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Three Pieces of a Stick Forming a Triangle

Date: 01/22/2007 at 23:33:08
From: Bob
Subject: Three Pieces of Stick Triangle problem

If you break a straight stick into three pieces, what is the 
probability that you can join the pieces end-to-end to form a 
triangle?



Date: 01/23/2007 at 04:19:47
From: Doctor Greenie
Subject: Re: Three Pieces of Stick Triangle problem

Hi, Bob -

After seeing your message, I sat down and figured out what I thought 
was a good answer.  Then I used Google to search for "break a stick 
into three pieces" and found several links which gave contradictory 
answers.  Many of the sites, however, showed the same answer I got, 
which is 1/4.

You can try a search of the web yourself to try to find information 
on this topic.  All of the formal and informal arguments I found on 
any of the pages I visited were far more complicated than the method 
I used to get my answer....

Here is the key idea for my analysis of the problem:

  If we consider the original stick to be of unit length, then 
  we can form a triangle whenever the longest stick is less than 
  a half unit long.

This idea is based on the Triangle Inequality Theorem, which says that
in a triangle the sum of any two sides must be greater than the third
side.  Having one piece that is equal to or more than half a unit
would leave the other two pieces to sum to exactly or less than half a
unit, violating the theorem.

For my analysis, we consider different cases where we make the first 
break and then, based on where that first break is, determine where 
the second break can be in order for the three resulting pieces to be
of lengths such that they can form a triangle.  So imagine the 
original stick marked like a ruler from 0.000 to 1.000.  Pick a few 
points for the first break and determine the range where the second 
break can be made so that the three pieces can form a triangle.  This
determination is based solely on the key idea stated above.

Suppose we make the first break at 0.4.  We must make the second break
somewhere after 0.5, because the "last" piece can't be more than a
half unit long.  And for the "middle" piece to be less than a half
unit long, we need to make the second break before 0.9 (that is,
before (0.4 + 0.5).  So, for a first break at 0.4, the range over
which we can make the second break is from 0.5 to 0.9, which is a
range of 0.4.

If you pick any other point "x" less than 0.5 for the first break,
then you will get a similar analysis; the second break will then have
to be made between 0.5 and (x + 0.5), which is a range of x.

Now let's pick a first break at a point past 0.5; suppose we make it 
at 0.85.  Then we will need to make the second break before 0.5 (so 
that the "first" piece is not more than a half unit long); and we will
need to make the second break no farther than a half unit from the
first break--i.e., at a point no less than 0.35.  So for a first break 
at 0.85, the range where we can make the second break and get three
pieces which can form a triangle is from 0.35 to 0.5--a range of 0.15.

If you pick any other point "x" greater than 0.5 for the first break,
then you will get a similar analysis; the second break will then have
to be made between (x - 0.5) and 0.5, which is a range of (1 - x).

Now if we make a graph of the range of values for which the second 
break can be made as a function of where the first break is made, the 
graph increases uniformly from a value of 0 at x=0 to a value of 0.5 
at a value of x=0.5; then it decreases uniformly to a value of 0 at 
x=1.0.  By a simple geometric argument, the average value of that 
graph over the entire interval is 1/4; therefore, the probability that 
the three pieces of the stick will form a triangle is 1/4.

I hope this helps.  Please write back if you have any further 
questions about any of this.

- Doctor Greenie, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
College Probability
College Triangles and Other Polygons
High School Probability
High School Triangles and Other Polygons

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