22. Given a regular pentagon. $M$ is an arbitrary point inside it (or on the boundary). Let's number the distances from point $M$ to the sides of the pentagon in ascending order: $r_{1} \leqslant r_{2} \leqslant r_{3} \leqslant$ $\leqslant r_{4} \leqslant r_{5}$. Find all positions of point $M$ for which the length $r_{3}$ becomes the smallest.
22. Let $A B C D E$ be a regular pentagon. Draw radii from its center to all vertices and perpendiculars to the sides. Then the pentagon will be divided into 10 equal triangles. It is sufficient to consider the case when point $M$ lies inside or on the side of one of these triangles (for example, $O A K$, where $K$ is the midpoint of $A B$). Note that the distance of a point from the side of an angle depends on its position relative to the bisector of this angle. Using this, we can write the distances of point $M$ from the sides of the angle as

$$
r_{A B} \leqslant r_{A E}, \quad r_{A E} \leqslant r_{B C}, \quad r_{B C} \leqslant r_{D B}, \quad r_{D B} \leqslant r_{D C}
$$

Let $r_{1}=r_{A B}, \quad r_{2}=r_{A E}, \quad r_{3}=r_{B C}, \quad r_{4}=r_{D E}, \quad r_{5}=r_{D C}$.

Obviously, the length $r_{B C}$ is the smallest when point $M$ coincides with point $K$. By symmetry, we conclude that $r_{3}$ is the smallest when point $M$ coincides with the midpoint of one of the sides of the pentagon.