A Limit Approach to a Classic Geometric Probability Problem

One of the common problems from probability texts is : "A stick is broken into three parts. What is the chance that the sum of the lengths of eveiy two is greater than the length of the third that is that the three parts will form a triangle?"

The traditional method for an area approach for pre-calculus students is to begin by assuming the stick is of unit length, and the breaks occur at distances of x, and x+y from one end, giving three pieces of length x, y, and 1-x-y. Then a graph of all possible outcomes can be found by graphing the lenghts x and y on the coordinate plane. The regiong bound by x>0, y>0 and x+y<1 defines a triangle containing all possible solutions. To find the subset which will fit together to form a triangle, we think about the basic triangle inequality from plane geometry; "any two sides must be greater than the third side." That restriction means x <1/2, (otherwise, the other two sides will be less than x) and similarly y<1/2. Finally, since the remaining part, 1-x-y must also be less than 1/2, x+y > 1/2. By adding these three lines to our graph, we see that solutions which make a triangle form must fall into the smaller triangle, which is 1/4 of the total sample space of outcomes.


I was reminded of this problem by an article in the Mathematical Spectrum, Vol 2, #2. by L. Rade. He begins with a finite version of the problem. A stick of length five units is cut into five equal spaces with six endpoints at A₀ to A₅.

Two of the four interior points are selected at random and the stick is broken at those two points. What is the probability that the broken pieces form a triangle?


He shows that there are $\dbinom{4}{2}$ , or 6 ways to break the stick, and only three of those will form a triangle. He then uses the lengths of the first two sticks (from the left) to plot a point on the xy plane to show the sample space of all solutions, and then a subset of those that form a triangle.
Then he proceeds to show that for five, six, seven, and eight points equally spaced on the stick, the probabilities of getting a triangle are 1/2, 1/10, 2/5, and 1/7 respectively. Finally he generalizes the previous graph to n points


and shows that the total number of points in the sample space is the triangular number for n-2, $\sum_{i=1}^{n-2}i = \frac{n^2-1}{8}$ .
Then he shows that the number of points in the sample space is given by a different formula if n is even or odd, and produces the probability of forming a triangle as [and guess who is doing piece-wise functions right now in pre=calc]
Finally, we derive the answer for an infinite number of possible points by letting n go to infinity and examine the graph of P(n)
I think this might be a very effective way to present this problem, and probably end up with a little less of the glassy eyed stare when you finish; and it might also provide a good model for them to attack geomtric probability problems in the future.

A history note:
Dr. David Singmaster's web page on the chronology of recreational mathematics gives "1873 Lemoine considers Probability that Three Lengths Form a Triangle." However, I have found the problem presented in the 1865 "A treatise on infinitesimal calculus" by Bartholomew Price. . I have asked Dr. Singmaster to comment, and if he has additional detail, I will include it here in a future edit.
and a followup question Given that the broken stick forms a rectangle, what is the probability that the triangle is acute? I just noticed after I had typed all this, that the followup question was sent to the Mathematical Spectrum by the same Dr. David Singmaster I have referred to above.

And a followup history note: J. J. Sylvester was the first to pose this question in 1865, according to David Singmaster's chronology.


just ... fiddling around.. , this is me trying to write piecewise functions in LaTeX
\$f(x)=\begin{cases}
\frac{n-4}{4(n-1}&\text{for n even } \\
\frac{n+1}{4(n-2)}&\text{for n odd} \end{cases} \$