In what ratio does the angle bisector of one of the acute angles of an isosceles right triangle divide its area?
Let's draw the angle bisector $AD$ in the right-angled triangle $ABC$ $\left(\angle B = 90^{\circ}\right)$, then

$$
\Delta ABD : \Delta ADC = AB \times BD : AB \times DC = BD : DC
$$

but

$$
BD : DC = AB : AC = 1 : \sqrt{2}
$$

therefore

$$
\Delta ABD : \Delta ADC = 1 : \sqrt{2}
$$

(Zoltán Harsányi, Eger.)

The problem was also solved by: Brachfeld S., Deutsch I., Erdélyi I., Freund R., Haar A., Kertész G., Kürti I., Ligeti P., Raab R., Ragány B., Szávay Z.