98****. A rod is broken into three parts; two break points are chosen at random. What is the probability that the three resulting pieces can form an acute triangle?
98. This problem considers the same experiment as problem 93. As in the second solution of problem 93, all possible outcomes of this experiment will be characterized by two numbers: $A K=x$ and $K L=z$ (see diagram $80, a$). In this case, the set of all outcomes is represented by the set of points of the triangle $O S T$, bounded by the coordinate axes and the line $x+z=l$ (diagram 86; see also diagram 81), and the probability of any event will be equal to the ratio of the area of the part of the triangle corresponding to favorable outcomes for this event to the area of the entire triangle.

We need to determine for which values of $x$ and $z$ it is possible to form an acute-angled triangle from three segments of lengths $x, z$, and $l-x-z$. As shown in the solution to the previous problem, for this to be true, the square of the length of each of these segments must be less than the sum of the squares of the lengths of the other two. Thus, unfavorable outcomes of the experiment will correspond to points of the triangle $O S T$ for which one of the following three inequalities is satisfied:

$$
\left.\begin{array}{c}
x^{2} \geqslant z^{2}+(l-x-z)^{2}, \quad z^{2} \geqslant x^{2}+(l-x-z)^{2} \\
(l-x-z)^{2} \geqslant x^{2}+z^{2}
\end{array}\right\}
$$

It is clear that no two of the listed inequalities can be satisfied simultaneously (since from each of them it follows that the length of the segment in the left part of the inequality is greater than the sum of the two other lengths; see footnote on page 326); therefore, no two of the three regions defined by these inequalities will intersect. Furthermore, when the rod is randomly divided into three pieces, the probability that the square of the length of one specific piece will exceed the sum of the squares of the lengths of the other two is the same for all three pieces of the rod ${ }^{1}$); hence, the areas of the three regions of the triangle $O S T$ defined by the three inequalities (*) must be the same. Therefore, to find the area of the region corresponding to all unfavorable outcomes of the experiment, it is sufficient to calculate the area of the region defined by any one of the inequalities (*) and multiply it by 3.

Now let's find the area of that part of the triangle $O S T$ for which the inequality

$$
(l-x-z)^{2} \geqslant x^{2}+z^{2}
$$

is satisfied. Expanding the brackets, this inequality can be rewritten as

$$
l^{2}+2 x z-2 l x-2 l z \geqslant 0
$$

Adding $l^{2}$ to both sides of the inequality and dividing by 2, we get:

$$
(l-x)(l-z) \geqslant \frac{l^{2}}{2}
$$

The equation $(l-x)(l-z)=\frac{l^{2}}{2}$, obviously, defines a hyperbola (graph of an inverse proportional dependence) passing through the points $x=\frac{l}{2}, z=0$ (point $N$) and $x=0$, $z=\frac{l}{2}$ (point $M$); the coordinate axes should be taken as the lines $O_{1} T$ and $O_{1} S$ (i.e., the lines $l-z=0$ and $l-x=0$; see diagram 86). The inequality $(l-x)(l-z) \geqslant \geqslant \frac{l^{2}}{2}$ will be satisfied for all points of the curvilinear figure $O N R M$, shaded in diagram 86, and only for them.

It remains to determine the area of the figure ONRM. Let's take the lines $O_{1} T$ (axis $O_{1} x^{\prime}$) and $O_{1} S$ (axis $O_{1} z^{\prime}$) as the new coordinate axes; in this coordinate system, our hyperbola will have the equation $x^{\prime} z^{\prime}=\frac{l^{2}}{2}$. Now, let's introduce a new unit of length, equal to $\frac{l}{\sqrt{2}}$; then this equation will be written as $x^{\prime} z^{\prime}=1$. In this new unit, the abscissa $O_{1} K$ of point $N$ will be $\frac{\frac{l}{2}}{\frac{l}{\sqrt{2}}}=\frac{\sqrt{2}}{2}$, and the abscissa $O_{1} T$ of point $M$ will be $\frac{l}{l}=\sqrt{2}$. By the result of problem 152 $\sqrt{2}$

it follows that the area of the curvilinear trapezoid $K N M T$ in the new square units will be $\ln \frac{\sqrt{2}}{\frac{\sqrt{2}}{2}}=\ln 2$, i.e., the natural logarithm of the number 2 (logarithm with base $e=2,718 \ldots$). Converting back to the old units of area, we find that the area of the curvilinear trapezoid

KNMT is $\ln 2 \cdot \frac{l^{2}}{2}=\frac{l^{2} \ln 2}{2}$. Therefore, for the area of the figure ONRM, shaded in diagram 86, we get the expression

$$
\frac{l^{2}}{2}-\frac{l^{2} \ln 2}{2}=\frac{l^{2}}{2}(1-\ln 2)
$$

The total area of the part of the triangle $O S T$ corresponding to unfavorable outcomes of the experiment will be three times this expression, and the area of the part corresponding to favorable outcomes will be equal to

$$
\frac{l^{2}}{2}-3 \frac{l^{2}}{2}(1-\ln 2)=\frac{l^{2}}{2}(3 \ln 2-2)
$$

Thus, the probability sought in the problem has the following value:

$$
\frac{l^{2}}{2}(3 \ln 2-2): \frac{l^{2}}{2}=3 \ln 2-2=0,082 \ldots
$$