TITLE: The vertices of an equilateral triangle are shrinking towards each other
QUESTION [16 upvotes]: For an equilateral triangle ABC of side $a$  vertex A is always moving in the direction of vertex B, which is always moving the direction of vertex C, which is always moving in the direction of vertex A. The modulus of their "velocity" is a constant. When and where do they converge. 
Attempt. Found the "when" using a physics style approach by "fixing the frame" on one of the vertices. (From this frame, other two vertex are moving towards origin in a straight line and components of their speed along this line can be used to find when the three meet at origin) For the "where" it is difficult using above approach as this is some kind of rotating and shrinking triangle which is difficult to translate. 

@all Apologies for bumping this question. I wished to give an answer the bounty but it wont let me until the next 23 hours. For the record: I am not seeking new answers.
Update: A cool example of PSTricks package of $\LaTeX$, for anyone who finds this question later. 

Link to code (a .tex file)
And using Pgf/TikZ

Source Page

REPLY [2 votes]: The following are the changes required to Will Jagy's solution to answer the  question (original problem statement).
$$
\begin{pmatrix}
\dot{x} \\
\dot{y}
\end{pmatrix} =
\frac{v}{2 s}
\begin{bmatrix}
-3 & -\sqrt{3} \\
\sqrt{3} & -3
\end{bmatrix}
\begin{pmatrix}
x \\
y
\end{pmatrix} =
\frac{v}{2 \sqrt{3}\sqrt{x^2+y^{2}}}
\begin{bmatrix}
-3 & -\sqrt{3} \\
\sqrt{3} & -3
\end{bmatrix}
\begin{pmatrix}
x \\
y
\end{pmatrix}.
$$
Here $s = \sqrt{3}\sqrt{x^{2} + y^{2}}$ is the edge length of the equilateral triangle at time $t$. Note that the velocity vector has a magnitude $v$.
Now, the vertices of triangle approach each other with velocity $v - v\cos(120^{\circ})$. So the edge length at time $t$ is given by
$$
s = l - \int^{t}_{0} \frac{3}{2}v(\tau) \;\text{d}\tau.
$$
where $l$ is the initial edge length of the equilateral triangle.
If the velocity $v$ remains constant, then $s = l - (3/2)vt$.
If $(x_{0}, y_{0})$ is the initial position of $(x, y)$, then the solution to the system of equations is
$$
\begin{pmatrix}
x \\
y
\end{pmatrix} = \exp\left(A \int_{0}^{t} \frac{v}{s} \;\text{d}\tau\right)
\begin{pmatrix}
x_{0} \\
y_{0}
\end{pmatrix}, \quad
A := \frac{1}{2}
\begin{bmatrix}
-3 & -\sqrt{3} \\
\sqrt{3} & -3
\end{bmatrix}.
$$
Observe that the shape of the curve traced by $(x,y)$ in the 2D plane is the same irrespective of the time integral inside the exponential. This integral is a parameter which only affect where on the fixed curve the point will be at time $t$. This confirms Steven Stadnicki's statement: " I believe this doesn't change the solution curve, but it does definitely change the parametrization of it...".
Setting $s = 0$ we get the time when the points meet. In other words we have to solve for $t$ in the equation
$$
\int^{t}_{0} \frac{3}{2}v(\tau) \;\text{d}\tau = l.
$$
If the velocity $v$ is constant, then the equilateral triangle will collapse in $2l/3v$ seconds. This is the question asked in I.E. Irodov's book ("Problems in General Physics", problem 1.12, page 13, MIR publishers Moscow) as pointed by Deepak. Following this line, we have
\begin{gather}
s(t) = l - \frac{3}{2}vt,\quad
t_{\text{c}} := \frac{2l}{3v},\quad
A \int^{t}_{0} \frac{v}{s} \;\text{d}\tau = -\frac{1}{3}\log\left(1-\frac{t}{t_{\text{c}}}\right)
\begin{bmatrix}
-3 & -\sqrt{3} \\
\sqrt{3} & -3
\end{bmatrix}, \\
\theta(t) := \frac{\log(1-t)}{\sqrt{3}},\quad
P(t) := (1-t)
\begin{bmatrix}
 \cos \theta(t) & \sin \theta(t) \\
-\sin \theta(t) & \cos \theta(t)
\end{bmatrix}, \\
\exp\left(A \int_{0}^{t} \frac{v}{s} \;\text{d}\tau\right) = 
P\left(\frac{t}{t_{\text{c}}}\right),\quad
\begin{pmatrix}
x \\
y
\end{pmatrix} = P\left(\frac{t}{t_{\text{c}}}\right)
\begin{pmatrix}
x_{0} \\
y_{0}
\end{pmatrix}.
\end{gather}
As $\cos$ and $\sin$ are bounded functions, the elements of $P(t/t_{\text{c}}) \to 0$ when $t \to t_{\text{c}}$. So the points will collide at the origin (centroid) at time $t_{\text{c}}$. Note that $\theta(t/t_{\text{c}}) \to -\infty$ when $t \to t_{\text{c}}$, which means that before the points collide they would have circled infinitely around the origin.
In Will Jagy's equation $v = \sqrt{3}\sqrt{x^2+y^{2}}$. The vertices trace the same curve but will never meet in finite time as the velocity tends to zero as the vertices are near collision.
By the way, this problem was first posed by Édouard Lucas (the problem of three dogs chasing each other).
Question 251, Nouvelle correspondance mathématique 3, 175--176, 1877.
and the solution was given by Henri BROCARD and M. Laisant which can be found on page 280 of the same journal (all these are in French).