<pre>
Q: If a line is cut in two independent random places, what are the odds
that the pieces can form a triangle?


A: 1/4.  Two proofs, one algebraic, one geometric:

Algebraic proof:

Let the length be 1 unit.  To form a triangle, the longest piece must
be less than .5, meaning that the two cuts must be on opposite sides of
the .5 mark.  The probability of this is 1/2.

Now let the cut that is below the .5 mark = x in length.  The other cut
must be between .5 and .5 + x.  For all x between 0 and .5, this
probability averages 1/2 (assuming they are on opposite sides of the .5
mark).  Multiplying the two probabilities gives 1/4.

Geometric proof:

If you pick a random point (x,y) in a square and break the stick at
the points of both coordinates, this is equivalent to breaking a stick
in two random places.

When can you get a triangle from the remaining pieces?  If x and y are
both less than 1/2, then the third piece will be longer than 1/2; this
has probability 1/4.  If x and y are both more than 1/2, then the first
piece will be longer than 1/2; this has probability 1/4.  If 0<x<y-1/2
or 0<y<x-1/2, then the middle piece will be longer than 1/2; this has
probability 1/4.  (See the figure for the corresponding regions of the
square, which are all disjoint.)  The remaining probability is 1/4.

 +---+---+
 ||22/||111||
 ||2/.||111||
 ||/..||111||
 +---+---+
 ||333||../||
 ||333||./2||
 ||333||/22||
 +---+---+
</pre>
