Probability of Forming a Triangle

Date: 01/24/2001 at 22:35:45
From: Mike
Subject: Probability

I don't know where to begin. 

A rope one unit long is cut in two places. What is the probability 
that the three resulting pieces can be arranged to form a triangle?

Thank you.
Mike

Date: 01/25/2001 at 00:56:51
From: Doctor Pat
Subject: Re: Probability

Mike,

Here is one way, I got this as an answer from the nephew of the late 
Isaac Asimov, the Science writer, when I posed this problem on the 
Internet several years ago. Dan is a very good mathematician on his 
own, as you can see from his approach to the problem...

If we let T denote the planar triangle in 3-space whose vertices are 
(1,0,0), (0,1,0), and (0,0,1), then choosing a point p at random on T
(i.e. with the probability of p lying in a subset S of T being 
proportional to the area of S, or more precisely Prob(p is in S) = 
area(S) / area(T) will simulate the random cutting of the unit segment 
desscribed above, with resulting pieces of lengths x, y, and z 
respectively.
     
The condition that x, y, and z "will form a triangle" is equivalent to 
the three conditions:  x + y > z, x + z > y, and y + z > x.  The 
subset of T where (x,y,z) satisfy these three conditions turns out to 
correspond to the little triangle whose vertices are the midpoints of 
the edges of T. Therefore Prob(the 3 pieces will form a triangle) is 
one-fourth.

To which I added...

From the three conditions above, you can also deduce that the 
conditions  x + y > z, x + z > y, and y + z > x with the additional 
knowledge that x + y + z = 1 reduces to Max(x,y,z) < .5,  which offers 
another easy approach...

Hope that helps

- Doctor Pat, The Math Forum
  http://mathforum.org/dr.math/