Maximum Number of Regions in the Plane Using Zig Lines?

Suppose that you already have $n$ zig lines in place, for a total of $2n$ rays. You draw a line $\ell_1$ that intersects all $2n$ rays. Then you draw a line $\ell_2$ that intersects that line and all $2n$ rays. Let $p$ be the point of intersection of $\ell_1$ and $\ell_2$, and for $i=1,2$ let $\{p_{i,k}:k=1,\ldots,2n\}$ be the set of points of intersection of $\ell_i$ with the $2n$ original rays. The underlined statement means that the points $p_{1,k}$ must all lie on the same side of $p$ on the line $\ell_1$, and the points $p_{2,k}$ must all lie on the same side of $p$ on the line $\ell_2$. When we erase half of $\ell_1$ and half of $\ell_2$ to convert them to a zig line with its bend at $p$, we’ll keep the half of $\ell_1$ that contains the points $p_{1,k}$ and the half of $\ell_2$ that contains the points $p_{2,k}$.

I don’t think that it’s immediately clear that this ensures that $Z_n=L_{2n}-2n$ for each $n\in\Bbb N$, but we can show this by induction on $n$.

The line $\ell_1$ cut each of the original $2n$ rays, so it split $2n+1$ regions in two and therefore added $2n+1$ regions. Then $\ell_2$ cut each of the $2n$ original rays and $\ell_1$, so it split $2n+2$ regions in two and so added another $2n+2$ regions. However, when we erase the halves of $\ell_1$ and $\ell_2$ that don’t contain the points of intersection with the original rays, we collapse three regions into one: the region between the two rays that we’re erasing merges with the two regions adjacent two it across those erased rays, and of course that means that they merge into each other as well. (Note that the original $2n$ rays never enter the region between the two rays that we’re erasing, and exactly what they do outside of it is irrelevant to this point.)

Assuming as an induction hypothesis that $Z_n=L_{2n}-2n$, this shows that

$$\begin{align*} Z_{n+1}&=Z_n+(2n+1)+(2n+2)-2\\ &=L_{2n}-2n+(2n+1)+(2n+2)-2\\ &=L_{2n}+2n+1\\ &=L_{2n+1}\\ &=L_{2n+2}-(2n+2)\;, \end{align*}$$

and since $Z_0=1=L_0-2\cdot0\cdot1$, we have a proof by induction that $Z_n=L_{2n}-2n$ for each $n\in\Bbb N$.