$19 \cdot 39$ On the same plane, draw two convex polygons $P_{1}, P_{2}$ with $n_{1}, n_{2}$ sides respectively, where $n_{1} \leqslant n_{2}$. If $P_{1}, P_{2}$ do not have any overlapping segments, then the maximum number of intersection points between $P_{1}, P_{2}$ is 
(A) $2 n_{1}$.
(B) $2 n_{2}$.
(C) $n_{1} n_{2}$.
(D) $n_{1}+n_{2}$.
(E) None of the above.
(19th American High School Mathematics Examination, 1968)
[Solution] By the definition of a convex polygon, each interior angle is less than $180^{\circ}$.
For the convex polygon $P_{2}$, each side of $P_{1}$ can intersect with the sides of $P_{2}$ at most at two points, so the maximum number of intersection points is $2 n_{1}$.

Next, we appropriately construct $P_{2}$ to prove that the number of intersection points can reach $2 n_{1}$.

First, select any two points on each side of $P_{1}$ and mark them. Then connect the two marked points adjacent to each vertex of $P_{1}$, forming a "cutting line" and extend them:
If they intersect, the intersection point serves as a vertex of $P_{2}$ (such as $P, Q$);
If they do not intersect, take one point on each extended line as a vertex of $P_{2}$ (such as $S$, $R$), and connect these two points, which becomes a side of $P_{2}$. Additionally, when selecting these points, ensure that their connecting line is parallel to the corresponding side of $P_{1}$ (such as $S R \parallel A C$).

Thus, each "cutting line" is a side of $P_{2}$, and it has two internal points on the side of $P_{1}$.

Therefore, each vertex of $P_{1}$ corresponds to one "cutting line", so there are $n_{1}$ such lines, and $P_{1}$ and $P_{2}$ have exactly $2 n_{1}$ intersection points.

Moreover, each angle of $P_{2}$ is an interior or exterior angle of a triangle, so it is always less than $180^{\circ}$, making $P_{2}$ a convex polygon.
Therefore, the answer is $(A)$.