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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.)
I think this is a solution | Comment 16 of 26 |
We have a stick of finite length L. We break it into three pieces a, b, and c, such that a+b+c=L.

We can graph the function a+b+c=L on a three-dimensional axes. Since a, b, and c are all positive, we are confined to the
first octant. We find that the solution space is a finite equilateral triangular plane, with vertices at (L,0,0), (0,L,0), and (0,0,L), and sides of length sqrt(2)*L. Total area of plane = sqrt(3)/4*(sqrt(2)*L)^2=sqrt(3)/2*(L^2).

The components of every point on the solution space add to L, but the components of every point do not constitute the requirements to form a triangle.

The fraction of the area of the solution space that have coordinates that can be made into triangles corresponds to the probablity that three segments taken at random will form a triangle.

It can be seen that a triangle cannot be made if two of the sides sum to less than half the total length of the stick.

Using our three dimensional graph, we can find the projection onto the solution space in which a+b < (1/2)*L, b+c < (1/2)*L, a+c < (1/2)*L. Those are the areas in which the solution points do not constitute the neccessary conditions for a triangle.

The remaining area will encompass the points where triangles can be formed.

Relative to our triangular solution space, we find that the vertices get cut out, that is, there are smaller triangles at the vertices, inside which area the points taken will not form triangles, because two of the three components sum to less than (1/2)*L.

The sides of these smaller triangles are given by sqrt(2)*(1/4)*L. The 1/4 follows from the fact that two of the components are less than or equal to (1/4)*L, and therefore must sum to less than (1/2)*L. Again using the formula for the area of an equilateral triangle we get sqrt(3)/4*(sqrt(2)*1/4*L)^2=sqrt(3)/32*L^2 for the smaller areas. There are three of these areas so the total area that contains points that will not make triangles is 3*sqrt(3)/32*L^2.

Finally, the probability that a point picked at random does not fall on this space is

sqrt(3)/2*(L^2) - 3*sqrt(3)/32*L^2
_____________________________________

sqrt(3)/2*(L^2)



which reduces to 1-(3/16)=13/16. So you have a 13/16 probability that three random lengths will have the right properties that they can form a triangle.
  Posted by puzzlesrfun on 2003-12-09 21:09:45
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