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### Triangle Inequality

```
Date: 09/14/97 at 14:05:03
From: Joe
Subject: IMP Math program, word problem

I am totally stuck on this problem, which is a Problem Of The Week in
IMP II.

Suppose that you have a straight pipe cleaner of a given length.  The
two ends are points A and C.

Now suppose that the pipe cleaner is bent into 2 portions, with one
portion twice as long as the other.  Label the place where it's bent
as B, so that the segment from B to C is twice as long as from A to B.
Then a fourth point, X, is chosen at random somewhere on the longer
section, and the pipe cleaner is bent at that portion, too.

Can the 3 segments of the pipe cleaner be made into a triangle by
changing the angles at the two bends?  As you might suppose, the
answer depends on the location of point X.

My question to you is: if point X is chosen at random along the
section from B to C, what is the probability that the 3 segments of
the pipe cleaner can be made into a triangle?

Thanks.
```

```
Date: 09/14/97 at 18:48:00
From: Doctor Anthony
Subject: Re: IMP Math program, word problem

You use the property of a triangle that any two sides must be greater
than the third side.

|<--- 1 --->|<--------  2  -------->|
A-----------B----|------------------C
x   D       2-x

If you look at the above figure, suppose the extra bend is made at D,
a distance x from the point B, as shown. Then for a triangle to be
possible in this configuration with D closer to B than C we require

1 + x > 2-x

2x > 1    and so  x > 1/2

Similarly, if D is closer to C than B the condition for a triangle to
be possible is:

x < 1 + (2-x)

2x < 3   so  x < 3/2

Putting these two inequalities together we get

1/2 < x < 3/2

Now x can vary with uniform probability between 0 and 2, and its
permitted range, 1/2 to 3/2, is exactly half this range. So the
probability that a triangle can be formed is  1/2.

-Doctor Anthony,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Probability

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