# OK, I gots one

PBE6
Posers and Puzzles 16 Feb '05 20:46
1. PBE6
Bananarama
16 Feb '05 20:462 edits
As with every problem I propose, this may have been asked and answered on a previous post. Them's the breaks.

Four cock-a-roaches, two male and two female, are sitting on a table in a square formation like so:

M----F
|......|
|......|
F----M

Each cock-a-roach is facing the next one going counterclockwise around the figure, getting an eye-load of tail while they're at it. The distance between consecutive cock-a-roaches is 1 m.

Now, as they are creatures of lust, each cock-a-roach tries to run straight towards the next one in line so they can get it on! Each cock-a-roach runs at a speed of 1 m/s (just like when the lights go on).

The question is: how long will it take before the cock-a-roaches catch each other, and have a huge orgy in the middle of the square? For the purposes of this question, assume that the orgy takes place as soon as they meet (all cock-a-roaches are smoking roaches during the race).
2. 16 Feb '05 22:34
Quick guess - 11/14 of a second? (pi/4 sec)?

Don't know if this is correct, going by intuition and basic maths.
3. Acolyte
17 Feb '05 08:45
Originally posted by PBE6
As with every problem I propose, this may have been asked and answered on a previous post. Them's the breaks.

Four cock-a-roaches, two male and two female, are sitting on a table in a square formation like so:

M----F
|......|
|......|
F----M

Each cock-a-roach is facing the next one going counterclockwise around the figure, getting an eye-l ...[text shortened]... orgy takes place as soon as they meet (all cock-a-roaches are smoking roaches during the race).
The cock-a-roaches will always be in a square, so all we need to do is calculate the rate of change of the side length, as a function of that length.

Suppose the side length is x. Then 'after a short time' x^2 will change to (x-h)^2 + h^2, which in the limit as h -&gt; 0 gives a rate of change of -2x, so d(x^2)/dt = -2x.

=&gt; -1/2sqrt(x^2) * d(x^2)/dt = 1

=&gt; -x = t + c

at t = 0, x = 1. So c = -1. Hence the cock-a-roaches meet at time 1.
4. PBE6
Bananarama
17 Feb '05 15:50
Originally posted by Acolyte
The cock-a-roaches will always be in a square, so all we need to do is calculate the rate of change of the side length, as a function of that length.

Suppose the side length is x. Then 'after a short time' x^2 will change to (x-h)^2 + h^2, which in the limit as h -> 0 gives a rate of change of -2x, so d(x^2)/dt = -2x.

=> -1/2sqrt(x^2) * d(x^2)/dt = 1

=> -x = t + c

at t = 0, x = 1. So c = -1. Hence the cock-a-roaches meet at time 1.
Excellent as usual, Acolyte.

As a follow-up, what shape does each roach's path take (mathematically defined, of course)?
5. CalJust
It is what it is
28 Feb '05 14:08
Originally posted by Acolyte
The cock-a-roaches will always be in a square, so all we need to do is calculate the rate of change of the side length, as a function of that length.

at t = 0, x = 1. So c = -1. Hence the cock-a-roaches meet at time 1.
Wonderful maths, but the intuitive answer is 1 sec. Here's why:

The roaches are 1 m apart at the start. Since the 'front' one will always move at right angles to the line connecting it with the 'rear' one, no vector of its movement will reduce that distance. The only decrease in the distance is from the movement of the 'rear' one gaining. Since the 'rear' one moves at 1 m/s, and the beginning distance is 1 m, he catches her (or she catches him!) in one second flat!

QED

CJ

6. PBE6
Bananarama
01 Mar '05 22:19
Nobody gives a crap about this puzzle anymore, which is fine, but I'll just send it to the top of the comment list anyway.

The path that each cock-a-roach takes is a logarithmic spiral! That's sexy as all get out. The thing I think is neat about that is this: the change in angle w.r.t. time (d(theta)/dt) as the time approaches 1 second increases without bound. Imagine swinging your partner dosey-do at an infinite rate of revolution. Your arms would rip off!!! Wicked. That's a messed up square dance if ever I saw one.