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Home » 2\sqrtx^2 + y^2 + z^2 = 4 - z. - Ready Digital AB

2\sqrtx^2 + y^2 + z^2 = 4 - z. - Ready Digital AB

📅 March 7, 2026 👤 scraface

Mar 07, 2026

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📌 y + \frac{1}{y} \geq 2

📌 Equality occurs when $y = 1$, i.e., $\tan x = 1 \Rightarrow x = \frac{\pi}{4}$.

📌 Thus, the minimum value is:

📌 + 2 = \boxed{9}

📌 Question: A regular tetrahedron has three of its vertices at $(0,0,0)$, $(1,1,0)$, and $(1,0,1)$. Find the coordinates of the fourth vertex, assuming all coordinates are integers.

📌 Let the fourth vertex be $(x, y, z)$, and recall that in a regular tetrahedron, all edges have equal length.

📌 Compute edge lengths between the known points:

📌 \text{Distance between } (0,0,0) \text{ and } (1,1,0): \sqrt{(1)^2 + (1)^2 + 0} = \sqrt{2} \\

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