99***. A rod is broken into two parts at a randomly chosen point; then the larger of the two resulting parts is again broken into two parts at a randomly chosen point. What is the probability that a triangle can be formed from the three resulting pieces?
99. For simplicity, let's assume that the length of our rod is 2 (this is legitimate, as we can always take half the length

![](https://cdn.mathpix.com/cropped/2024_05_21_68334347b64402497ba5g-331.jpg?height=559&width=378&top_left_y=968&top_left_x=149)

Fig. 87. of the rod as the unit of length). Let \( AB \) be our rod, \( K \) and \( L \) the points of the first and second break, respectively. The outcome of the experiment in this problem is completely determined by the numbers \( x = AK \) and \( z = KL \), where \( x \) is chosen randomly between 0 and 1 ( \( x \) is the length of the smaller of the two parts formed after the first break, and thus \( x \leq 1 - x \)).

Since \( y = \frac{z}{2 - x} \), dividing the written inequalities by \( 2 - x \), we get that favorable outcomes will correspond to pairs of numbers \( (x, y) \) such that \( y \geq \frac{1 - x}{2 - x} \), i.e.,

\[
y \geq 1 - \frac{1}{2 - x}
\]

We now need to visualize the region on Fig. 88 defined by these inequalities.

Introduce instead of coordinates \( (x, y) \) coordinates \( \left(x', y'\right) \), where

\[
y' = y, \quad \text{and} \quad x' = 2 - x
\]

Such a change of coordinates is equivalent to moving the origin to point \( O' \), located on the x-axis at a distance of 2 from the old origin and changing the direction of the x-axis (see Fig. 88). In the new coordinates, the inequalities (*) will transform as follows:

\[
y' \geq 1 - \frac{1}{x'}
\]

The first of these inequalities indicates that all points corresponding to favorable outcomes will be located below the hyperbola

\[
y' = \frac{1}{x'}
\]

Now let's construct the curve

\[
\begin{gathered}
y' = 1 - \frac{1}{x'} = \frac{1}{2} - \left(\frac{1}{x'} - \frac{1}{2}\right) \\
y' - \frac{1}{2} = -\left(\frac{1}{x'} - \frac{1}{2}\right)
\end{gathered}
\]

This curve will obviously be symmetric to the curve \( y' = \frac{1}{x'} \) relative to the line \( y' = \frac{1}{2} \) (see Fig. 88). The second inequality (**) shows that all points corresponding to favorable outcomes are located above the curve

\[
y' = 1 - \frac{1}{x'}
\]

Thus, favorable outcomes correspond to points filling the shaded area on Fig. 88. We need to determine the area of this region.

First, note that due to the symmetry of the curves

\[
y' = \frac{1}{x'} \text{ and } y' = 1 - \frac{1}{x'}
\]

relative to the line \( y' = \frac{1}{2} \), the area of the curvilinear triangle \( OTW \) is equal to the area of the curvilinear triangle \( TUV \); thus, the desired area is \( 1 - 2 \) area \( TUV \). But due to problem 152, the area of the curvilinear trapezoid \( OTVW \) is \( \ln 2 \) (where \( \ln \) denotes the natural logarithm, i.e., the logarithm with base \( e = 2.718 \ldots \)). From this, it follows that the area \( TUV = 1 - \ln 2 \) and, consequently, the area of the curvilinear triangle \( TVW \) is

\[
1 - 2(1 - \ln 2) = 2 \ln 2 - 1 = 0.388 \ldots
\]

This expression is also equal to the probability that the segments \( x, z \), and \( 2 - x - z \) can form a triangle.