SOLUTION: Two points are chosen at random on a line segment whose length is a > 0. Find the probability that the 3 line segments thus formed can be the sides of a triangle.

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Question 397657: Two points are chosen at random on a line segment whose length is a > 0. Find the probability that the 3 line segments thus formed can be the sides of a triangle.
Found 2 solutions by robertb, Edwin McCravy:
Answer by robertb(5830)   (Show Source): You can put this solution on YOUR website!


Let two of the segments have lengths x, y. Then the third segment must have length a - x - y. Use the triangle inequality on 3 different instances:
(i)x + y > a - x - y <==> x + y > a/2.
(ii) x +a - x - y > y <==> a/2 > y
(iii) Similarly , from y +a -x - y > x we get a/2 > x.

The initial conditions are x >0, y >0, and a - x -y >0, or a > x+y. In the Cartesian plane, this feasibility region is an (open) isosceles triangle with vertices at (0,0), (0, a), and (a,0). It has area .
Incidentally, the region defined by the instances (i), (ii), and (iii) above is also an open isosceles triangle with vertices (a/2, 0), (a/2, a/2), and (0, a/2). This triangle has area .
Therefore the probability that the 3 line segments can be the sides of a triangle is .
Answer by Edwin McCravy(20054)   (Show Source): You can put this solution on YOUR website!


The other tutor is right! Here's a bit more geometrical drawings:  

Suppose this is the line segment.

0------------x----------y------------a

Then the sides of the triangle are x, y-x, and a-y

The sum of any two sides must be greater than the third side.

So

x + (y-x) > a-y,    x + (a-y) > y-x,     (y-x) + (a-y) > x      y > x
x + y - x > a-y     x + a - y > y-x            y-x+a-y > x 
        y > a-y       2x - 2y > -a                -x+a > x   
       2y > a        -2x + 2y < a                    a > 2x
        y > a/2       2y - 2x < a                  a/2 > x   
                       2(y-x) < a                    x < a/2   
                          y-x <           

We also know that y < a  and x > 0

So we have a system of inequalities:



We draw the boundary lines:




and find that the solution to the system of inequalities is the triangle
marked "SOLUTION".  It's area is 



So the numerator of the probability is the area of that triangle, for
the coordinates of any point within that triangle will represent points
on the original line:


0------------x----------y------------a

that satisy the system of inequalities.

The denominator of the probability is the area of this triangle:


 
Because the coordinates of any point inside this triangle represents 
a pair of values for x and y that could have been picked on this line:

0------------x----------y------------a
 
The area of this triangle is 

So the desired probability is  





Edwin
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