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The title says it all: Is there a (regular) icosahedron containing a rational point on each of its faces?

For other Platonic solids, the affirmative answer is easy. Indeed, regular tetrahedra, cubes, and octahedra may have all their vertices rational. A dodecahedron cannot, but 8 of its vertices form a cube, and each face contains an edge of that cube as its diagonal.

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    $\begingroup$ Two different (mis)spellings of icosahedron in one question! Nice question anyway. $\endgroup$
    – Gerald Edgar
    Commented Aug 16, 2023 at 10:48
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    $\begingroup$ @KentaSuzuki No, you cannot find a regular pentagon in the integer lattice (in any dimension). $\endgroup$
    – Ilya Bogdanov
    Commented Aug 16, 2023 at 14:36
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    $\begingroup$ I'm curious: is there a pentagon with a rational point on each edge? $\endgroup$
    – Yaakov Baruch
    Commented Aug 17, 2023 at 10:23
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    $\begingroup$ Here's an example that doesn't work but comes close. The icosahedron with vertices $(\pm (\phi-2), \pm (\phi-1), 0)$, $(\pm (\phi - 1), 0, \pm (\phi - 2))$, $(0, \pm (\phi-2), \pm (\phi-1))$ has the property that each point in the set $\{ (1,1,-1), (1,-1,1), (1,-1,-1), (-1,1,1), (-1,1,-1), (-1,-1,1) \}$ lies on six of the extended faces of the icosahedron (with each extended face having at least one). However, one can show that for 12 of the faces of this icosahedron, any rational point lies a distance at least $\sqrt{2}$ from the origin, while the circumradius is $\sqrt{7-4\phi} < \sqrt{2}$. $\endgroup$
    – Jeremy Rouse
    Commented Aug 17, 2023 at 16:10
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    $\begingroup$ Curious side question: is it possible that the answer depends on the dimension? (e.g., that such an icosahedron exists in $\mathbb{R}^4$ but not $\mathbb{R}^3$) $\endgroup$
    – Steven Stadnicki
    Commented Aug 20, 2023 at 3:12

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