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.
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|22/|111| |2/.|111| |/..|111|
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|333|../| |333|./2| |333|/22|
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