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Partitions and Triangle Inequality


Date: 26 May 1995 09:29:06 -0400
From: william parker school
Subject: Number of Triangles with same Perimeter

How is the number of different triangles with integer sides related to the
perimeter of the triangle?


Date: 28 May 1995 18:54:00 -0400
From: Dr. Ken
Subject:Number of Triangles with same Perimeter

Hello there!  

This is quite a neat problem you've sent us.  First let's look at one
specific case, the case where the perimeter is 15.  This is a question that
I answered a couple of weeks ago from another student.  
                                        
                                  
Date: Sat, 6 May 1995 21:52:31 -0400 (EDT)

Hello Dana!

What a neat problem!  I think the first thing to do in a problem like this
is to take a survey on what our tools are, and how we should attack the
problem.  The way I see it, our tools are the Triangle Inequality and the
partitions of 15.  (just to make sure you're still with me, the Triangle
Inequality says that the sum of the lengths of any two sides must be greater
than the length of the third side; this rules out possibilities like
1,1,13.  The partitions of 15 are simply all the combinations of positive
integers that add up to 15, like 1,1,13 or 2,13.  Obviously, we'll only be
using the partitions that use 3 positive integers)

So let's find a method.  First think about how many triangles we can make
using a side of length 1.  Well, since the Triangle Inequality says that 1
plus the second side has to be more than the third side and vice versa, the
only triangle that will work with a side of length 1 is the triangle 1,7,7.
The ONLY one.  So from now on, when we're looking for triangles, we don't
have to worry about sides of length 1.

Now how many triangles can we make that have a side of length 2?  Again, use
the Triangle Inequality to tell you that 2 plus the second side is greater
than the third side, and vice versa.  The only numbers that will fulfill
that requirement are 2,6,7 and 2,7,6, and since you sent in the problem, you
get to decide whether or not those are the same triangle or two different
congruent triangles.  

Then you can keep going like this, and each time you deal with a side
length, you get to throw it out and not consider it for any of the rest of
the problem, since you've gotten ALL triangles that have it for a side
length.  One more hint: what is the _longest_ a side can be in one of these
triangles?

Now let's look at what happens when the perimeter is small, like for example 
5.  These are the only partitions of 5 that use three parts: 1,1,3 & 1,2,2.
Note that 1,1,3 doesn't fulfill the triangle inequality because 1+1<3.  So
the only triangle that works with perimeter 5 is 1,2,2.  This is typical of
small perimeters; there aren't too many triangles you can make.  What
happens with perimeter 3?  What about 4?  I think that's kind of neat.

Now look at a large perimeter, like 4567.  There are LOTS of triangles you
can make.  Think of it this way:  you've got a stick with 4566 hinges in it,
and you can bend two of them to make a triangle.  If you tried this, you'd
quickly realize that you could make a bunch of different triangles!  In
fact, the larger the perimeter the closer you'll come to having hinges
everywhere.

Hope this helps!

-K
    
Associated Topics:
High School Discrete Mathematics

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