What is the area of a regular dodecagon inscribed in a circle, if the radius of the circle is $r$?
Let the three consecutive vertices of a regular dodecagon be $A, B, C$ and its center $O$. Since $AC = AO = OB = r$, the area of the kite $OABC$ is

$$
t=\frac{r r}{2}=\frac{r^{2}}{2}
$$

The regular dodecagon is formed by six such kites, so its area is

$$
T=6 t=3 r^{2}
$$

(Endre Makó, Eger.)

Number of solutions: 17.