Remember

"Four Bugs" or

"Three Bugs"?

In this problem, the six bugs start at the corners of a regular hexagon (with side length=10 inches).

Again, the bugs travel directly towards their neighbor (counter-clockwise). And, again, each bug homes in on its target, regardless of its target's motion. So, their paths will be curves spiraling toward the center of the hexagon, where they will meet.

What distance will the bugs have covered by then, and how did you determine it?

(In reply to

answer by K Sengupta)

Let the regular hexagon be denoted by ABCDEF. At any given instant, the three bugs located at each of the vertices A, B, C, D, E and F of the hexagon which shrinks and rotates as the six bugs move closer together. Thus, the six bugs will describe six congruent logarithmic spirals that will meet at the center of the hexagon.

Since each angle of a regular hexagon is 120 degrees, it follows

that the path of the pursuer and the pursued will describe an

angle of 120 degrees between them and consequently, each bug’s motion will have a component equal to cos 120 = - 0.5 times its velocity that will carry it towards its pursuer. Thus, these two bugs will have a mutual approach speed of 1 - 0.5 = 0.5 times the velocity of each bug.

Consequently, the distance traversed by each of the six bugs

until they meet at the center of the regular hexagon triangle

= 10/(0.5)

= 20 inches