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Home » "The Long-Awaited Voltron Movie Finally Dropped—Are You Ready for the Epic Reveal?! - Get link 4share

"The Long-Awaited Voltron Movie Finally Dropped—Are You Ready for the Epic Reveal?! - Get link 4share

📅 March 10, 2026 👤 scraface

Mar 10, 2026

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📌 $k=2$: $\text{cis}(2\pi/3) = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$ → $(-0.5, \approx 0.866)$

📌 $k=3$: $\text{cis}(\pi) = -1 + 0i$ → $(-1, 0)$

📌 $k=4$: $\text{cis}(4\pi/3) = -\frac{1}{2} - i\frac{\sqrt{3}}{2}$ → $(-0.5, \approx -0.866)$

📌 $k=5$: $\text{cis}(5\pi/3) = \frac{1}{2} - i\frac{\sqrt{3}}{2}$ → $(0.5, \approx -0.866)$

📌 The node **opposite to $A = (1, 0)$** is the one diametrically opposite, i.e., rotated $180^\circ$, so $k = 3$:

📌 z = -1 + 0i = (-1, 0)

📌 But the question says "the node opposite to $O$", i.e., opposite the origin. That wording is misleading — $O$ is at the origin. But in the hexagon, the node farthest from $O$ is at $(-1, 0)$, which is at distance 1 — same as all others. But in a **regular hexagon**, all vertices are equidistant from center, so **no node is farther than others**.

📌 However, in a **scaled hexagonal lattice** used for energy grid modeling, nodes may be placed at **integer-coordinate projections** using vectors $\vec{v}_1 = (1, 0)$, $\vec{v}_2 = \left(\frac{1}{2}, \frac{\sqrt{3}}}{2}\right)$, but these are not integer.

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