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
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