Vrhovi kocke u prostornom koordinatnom sustavu s ishodištem OO su u točkama A(1,1,1)A(1,1,1), A(1,1,1)A'(-1,-1,-1), B(1,1,1)B(-1,1,1), B(1,1,1)B'(1,-1,-1), C(1,1,1)C(-1,-1,1), C(1,1,1)C'(1,1,-1), D(1,1,1)D(1,-1,1), D(1,1,1)D'(-1,1,-1). Točka OO je središte kocki opisane sfere. Neka točka TT nije na toj sferi i d=OTd = |OT|. Označimo ss α=ATA\alpha = \measuredangle ATA', β=BTB\beta = \measuredangle BTB', γ=CTC\gamma = \measuredangle CTC' i δ=DTD\delta = \measuredangle DTD'. Dokažite da je tg2α+tg2β+tg2γ+tg2δ=32d2(d23)2.\mathrm{tg}^2 \alpha + \mathrm{tg}^2 \beta + \mathrm{tg}^2 \gamma + \mathrm{tg}^2 \delta = \frac{32d^2}{(d^2 - 3)^2}.