Randomly cut a line segment into three pieces.What is the probability that these three pieces can form a triangle?

Question

Sigiloso · Accepted Answer

1/8. let x and y be lengths of 2 segments, 1-x-y length of the 3rd segment. twe following inequations should be true: x + y > 1-x-y, x + 1-x-y > y, y + 1-x-y > x. therefore x + y > 1/2, x  0, y > 0)

Anh Le · Answer

I modify the above answer a little bit:

Let x and y be lengths of 2 segments, 1-x-y length of the 3rd segment. 

We have triangle inequalities: 

x + y > 1-x-y

x + 1-x-y > y

y + 1-x-y > x

Therefore, we have:  x + y > 1/2, 1> x 0, 1> y > 0. The probability is the ratio between two areas of the sub trapezoid and the unit triangle.

P = area(trapezoid) / area(triangle) = 1 - area(sub-triangle) / area(triangle) = 1 - 1/4 = 3/4

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