Question #381

The stick of the length L is broken in two randomly selected points. With what probability it is possible to make a triangle with its parts?

Expert's answer

If we take x and y as the distance from the left end of the stick to the points of break, the set of all outcomes can be shown on a square of the side l, one side of which lies on the coordinate axis x and the other – on the coordinate axis y. If we take the condition y>x, then the set of outcomes displayed on the triangle OBC. The area of this triangle is L2/2. The parts of the stick will have the lengths: x, y-x, and L-y.

Now remember geometry. We can form a triangle with the three segments if and only if the length of each segment is less than the sum of the lengths of the other two segments. This condition in our case leads to a system of three inequalities:

x < y-x+l-y

l-y < x+y-x

y-x < x+l-y

The first inequality is transformed to the form x<(L/2), the second - to the form y>(L/2), and the third inequality – to the form y<(L/2). The set of pairs of numbers x and y is a solution of inequalities shown in the shaded triangle. The area of this triangle is 4 times smaller than the area of a triangle OBC. Thus, the answer is 1/4.

Now remember geometry. We can form a triangle with the three segments if and only if the length of each segment is less than the sum of the lengths of the other two segments. This condition in our case leads to a system of three inequalities:

x < y-x+l-y

l-y < x+y-x

y-x < x+l-y

The first inequality is transformed to the form x<(L/2), the second - to the form y>(L/2), and the third inequality – to the form y<(L/2). The set of pairs of numbers x and y is a solution of inequalities shown in the shaded triangle. The area of this triangle is 4 times smaller than the area of a triangle OBC. Thus, the answer is 1/4.

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