# [05AB1E](https://github.com/Adriandmen/05AB1E/wiki/Commands), <s>31</s> 25 [bytes](https://github.com/Adriandmen/05AB1E/wiki/Codepage)

 ®X‚D3ãsãvy5t<;D>‚*¾š2Ý._«

Uses \$\sqrt{3}\$ as the distance of each vector from the origin \$\{0,0\}\$. Will output the following 20 vertex-coordinates:

$$[\{-1,-1,-1\},\{-1,-1,1\},\{-1,1,-1\},\{-1,1,1\},\{1,-1,-1\},\{1,-1,1\},\{1,1,-1\},\{1,1,1\},\{0,-\phi,-\phi-1\},\{-\phi,-\phi-1,0\},\{-\phi-1,0,-\phi\},\{0,-\phi,\phi+1\},\{-\phi,\phi+1,0\},\{\phi+1,0,-\phi\},\{0,\phi,-\phi-1\},\{\phi,-\phi-1,0\},\{-\phi-1,0,\phi\},\{0,\phi,\phi+1\},\{\phi,\phi+1,0\},\{\phi+1,0,\phi\}]$$

[Try it online.](https://tio.run/##yy9OTMpM/f//0LqIRw2zXIwPLy4@vLis0rTExtrFDiiidWjf0YVGh@fqxR9a/f8/AA)

**Explanation:**

<!-- language-all: lang-python -->

 ®X‚ # Push pair [-1,1]
 D # Duplicate it
 3ã # Get all possible triplets using the cartesian power of 3
 s # Swap so the [-1,1] pair is at the top again
 ã # Also get all possible pairs using the cartesian power of 2
 vy # Loop over each pair of [±1,±1]:
 5t<; # Push the golden ratio: (sqrt(5)-1)/2
 D # Duplicate it
 > # Increase the copy by 1
 ‚ # Pair it together with the (sqrt(5)-1)/2
 y * # Multiply the values in the pair to pair [±1,±1] at the same positions
 ¾š # Prepend a 0 to this pair
 2Ý # Push list [0,1,2]
 ._ # Rotate the triplet that many times to the left
 « # Merge it to the original list of [-1,1]-triplets
 # (after which the list of 20 triplets is output implicitly as result)

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I started with a port of [this SO C# answer](https://stackoverflow.com/a/10462220/1682559), but noticed (for \$r=\sqrt{3}\$ at least):

- \$±\frac{1}{\sqrt{3}}\times\phi\times\sqrt{3}\$ is simply \$±\phi\$
- \$±\frac{\frac{1}{\sqrt{3}}}{\phi}\times\sqrt{3}\$ is simply \$±(\phi+1)\$

[Try it online.](https://tio.run/##yy9OTMpM/W9aYmMd@t@4pCpCy7hES4crQoeLC8TTh/DsdP7/BwA)