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### Probability of Making Triangle with Three Stick Pieces

```Date: 02/24/2007 at 03:26:40
From: Kiran
Subject: Find probability that 3 parts of a stick form a triangle.

Take a stick of unit length and break it into two pieces, choosing
the break point at random.  Now break the longer of the two pieces at
a random point.  What is the probability that the three pieces can be
used to form a triangle?

I'm not sure what to do after taking the initial condition.  Let a be
the point of breakage and its distance from one end is x.  Then
x > 1/2.  The second probability gives y > 1/4.  How to proceed from
there and get the probability?

```

```

Date: 02/24/2007 at 09:37:07
From: Doctor Greenie
Subject: Re: Find probability that 3 parts of a stick form a triangle.

Hello, Kiran -

I made a rather extensive search of the Internet and the Dr. Math
archives and found no satisfactory discussion of this problem.  There
are several sites I found which discuss the similar problem where the
stick is broken randomly in two places--unlike your problem, where the
first break is made and then the second break is made in the longer
piece.  Even in that other case, I found disagreement about the
correct answer.  And all of the sites I found used methods of solving
the problem which seem more difficult than mine.

So here is my solution to your problem....

To make a triangle out of three pieces of a stick of unit length, the
longest piece must have a length which is less than half a unit.

Let's make our first break and let x be the length of the shorter
piece.  x can vary between 0 and 0.5.  Let's pick values in this range
and consider the probability that we can make a triangle for that
value of x.

Suppose x = 0.2; then the long stick has length 0.8.  The break on
this stick must be made so that the longest piece has length less than
0.5.  This means the break must be made between 0.3 and 0.5.  So the
probability of being able to make a triangle with the three pieces if
the first break is at 0.2 is 0.2/0.8.

Now let's pick one more value for where we make the first break.
Suppose x = 0.4; then the long stick has length 0.6.  Again, the break
on this stick must be made so that the longest piece has length less
than 0.5.  This means the break must be made between 0.1 and 0.5.  So
the probability of being able to make a triangle with the three pieces
if the first break is at 0.4 is 0.4/0.6.

Using our two specific examples, we can see that, in the general case,
the probability of being able to make a triangle out of the three
pieces, as a function of the length x of the short piece we get on our
first break, is

x/(1 - x)

Our variable x can range from 0 to 0.5, so the overall probability
that we can make a triangle with the three pieces of the stick, given
your statement of the problem, is

integral from 0 to 0.5 of (x/(1 - x))
---------------------------------------
0.5 - 0

This turns out to be

2(-.5 - ln(.5)) = 2(.19314718....) = .38629436...

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

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

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