Problem about points on an equilateral triangle Suppose that $A$, $B$, and $C$ are three points in a plane, such that $AB = AC = BC = 1$.
At each point in time, $A$ is moving toward $B$, $B$ is moving toward $C$, and $C$ is moving
towards $A$, all with speed $v = 50$.
At what time $T$ will all the points reach the center of the triangle?

 A: In rotating frame of reference it is clear that points should walk $L=a\cdot\frac{1}{\sqrt3}$ — it is the distance from every vertice to the point of meeting.
Because of rotating only the radial component of velocity makes sense. And it is $V=\frac{v\sqrt3}{2}$.
So, $T=\frac LV=\frac{2a}{3v}=\frac{1}{75}$.

OK, more formal proof.
We have three points: $A_1(R, 0),\, A_2(-R\cos\frac{2\pi}{3}, R\sin\frac{2\pi}{3}),\, A_3(R\cos\frac{4\pi}{3}, R\sin\frac{4\pi}{3})$
We will calculate the law of motion of the first point in polar coordinates $(r(t),\phi(t))$.
Initial conditions are $r(0)=R, \ \phi(0)=0$.
The second point has law of motion (due to the symmetry):  $\tilde r(t)=r(t),\ \tilde \phi(t)=\phi(t)+\alpha$, where $\alpha=\frac{2\pi}{3}$.
Generally we have $$\dfrac{\dot y}{\dot x}=\dfrac{\tilde y-y}{\tilde x-x}.$$
In polar $$\dfrac{(r\sin\phi)_t}{(r\cos\phi)_t}=\dfrac{r\sin(\phi+\alpha)-r\sin\phi}{r\cos(\phi+\alpha)-r\cos\phi}$$
From it we obtain $$\dot r+r\dot\phi\tan\frac\alpha 2=0.$$
Also points move with constant speed: $$\dot r^2+r^2\dot\phi^2=const=v^2$$
From two last equations we have $$\dot r=\pm v\sin\frac\alpha 2$$
So $$r(t)=R-vt\sin\frac\alpha 2 $$
And $$\dot\phi=\dfrac{v\cos\frac\alpha 2}{R-vt\sin\frac\alpha 2},\quad \phi(t)=-\dfrac{1}{\tan\frac\alpha 2}\ln\left(1-\dfrac{vt\sin\frac\alpha 2}{R}\right).$$
We also could obtain $r(\phi)$: $$\phi=-\dfrac{1}{\tan\frac\alpha 2}\ln\dfrac{r}{R}$$
And finally $$r(\phi)=Re^{-\phi\tan\frac\alpha 2}.$$  
Time of meeting could be obtained by pointing $r(T)=0$, $T=\dfrac{R}{v\sin\frac\alpha 2}$
A: Hint:


*

*Calculate the relative velocity of $A$ to $B$ and observe that it does not change.

*Try to solve a similar question about a square. In such case use the fact, that velocity vector of $B$ is perpendicular to velocity vector of $A$, hence it does not matter – the travel time (and therefore distance too) is just as it would take to go straight to the other corner.

*In case of a triangle, adjust for the non-orthogonal vectors. The relative velocity of $A$ to $B$ is $1\cdot|v_A|+\frac{1}{2}|v_B|$ and so the time and traveled distance would be a base multiplied by $\frac{2}{3}$.


I hope this helps $\ddot\smile$
