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.

enter image description here

Link to code (a .tex file)

And using Pgf/TikZ

enter image description here

Source Page

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    $\begingroup$ Well, by symmetry they should obviously meet at the centre (e.g. use triangular/barycentric coordinates and argue that all three coordinates must be the same). $\endgroup$ Jun 12 '11 at 3:01
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    $\begingroup$ Martin Gardner had a column in Scientific American where he showed ladybugs on a square doing the same thing. It made a pretty picture. $\endgroup$ Jun 12 '11 at 3:28
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    $\begingroup$ Very apropos... (See the Mathematica notebook for a generalization to regular polygons...) $\endgroup$ Aug 1 '11 at 5:41

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.

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    $\begingroup$ Your analysis looked interesting. It would be great if you found the time to finish it sometime later. +1 in anticipation :) $\endgroup$
    – kuch nahi
    Jun 13 '11 at 2:47
  • $\begingroup$ Actually, now that I (re-) look at this, I think there's a problem with it - you say 'the easiest vector length would seem to be the length of the edge being copied', but that's different from the problem's statement (that the velocity remain constant), which would want a unit vector at each point. I believe this doesn't change the solution curve, but it does definitely change the parametrization of it, and in particular affects how long it takes to reach the origin (which AFAIK still hasn't been answered)... $\endgroup$ Aug 19 '11 at 17:48

Here's an animation showing the shrinking triangle:

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.

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    $\begingroup$ Very nice animation. How was it made? $\endgroup$ Jun 12 '11 at 16:09
  • $\begingroup$ @Shreevatsa: Thanks. I made the animation in Mathematica and then exported as an animated GIF. $\endgroup$
    – Jim Belk
    Jun 12 '11 at 18:09
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    $\begingroup$ @Jim could you post the mathematica code? $\endgroup$
    – kuch nahi
    Jun 12 '11 at 18:40
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    $\begingroup$ @Jim, This is really neat! Your Mathematica skills are impressive. $\endgroup$ Jun 14 '11 at 13:18
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    $\begingroup$ @yayu: Many thanks for the bounty! $\endgroup$
    – Jim Belk
    Jun 15 '11 at 7:58

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).


At time $0$, position of the bodies are $A,B,C$ respectively, After infinitesimal time $dt$, $A',B',C'$ are the respective positions such that $AA',BB',CC'$ all equal to $dx$, as velocity is same for all.

Relative displacement of $B$ with respect to $A$ is,


Solving by ignoring higher powers of $dx$ and $dx'$, we get, $dx'=1.5dx$

Thus, $v'=1.5v$

Total time $= \dfrac{L}{v'} = \dfrac{0.667L}v$

Also, total distance of spiral path covered is $0.667L$.

  • 1
    $\begingroup$ We generally require images and such to be in the body of the answers. Also, please use LaTeX/MathJax formatting, if you could! $\endgroup$
    – The Count
    Jan 23 '17 at 19:10
  • $\begingroup$ Now challenge is to find the length of that spiral $\endgroup$ Jul 13 '20 at 14:01

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