Since the dogs are moving symmetrically, the distance saperating them will decrease evenly as they case each other, so the dogs should catch each other in the center of the room. Their path will be curved, but for sake of approximation, lets se it’s a straigt line to the center.

Assuming that the dogs run at 1m/sec, and room is y x y in diamater.

h^2 = (y/2)^2 + (y/2)^2

h^2 = 2(y/2)^2

h = sqrt(2) * y/2

h = y/sqrt(2) meters

thus it would take the dogs approximtly y/sqrt(2) seconds to catch each other. This is a lower bounds, as the dog will not be traveling in a straight line, but along a curve.

Some Other & Good Aproaches are

Each dog starts at the corner and moving symmetrically. So each dogs start moving perpendicular to the adjacent dogs. Lets assume v.

So each one start moving with v towards the next dog.

If we see realtive speed of the dog1 (v1), w.r.t dog 2, it changes perpendicularly. So it’ll not affect the time taken along the direction of the dog1 to dog2 and the speed will be v only always.

So it they have started at corners with the distance of the length of the square(s).

Time = s/v. They’ll meet at the center.

Another Similar Approach is

Let’s the dogs be A, B, C and D where A is chasing B, B is chasing C, C is chasing D and D is chasing A.

All the dogs will eventually meet in the center of the square. Since all the dogs move in symmetry, the only logical answer to the location of their meeting is the center of the square.

At any point in time, dog A is perpendicular to dog B and B perpendicular to C and so on. Dog A moves towards dog B but dog B does not move towards or away from dog A since it is moving perpendicular to dog A. Therefore, the distance that dog A needs to cover to reach dog B is the same as the original distance between them, one unit.

The speed of each of the dog towards the dog it is chasing is given by (1 + cos(t)) where t is the angle on each corner of the square.

Speed of dog = 1 + cos(90) = 1 + 0 = 1

Time needed = Distance/Speed = 1 / 1 = 1 unit.

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