theta/dt}{-dr/dt} = \tan \alpha, $$ 
where $\alpha$ is constant ratio.
$$ \frac{r\cdot d\theta}{-dr} = \tan \alpha $$ 
Integrating
$$ r= r_{o}e^{- \cot \alpha.\theta} $$
The angle $\alpha $ is half of equilateral triangle angle $60^{\circ}$ due to symmetry.  One of three curves spiralling inwards  plotted with $r_{o}=1$ is
$$ r= r_{o}e^{- \sqrt{3}\theta} $$ 

This is a logarithmic spiral.
It can go to the center $(r=0)$ only when $ \theta\rightarrow \infty $. 
We can twice polar rotate the curve about origin by $ 2\pi/3$ to see the other two loci. That is, the three points approach center arbitrarily close, but can never meet at a finite time... never to converge.
EDIT1:
If the velocity is directed between two vertices/along sides  of an equilateral triangle without rotation then the variable triangle becomes smaller and progressively becomes minimized at the sides midpoints (in-circle tangent points) and becomes bigger again without coming near each other. So this option can be ruled out.
