# probability and statistics

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(Right Triangle Problem) Randomly cut a line segment into three pieces. What is the probability that these three pieces can form a right triangle?

• probability and statistics -

What a neat question !

Some starting thoughts ....
the line would have to be cut into the ratio:
3:4:5 or
5:12:13
7:24:25
8:15:17
etc

using the ratio 2mn : m^2-n^2 : m^2 + n^2 , where
1. m > n
2. m and n are relatively prime
3. one of m and n is even, the other is odd
to get triples in lowest ratio terms that will form right-angled triangles

e.g. for 8:15:17 ---> 8^2 + 15^2 = 17^2

The problem is that there is an infinite number of such triples, and we can cut the line in an infinite number of ways. But intuitively we feel that that has to be more of the latter than the former, so there are different degrees of infinity ??
I suggest you Google "Aleph-null" or Aleph-naught" to find out more.

One might set up a set of empirical data by setting up some kind of computer program which randomly selects any 3 term ratio, then tests if the selected ratio satisfies the Pythagorean condition.

• probability and statistics -

I would encourage a you run a simulation of 1E6 simulations, and compute probability.

computer pseudo program:
set endnumber=1,000,000
set s=0, f=0
error= .0000000001
newtrial:
take L, cut a piece from it 0 to a
then cut L-a in to two pieces, b, and c.

if a^2-(b^2+c^2)<error or b^2-(a^2+c^2)<error or if c^2-(a^2 +b^2)<error then goto success
otherwise go to failure

success: S=S+1
Failure: F=F+1

when S+F= endnumber, stop, print s/(s+F)
otherwise new trial.

end

I suspect you will get zero, but if you make error say .1, you might get some successes.

I am reminded of a much easier triangle q: if one cuts a straw randomly into three pieces, what is the probability that the three pieces will make a triangle. Because I am so kind an generous, the answer is 1/4, but I leave it to you to figure it out why.

• probability and statistics -

Sorry, my computer was on the blink, and I am catching up!

If the line segment is "commensurable" in Euclid's terms, i.e. the original length is an integer in modern terms, then it is possible to calculate the probability.

The problem is one of partitioning an integer into three parts, in combinatorics terms.

If the integer (total length) is known, the probability can be calculated explicitly, since there is only a limited number of ways of partitioning an integer into three (positive) integers.

If the integer is not known, then the problem is more complex, namely we have to calculate the cumulative probability for lengths of 3 up to infinity, if there exists such a solution.

If the lengths are not commensurable, i.e. not rational, the probability is close to nil. We can see this by realizing that the number of irrational numbers between any two consecutive integers is infinite. So, like Reiny said, there are different degrees of inifinity!

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