We break a thin wooden stick into 3 pieces. What is the probability that we can form a triangle with these pieces?

[^0]:    ${ }^{1 *}$ We are happy to accept solutions to problems marked with a star from university students as well.
Divide the stick of length $2n$ into $2n$ equal parts and calculate the sought probability first in the case where the stick can only be broken at the generated division points.

Let $x$, $y$, and $z$ be the lengths of the three parts of the stick; for these parts to form a triangle, the following inequalities must hold:

$$
xn
$$

These inequalities are only satisfied by the following value pairs:

$$
\begin{aligned}
& \text { If } x=2, \quad \text { then } y=n-1, \\
& \text { If } x=3, \quad \text { then } y=n-1, \quad \text { or } n-2, \\
& \text { If } x=4, \quad \text { then } y=n-1, \quad \text { or } n-2, \quad \text { or } n-3, \\
& \ldots . . . \quad \ldots . . . \\
& \ldots . . . . . . . \\
& \text { If } x=n-1, \quad \text { then } y=n-1, \quad \text { or } n-2, \quad \ldots \text { or } 3, \quad \text { or } 2 .
\end{aligned}
$$

The number of favorable cases is therefore:

$$
1+2+3+\ldots+(n-2)=\frac{(n-2)(n-1)}{2}
$$

The total number of possible cases is as follows:

$$
\begin{gathered}
x=1 \text { and } y=1,2,3, \ldots, 2 n-2 \\
x=2 \text { and } y=1,2,3, \ldots, 2 n-3 \text { etc. }
\end{gathered}
$$

Thus, their number is:

$$
2 n-2+2 n-3+2 n-4+\ldots 2+1=\frac{(2 n-2)(2 n-1)}{2}
$$

And so the sought probability is:

$$
v=\frac{(n-2)(n-1)}{(2 n-2)(2 n-1)}
$$

If we now take $n$ and thus the number of division points to be $\infty$, meaning the stick can break at any point, then, since $v$ (dividing the numerator and the denominator by $n^{2}$) can also be written as:

$$
\frac{1-\frac{3}{n}+\frac{2}{n^{2}}}{4-\frac{6}{n}+\frac{2}{n^{2}}}
$$

the sought probability is: $\frac{1}{4}$.

(Dénes König, Budapest.)

The problem was also solved by: Aczél V., Bartók I., Bayer B., Lázár L., Póka Gy., Selényi M., Szmodics K.