= Solution
The vectors from the B atom to its three nearest N neighbours can be chosen as
$$
\boxed{\mathbf n_1=\left(\frac a{\sqrt3},0\right),\quad
\mathbf n_2=\left(-\frac a{2\sqrt3},-\frac a2\right),\quad
\mathbf n_3=\left(-\frac a{2\sqrt3},\frac a2\right).}
$$
Indeed $\mathbf n_1=\mathbf d_N-\mathbf d_B$, $\mathbf n_2=\mathbf n_1-\mathbf a_1$ and $\mathbf n_3=\mathbf n_1-\mathbf a_2$. All three have length $a/\sqrt3$. Repeating the two-site <crystal motif>[basis] over $\mathbf R=m\mathbf a_1+n\mathbf a_2$ produces the honeycomb pattern in the first panel of the original figure in <past-exam-of-the-condensed-matter-physics-major-option-course-of-the-university-of-oxford/2018/1/c/ii/solution>.
Solving $\mathbf a_i\cdot\mathbf b_j=2\pi\delta_{ij}$ gives
$$
\boxed{\mathbf b_1=\left(\frac{2\pi}{\sqrt3a},\frac{2\pi}a\right),\quad
\mathbf b_2=\left(\frac{2\pi}{\sqrt3a},-\frac{2\pi}a\right).}
$$
The <first Brillouin zone> is the Wigner–Seitz hexagon of this <reciprocal lattice>. Its vertices are $(0,\pm4\pi/(3a))$ and $(\pm2\pi/(\sqrt3a),\pm2\pi/(3a))$, with independent signs in the latter set. The centre is $\Gamma=(0,0)$; one inequivalent corner is $K=(0,4\pi/(3a))$, with $K'=-K$. Corners related by a <reciprocal lattice> translation are equivalent, leaving two inequivalent valleys.
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