
Re: probability of a triangle (SPOILER)
Posted:
Jun 7, 1996 5:28 AM


Dan,
A very pretty solution. My own approach was twodimensional but also used geometric probability. The three segments have length x, y, and 1xy so the domain is the right triangle D whose corners are (0, 0), (1, 0), and (0, 1) . . . any point inside this figure represents a possible breaking of the unit segment. The three triangle inequalities become x + y > 1/2, x < 1/2, y < 1/2 which define a right triangle formed by joining the midpoints of D. The probability of forming such a triangle is 1/4.
Alan Lipp

