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Intersection of right-angled parallelepiped (cuboid) and a surface is regular hexagon. Prove that the parallelepiped is actually a cube.

It's somehow obvious to me that it has to be regular cuboid (=cube) if the intersection is a regular hexagon, but I need formal proof.

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In the case of a cube, do you know how the intersection actually occurs? That might give you some hints as to why the side lengths must be the same. – Calvin Lin Feb 6 '13 at 23:53
Hexagon that comes out as intersection of surface and cube doesn't actually have to be regular, right? – Lazar Ljubenović Feb 6 '13 at 23:56
Yes, what I meant is, given a cube, do you know how to slice it to get a regular hexagon? – Calvin Lin Feb 7 '13 at 0:04
Sides of hexagon have to be $a\sqrt2/2$ if $a$ is side of cube. – Lazar Ljubenović Feb 7 '13 at 0:07

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