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### Triangle Perimeter

```
Date: 07/20/97 at 14:49:27
From: Jody Caraher
Subject: Triangle perimeter

How many triangles have sides whose lengths total 15 units?  I have
been browsing Web trying to discover the answer...
```

```
Date: 07/25/97 at 17:28:30
From: Doctor Terrel
Subject: Re: Triangle perimeter

Dear Jody,

There are many kinds of answers for your question.  Do you mean that
the lengths of the triangle's sides must be "whole numbers," or might
they be "real numbers" (i.e. "decimals" or fractions)?  In the
former case, there is only a limited number of possibilities - seven,
in fact (see below) - whereas in the latter case, there is an

To see why, let's take a simple example. If we had two sides of 7
units each, the third side would be 1 unit. This is a very "narrow"
isosceles triangle, resembling a sharp-pointed knife. But we could
also imagine another triangle, also isosceles, with these side
lengths: 7.1, 7.1, and 0.8. Or another: 7.05, 7.05, and 0.9. Or
another: 6.9, 6.9, and 1.2. And those are just some isosceles
versions. We could also form a scalene triangle like this: 7.1, 7.0,
and 0.9. When we begin admitting number values such as these, it
begins to be clear that there are many ways to do it.

To see this, take a loop of string of length fifteen units and note
that you can form many non-similar triangles out of it by holding
three fingers in the loop and pulling taut.

If we limit ourselves to positive integer values, however, the story
changes considerably. Essentially, we need to find how many solutions
there are for this equation:

a + b + c = 15

where a, b, and/or c may be distinct values or even equal to one
another. One solution (7, 7, 1) was just discussed. But even this is
not quite enough.  You see, (2, 5, 8) would be another solution to
the equation, but you couldn't form a triangle with those lengths.
With 8 as one side, the two sides of 2 and 5 (whose sum is 7) wouldn't
"meet" or connect.

So we need to add one more fact to our search, called the Triangle
Inequality Property. This says that the sum of the lengths of any two
sides must exceed the length of the third side. In our good example,
we have 7 + 1 > 7; but in our bad example, we have 2 + 5 < 8.

With a little patience, we can systematically form a list of
solutions:

1) 1-7-7
2) 2-6-7
3) 3-6-6
4) 3-5-7
5) 4-4-7
6) 4-5-6
7) 5-5-5

Note how we let "a" equal the smallest side, and kept it constant as
long as possible while looking for the lengths of "b" and "c".  This
is just to get organized. If you are concerned about obtaining all the
solutions, it helps to have a good, systematic way of searching for
them.  For example, you could let "a" be the length of the smallest
side... (etc.).

Does this do what you want?  I hope so.  If not, contact me again.

-Doctor Terrel,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```

```
Date: 07/25/97 at 17:14:06
From: Doctor Ceeks
Subject: Re: Triangle perimeter

In this case, it is the triangle inequality that contains the crux of
the problem, not the fact that a + b + c = 15. If you solve the
triangle inequality for a different given perimeter, you have
essentially solved this problem because the solutions to the two
problems are in bijection via scaling.

However, if you solve the equation a + b + c = 15, you have not begun

-Doctor Ceeks,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Basic Algebra
High School Geometry
High School Triangles and Other Polygons

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