Why should the particles meet at a common point? I saw a question in my physics book asking for the time when all the three particles (each at the corner of an equilateral triangle and each having constant velocity v along the sides of the triangle) meet at a common point.

I can't find the reason why these particles should meet at a common point.
What I think is that since they all have the same velocity and each travels the same distance so after some time ($ t = \frac{a}{v}$) ($a$ is the side of the triangle) their corners should be interchanged and this should continue all the time and they should never be at the same point.
But it is not the answer and the solution shows that they met at the centroid of the triangle.

Why should they follow a curved path? Shouldn't they just go on along the sides of the triangle?
 A: I think it's possible that you either misunderstood the problem or that it was badly phrased. I think you're right that if the particles were simply moving in straight lines with a constant velocity, they would never meet.
I believe, however, that the problem was probably intended as a Pursuit Curve-type problem: the particles are initially at the vertices of a triangle, but each particle "pursues" the other, with their velocity directed towards the particle they're pursuing. $A$ pursues $B$, $B$ pursues $C$, and $C$ pursues $A$. (In other words, the points $A$, $B$, and $C$ represent the particles, and not fixed vertices of some triangle.)
A: The question says that the particles always point towards each other. This is a very famous question in India for Jee's preparation for kinematics. Here is the question:

Three particles A, B, and C are situated at the vertices of an equilateral triangle ABC with sides d at t = 0. Each of the particles moving with constant speed always has its velocity along AB, B along BC and C along CA. At what time will the particles meet each other?

With the associated diagram which is exactly what you have given. Here the triangle always  refers to the traingle made by the particles as its vertices.
So in this case obviously the particles cant move in a straight path.
Another challenge that my teacher had proposed to me was to find the equation trajectory of the particles. Good Luck!
A: If the velocities are all equal and always directed towards the position of the other particles, the velocity vectors will start rotating. Each particle at an equal rate. So the velocities will converge to the point in the middle.
What will happen to the particles after they've reached that point of coincidence? You can probably imagine yourself. The particles still have the same velocity (if the collision is elastic) but opposite in direction. So they will start diverging back to the triangle after which the process repeats itself. A Perpetuum Mobile! (Which, of course, in reality, is impossible).
A: Another way to see this, is by leveraging symmetry, since the whole system is invariant under a rotation of $ 120^{\circ} $ or integer multiple of that.  Suppose they do not meet at the center, but some nearby point P.
Now since the answer should also not depend on whether or not you rotate the whole system by certain angles, meeting at point P is obviously wrong, because the position of point P is not invariant under such rotations. We are left to conclude that they must meet at the center of the triangle.
A: If the speed is constant then the acceleration must be centripetal, which determines the curvature of the arcs which meet at the center of the triangle. The acceleration is determined by the speed of the particle at the apex ahead. From the point of view of particle (1), the particle (2) ahead moves through an arc: v cos($30^o$)dt/L (where L is the instantaneous length of the side of the triangle). In that same short time, the velocity vector of particle (1) rotates through that same angle:  dv/v.  Equate these to find the acceleration: dv/dt.  Note that as L gets smaller, the acceleration and curvature increase.
