1. Find all triangles $ABC$ for which the equality

$$
|BC| \cdot|AX|=|AC| \cdot|BY|,
$$

holds, where point $X$ is the intersection of the angle bisector of $\angle BAC$ with side $BC$ and point $Y$ is the intersection of the angle bisector of $\angle ABC$ with side $AC$. (P. Černek)
Solution: We rewrite the given equality in the form $|B C|:|B Y|=|A C|:|A X|$ and express both ratios using the sine rule for triangles $B C Y$ and $A C X$ (in which, with the usual notation for the interior angles of triangle $A B C$, it is clear that $|\angle B Y C|=\alpha+\frac{1}{2} \beta$ and $\left.|\angle A X C|=\beta+\frac{1}{2} \alpha\right):$

$$
\frac{|B C|}{|B Y|}=\frac{\sin \left(\alpha+\frac{1}{2} \beta\right)}{\sin \gamma} \quad \text { and } \quad \frac{|A C|}{|A X|}=\frac{\sin \left(\beta+\frac{1}{2} \alpha\right)}{\sin \gamma} .
$$

We are therefore looking for triangles for which $\sin \left(\alpha+\frac{1}{2} \beta\right)=\sin \left(\beta+\frac{1}{2} \alpha\right)$. Since both arguments lie between $0^{\circ}$ and $180^{\circ}$, the equality of their sines will occur only if $\alpha+\frac{1}{2} \beta=$ $=\beta+\frac{1}{2} \alpha$, or $\left(\alpha+\frac{1}{2} \beta\right)+\left(\beta+\frac{1}{2} \alpha\right)=180^{\circ}$. The first condition means $\alpha=\beta$, the second $\alpha+\beta=120^{\circ}$, or $\gamma=60^{\circ}$.

A small variation of the first part: If we compare the given equality with the generally valid equality $|B C| \cdot|A P|=|A C| \cdot|B Q|$, where $A P$ and $B Q$ are the altitudes of the given triangle, we get the equivalent condition in the form $|A P|:|A X|=|B Q|:|B Y|$. From the right-angled triangles $A P X$ and $B Q Y$, we again get the equality $\sin \left(\alpha+\frac{1}{2} \beta\right)=\sin \left(\beta+\frac{1}{2} \alpha\right)$.

ANSWER: The sought triangles are precisely those for which $|A C|=|B C|$ or $|\angle A C B|=60^{\circ}$.