Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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
add comment

Know someone who can answer? Share a link to this question via email, Google+, Twitter, or Facebook.

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.