Three Pieces of a Stick Forming a Triangle
Date: 01/22/2007 at 23:33:08 From: Bob Subject: Three Pieces of Stick Triangle problem If you break a straight stick into three pieces, what is the probability that you can join the pieces end-to-end to form a triangle?
Date: 01/23/2007 at 04:19:47 From: Doctor Greenie Subject: Re: Three Pieces of Stick Triangle problem Hi, Bob - After seeing your message, I sat down and figured out what I thought was a good answer. Then I used Google to search for "break a stick into three pieces" and found several links which gave contradictory answers. Many of the sites, however, showed the same answer I got, which is 1/4. You can try a search of the web yourself to try to find information on this topic. All of the formal and informal arguments I found on any of the pages I visited were far more complicated than the method I used to get my answer.... Here is the key idea for my analysis of the problem: If we consider the original stick to be of unit length, then we can form a triangle whenever the longest stick is less than a half unit long. This idea is based on the Triangle Inequality Theorem, which says that in a triangle the sum of any two sides must be greater than the third side. Having one piece that is equal to or more than half a unit would leave the other two pieces to sum to exactly or less than half a unit, violating the theorem. For my analysis, we consider different cases where we make the first break and then, based on where that first break is, determine where the second break can be in order for the three resulting pieces to be of lengths such that they can form a triangle. So imagine the original stick marked like a ruler from 0.000 to 1.000. Pick a few points for the first break and determine the range where the second break can be made so that the three pieces can form a triangle. This determination is based solely on the key idea stated above. Suppose we make the first break at 0.4. We must make the second break somewhere after 0.5, because the "last" piece can't be more than a half unit long. And for the "middle" piece to be less than a half unit long, we need to make the second break before 0.9 (that is, before (0.4 + 0.5). So, for a first break at 0.4, the range over which we can make the second break is from 0.5 to 0.9, which is a range of 0.4. If you pick any other point "x" less than 0.5 for the first break, then you will get a similar analysis; the second break will then have to be made between 0.5 and (x + 0.5), which is a range of x. Now let's pick a first break at a point past 0.5; suppose we make it at 0.85. Then we will need to make the second break before 0.5 (so that the "first" piece is not more than a half unit long); and we will need to make the second break no farther than a half unit from the first break--i.e., at a point no less than 0.35. So for a first break at 0.85, the range where we can make the second break and get three pieces which can form a triangle is from 0.35 to 0.5--a range of 0.15. If you pick any other point "x" greater than 0.5 for the first break, then you will get a similar analysis; the second break will then have to be made between (x - 0.5) and 0.5, which is a range of (1 - x). Now if we make a graph of the range of values for which the second break can be made as a function of where the first break is made, the graph increases uniformly from a value of 0 at x=0 to a value of 0.5 at a value of x=0.5; then it decreases uniformly to a value of 0 at x=1.0. By a simple geometric argument, the average value of that graph over the entire interval is 1/4; therefore, the probability that the three pieces of the stick will form a triangle is 1/4. I hope this helps. Please write back if you have any further questions about any of this. - Doctor Greenie, The Math Forum http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.