20. Given a finite number of polygons in the plane, if for any two of them, there is a line passing through the origin that intersects both, then these polygons are said to be properly placed.

Find the smallest positive integer $m$, such that for any set of properly placed polygons, it is possible to draw $m$ lines passing through the origin, so that each of these polygons intersects at least one of these $m$ lines.
20. First, it is easy to make 3 properly placed quadrilaterals such that any line passing through the origin intersects at most two of them (the drawing is omitted), so the required minimum $m \geqslant 2$. Secondly, since there are only a finite number of polygons, there must be one with the smallest angle subtended at the origin, denoted as $M$, with the angle $\alpha$, and the lines containing the sides of the angle $\alpha$ subtended by $M$ at the origin are denoted as $l_{1}$ and $l_{2}$. It is sufficient to prove that any polygon must intersect at least one of $l_{1}$ or $l_{2}$, then the required minimum $m=2$. In fact, let $M^{\prime}$ be any polygon different from $M$. By the problem's condition, there must be a line $l$ passing through the origin that intersects both $M$ and $M^{\prime}$. Clearly, $l$ is either within the angle $\alpha$ or coincides with one of $l_{1}, l_{2}$. If $l$ coincides with $l_{1}$ or $l_{2}$, the proof is complete. If $l$ is within the angle, since the angle subtended by $M^{\prime}$ at the origin $\geqslant \alpha$, there must be a line $l^{\prime}$ passing through the origin that intersects $M^{\prime}$, and $l^{\prime}$ is either outside the angle $\alpha$ or coincides with one of $l_{1}, l_{2}$. If $l^{\prime}$ coincides with one of $l_{1}, l_{2}$, the problem is solved. If $l^{\prime}$ is outside $\alpha$, it precisely means that one of $l_{1}$, $l_{2}$ lies between $l$ and $l^{\prime}$, hence it must intersect $M^{\prime}$, proof complete.