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Home » The solutions are $ \boxed{2 + \sqrt7} $ and $ \boxed{2 - \sqrt7} $. - Treasure Valley Movers

The solutions are $ \boxed{2 + \sqrt7} $ and $ \boxed{2 - \sqrt7} $. - Treasure Valley Movers

📅 April 16, 2026 👤 scraface

Apr 16, 2026

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Let’s try integer values. Try $x = 0$:
+ 2y^2 = 2 \Rightarrow y^2 = 1 \Rightarrow y = \pm1 \Rightarrow z = \p...
So possible coordinates: $(0,1,1)$ or $(0,-1,-1)$.
Check distances:
From $(0,1,1)$ to $(0,0,0)$: $\sqrt{0 + 1 + 1} = \sqrt{2}$
To $(1,1,0)$: $\sqrt{1 + 0 + 1} = \sqrt{2}$
To $(1,0,1)$: $\sqrt{1 + 1 + 0} = \sqrt{2}$
All distances match.
Thus, the fourth vertex is $\boxed{(0,1,1)}$.
Question: Find all angles $z \in [0^\circ, 360^\circ]$ that satisfy

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