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I've a broken stick (Posted on 2003-12-07) Difficulty: 5 of 5
I've a straight stick which has been broken into three random-length pieces.
What is the probability that the pieces can be put together to form a triangle?
If you can answer this at this point, please do.
If not, perhaps this will help: here are several methods to break the stick into the three random length pieces:
  1. I select, independently, and at random, two points from the points that range uniformly along the stick, then break the stick at these two points.
  2. I select one point, independently, and at random (again uniformly), and break the stick at this point. I then randomly (with even chances) select one of the two sticks and randomly select a point (again uniformly) along that stick, and break it at that point.
  3. I select one point, independently, and at random (again uniformly), and break the stick at this point. I then select the larger stick, and randomly select a point (again uniformly) along that stick, and break it at that point.
If this clarifies the problem, please show how this affects your work.

No Solution Yet Submitted by SilverKnight    
Rating: 3.6000 (5 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution solution?? | Comment 25 of 26 |
<BLOCKQUOTE>Let the points be A and B, so 0<=A<=1 and 0<=B<=1. For the three line segments to form a triangle, the sum of the lengths of any
two must be greater than or equal to the length of the third.

Case 1: If 0<=B<=A, the three line segments have lengths B, A-B, and 1-A. For these to form a triangle, the following three conditions must be true:

</BLOCKQUOTE> <BLOCKQUOTE>    B + (A-B) >= 1-A    ==>    A >= 1/2
    (A-B)+(1-A) >= B    ==>    B <= 1/2
    B + (1-A) >= A-B    ==>  A-B <= 1/2

Case 2: If A<=B<=1, the same conditions must apply as in case 1, except with A substituted for B and B subsituted for A, that is:

    A + (B-A) >= 1-B    ==>    B >= 1/2
    (1-B)+(B-A) >= A    ==>    A <= 1/2
    A + (1-B) >= B-A    ==>  B-A <= 1/2

Plotting these cases on a graph with horizontal axis for A and vertical axis for B, we get:

 1  +--+--+
    | /|  |
    |/t|  |
1/2 +--+--+
    |  |t/|
    |  |/ |
 0  +-----+
    0 1/2 1
 

The two triangles labeled t are the only regions of this unit square where A and B create three line segments that agree with the conditions for case 1 (the lower right triangle) or case 2 (the upper left triangle).  They total 1/4 of the area of the unit square, so the probability of two randomly chosen points craeting line segments that form a triangle is 1/4.

</BLOCKQUOTE>
  Posted by lee kian keong on 2004-07-08 02:31:03
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