The answer to this problem is either really obvious, or it’s not.
Say there are four bugs standing at the corners of an imaginary 1m * 1m square. Each of the bugs is facing the bug that is on the adjacent, clockwise corner from where they are.
At the same instant, each of the four bugs starts walking directly towards the bug they are facing at 1m per hour, and keeps walking directly towards that bug as that bug itself starts moving. So the four bugs spiral towards the center of the square in a clockwise direction.
How long does it take for the four bugs to meet?
Answer
There are two different insights one could have in thinking about this problem.
The first insight – let’s call it the “square formation” insight – is that the bugs will effectively always stay in a square formation in their path to the center. A shrinking and spiraling square, but a square that is always centered on the original center. This means that regardless of where the bugs are in their spiral towards the center, they will always be moving 45 degrees to a line connecting where they are to the center.
So one can calculate their effective velocity towards the center:
V_{center} = \cos\;45^\circ\;*\; V_{absolute}\\ V_{center} = \frac{1}{\sqrt(2)}\;m/hrSince the total distance towards the center from the starting position is \frac{1}{\sqrt(2)}\;m, it will take the bugs 1hr to meet in the center.
Those that have the second insight – let’s call it the “perpendicular trajectory” insight – will think this problem is so obvious they’ll wonder what the fuss is about. A bug’s movement towards the bug it is following is always perpendicular to the motion of the followed bug. So in effect the movement of the followed bug has no effect on how far the follower has to travel to reach it or how long it will take, so the follower has to travel 1 meter, and will catch up to the followed bug in 1 hr. A-ha!
What if the 3 bugs were in an equilateral triangles with sides 1m?
Same principle as the above: they stay in formation as they spiral to the center. One bug is heading towards the other at 1 meter per hour. In this case, the followed bug is making effective progress towards the follower at cos(60), or .5 meters per hour. So they have a combined speed of 1.5 meters per hour towards each other, so they should cover the 1 meter between them in 40 minutes.