The Angle of Symmetry of a Different Kind

To get the regular hexagon, the cut must connect midpoints of sides of the cube.

Let's set the length of the edge of the original cube to $AD=4$

(Since we are only interested in angles, actual size is irrelevant and can be set arbitrarily.)

The lower image shows the cube, and the cut, as viewed from above, with a unit grid.

We can see that $AB=\sqrt2$

$DC=3\sqrt2$

and $EC=2\sqrt2$ .

We can now get the $\measuredangle ECB$ from $\triangle BEC$

$\measuredangle ECB=\arctan \frac{EB}{EC}=\arctan \frac4{2\sqrt2}\approx\boxed{54.74^\circ}$ .

We use the idea of vector to solve this problem.

Let the base of the cube (of the length $a$ ) on the x-y plane.

In order in have a regular hexagon as the intersecting (between the cube and the plane), the plane (or the cut) is to the segment joining the points $(0,0,0)$ and $(a,a,a)$ . This means that the equation of the plane is $x+y+z=d$ , for some $d$ .

Let the required angle be $\theta$ . Now $\begin{pmatrix}1 \\ 1\\ 1\end{pmatrix} \cdotp \begin{pmatrix}0 \\ 0\\ 1\end{pmatrix} = \left| \begin{pmatrix}1 \\ 1\\ 1\end{pmatrix} \right| \left| \begin{pmatrix}0 \\ 0\\ 1\end{pmatrix} \right| \cos \theta$ implies that $\cos \theta =\frac{1}{\sqrt{3}}$ and thus $\theta \approx \boxed{54.74 ^{\circ}}$ .