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

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).