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📅 May 4, 2026 👤 scraface

May 04, 2026

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\sqrt{x^2 + y^2 + z^2} = 1 \quad \Rightarrow \quad x^2 + y^2 + z^2 = 1
The distance from \((1, 0, 0)\) to \((x, y, z)\) is:
\sqrt{(x-1)^2 + y^2 + z^2} = 1 \quad \Rightarrow \quad (x-1)^2 + y^2 +...
x^2 - 2x + 1 + y^2 + z^2 = 1 \quad \Rightarrow \quad x^2 + y^2 + z^2 -...
Since \(x^2 + y^2 + z^2 = 1\), we have:
- 2x + 1 = 1 \quad \Rightarrow \quad 2 - 2x = 1 \quad \Rightarrow \qua...
Since \(x\) must be an integer, this is not possible. Re-evaluate usin...
Try \((\frac{1}{2}, \frac{1}{2}, \frac{\sqrt{2}}{2})\) scaled to integ...
Verify distances:
\sqrt{\left(1 - 0\right)^2 + \left(1 - 0\right)^2 + \left(1 - 0\right)...

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