The angles of a triangle are $\alpha, \beta, \gamma$. How does the area of this triangle compare to the area of the circle passing through its vertices?
If $R$ is the radius of the circle passing through the vertices of a triangle, then the area of the circle is $R^{2} \pi$, and the area of the triangle is $\frac{a b}{2} \sin \gamma = \frac{4 R^{2}}{2} \sin \alpha \sin \beta \sin \gamma$, so the area of the triangle is in the ratio to the area of the circle as

$$
2 \sin \alpha \sin \beta \sin \gamma: \pi
$$

(Popoviciu Miklós, VII. o. t., Lugos.)

The problem was also solved by: Brachfeld S., Bagyinka M., Déri Zs., Deutsch I., Haar A., Hirschfeld Gy., Kelemen M., Kertész G., Kohn L., Neidenbach E., Osztián K., Raab R.