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Consider placing $n$ lines in $d$ dimensional space in a way that the angles between any two pairs of lines is always the same (and they all pass through the origin). When $d=3$, we get configurations of lines with this property if we take any Platonic solid with triangular faces, mark the center of mass as the origin and draw lines from there to each of the vertices. Doing this with an Octahedron produces the coordinate system. Doing it with a Tetrahedron produces four lines where any pair is at $60^{\circ}$. Doing it with an Icosahedron produces $6$ lines where any pair is at about $97^{\circ}$.

Doing this with Platonic solids like the cube where the faces aren't triangles doesn't produce the same effect. This can't be a coincidence. Can we prove this will always be the case? And can the observation we made in $3$ dimensions be extended to $d$ dimensions?

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    $\begingroup$ As far as I can tell all the angles you get using a cube are in fact the same. In any case, you can very easily explain this by just looking for a symmetry of the Platonic solid that sends any angle to any other angle. $\endgroup$ Aug 7, 2021 at 1:59
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    $\begingroup$ The angles between lines OA and OB and between lines OA and OC are the same (note that a pair of lines forms two different angle measures, unless they are orthogonal). You can see this because A is adjacent to the opposite point to C on the cube, which is also on line OC. $\endgroup$ Aug 7, 2021 at 2:41
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    $\begingroup$ Angles AOB and AOC aren't the same, but AOB and AOE are, where E is the point diametrically opposite C (hence on the same line). Any two intersecting lines have two supplementary angles between them, and AOB is the acute one while AOC is the obtuse one. They're still the same relative line arrangements. $\endgroup$ Aug 7, 2021 at 18:57
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    $\begingroup$ In the case of the regular tetrahedron the angle between any pair of the four lines is not $60^{\circ}$" but $70.5288^{\circ}$ or more accurately $\arccos\left(\frac{1}{3}\right)$ $\endgroup$ Aug 7, 2021 at 20:03
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    $\begingroup$ I agree the dodecahedron doesn't have it - a necessary condition for this to occur is that any two vertices on the polyhedron are at most $3$ edges apart (i.e there are at most $4$ distances from a given vertex), which is true of all platonic solids except the dodecahedron. $\endgroup$ Aug 7, 2021 at 23:32

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This does not extend to $d$ dimensions; the $600$-cell has triangular faces and tetrahedral cells, but its $60$ axes are not all at equal angles to each other. You can see this by fixing one axis and considering the possible "altitudes" of vertices relative to the "north pole" at the top of this axis - if all lines were at the same angles to one another, there could only be two such altitudes besides the north and south poles, but in fact there are $7$ (in groups of size $12,20,12,30,12,20,12$ from north to south).

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  • $\begingroup$ Maybe triangular faces are required in 3 dimensions, but when you get to 4 dimensions, you need square faces for the property to hold. $\endgroup$ Aug 7, 2021 at 21:54
  • $\begingroup$ It's not true of the tesseract either, though - the line from the origin to $(1,1,1,1)$ has a different angle to the line passing through $(1,1,1,-1)$ than to the line passing through $(1,1,-1,-1)$. $\endgroup$ Aug 7, 2021 at 23:30

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