Date: Jun 7, 1996 5:28 AM
Author: Alan Lipp
Subject: Re: probability of a triangle (SPOILER)

Dan,

A very pretty solution. My own approach was two-dimensional but also used
geometric probability. The three segments have length x, y, and 1-x-y 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