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Four turtles/bugs puzzle

Starting from the corners of a square of side a, each bug chases the one clockwise from it. If they all start at the same time and run at the same speed, how far has each bug run by the time they all collide at the center of the square?

I know that this can be solved with or without calculus. Can anyone help me please? Thank you.

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marked as duplicate by Ross Millikan, Lukas Geyer, Rudy the Reindeer, userNaN, Chris Eagle Nov 18 '12 at 9:24


The answer is given in this previous question. –  Rahul Nov 18 '12 at 5:52
@RahulNarain: Good on you for finding it –  Ross Millikan Nov 18 '12 at 5:59

2 Answers 2

By symmetry, the bugs are always at corners of a square as they spiral in. A quasi-calculus solution is to recognize that the path of one bug is (infinitessimaly) directly toward the next one, and the path of the next one is perpendicular, so the distance shortens with the speed of the bug. The distance traveled is then the side of the square.

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Let $A(t)$ and $B(t)$ be two of these bugs, $A$ chasing $B$. Note that $A$ always moves towards $B$ in a direction perpendicular to the motion of $B$. Thus, it makes no difference how $B$ is moving, as the time it takes $A$ to reach $B$ remains the same in any event.

As a result, the time it takes is the original distance between $A$ and $B$, written $|A(0) - B(0)|$, divided by the speed of bug $A$, and the distance each bug has run is $|A(0) - B(0)|$.

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