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?

(In reply to

answer by K Sengupta)

At any given instant, the four bugs located at each of the vertices A, B, C and D of the square which shrinks and rotates as the four bugs move closer together. Thus, the four bugs will describe four congruent logarithmic spirals that will meet at the center of the square.

Since each angle of an square 90 degrees, it follows that the path of the pursuer and the pursued will describe an angle of 90 degrees between them and consequently, each bug’s motion will have a component equal to cos 90 = 0 times its velocity that will carry it towards its pursuer. Thus, these two bugs will have a mutual approach speed of 1 + 0 = 1 times the velocity of each bug. In other words, the mutual approach speed would precisely equal the velocity of any given bug.

Consequently, the distance traversed by each of the four bugs

until they meet at the center of the triangle = 10 inches