We divide a segment into four parts with three randomly chosen points. What is the probability that these four sub-segments can be the four sides of some quadrilateral?
I. solution: A necessary and sufficient condition for the existence of a quadrilateral is that any side is smaller than the sum of the other three sides. Let the sides of the quadrilateral be denoted by $x, y, z, u$, and let the perimeter be $x+y+z+u=2 s$. According to the above,

$$
x < y + z + u
$$

Similarly, the other sides must also satisfy this condition. If we consider the inequality $x + y + z + u < 2(x + y)$, it simplifies to $z + u < x + y$. This inequality is equivalent to the condition that the point $(x, y)$ lies on the opposite side of the line

$$
x+y=\frac{1}{2}
$$

from the origin.

Thus, the coordinates of the points lying inside the triangle defined by the points $C\left(\frac{1}{2} ; 0\right), D\left(0 ; \frac{1}{2}\right)$, and $E\left(\frac{1}{2} ; \frac{1}{2}\right)$ (and only these) satisfy both (1) and (a). If (a) is satisfied, then

$$
z+u \leq \frac{1}{2}
$$

and thus $z < \frac{1}{2}$. Now, if we perform the previous considerations in a coordinate system with $z$ and $u$ axes, we similarly find that

$$
v_{b}=\frac{1}{4}
$$

Therefore, the sought probability is

$$
v=v_{a}+v_{b}=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}
$$

(Kálmán Györy, Ózd, József Attila Grammar School III. class)

II. solution: A simple and elegant solution can be obtained if we use the following lemma: The sum of the distances from any interior point of a regular tetrahedron to its four faces is equal to the height of the tetrahedron.

Indeed, let the height of the tetrahedron be $m$, the area of one face be $t$, and the distances from the point $P$ to the faces be $m_{1}$, $m_{2}$, $m_{3}$, and $m_{4}$. The point $P$ defines a triangular pyramid with each face. The sum of the volumes of these 4 sub-pyramids is equal to the volume of the original tetrahedron. Thus,

$$
\frac{t m_{1}}{3}+\frac{t m_{2}}{3}+\frac{t m_{3}}{3}+\frac{t m_{4}}{3}=\frac{t m}{3}
$$

from which

$$
m_{1}+m_{2}+m_{3}+m_{4}=m
$$

Let the given segment be $m$, and consider a regular tetrahedron of height $m$. The distances $m_{1}, m_{2}, m_{3}, m_{4}$ from any interior point of this tetrahedron to the faces, according to the lemma, represent a possible partition of $m$.

Favorable points are only those for which $m_{i}<\frac{m}{2}(i=1,2,3,4)$, as we saw in the first solution. Therefore, the favorable points are inside the octahedron defined by the midpoints of the tetrahedron's edges as vertices. The sought probability is thus the ratio of the volume of the octahedron to the volume of the tetrahedron.

Let the volume of the tetrahedron be $K$. The octahedron can be formed by removing 4 half-edge-length regular tetrahedra, cut by the 4 mid-edge planes, from the original tetrahedron. The volume of each of these smaller tetrahedra is $\left(\frac{1}{2}\right)^{3} \cdot K=\frac{K}{8}$, and thus the total volume of the 4 removed tetrahedra is $\frac{K}{2}$. The volume of the remaining octahedron is therefore $K-\frac{K}{2}=\frac{K}{2}$, and the sought probability is

$$
v=\frac{\frac{K}{2}}{K}=\frac{1}{2}
$$

(Gyula Argay, Balassagyarmat, Balassi Grammar School IV. class)