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Probability to form a Triangle by breaking a Rod

Updated: Feb 8

It is a very popular question throughout the web. Even in some job applications, it is being asked to the candidates so as to test their practical problem solving skills. Here is the question:

You are given a rod with length L, and you are asked to break it into three pieces. What is the probability that these three pieces can form a triangle?





SOLUTION:


The triangle inequality will help us solve this problem. It simply states that the length of any side of a triangle is always greater than the difference between the lengths of the other two sides, and less than their sum. Mathematically speaking,


| b – c | < a < b + c (1a)

| a – c | < b < a + c (1b)

| a – b | < c < a + b (1c)


where a, b and c are the lengths of the sides of the triangle. Unless these inequalities hold, a triangle can not be formed.


In our problem, we know that the total length of the given rod is given as L. Therefore,

after it has been broken into three pieces, we will have the relation


a + b + c = L (2)


where a, b and c are now the lengths of the pieces, respectively. In order to solve this problem, we will try to determine how many different outcomes can occur regarding the

formation of a triangle with these three pieces. We will first consider the option for which a triangle can not be formed.



What is the condition for not being able to form a triangle?


From the triangle inequalities, we know that if one of the lengths (let's say a) is larger than the sum of the other two lengths, a triangle can not be formed. Then, we will have the condition


a > b + c ––> a > L – a ––> a > L/2 (3)


Then, Eq. (3) suggests that if one of the lengths is larger than half of the length of the original rod, a triangle can not be formed. Here, we have taken the side length a to be the largest, but b or c could have been chosen to be the largest as well. In addition, if we know that one of the lengths is larger than L/2, then each one of the other two lengths will be simply less than L/2.



What is the condition for being able to form a triangle?


Based off of Eq. (3), we will now check if each of the lengths being less than L/2 is a sufficient condition to form a triangle, i.e.


a < L/2 and b < L/2 and c < L/2 (4)


Assuming Eq. (4) is correct, let's check if the triangle inequalities would hold for the side with length a:


  • Check #1: (Upper Limit) Does "a < b + c" hold?

a < b + c ––> a + a < a + b + c ––> 2a < L ––> a < L/2 (5)


Since, a < L/2 is a true statement based on Eq. (4), the inequality a < b + c simply holds.


  • Check #2: (Lower Limit) Does "| b – c | < a" hold?

Here, if we choose to name b as the larger side among b and c, the triangle inequality for the lower limit would read


bc < a ––> b < a + c ––> b + b < a + b + c ––> 2b < L ––> b < L/2 (6)


Since, b < L/2 is also a true statement due to Eq. (4), the lower limit in the triangle inequality also holds.


Hence, while we were doing the above checks, none of the side lengths takes precedence over the others. The names are interchangeable meaning that a can be named as b or c can be named as a, so for and so forth. In short, we can conclude that


a < L/2 and b < L/2 and c < L/2 (see 4)


is the necessary and sufficient condition to form a triangle with side lengths a, b and c, with the requirement that a + b + c = L.



Then, the number of possible outcomes after breaking a rod with length L into three pieces (within the context of forming a triangle) is 4 as summarized in the below table:


Table 1: When a rod with length L is broken into three random pieces with lengths a, b and c, respectively, we end up with 4 possible outcomes one of which allows the formation a triangle by using these three pieces. If any of a, b or c is larger than the half of the length of the original rod, a triangle by using these three pieces can not be formed.



Only one of these options (Outcome #4) forms a triangle. Due to the symmetric nature of the problem, one might intuitively think that all these outcomes have the same likelihood to occur. Then, the probability for forming a triangle by breaking a rod into three pieces is found to be:


Probability = # of outcomes forming a triangle / total # of outcomes = 1 / 4 = 25%


However, a thorough calculation is worth being given to prove that all possible outcomes given in Table 1 have the equal likelihood to occur. Here is the proof:



Problem:


Prove that the outcomes given in Table 1 are equally likely to occur.


Now, we would like to prove that the outcomes presented in Table 1 are all equally likely to occur. As is widely used in solving various probability problems, we will follow a geometrical approach for the proof. Each parameter in the problem will now be treated as a dimensional degree of freedom. This, then, allows us to define a three dimensional parameter space with mutually orthogonal axes denoted as a, b and c (see Fig. (1)). Unless there was any condition on the values of these parameters, we would simply have an indefinitely large sample space. However, according to our problem, we know that a, b and c should satisfy


a + b + c = L (see 2)


and none of these parameters can be negative. These requirements, then, suggest that we restrict ourselves to the triangular plane (this is nothing but the plot of Eq. (2) in abc-space) shown in Fig (1) with a light blue color. The area of this plane is composed of points which give all possible (a,b,c) values satisfying Eq. (2). So, this triangular plane will basically be our sample space containing all possible outcomes based on the above given requirements. Please note that, in Fig. (1), we show the sample space from different viewing angles so as to make the reader better visualize the picture. Hence, the sample space crosses the axes at a=L, b=L and c=L points, and it simply forms an equilateral triangle.



Figure 1: The plot of Eq. (2) in (abc)-parameter space. The shaded area defines the sample space used in the solution to the problem. The shaded region crosses the axes at a=L, b=L and c=L points.


Now, the problem has been reduced to finding the areas on this sample space representing each of the outcomes given in Table 1. Then, the probability for a given outcome to occur would simply be the ratio of the area representing this outcome to the whole area of the sample space. For example, let's consider the first outcome, i.e. the case when a>L/2, b<L/2 and c<L/2. In order to find the area which represents this choice, we simply draw the a = L/2 plane as shown in Figure (2) and determine the part on the sample space which intersects with the points in the region where a > L/2. We already know that if a > L/2 both b and c will be less than L/2. So, the area (red equilateral triangle) represents all possible values for (a,b,c) for Outcome #1 in Table 1.




Figure 2: The plane a=L/2 is used to find the part on the sample space given in Fig. (1) which represents all possible values for (a,b,c) satisfying the condition for the Outcome #1 to occur.



Now, from the geometry one can easily show that the red area turns out to be the one forth of the area of the sample space, leading the probability for the Outcome #1 to be 25%.


P (#1) = red area / area of the sample space = 1 / 4 (7)


Similarly, the other probabilities can be calculated following a similar approach. The planes b = L/2 and c = L/2 can be used to determine the parts on the sample space which represent the Outcome #2 and Outcome #3, respectively. Fig. (3) summarizes our results:




Figure 3: All parts on sample space representing the Outcomes in Table 1. The light blue region in the middle contains all the (a,b,c) values which satisfy the condition in Outcome #4; a < L/2, b < L/2 and c < L/2, and this represents the values of the parameters which would result in the formation of a triangle as asked in the original problem.


As is obvious here, all parts which represent the outcomes in Table 1 possess equal areas on the sample space, and they also cover the entire sample space with no overlaps. Thus, the outcomes in Table 1 are statistically independent and are all equally likely to occur. Then, we can conclude that


P(#1) = P(#2) = P(#3) = P(#4) = 25% (8)


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