A problem related to Vectors.

Hi, I'm new here. Please excuse me if I break some rule of this website.

Three particles $A,B$ and $C$ are at the vertices of an equilateral trianagle $ABC$. 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.

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Comments

I dont know what you're doing, I just know you said all particles move cyclically towards one another [A to B, B to C & C to A] in an equilateral right? thereby the sum of their velocities at every instant is zero..., no collision, no interaction, nothing... therefore momentum conservation reveals centre of mass remains at same location..., to correlate centre of mass with centroid I assume m1=m2=m3, which says that Centre of mass, or the centroid, wont move.

Jayant Kumar - 7 years ago

Look, try to picture this in your head. If we have an equilateral triangle, and we try to imagine that all the sides decreases at an equal and uniform rate, its center of gravity will be unchanged.

If you are confused, just think about the fact that if the rates at which the sides change are not equal, then it will no longer be an equilateral triangle. Then the centroid WILL CHANGE.

I hope this helps.

Awnon Bhowmik - 6 years, 12 months ago

I think I got it. Since all the sides decreases at the same rate so it should always be an equilateral triangle. Thankyou

Anupam Khosla - 6 years, 11 months ago

It looks likes that you are a "shady"fan & are you really 106 !

Archiet Dev - 7 years ago

Do you know how to edit "age" in the profile?

Anupam Khosla - 7 years ago

No, I'm 20. Haven't got familiar with this website yet.

yes we can simply prove it.take some value of the sides of the equilateral triangle.now we know the velocities each of them and again we know their directions,so now we can calculate how much time it will take to meet each other.now take a time less than that,and get the co -ordinates of position of that time,now take a suitable co-ordinate system of your own and place the value then simply you will prove that you want to.

Arpan Misra - 6 years, 11 months ago

It resembles a cyclic movement.... position of vertices change but the centre point remains same.... looks like as if the triangle is rotating..

Debanjan Dey - 6 years, 11 months ago

what if the triangle is scalene???

Arghadip Koner - 6 years, 10 months ago

A solved example of Concepts of Physics by H.C.Verma

Akshit Vashishth - 6 years, 11 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$