If the wire of length $\ell $ is cut into three pieces then find the probability that the three pieces form a triangle

Question

Question: If the wire of length is cut into three pieces then the probability that the three pieces form a triangle is ____

My approach is as follow

For a triangle to exist the sum of two sides should be greater than the third side

Hence $x + y + z = \ell \Rightarrow z = \ell - \left( {x + y} \right) > 0$

$x > 0,y > 0,z > 0$

$x + y > z \Rightarrow x + y > \ell - \left( {x + y} \right) \Rightarrow x + y > \frac{\ell }{2}$

$y + z > x$ & $y + z > \frac{\ell }{2}$

$z + x > y$ & $z + x > \frac{\ell }{2}$, therefore $x + y + z > \frac{{3\ell }}{4}$

How do we proceed further

Answer 1

Solution 1 :

Let $x$ be the first cut , then $y$ be the second cut.
Then $z$ is remaining.

This will not form a triangle when $z$ is longest with $x+y$ less than half.

We can see with the lines $x+y=1$ (Purple) & $x+y=0.5$ (Light Gray) , that the Probability (Dark Gray triangular Area) is $1/4$.

Likewise , we do not get a triangle when $y$ is longest with $z+x$ less than half & we do not get a triangle when $x$ is longest with $y+z$ less than half , hence that Probability is $3 \times 1/4 = 3/4$

The Probability that we will get the triangle is $1-3/4=1/4$

Solution 2 :

In 3D Coordinate System , with $(x,y,z)$ on the 3 Axes , we can see that every Point on the triangle between $(1,0,0)$ & $(0,1,0)$ & $(0,0,1)$ will have $x+y+z=1$ & hence that can be a way to cut the wire.

That triangle in 3D is like this :

The Central Area is where we get a triangle , while the Corners will not give triangles.

Hence Probability is $1/4$

Answer 2

You’d have to describe the process of “cutting in pieces” precisely.

I cut the wire at a random point with linear distribution. If I then cut the smaller piece, it’s impossible to get a triangle. If I cut the longer piece then I get some probability, if I first mark the complete wire in two places and cut at the two marks I get a different probability.