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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?

 Mar 29, 2018
 #1
avatar+748 
+2

Suppose the points where we break the line are x and y, where 0 < x < y < 1. So our "possible" region is the following triangle on the plane:



The three sides have lengths x, y-x, and 1-y. We must satisfy the Triangle Inequality on all three sides, meaning that the sum of any two of these lengths must be greater than the third. This gives us

1.  : This gives y>1-y, so y> 1/2.

2. : This gives 1-x> x, so x<1/2.

3.  : This gives  , so  .

Adding these lines to the diagram gives us the darker "success" region.



The darker region makes up 1/4 of the total region, so the probability is 1/4.

 Mar 29, 2018
 #2
avatar+111329 
+1

Very nice, MIRB16  !!!

 

 

cool cool cool

 Mar 29, 2018
 #3
avatar+109520 
+2

 

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?

 

I would do this with a boolean contour map.

The line segment can be any length so let it be 1

Let the segment by cut into segments that are    x,  1-y and   y-x    units long.

(If you add these you can see that the total length is 1 unit.)

 

No each of these lengths must be between 0 and 1 unit long.

so

\(0 x \quad (3) \qquad and \qquad y

 

I need to plot these to get the full sample space.  The dark triangle is the sample space.

 

 

Now in order for the 3 peices to from a triangle, the sum of the two little ones must be greater than the length of the long one.

Remembering that the  segments  are    x,  1-y and   y-x    units long.

 

 

\(x+(1-y)>y-x\\ 2x+1>2y\\ y

 

AND

 

\(x+(y-x)>1-y\\ y>1-y\\ 2y>1\\ y>0.5 \qquad(6)\)

 

AND

 

\((1-y)+(y-x)>x\\ 1-x>x\\ 1>2x\\ x<0.5 \qquad (7)\)

 

 

So a quarter of the sample space is the area of the desired region.

 

P(the three pieces form a triangle) = 0.25

 Mar 29, 2018
 #4
avatar+109520 
+1

I see you beat me to the finish line MIRB16.      laugh

At least we got the same answer, always a good sign :)     

 Mar 29, 2018

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