Kinematics Problem -- 3 particles at the vertices of an equilateral triangle

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Three particles A, B, and C are positioned at the vertices of an equilateral triangle and move with constant speed towards each other along the sides. Despite being restricted to their respective sides, they will meet at the center of the triangle due to their continuous adjustment in direction towards one another. The motion is not fixed to the original triangle's vertices but dynamically changes as they approach each other. This understanding clarifies the apparent contradiction of how they can meet while remaining on the sides. The particles will converge at a single point over time.
RandomGuy1
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Here's the question:

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

And here's what I don't understand - If each particle is restricted to its corresponding side of the triangle, how exactly will they "meet"?
 
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The sides of the triangle are getting smaller.
 
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RandomGuy1 said:
The particle at A always has its velocity along AB, B along BC and C along CA.
It's a bit oddly worded. Don't think in terms of fixed directions based on the original positions. As the particles move and change position: A always moves toward B, B always moves toward C, C always moves toward A.
 
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Doc Al said:
It's a bit oddly worded. Don't think in terms of fixed directions based on the original positions. As the particles move and change position: A always moves toward B, B always moves toward C, C always moves toward A.

You were right. That cleared a lot up. Thanks!
 
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