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 unlimited, infinite number of answers. 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 to address the essence of this problem, which is about triangles. -Doctor Ceeks, The Math Forum Check out our web site! http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.