5. (8 points) Through the sides of a regular $2 n$-gon, lines are drawn. Into how many parts do these lines divide the plane
Solution: We will make two remarks: 1) no three lines pass through one point; 2) lines will be parallel if and only if they contain opposite sides of a regular $2 n$-gon.

Lemma: Suppose several lines are drawn on a plane, and no three of them intersect at one point. Draw another line that intersects $m$ already drawn lines at $m$ different points, then the number of parts into which the plane is divided by these lines will increase by $m+1$.

$\triangleright$ Denote the new intersection points on our line as $L_{1}, L_{2}, \ldots, L_{m}$ (the numbering coincides with the arrangement of the points on the line). Consider all parts of the plane that were formed before drawing the new line. If a part contains the intervals $\left(-\infty L_{1}\right),\left(L_{1} L_{2}\right), \ldots,\left(L_{m-1} L_{m}\right),\left(L_{m} \infty\right)$, then it will be divided into two, otherwise it will remain unchanged. Therefore, since we have $m+1$ intervals, the number of parts will increase by $m+1$.

Let's return to solving our problem. We will draw lines sequentially through the sides of a regular $2 n$-gon. The first side forms 2 parts; the second will intersect only the first, so it will add 2 more parts; and so on, each subsequent one will add one more part than the previous one. We will continue this operation up to the $n$-th line. Starting from the $(n+1)$-th line, parallel lines will begin to form: the $(n+1)$-th line will intersect $n$ previous ones; the $(n+2)$-th - will intersect $n+1$ previous ones; and so on, each subsequent one will intersect one more than the previous one. In the end, we get the following answer

$$
\begin{gathered}
2+2+3+\ldots+n+n+(n+1)+\ldots+(2 n-1)= \\
=1+n+\frac{2 n(2 n-1)}{2}=2 n^{2}+1
\end{gathered}
$$

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