As far as I understand, the construction of Brillouin zones stems from the relation$$
2 \vec{k}\cdot \vec{G} +G^2 = 0
\,,$$where $\vec{k}$ is the wave vector and and $\vec{G}$ is the reciprocal lattice vector. This condition is supposed to be fulfilled when $\vec{k}$ terminates on a line normal to $\vec{G}$ at half of the length of $\vec{G}.$
If so, why does the third Brillouin zone take the form of Figure 1, rather than Figure 2, for a quadratic lattice? The second figure shows the area enclosed by lines situated half of $2b\hat{x}$ and $2b\hat{y}$ from the origin, perpendicular to the $x$- and $y$- directions respectively.
$\hspace{50px}
{\def\place#1#2#3{\smash{\rlap{\hskip{#1px}\raise{#2px}{#3}}}}}
\place{0}{220}{\begin{array}{c}\textbf{Figure 1} \\ \hspace{250px} \end{array}}
$
$\hspace{50px}{\def\place#1#2#3{\smash{\rlap{\hskip{#1px}\raise{#2px}{#3}}}}}
\place{0}{220}{\begin{array}{c}\textbf{Figure 2} \\ \hspace{250px} \end{array}}$
Best Answer
The first BZ should contain one lattice point, the second BZ two, and the third BZ three, etc. Figure 1 is correct because it is exactly the case. The third BZ in Figure 2 contains 4 lattice points, so it is not correct.