Let $A B C$ be the sides of a triangle; we draw a line parallel to $B C$ that intersects sides $A B$ and $A C$ at points $X$ and $Y$. What is the length of $X Y$, if the perimeter of triangle $A X Y$ is equal to the perimeter of trapezoid $X B C Y$?
From the similarity of triangles $A X Y$ and $A B C$, it follows that

$$
A X: c = X Y: a \text{ and } A Y: b = X Y: a
$$

from which

$$
A X = \frac{c}{a} \cdot X Y \text{ and } A Y = \frac{b}{a} \cdot X Y
$$

thus

$$
A X + A Y = \frac{b+c}{a} X Y
$$

But $A X + A Y$ is, according to the problem, the semi-perimeter of the triangle $(s)$, so

$$
X Y = s \cdot \frac{a}{b+c}
$$

(Perl Gyula.)

The problem was also solved by: Frankl I., Freibauer E., Goldberger M., Groffits G., Kiss A., Krisztián Gy., Lukhaub Gy., Oltay K., Prohászka J., Weisz Á., Weisz J.