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A line segment is broken at two random points along its length. What is the probability that the three new segments can be arranged to form a triangle?

Nov 24, 2018

#1
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Thinking about this for a bit it seems to me the only way you can't form a triangle with the 3 segments is if one of those segments has length greater than or equal to half the length of the original segment.

Suppose our original segment is length 1.

We can get a length 1/2 or greater segment in 3 ways

a) both points are greater than 1/2.

b) both points are less than 1/2.

c) the difference between the two points is greater than 1/2

The region satisfying the union of these constraints is as shown below

And it's easy to eyeball that given underlying uniform distributions that the region of interest (beige)

has an area of 3/4 and thus

$$P[\text{triangle cannot be made from 3 segments formed by 2 random points}]=\dfrac 3 4\\ P[\text{triangle can be made}] = 1 - \dfrac 3 4 = \dfrac 1 4$$

Another set of eyes on this would be helpful.

Nov 24, 2018
edited by Rom  Nov 24, 2018
edited by Rom  Nov 24, 2018
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Why do you want another set of eyes Rom?

Melody  Nov 24, 2018
#3
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this is correct then?

Rom  Nov 24, 2018
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Don't know, I've answered this question before, more then once, but I never find it  simple....

mmmmm  I could just chase down an old answer .... would that be cheating?  That is rhetorical,    LOL.

Melody  Nov 24, 2018
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Here is my take.

A line segment is broken at two random points along its length. What is the probability that the three new segments can be arranged to form a triangle?

Let the length of the line be 1 unit.

Let the lengths of the segments be  x, 1-y, and y-x   units.      (if you add these they equal 1)

Now all these segments must be between 0 and 1 unit

So I will graph this and the area will give the full population of possible cuts.

Represented on this contour map in dark green

Now lets see how much of this is included if the peices can form a triangle.

To make a triangle the sum of the smaller peices has to be bigger than the length of the longest one.

so all three of the following conditions must be met.

$$x+1-y>y-x, \qquad x+y-x>1-y, \qquad 1-y+y-x>x\\ \text{These simplify to }\\ y 0.5, \qquad x<0.5$$

So now I can plot these over the population and see what happens.

I can see that a quarter of the population triangle is included so

The probablility that the peices will be able to form a triangle is 1/4

(I keep editing this because some of what I write keeps getting deleted. There must be more elements here than web2.0 can happily cope with. I will see what happens this time but if it is not complete you will have to join the dots for youselves :/)

Nov 24, 2018
edited by Melody  Nov 24, 2018
edited by Melody  Nov 24, 2018
edited by Melody  Nov 24, 2018