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I have seen several interactive models showing that a plane slice of the cube can reveal a cross section that is a regular hexagon. I know that it has to be a plane passing through the origin and normal to (1,1,1) or in general I guess just perpendicular to a the main diagonal. Is there a way to verify algebraically that this is indeed a regular hexagon I can't think of any obvious way to start.

Note: Oh and sorry I am considering a cube centered at (1/2,1/2,1/2) with sides of length on Vertices at (0,0,0) (1,0,0) (0,1,0) (1,1,1) etc.

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Why algebraically? You can see it geometrically. There are several such planes, but all symmetric with respect to the symmetries of the cube. Basically, take a main diagonal of the cube (diagonal connecting diametrically opposite vertices) and draw the plane passing through the center of the cube (i.e. the midpoint of the diagonal) and perpendicular to that diagonal. The intersection of the plane with the cube is a regular hexagon. If you rotate the cube around the diagonal $120^{\circ}$ in one direction, the cube is mapped to itself, the plane perpendicular to the diagonal is also mapped itself and thus the hexagon is mapped to itself too. So it has $120^{\circ}$ rotational symmetries. Moreover, if you take the symmetry that maps each point on the cube to its diametrically opposite, you get again that both the cube and the plane are mapped to themselves, giving you a cantral, i.e. $180^{\circ}$ symmetry of the hexagon. The combination of the $120^{\circ}$ rotational symmetry and the $180^{\circ}$ degree yields a $60^{\circ}$ of the hexagon, yielding that it is a regular hexagon.

Edit 1. For instance, let me for simplicity look at the cube with edges of length one spanned by the coordinate vectors $(1,0,0),\,(0,1,0),\, (0,0,1)$. Notice that the center of this cube is $(1/2, 1/2, 1/2)$. The triangle formed by the points with coordinates $(1,0,0),\,(0,1,0),\, (0,0,1)$ (they are vertices of the cube, the three neighbors of $(0,0,0)$) is equilateral and its plane is orthogonal to the diagonal spanned by the vector $(1,1,1)$. Now, if you move the plane of that equilateral triangle to a parallel plane but bit further along the diagonal, so that it passes through the midpoint of the diagonal, which is center of the cube $(1/2, 1/2, 1,2)$ you have exactly the plane you want. Equation $x+y+z = \frac{3}{2}$.

Edit 2. The vertices of the hexagon are the midpoints of the six edges that the plane $x+y+z = \frac{3}{2}$ intersects (the plane passing through the center of the cube and orthogonal to one main diagonal). You can see that the length of each edge of the hexagon is one half from the length of a face diagonal of the cube. Three alternating edges of the hexagon are parallel and one half of the edges of the equilateral triangle with vertices $(1, 0, 0), \, (0, 1, 0), \, (0, 0, 1)$ and the other three alternating edges of the hexagon are parallel and one half of the edges of the equilateral triangle with vertices $(1, 1, 0), \, (1, 0, 1), \, (0, 1, 1)$. The two equilateral triangles are congruent and one is a sixty degree rotation of the other and then moved by a translation to its place.

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  • $\begingroup$ Hm interesting a little confusing to me but I'm going to spend some time thinking about this. Oh and sorry I am considering a cube centered at (1/2,1/2,1/2) with sides of length on Vertices at (0,0,0) (1,0,0) (0,1,0) (1,1,1) etc. $\endgroup$ – numtheory1555242 Sep 28 '16 at 16:41
  • $\begingroup$ @numtheory1555242 I edited my post, made it more precise and added some extra description. $\endgroup$ – Futurologist Sep 28 '16 at 17:55
  • $\begingroup$ @numtheory1555242 Another edit for better visualization :). $\endgroup$ – Futurologist Sep 28 '16 at 19:05
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First, find the intersections between the plane and the edges of the cube. Those are the vertices of the hexagon.

Next, find the distances between them. That should tell you the sides are all the same.

Finally, get the angles between them (probably using the law of cosines). That should tell you that all the angles are the same.

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  • $\begingroup$ Okay cool this seems like the simplest way for me at my level to do this algebraically but I should probably investigate some of these other methods too. Appreciate the advice! $\endgroup$ – numtheory1555242 Sep 28 '16 at 16:43
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Take the center at the origin of 3-space, and let the points $A,B,C,D,E,F$ [in that order] be $(1,-1,0),(1,0,-1),(0,1,-1),(-1,1,0),(-1,0,1),(0,-1,1).$ Then using dot product one can check that all the angles AOB etc formed by two adjacent vertices and the center are 60 degrees, and all the lengths are the same.

Also all the points lie on plane $x+y+z=0.$

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Let the vertices of the cube be at $(\pm 1, \pm 1, \pm 1)$, so each of the six faces has one coordinate set to $\pm 1$ and the other two varying over $[-1,1]$; and each of the twelve edges has two coordinates set to $\pm 1$ and the third varying over $[-1,1]$. Now consider the plane $x+y+z=0$. This meets just those edges where the two fixed coordinates have opposite signs, and the third coordinate in each case is $0$. I.e., it hits the points $(1,0,-1)$, $(1,-1,0)$, $(0,-1,1)$, $(-1,0,1)$, $(-1,1,0)$, and $(0,1,-1)$. Each pair of neighboring points is on the same face, and each segment has length $\sqrt{2}$.

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