97**. A piece is randomly broken off from each of three identical rods. What is the probability that these three pieces can be used to form an acute triangle?
97. In this problem, we consider the same experiment as in the previous one. As before, the set of all possible outcomes of the experiment is represented by the set of points of the cube \(O A B C D E F G\) with side length \(l\); the probability of an event is equal to the ratio of the volume of the part of the cube corresponding to favorable outcomes of the experiment to the volume of the entire cube. Thus, we only need to determine which points of the cube correspond to favorable outcomes of the experiment.

According to the cosine theorem in an acute triangle, the square of each side is less than the sum of the squares of the other two sides, while in an obtuse triangle, the square of the side opposite the obtuse angle is greater than the sum of the squares of the other two sides, and in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the legs. Therefore, for three segments to form an acute triangle, the square of each of these segments must be less than the sum of the squares of the other two \({ }^{1}\). Thus, the difference between the solution of this problem and the previous one is that now the favorable outcomes of the experiment will be determined not by inequalities (*) (see the solution to problem 96), but by the inequalities

$$
x^{2}+y^{2}>z^{2}, \quad x^{2}+z^{2}>y^{2}, \quad y^{2}+z^{2}>x^{2}
$$

The equality \(x^{2}+y^{2}=z^{2}\) means that the distance \(M Q=\sqrt{x^{2}+y^{2}}\) from the point \(M\) with coordinates \(x, y, z\) to the \(O z\) axis is equal to the segment \(O Q\) on the \(O z\) axis (Fig. 85, a). From this, it is clear that all points for which \(x^{2}+y^{2}=z^{2}\) will lie on the surface of a cone with the axis being the line \(O D\) and the angle between the axis and the generatrix being \(45^{\circ}\); points for which \(z^{2}>x^{2}+y^{2}\) (these points correspond to unfavorable outcomes of the experiment) will be located inside this cone. Similarly, the inequalities \(y^{2}>x^{2}+z^{2}\) and \(x^{2}>y^{2}+z^{2}\) define two cones with axes being the lines \(O C\) and \(O A\) (Fig. 85, b). Inside the cube \(O A B C D E F G\), one quarter of each of these three cones is placed; they do not intersect and the height of each is equal to the length of the cube's edge \(l\), and the radius of the base is also equal to \(l\) (since the angle between the axis and the generatrix is \(45^{\circ}\)). The volume of one quarter of such a cone is obviously equal to

$$
\frac{1}{4} \cdot \frac{1}{3} \pi l^{2} \cdot l=\frac{\pi l^{3}}{12}
$$

Therefore, the volume of the part of the cube corresponding to unfavorable outcomes of the experiment is \(3 \frac{\pi l^{3}}{12}=\frac{\pi l^{3}}{4}\), and the volume of the part of the cube corresponding to favorable outcomes is \(l^{3} - \frac{\pi l^{3}}{4}\).

From this, it follows that the probability sought in the problem is

$$
\frac{l^{3}-\frac{\pi l^{3}}{4}}{l^{3}}=1-\frac{\pi}{4}=0.2146 \ldots
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_68334347b64402497ba5g-328.jpg?height=382&width=872&top_left_y=465&top_left_x=134)

Fig. 85.