$\mathrm{Az} A O B$ arc, draw a tangent circle inside it. What is the area of this circle if $A O=B O=r$ and $A O B \varangle=60^{\circ}$?
We draw the angle bisector of $\angle AOB$, which intersects the circumference at point $C$. At point $C$, we draw a tangent to the circle, which intersects $OA$ and $OB$ at points $D$ and $E$. The triangle $ODE$ is equilateral because $\angle AOB = 60^{\circ}$. The radius of the circle inscribed in the triangle $ODE$ - as is well known - is $\frac{r}{3}$, and thus the circumference of the circle is $\frac{r^{2}}{9} \pi$.