(Right Triangle Problem) Randomly cut a line segment into three pieces. What is the probability that these three pieces can form a right triangle?
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.
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.
Sorry, my computer was on the blink, and I am catching up!
Here are my comments.
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!
To solve this problem, we need to consider the conditions for the three line segments to form a right triangle.
A right triangle is a triangle in which one of the angles is a right angle (90 degrees). For the three pieces of a line segment to form a right triangle, the Pythagorean theorem must hold true, which states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the other two sides.
Let's denote the three line segments as a, b, and c, with a being the longest segment (hypotenuse) and b and c being the other two segments.
Given that the three line segments are randomly cut, we can assume that each of them has a uniform probability distribution. Now, we'll analyze the possible ranges of values for a, b, and c.
Since a is the longest segment (hypotenuse), it should be greater than both b and c. Hence, let's assume that 0 < c < b < a.
Given these conditions, we can visualize the possible values geometrically. The lengths of b and c can vary in the range of 0 to a, and a can vary from a minimum length if the two smaller segments are zero, to infinity if b approaches a and c approaches 0.
Now, to find the probability of forming a right triangle, we need to determine the valid values of a, b, and c that satisfy the Pythagorean theorem.
The Pythagorean theorem is given by:
a^2 = b^2 + c^2
Given the range of possible values for a, b, and c, we can calculate the probability by integrating over the valid region in the a-b-c space. The valid region will be determined by the inequality a > b + c.
This integration can be quite complex, involving double integrals to determine the probability density function. However, I will provide a simplified approach to estimate the probability.
We can consider an equilateral triangle with a side length of 1 as a reference. In this case, the hypotenuse (a) can range from 1 to √2, while b and c can vary from 0 to 1.
The valid region in this equilateral triangle is the area under the line a = b + c, which represents the condition for a right triangle. This forms a smaller triangle within the equilateral triangle.
By calculating the areas of the smaller triangle and the entire equilateral triangle, we can estimate the probability using the formula:
Probability = Area of smaller triangle / Area of entire equilateral triangle
Now, applying the formula, we find:
Area of entire equilateral triangle = (1/2) * base * height = (1/2) * 1 * (√3/2) = sqrt(3)/4
For the smaller triangle, we can find its area by integrating as follows:
Area of smaller triangle = ∫[0 to 1] ∫[0 to 1−c] (1−c−b) db dc = 1/24
Therefore, the probability that three randomly cut pieces can form a right triangle is:
Probability = Area of smaller triangle / Area of entire equilateral triangle
= (1/24) / (sqrt(3)/4)
= 1 / (6 * sqrt(3))
≈ 0.096225
Hence, the probability is approximately 0.096225 or about 9.6225%.