TITLE: Cross section is a regular hexagon.Is it a cube?
QUESTION [7 upvotes]: One of the cross sections in a rectangular box is a regular hexagon.Prove that the box is a cube
I tried to prove that certain lengths were equal by showing that certain triangles are congruent but that was of no use.Could someone help me with this one?

REPLY [3 votes]: Since the lines on opposite faces are parallel, and every diagonal is the same length, we have $b^2+c^2 = d^2+e^2 = f^2+a^2$.
For the hexagon to be in a plane, we need
$det \begin{pmatrix}b&0&-a\\c&d&0\\0&e&f\end{pmatrix} = 0$,
hence $bdf = ace$
Finally, we need the angles of the hexagon to be the same, so looking at the scalar products, you get $cd=ef=ab$.
This makes $5$ equations. So proving $a=b=c=d=e=f$ might be possible from there :
$(b-a)((a+b)bd+abc) = bd(b^2-a^2)+(ab)bc-(ab)ac = bd(b^2-a^2)+(cd)bc-(ef)ac \\= bd(b^2+c^2-a^2) - (ace)f = bdf^2 - bdf^2 = 0$
Since $(a+b)bd+abc > 0$, we get $a=b$.
Similarly, $c=d$ and $e=f$, and $cd = ef = ab$ becomes $c^2 = e^2 = a^2$, hence $a=b=c=d=e=f$