Cross section is a regular hexagon.Is it a cube?

Question

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?

Answer 1

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 $a=b=c=d=e=f$

Answer 2

Given $a$, $b$ and $c$ lengths of block(rectangular box) and $u$ length of edge of hexagon. Now we insert a line $u$ into a rectangle so that the side $a$ will divided to sections $x_a$ and $a-x_a$ and side $b$ will divided to $x_b$ and $b-x_b$. Inside this rectangle is a triangle for which the $x_a^2+x_b^2 = u^2$(see image)
, for rectangle $a\cdot c$ analogously $x_a^2+x_c^2=u^2$ and for rectangle $a\cdot c$ we get $x_a^2+x_c^2=u^2$.
Now
$x_a^2+x_c^2 = u^2 = x_b^2+x_c^2 \implies x_a=x_b$,
$x_a^2+x_b^2 = u^2 = x_b^2+x_c^2 \implies x_a=x_c$.
($x_a^2+x_b^2 = u^2 = x_a^2+x_c^2 \implies x_b=x_c$ is redundant).
All these equalities are valid for each the side and his opposite side, this mean for $a\cdot b$ and also for oposite side $a\cdot b$. And therefore, $x_a = a-x_a \implies x_a = \frac{a}{2}$, $x_b = b-x_b \implies x_b = \frac{b}{2}$ and $x_a = a-x_a$, $x_c = c-x_c \implies x_c = \frac{c}{2}$.

And from these and previous equalities follows $x_a = x_b = x_c$ and also $a = b = c \equiv v$, therefore, that hexagon can be entered only in the cube.
If length of edge of regular hexagon equals to $u$, the length of edge $v$ of this cube will $v = 2 \sqrt u$.