A few days ago I posted an answer to a question on Phys.SE.
The question is:
Three particles $A,B$ and $C$ are at the vertices of an equilateral trinagle $ABC$. Each of the particle moves with constant speed $v$. $A$ always has its velocity along $AB$, $B$ along $BC$ and $C$ along $CA$. They meet each other at the centroid. At any instant, the component of velocity of B along BA is $v\cos60^∘$.
I know that by symmetry $ABC$ will always make an equilateral triangle. But can we prove this mathematically, that is
- Can we mathematically show that at any instant of time $t$ the points $ A(t),B(t),C(t)$ will make an equilateral triangle whose centroid is the same as that of the triangle made by the points $A(0),B(0),C(0)$?
My own understanding:
Since the all the distances $AB,BC$ and $CA$ decreases at the same rate the points $ABC$ must meet coincidentally and at any instant $ABC$ should make a equilateral triangle. Now I have to prove that the centroid must remain constant. For this I am thinking of two methods
(1) I choose the initial centroid as the origin and label the points $A,B$ and $C$ as $\vec r_A,\vec r_B, \vec r_C$ respectively now $ |\vec r_A(0)|=|\vec r_B(0)|=|\vec r_C(0)|$ and $\dfrac {d|(\vec r_B- \vec r_A)|}{dt}=const$...
(2) For simplification I choose the point $A$ as the centre of the coordinate system and.....
As the things were going tedious I thought it would be better to ask on Math.SE. Sorry for the homework question $-$ could anyone tell me how should I proceed?