Question 397657



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 {{{a^2/2}}}.

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 {{{(1/2)*(a/2)*(a/2) = a^2/8}}}.

Therefore the probability that the 3 line segments can be the sides of a triangle  is {{{(a^2/8)/(a^2/2) = 2/8 = 1/4}}}.