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Break a dowel to form a triangle

Date: 8 Mar 1995 17:04:44 -0500
From: Chianne Chen
Subject: question

Hi, I have a problem:

A wooden dowel is randomly broken in 2 places.  What is the 
probability that the 3 resulting fragments can be used to form 
the sides of a triangle?


Date: 9 Mar 1995 17:55:17 -0500
From: Stephen Weimar
Subject: Re: question

I'd say it makes a difference how it's broken and what you mean 
by random.  For instance, is the stick broken once and then one 
of two pieces is selected at random to be broken again ("pick one 
and break again")?  Or do you pick two spots on the dowel at 
random where the breaks will be made (two-at-once)?

Let's look at the second situation because the first seems similar 
but a little harder to calculate.

Two-at-once is like "pick one and break again" except that with 
the latter we have a fifty percent chance of choosing the bigger 
piece for the second break; whereas for two-at-once, since 
the location of the second break is made with reference to the 
whole stick, you have a greater than fifty percent chance of 
making the next break in the bigger piece since the bigger piece 
is more than fifty percent of the whole stick.  So the probability 
of "two-at-once" resulting in a triangle should be better than 
that of "pick one and break again."

Now, what does it take for three sticks to be able to make a 
triangle? Will any three sticks do?  Once you express the 
minimum conditions the broken pieces must satisfy in relation 
to each other if they are to make a triangle, then you can make 
3 statements, in this case, 3 inequalities to solve. Try to figure 
this out before reading the "spoiler" below.  When I set up my 
equations, I let x be the distance from the left side of the dowel 
to the first break. Let y be the distance from the left side to the 
second break. Now we have two sets of inequalities. When y 
is to the right of x, we have one set. When y is to the left of x, 
we have another.  But we can see that these two situations are 
equally likely, in fact the probability in each case should be the 
same, so once we have calculated one the total will be easy to 

"y to the right of x"


The lengths of the pieces are x, 1-y, y-x. So we can set up 
three inequalities since any two sides added together have to 
be longer than the third.

What happens if you set up these inequalities and solve them?  
I'll leave it to you to write out the inequalities.

When doing problems such as this I like to make a 1 by 1 square, 
the area of which represents the total probability (1). Let's say 
the length of the unbroken stick is one unit. The horizontal side 
of our square could represent the first break which is anywhere 
from 0 to 1. The vertical side of our square will be the second 
break, y.

If you graph the inequalities you get a triangular area occupying 
the lower right hand half of the upper left quadrant. I don't know 
if this picture comes out on your screen, you may have to change 
fonts. The area of that triangle is equal to the probability of "y 
to the right of x". 

     |   /|
     |  / |
     | /  |
     |/_ _|
     |_ _ _ _ _ _
     0    1/2    1

Don't forget about "y to the left of x".

Once you have completed this version of the problem you might 
want to go back and see if you can set up the inequalities for 
"pick one and break again".  The idea is the same but the math 
can be harder.

-- Dr. Steve
Associated Topics:
High School Geometry
High School Probability
High School Triangles and Other Polygons

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