# What is the probability that the resulting four line segments are the sides of a quadrilateral?

Question: Divide a given line segment into two other line segments. Then, cut each of these new line segments into two more line segments. What is the probability that the resulting line segments are the sides of a quadrilateral?

I am stuck on this problem. I think I am close, but I am not sure if it is correct. Any help or conformation on this would be helpful.

My thoughts on the problem:

Let us say that the line segment is of length 1. The only restriction for these for line segments to form a quadrilateral is that no one side > .5 (correct me if I am wrong).

With our first cut we have two smaller line segments, one larger than the other. We only need to look at the longer on of these two line segments. Let us call $y$ the smaller line segment and $x$ the larger one. $x$ will be between 0.5 and 1.

When we cut each of these new line segments we need to find where it does not work for it to be a quadrilateral. We only have to look at $x$. Let us call $a$ the length that we cut.

If we use the example $x=0.6$ we can see that $a$ cannot be less than 0.1 or greater than 0.5.

We can generalize this for any $x$. $a$ cannot be less than $(x-1/2)$ or greater than $1/2$.

This is where I get stuck.

I believe that the probability for any $x$ value that these four line segments will not be a quadrilateral is $$\frac{2(x-1/2)}{x}$$

If this is correct is the total probability that it cannot be a quadrilateral $$\int\limits_{1/2}^1\frac{2(x-1/2)}{x}\mathrm{d}x?$$

Any help is much appreciated. Thank you

• +1 for showing your work. In the very last integral, $\mathrm dx$ should be $2\mathrm dx$. Apart from that, your reasoning looks fine. To compute the last integral, use $\int(4-2/x)\mathrm dx=4x-2\log x$, which yields the numerical value $2-2\log2$. – Did Jan 9 '13 at 6:06
• Why should it be $2dx$? – yousuf soliman Jan 9 '13 at 6:09
• Because the total mass of $\mathrm dx$ on $[1/2,1]$ is $1/2$ and you want a total mass of $1$. – Did Jan 9 '13 at 6:15
• I am sorry, I'm not sure I entirely understand this. I am fairly new to Calculus in general and have never come across a situation like that. Could you please explain a little more in depth? Thank you! – yousuf soliman Jan 9 '13 at 6:19
• @MuadDib42 Another reason why there is a 2, is because you did a symmetric argument and only considered the case where the break occurred in $[0.5, 1]$, instead of when the break occurred in the entire domain $[0,1]$. – Calvin Lin Jan 9 '13 at 7:07

## 1 Answer

In your very last integral, $\mathrm dx$ should be $2\mathrm dx$ $(*)$. Apart from that, your reasoning looks fine. To compute the last integral, use $\displaystyle\int(2−1/x)\mathrm dx=2x−\log x$, which yields the numerical value $2−2\log2\approx61\%$ for the probability of no quadrilateral.

$(*)$ To understand why, note that the total mass of $\mathrm dx$ on $[1/2,1]$ is $1/2$ and that one wants a total mass of $1$. This indicates that the probability density function of $X$ the longest of the two first lengthes is $2$ on the interval $[1/2,1]$. Once $X=x$ is known, the probability of no quadrilateral is $p(x)=2(x-1/2)/x$, as you aptly showed, hence the total probability of no quadrilateral is $$\int_{1/2}^1p(x)\,(2\mathrm dx)=2\int_{1/2}^1(2-1/x)\mathrm dx.$$