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- between $(0,0,0)$ and $(1,0,1)$: $\sqrt{2}$,
- between $(0,0,0)$ and $(0,1,1)$: $\sqrt{2}$,
- and between $(1,1,0)$ and $(1,0,1)$: $\sqrt{(0)^2 + (-1)^2 + (1)^2} = \sqrt{2}$, etc. All edges $\sqrt{2}$. Thus, the fourth vertex is $(0,1,1)$.
- Question: Find all angles $z \in [0^\circ, 360^\circ]$ satisfying $2\sin^2 z - 3\sin z + 1 = 0$.
- Solution: Let $u = \sin z$. The equation becomes $2u^2 - 3u + 1 = 0$. Solving:
- u = \frac{3 \pm \sqrt{9 - 8}}{4} = \frac{3 \pm 1}{4} \implies u = 1 \text{ or } u = \frac{1}{2}.
