The vertices of an equilateral triangle are shrinking towards each other

Here's an animation showing the shrinking triangle:

As is required by the statement of the problem, the sides of the triangle are everywhere tangent to the blue curves.

The blue curves are logarithmic spirals. The curve on the right has endpoints $(0,0)$ and $(1,0)$, and is defined by the equation $$ r \;=\; \exp\left(-\sqrt{3}\;\theta\right). $$ Note that the triangle actually undergoes an infinite number of rotations as it shrinks towards the origin.

The rate at which the sides of the triangle shrink is equal to 3/2 of the speed at which the points move. (This follows from the fact that one endpoint of each edge has velocity tangent to the edge, while the other has a velocity component of $\sin(30^\circ)=1/2$ in the direction of the edge.) As a consequence, the length of each blue curve is $2/3$ of the side length of the large triangle.

Edit: Here is the Mathematica code for the animation above:

PolarToRectangular[{r_, theta_}] := {r*Cos[theta], r*Sin[theta]}
tmax = 2/Sqrt[3];
PolarCurve[t_] := {1 - t/tmax, -Log[1 - t/tmax]/Sqrt[3]}
f1[t_] := PolarToRectangular[PolarCurve[t]]
f2[t_] := PolarToRectangular[PolarCurve[t] + {0, 2 Pi/3}]
f3[t_] := PolarToRectangular[PolarCurve[t] + {0, 4 Pi/3}]
spirals = ParametricPlot[{f1[t], f2[t], f3[t]}, {t, 0, tmax},
    Axes -> None, ImageSize -> 300, PlotStyle -> Darker[Blue],
    PlotRange -> {{-0.7, 1.1}, {-0.9, 0.9}}];
triangle[t_] := Graphics[{Opacity[0], EdgeForm[Black],
    Polygon[{f1[t], f2[t], f3[t]}]}]
points[t_] := Graphics[{PointSize[Large], Point/@{f1[t],f2[t],f3[t]} }]
dt = tmax/75;
myframes = Table[Show[spirals, triangle[t], points[t]], {t, 0, 75*dt, dt}];
Export["ShrinkingTriangle.gif", myframes, 
    "DisplayDurations" -> {1}~Join~ConstantArray[0.04, 74]~Join~{1}]

This code exports the animation as a GIF. If you want to see the animation from within Mathematica, the last command would be ListAnimate[myframes] instead.

This is easier than a general case without symmetries. Fix the origin as the center of the triangle. Define a vector field on all of $\mathbb R^2$ such that, at every point $p,$ the vector points at the rotation of $p$ by exactly 120 degrees, taking a consistent rotation direction, say counterclockwise. The easiest vector length would also seem to be the length of the edge being copied, which is then a constant multiple of the distance of $p$ from the origin. Finally, find the integral curve through your favorite vertex on the original triangle.

EDIT: $$ \left( \begin{array}{c} \dot{x} \\ \dot{y} \end{array} \right) \; = \; \left( \begin{array}{cc} \frac{-3}{2} & \frac{-\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{-3}{2} \end{array} \right) \left( \begin{array}{c} x \\ y \end{array} \right) $$

With eigenvalues $\frac{-3}{2} \pm \frac{i \sqrt 3}{2},$ what we do is diagonalize (or, if necessary, the Jordan normal form), at which point the esxponential of the matrix multiplied by the variable $t$ is evident. Then change back, the result is called the fundamental matrix. Often called $\Phi,$ it solves $\Phi' = A \Phi,$ where, in this case, $$ A \; = \; \left( \begin{array}{cc} \frac{-3}{2} & \frac{-\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{-3}{2} \end{array} \right) $$

I get the fundamental matrix as

$$ \Phi(t) \; = \; e^{-3t/2} \left( \begin{array}{cc} \cos \left( \frac{\sqrt{3}}{2} \; t\right) & - \sin \left( \frac{\sqrt{3}}{2} \; t\right) \\ \sin \left( \frac{\sqrt{3}}{2} \; t\right) & \cos \left( \frac{\sqrt{3}}{2} \; t\right) \end{array} \right) , $$

Any solution of the system is found by starting with a constant column vector, and the curve is $\Phi$ multiplied by that column vector, and goes through that at $t=0.$

For those who have not taken a differential equations course, we have the equation
$$ \Phi(t) = e^{tA}= I + t A + \frac{t^2 A^2}{2} + \frac{t^3 A^3}{6} + \cdots$$

As you can see, any such curve actually never reaches the origin, and wraps around the origin infinitely many times. I am beginning to think that this is the same as the logarithmic spiral answer, in which case I have shown you the rudiments of solving the relevant ODE system.

I've seen this on IE Irodov's physics problems. My view of the problem was we know the initial velocity vector in the direction of the center from the starting vertex is $V \cos 30$. As it moves towards the center the velocity vector attains the actual velocity V. We know the distance traveled as $a/\sqrt{3}$. So using the equations of motion it came to $4\frac{2-\sqrt{3}}{\sqrt{3}} \times \frac{a}{v}$. Apparently the solution is even simpler : $\frac{2}{3} \times \frac{a}{v}$. http://exir.ru/1/resh/1_12.htm