Four bugs are located in the corners of a square, 10 inches on the side. They are arranged like this:
A---B
| |
D---C
As the clock starts,
A begins crawling directly toward
B, which goes to
C,
C goes to
D and
D to
A.
Each bug will home in exactly on its target, reguardless of the target's motion, so their paths will be curves spiraling toward the center of the square where they will meet.
What distance will each of the bugs have covered by then?
10 inches. First thing to notice is the symmetry. Each bug will home in on its target at exactly the same speed, and each will approach the center at exactly the same speed. This means that the four bugs will always form a square. Since they form a square, their travel paths will always be at right angles to one another. That means that at any given instant, Bug B's motion does not take it any closer or any further away from Bug A. Hence, Bug B's motion does not increase or decrease the distance Bug A must travel. So Bug A must travel 10 inches. Likewise for the other bugs.
Solution
