Question

Take a stick of unit length and break it into two pieces, choosing the break point at random. Now break the longer of the two pieces at a random point. What is the probability that the three pieces can be used to form a triangle?

Answer 1

Answer:

≈ 38.6%

Step-by-step explanation:

For three lengths to form a triangle, the sum of the shortest two must be greater than the longest.

a + b > c

We start with a stick of length 1.  Let's say the shorter length after the first break is a, such that 0 < a < 0.5.  The remaining length is 1−a.

On the second break, we get two pieces, b and c.  Each piece can be anywhere between 0 and 1−a, but in order to form a triangle, neither can be more than 0.5.  If b = 0.5, then:

c = 1−a−b

c = 1−a−0.5

c = 0.5−a

So each b and c must be between 0.5−a and 0.5.  The width of that range is 0.5−(0.5−a) = a.  So the probability that they can form a triangle is a / (1−a).

a can be anywhere between 0 and 0.5, so the cumulative probability is:

$\frac{\int\limits^\frac{1}{2} _0 {\frac{x}{1-x} } \, dx}{\frac{1}{2}-0}\\2\int\limits^\frac{1}{2} _0 {\frac{x}{1-x} } \, dx$

Using u substitution:

u = 1−x, du = -dx

$-2\int {\frac{1-u}{u} } \, du\\-2\int {\frac{1}{u} - 1} \, du\\-2(ln|u|-u)+C\\-2(ln|1-x|-(1-x))+C\\-2(ln|1-x|-1+x)+C\\-2ln|1-x|+2-2x+C\\-2ln|1-x|-2x+C$

Evaluating this between x=0 and x=0.5:

[ -2 ln|1−0.5| − 2(0.5) + C ] − [ -2 ln|1−0| − 2(0) + C ]

[ -2 ln|0.5| − 1 + C ] − [ C ]

-2 ln|0.5| − 1

ln|4| − 1

≈ 0.386

Answer 2

The probability that the three pieces can be used to form a triangle is 0.3863

Represent the lengths of the stick with x, y and z.

Where z represents the longest side of the triangle, and x represents the shortest side length

By triangle inequality theorem, we have:

$z < x + y$

Assume the length of the stick is a unit length, the following equation must be true:

$x + y + z = 1$

Make z, the subject

$z = 1 - x - y$

Where:

$0 < x < y$

We can assume that y is at the 0.5 position.

So, we have:

$0 < x < 0.5$

and

$z=1 -x -0.5$

Collect like terms

$z=1 -0.5-x$

Evaluate

$z=0.5-x$

The probability that the three pieces can form a triangle is represented as:

$p = \int\limits^{y}_0 {\frac{x}{1 - x}} \, dx \div y$

So, we have:

$p = \int\limits^{0.5}_0 {\frac{x}{1 - x}} \, dx \div 0.5$

Divide

$p = 2\int\limits^{0.5}_0 {\frac{x}{1 - x}} \, dx$

Let

$u = 1- x$

So, we have:

$du = -dx$

The equation becomes

$p = -2\int\limits^{0.5}_0 {\frac{x}{u}} \, du$

This gives

$p = -2\int\limits^{0.5}_0 {\frac{1 - u}{u}} \, du$

Split

$p = -2\int\limits^{0.5}_0 {\frac{1}{u} - 1} \, du$

Integrate

$p = -2*[ {\ln(u) - u]\limits^{0.5}_0$

Substitute $u = 1- x$

$p = -2*[ {\ln(1 - x) - 1 + x]\limits^{0.5}_0$

Expand

$p = -2*[ \ln(1 - 0.5) - 1 + 0.5] -[ \ln(1 - 0) - 1 + 0]]$

$p = -2*[ \ln(1 - 0.5) -0.5] -[ \ln(1 - 0) - 1]]$

$p = -2*[ \ln(0.5) -0.5] -[ \ln(1) - 1]]$

Open the bracket

$p = -2*[ \ln(0.5) -0.5 - \ln(1) + 1]$

Evaluate the like terms

$p = -2*[ \ln(0.5)- \ln(1) + 0.5]$

Evaluate the expression

$p = 0.38629436112$

Approximate

$p = 0.3863$

Hence, the probability that the three pieces can be used to form a triangle is 0.3863

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