7.44 A finite set of polygons in the plane is said to be properly placed if for any two of them, there is a line through the origin that intersects both.

Find the smallest natural number $m$ such that for any properly placed set of polygons, one can draw $m$ lines through the origin so that each polygon in the set intersects at least one of these $m$ lines.
［Solution］In the case of 3 polygons shown in the right figure, any line passing through the origin intersects at most two of them, hence we know that $m \geqslant 2$.

Since there are only a finite number of polygons, there must be one polygon that subtends the smallest angle at the origin. Let this polygon be $M$ and the angle $\alpha$ subtended by $M$ at the origin be bounded by two lines $l_{1}$ and $l_{2}$. Then, any polygon in the group must intersect at least one of these two lines. Indeed, let $M^{\prime}$ be any polygon different from $M$. By the given condition, there must be a line $l$ passing through the origin that intersects both $M$ and $M^{\prime}$. Thus, the line $l$ is either between $l_{1}$ and $l_{2}$ (i.e., intersects the interior of $\angle \alpha$) or coincides with one of $l_{1}$ or $l_{2}$. If the latter is the case, the problem is already solved. If $l$ is indeed between $l_{1}$ and $l_{2}$, then since the angle $\beta$ subtended by $M^{\prime}$ at the origin is not less than $\alpha$, there must be a line $l^{\prime}$ passing through the origin that intersects $M^{\prime}$ but does not intersect the interior of $\angle \alpha$. If $l^{\prime}$ coincides with one of $l_{1}$ or $l_{2}$, the problem is solved. If $l^{\prime}$ is outside $\angle \alpha$, this precisely means that one of $l_{1}$ or $l_{2}$, say $l_{1}$, lies within the angle $\beta$ between $l$ and $l^{\prime}$. Thus, both polygons $M$ and $M^{\prime}$ intersect the line $l_{1}$. Therefore, the smallest natural number $m=2$.