pick-up sticks

[BMAD 64]

Suppose that you have 3 rods that have a radius of 2in and a length of 6in.  If they were dropped in a square-based box with a length of 7in, what is the probability that none of the rods touch?

 

You can assume that the two rods would not roll once they came in contact with the box and that the rods always landed flat in the box.

[BASKARAN R 0]

0 (zero)

[telesphorelalancette 0]

Standing up

[bonanova 84]

I don't have an answer yet, but here is how I would attack the problem

Spoiler

1. Define a random position for a rod. One might assume the (x, y) coordinates of the rod's center point are uniformly distributed on the interval [0,7], inclined at an angle a uniformly distributed on the interval [0,180^o). We would discard a result if the endpoints of the rod lay outside the box.
    
2. Set up three pairwise integrals over x₁y₁a₁x₂  y₂a₂ to check for intersections. Or,
    
3. Run a simulation for a million cases of three random rods and count the cases where an intersection occurs.

[bonanova 84]

14 hours ago, bonanova said:

I don't have an answer yet, but here is how I would attack the problem

  Reveal hidden contents

1. Define a random position for a rod. One might assume the (x, y) coordinates of the rod's center point are uniformly distributed on the interval [0,7], inclined at an angle a uniformly distributed on the interval [0,180^o). We would discard a result if the endpoints of the rod lay outside the box.
    
2. Set up three pairwise integrals over x₁y₁a₁x₂  y₂a₂ to check for intersections. Or,
    
3. Run a simulation for a million cases of three random rods and count the cases where an intersection occurs.

After re-reading the problem, I'd agree with Baskaran. In fact,

Spoiler

No two such rods could escape touching each other.

 

[BMAD 64]

8 hours ago, bonanova said:

After re-reading the problem, I'd agree with Baskaran. In fact,

  Hide contents

No two such rods could escape touching each other.

 

lol, i meant to make the rods' diameter 2in not the radii

 

[Phr0stByt3 0]

radius of 2 in = diameter of 4in.  There is no possibility to even fit 2 rods in the box flat.  Even if you allow for the rods to be standing up in a 7in cubed box, you cant fit 3 in there.  I think you asked the wrong question ;-).

But, if you ask the same question with only 2 rods then the entire riddle hangs on the fact that someone does not know this simple fact = diameter = radius x 2 and you may be able to trick them.

 

You could make this a little more tricky like this (if you require them to be laying flat):

Suppose that you have 3 rods that have a radius of 1in and a length of 4in.  If they were dropped in a square-based box with a length of 5in, how many ways can they land that none of the rods touch.

Answer is still 0... some might forget that simple fact above and try to either place them side-by-side, or 2 side by side and then one on top, but a 2in diameter prevents that (2x3 > 5, and 2+4 > 5 respectively).  Finally, going with 1 on top and 2 diagonally, the math would be a bit more tricky, except that this simple formula proves it cant be done:  2x4 = 8 (2-dimensional area of cylinder from top).  8x3 = 24 (area of 3 such cylinders).  While this is short of the 25in area of the box, any diagonal positioning of the cylinders will waste a lot more than 1 square inch (in that a wedge shaped area becomes unusable).