= Solution
For one <electron> per <primitive unit cell>, the areal <number density> is $n_s=1/(ab)$. The <Fermi circle> has radius $k_F=\sqrt{2\pi/(ab)}$. The nearer first-zone boundary is $k_x=\pm\pi/a$, because $a>b$. The circle protrudes precisely when
$$
k_F>\frac\pi a
\quad\Longleftrightarrow\quad
\frac{2\pi}{ab}>\frac{\pi^2}{a^2}
\quad\Longleftrightarrow\quad
\boxed{a>\frac{\pi b}{2}.}
$$
At equality the <Fermi surface> only touches the boundary. Crossing an extended-zone boundary alone does not make a gap: the <free electron gas> remains metallic when the ionic <periodic potential> is neglected.
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