$19 \cdot 45$ a “$n$-pointed star” can be drawn as follows: take a convex polygon, sequentially label its sides as $1, 2, \cdots, n (n \geqslant 5)$, and assume that for any integer $k$ between 1 and $n$, side $k$ is not parallel to side $k+2$ (side $n+1$ and side $n+2$ are considered as side 1 and side 2, respectively). Now, for all such $k$, extend side $k$ and side $k+2$ until they intersect (the figure below is an example for $n=5$). Let $S$ be the sum of the interior angles of the $n$ points, then $S$ equals
(A) $180^{\circ}$.
(B) $360^{\circ}$.
(C) $180^{\circ}(n+2)$.
(D) $180^{\circ}(n-2)$.
(E) $180^{\circ}(n-4)$.
(17th American High School Mathematics Examination, 1966)
[Solution 1] Let the $n$ angles of the $n$-pointed star be $\beta_{1}$, $\beta_{2}$, $\beta_{3}$, ..., $\beta_{n}$, and the $n$ angles of the polygon be $\alpha_{1}$, $\alpha_{2}$, $\alpha_{3}$, ..., $\alpha_{n}$. Then we have
$$
\begin{array}{l} 
\beta_{1}=180^{\circ}-\left(180^{\circ}-\alpha_{1}\right)-\left(180^{\circ}-\alpha_{2}\right) \\
=\alpha_{1}+\alpha_{2}-180^{\circ}, \\
\beta_{2}=180^{\circ}-\left(180^{\circ}-\alpha_{2}\right)-\left(180^{\circ}-\alpha_{3}\right) \\
=\alpha_{2}+\alpha_{3}-180^{\circ}, \\
\ldots \ldots \ldots \ldots \ldots
\end{array}
$$
$$
\beta_{n}=180^{\circ}-\left(180^{\circ}-\alpha_{n}\right)-\left(180^{\circ}-\alpha_{1}\right)=\alpha_{n}+\alpha_{1}-180^{\circ} \text {. }
$$

Adding all the above equations, we get
$$
S=2 \sum_{k=1}^{n} \alpha_{k}-180^{\circ} \cdot n,
$$

The sum of the interior angles of a convex $n$-sided polygon is $(n-2) \cdot 180^{\circ}$, so $S=(n-4) \cdot 180^{\circ}$. Therefore, the answer is $(E)$.
[Solution 2] From problem $19 \cdot 40$, we have
$$
S=n \cdot \frac{(n-4) 180^{\circ}}{n}=(n-4) \cdot 180^{\circ} .
$$

Therefore, the answer is $(E)$.