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 A

_{0}to A

_{5}.

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} \$

## No comments:

Post a Comment