Consider three time dependent points $P_1(t)$, $P_2(t)$ and $P_3(t)$ in the plane($\mathbb{R}^2$), such that:




This system should look like a triangle rotating but changing its shape.

Check if the images $P_1(\mathbb{R})=P_2(\mathbb{R})=P_3(\mathbb{R})$ and they are an ellipse.

I have simulated this on my computer some time ago, and the trajectories seem to be one and the same ellipse. I don't know how to solve the differential equations to check it directly. However, I know for sure that the area of the triangle formed by the points is constant, since the points move along the base line.


To avoid a proliferation of subscripts, I’m going to rename the points $P$, $Q$ and $R$. The system of equations can be expressed in matrix form as $$\begin{bmatrix}\dot P \\ \dot Q \\ \dot R\end{bmatrix} = \begin{bmatrix}0&-1&1\\1&0&-1\\-1&1&0\end{bmatrix} \begin{bmatrix}P\\Q\\R\end{bmatrix}.$$ It’s not terribly difficult to work out the solution to this system using the usual methods: the eigenvalues are $0$ and $\pm\sqrt3i$, and so the coefficient matrix is similar to $$J=\begin{bmatrix}0&0&0\\0&0&-\sqrt3\\0&\sqrt3&0\end{bmatrix}$$ with exponential $$e^{tJ} = \begin{bmatrix}1&0&0\\0&\cos{\sqrt3t}&-\sin{\sqrt3t}\\0&\sin{\sqrt3t}&\cos{\sqrt3t}\end{bmatrix}.$$ A corresponding eigenbasis is easily computed (in fact one can find an orthonormal basis—the coefficient matrix is the “cross-product matrix” of $(1,1,1)^T$), and setting $C = \frac13(P_0+Q_0+R_0)$, the centroid of the initial configuration of points, gives for the general solution $$P(t)=C+(P_0-C)\cos{\sqrt3t}+(Q_0-R_0)\frac1{\sqrt3}\sin{\sqrt3t} \\ Q(t) = C+(Q_0-C)\cos{\sqrt3t}+(R_0-P_0)\frac1{\sqrt3}\sin{\sqrt3t} \\ R(t) = C+(R_0-C)\cos{\sqrt3t}+(P_0-Q_0)\frac1{\sqrt3}\sin{\sqrt3t}.$$ Now, set $P_0=(1,0)$, $Q_0=\left(-\frac12,\frac{\sqrt3}2\right)$ and $R_0=\left(-\frac12,-\frac{\sqrt3}2\right)$, the vertices of an equilateral triangle centered on the origin. The parametric equations of the three curves simplify to $$P(t) = (1,0)\cos{\sqrt3t}+(0,1)\sin{\sqrt3t} \\ Q(t) = \left(-\frac12,\frac{\sqrt3}2\right)\cos{\sqrt3t}+\left(-\frac{\sqrt3}2,-\frac12\right)\sin{\sqrt3t} \\ R(t) = \left(-\frac12,-\frac{\sqrt3}2\right)\cos{\sqrt3t}+\left(\frac{\sqrt3}2,-\frac12\right)\sin{\sqrt3t}, $$ which are all obviously parameterizations of the unit circle. Any other initial point configuration can be obtained from this one by an affine transformation, which also maps the three identical unit circles onto three identical ellipses.

With the above in mind, you could with a bit of tedious trigonometric manipulation verify that the three trajectories in the general solution are phase-shifted versions of each other, i.e., that $Q(t)=P(t+\delta)$, $R(t)=Q(t+\delta)$ and $P(t)=R(t+\delta)$, where $\delta = {2\pi\over3\sqrt3}$.

  • $\begingroup$ I could’ve gone directly to the unit-circle solution, but I thought seeing the general case would be instructive. The vectors involved have transparent geometric interpretations. $\endgroup$ – amd Jul 18 '18 at 23:14

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