We divide a given segment into three parts by two randomly chosen points. What is the probability that a triangle can be constructed from the three resulting segments? (L. 4, p. 103, Example 2). Solve the problem using an equilateral triangle as a model instead of a square.
Consider the given segment as a unit. It is known that in an equilateral triangle, the sum of the distances from any point $P$ to the sides is equal to the height of the triangle. Therefore, in a regular triangle with a unit height, any point $P$ inside the triangle divides the unit into three parts (see Figure 1):

$$
x+y+z=1
$$

where $x, y, z$ represent the distances from $P$ to the sides $a, b, c$, respectively.

![](https://cdn.mathpix.com/cropped/2024_05_02_e982117985c4d6892600g-1.jpg?height=390&width=442&top_left_y=431&top_left_x=836)

Conversely, any partition $x, y, z$ that satisfies the condition (1) corresponds to one and only one point $P$ inside the triangle. Therefore, the "possible area" is the area of $\triangle ABC$: $t$.

The segments $x, y, z$ can form a triangle if the sum of any two values is greater than the third, i.e.,

$$
x<\frac{1}{2}, \quad y<\frac{1}{2} \quad \text{and} \quad z<\frac{1}{2}
$$

The points $P$ for which the corresponding $x, y, z$ values satisfy the conditions (2) lie within the triangle $\triangle A_{1}B_{1}C_{1}$, defined by the midpoints of the sides of the equilateral triangle, as shown in the hatched area of the figure.

Thus, the favorable area is $\frac{t}{4}$, and the sought probability is

$$
v=\frac{\overline{4}}{t}=\frac{1}{4}
$$

Zsombok Zoltán (Bp., IV., Konyves Kalman g. I. o. t.)