What is the probability that a line cut randomly into the pieces will have length that can form a triangle?
Three pieces.
3-4-5 yes.
3-4-9 no, for example.
Three pieces.
3-4-5 yes.
3-4-9 no, for example.
Puzzling
Favorite Answer
Look at Problem 2 in the link below.
Answer:
1/4
Steve A
Thank yoj for the link to that elegant solution.
I got 1/4 on a spreadsheet, but it was not elegant, and certainly not a proof. Assume a line length 100 and segment x is the shortest of the three (or tied), and all segments are integers. The number of possible lengths of the other two segments = (100 -3x)/2.
Of those, x/2 form triangles from x = 1 to x = 25 at which point x > (100 -3x)/2 and that limits the triangles from 25 to 33. x cannot exceed 100/3.
That got me to Denominator = sum(1,33) (100 -3x)/2 and numerator = Sum(1,25) n/2 + Sum(26,33) (100-3x)/2
The /2 cancel. Use integrals.
Integral (0,.25) (n) + Integral(.25,1/3) (1 - 3x) all divided by Integral (0,1/3) (1-3x)x
Yours is far better