4 dogs chasing each other

[BMAD]

Four dogs are positioned at the corners of a square (d=1m), chase each other in clockwise direction with the same constant speed . As their target is moving, they will follow a curved path, eventually colliding in the center of the square.

 

Why is the total length of the path just 1m?

[bonanova]

Spoiler

The paths of each pursuing dog and his target are perpendicular. Because their speeds are equal, the dogs remain at the corners of a (shrinking) square. Although the target dog is not standing still, his motion does not increase or decrease the distance from his pursuer.

Another way to think of it is to rotate the square as the dogs run, so as to keep the sides of their square horizontal and vertical. The dog in the LR corner, then, would be seen to run vertically until he reached the dog in the UR corner, thus running exactly the length of the original square.

If three dogs were at the vertices of an equilateral triangle with sides of unit length, they would run less than 1 unit distance, because the velocity of their target dogs would have a component that pointed toward their pursuers, thus shortening the needed distance to be caught.  Five dogs on a pentagon would have to run farther than 1 unit distance. The deviations in each case are given by the sine of the angle at the vertices. For a square sin(90) = 0, and so on.