A circle with radius $r$ is divided into 12 equal parts, and a star-shaped figure is drawn inside it such that each straight line is the chord of a 150-degree arc. What is the area of the region between the star-shaped figure and the circumference of the circle?
I. solution. Let the vertices of the star inscribed in the circle be $A_{1}, A_{2}, A_{3}, \ldots A_{12}$. The area we are looking for is equal to the sum of the areas of 12 congruent triangles and 12 congruent circular segments. The sum of the areas of the circular segments, $t_{1}$, is obtained by subtracting the area of the regular dodecagon from the circumference of the circle; thus

$$
t_{1}=r^{2} \pi-3 r^{2}=r^{2}(\pi-3)
$$

Now let's calculate the sum of the areas of the triangles, $t_{2}$. Let the intersection point of the chords $A_{2} A_{9}$ and $A_{1} A_{6}$ be $C$. Then the triangle $A_{1} A_{2} C$ is equilateral, because $A_{2} A_{1} C \varangle=A_{1} A_{2} C \varangle=60^{\circ}$ (the central angle corresponding to the arc $A_{2} A_{6}$ is indeed $\left.120^{\circ}\right)$. Therefore, the area of the triangle $A_{1} A_{2} C$ is

$$
\frac{{\overline{A_{1} A_{2}}}^{2}}{4} \sqrt{3}=r^{2} \sin ^{2} 15^{\circ} \sqrt{3}=\frac{r^{2}}{2}\left(1-\cos 30^{\circ}\right) \sqrt{3}=\frac{r^{2}}{4}(2 \sqrt{3}-3)
$$

Thus, the sum of the areas of the 12 triangles is:

$$
t_{2}=3 r^{2}(2 \sqrt{3}-3)
$$

Therefore, the area we are looking for is:

$$
T=t_{1}+t_{2}=r^{2}(\pi-3)+3 r^{2}(2 \sqrt{3}-3)=r^{2}(\pi+6 \sqrt{3}-12)=1.53389 r^{2}
$$

(Riesz Marcell, Györr.)

II. solution. The area of the star is equal to the sum of the areas of 24 isosceles triangles. The base of each triangle is equal to $r$, the radius of the circle; the angles at the base are each $15^{\circ}$, and the height is $\frac{r}{2} \operatorname{tg} 15^{\circ}$. Thus, the area of each triangle is

$$
\frac{r^{2}}{4} \operatorname{tg} 15^{\circ}
$$

and the area of the entire star is

$$
6 r^{2} \operatorname{tg} 15^{\circ}
$$

so the area we are looking for is:

$$
T=r^{2} \pi-6 r^{2} \operatorname{tg} 15^{\circ}=r^{2}\left(\pi-6 \operatorname{tg} 15^{\circ}\right)
$$

but

$$
\operatorname{tg} 15^{\circ}=2-\sqrt{3}
$$

and thus again

$$
T=r^{2}(\pi+6 \sqrt{3}-12)
$$

(Ádámffy Elek, Eger.)

The problem was also solved by: Dömény I., Erdős V., Földes R., Haar A., Harsányi Z., Heimlich P., Jánosy Gy., Kertész G., Kräuter F., Kürti I., Messer P., Pám M., Rássy P., Rosenberg J., Ruvald S., Schuster Gy., Schwarz Gy., Sonnenfeld J.