Home > Mathematica, Mathematics, Uncategorized > Breaking Sticks

Breaking Sticks

This post is based on a probability puzzle. What is the probability that breaking a straight stick at 2 random points will result in 3 segments that can form a triangle.

Consider the resulting segment lengths: x_i, i \in \{1,2,3\}. Let the stick length =1.

Now,

\sum_{j=1}^3 x_j=1

To form a triangle

 x_i+x_j>x_{6-i-j}, i\neq j for all {i,j}\in \{1,2,3\}

These constraints reduce to: x_1+x_2>1/2 , x_1<1/2,x_2<1/2

You can visualize this as follows. As there are only 2 degrees of freedom, consider only space of x_1,x_2. The space of allowable x_1, x_2 is the region of the unit square: x_1+x_2<1 . Within this space the values of x_1,x_2 that form a triangle with lengths (x_1,x_2,1-x_1-x_2) is the region satisfying the constraint above.

The probability is then: (1/8)/(1/2)=1/4.

 

Just for fun:

 

msestick

 

The light blue triangle is the region of x_1,x_2 that can form triangle and the pink triangle encloses the allowable stick segments.

The triangles on the right have been rescaled to be circumscribed by the unit circle. The relative lengths pertain not the absolute lengths.

Advertisements
  1. No comments yet.
  1. No trackbacks yet.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: