# Four turtles/bugs puzzle

I was reading about the the four turtles/bugs math puzzle Four bugs are at the four corners of a square of side length D. They start walking at constant speed in an anticlockwise direction at all times directly towards the bug ahead of them. How far does each bug walk before they meet with each other?

and the a-ha answer given is that the turtles are always in the shape of a square, and since each one is always walking perpendicular to the direction of his pursuer's approach, he neither hinders nor helps the pursuer in reaching him. Therefore the time to the center is the same as the time it takes one turtle to walk across a side of the square.

Since each creature is running directly to its target, the paths of a chaser and its chasee are always at right angles. That is, the distance between the chaser and its chasee depends only on the movement of the chaser, not its chasee. Therefore, the distance that the chaser runs is the same as if the chasee did not run at all.

Could you please explain how this happens?

• This is a great explanation. Where did you read it? And what part do you have a problem with? Do you accept that they are in the shape of a square? – Phira Jun 8 '12 at 0:06
• This was in a Martin Gardner column many years ago. – Ross Millikan Nov 18 '12 at 5:58
• The idea is, if you're the chasee, and your direction is always perpendicular to your chasers walk, then you don't change the distance between you, only the direction (why ? see by projection on the line between you and the chaser). Therefore, only the chaser changes this distance, hence the time for him to reach you is the same as if you didn't move at all. – Jean-Claude Arbaut Mar 19 '13 at 11:53

## 1 Answer

If we accept the symmetry that the bugs form a square that shrinks, the problems becomes easier if we look at it as seen from the centre O of the initial square: if you draw a line from O to B1 (where the first bug is), its velocity v has two components: $v/\sqrt 2$ perpendicular to the ray OB1 and $v/\sqrt 2$ along the ray, towards O! The same for all bugs. Now you see that the motion of the bug B1 is composed by two motions: one of rotation around O with constant speed (given above) and one along the radial line from O, towards O. You can calculate the trajectory of one bug: $\dot\rho = -\frac{v}{\sqrt 2}$ and $\rho\dot\phi=\frac{v}{\sqrt 2}$, where $\rho, \phi$ are polar coordinates of B1, as seen from O. Initial conditions are known, now you can easily calculate the time to impact, the trajectory length, etc.

• Could you create a full derivation? – Royi Aug 12 '15 at 5:32