= Solution
The <body-centered tetragonal lattice> has points at $(0,0,0)$ and $(1/2,1/2,1/2)$ in the conventional cell. Add the two-atom motif to each. Reducing coordinates modulo integer translations gives
$$
\boxed{(0,0,0),\quad(0,1/2,1/4),\quad
(1/2,1/2,1/2),\quad(1/2,0,3/4).}
$$
Thus the cell contains \b[four atoms], not merely the two lattice points. In projection along $[001]$, label each projected point by its fractional height $z/c$; the four sites do not all lie in the same plane.
\Image[/past-exam-of-the-materials-science-course-of-the-university-of-cambridge/2018/paper-1-tin-projection.png]
{title=Original calculated $[001]$ projection of two-by-two conventional beta-tin cells; colors and labels give fractional height, not different chemical species}
The plotted repeat is $a=0.5831\,\mathrm{nm}$ in both horizontal directions. Atoms on the outer drawing boundary are repeated images needed to show neighboring cells, rather than additional unshared atoms in the four-cell counting volume.
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