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=∣OT∣d = |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(d2−3)2.\mathrm{tg}^2 \alpha + \mathrm{tg}^2 \beta + \mathrm{tg}^2 \gamma + \mathrm{tg}^2 \delta = \frac{32d^2}{(d^2 - 3)^2}.