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June 16th, 2017, 06:51 PM   #1
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Polygons in a cube

Hello,

the situation is the following: A cube is intersected by a plane. Depending on the coordinates of the plane, different polygons can be created by the intersection.
How can I prove (with analytic geometry) that these polygons can be triangles, rectangles, pentagons or hexagons?

Intuitively (and by drawings) I know that this must be the case but how can I prove it? I thought about showing that the sums of the angles are (n-2)*pi for n from 3 to 6 but I didn't manage to calculate the angles with an arbitrary plane.

Does someone have an idea?
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June 16th, 2017, 08:30 PM   #2
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I do have an idea, but I have not worked it out.

Set up your cube with one corner at the origin of a three dimensional Cartesian grid. You can do this because the orientation of the grid is free with respect to the cube.

Now any plane that is parallel to the xy-plane and intersects the cube will form a rectangle.

Now alter the orientation of the intersecting plane with respect to the xy-plane without making it parallel to the yz-plane or or xz-plane. Presumably there will be only a small number of ways in which such an alteration can be done, and each way should generate a small number of patterns of intersection.
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June 17th, 2017, 05:41 AM   #3
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Ok, thanks for the idea!,

I tried to set up the plane like this:
a is the length of the cube,
I chose (a|a|d) as the anchor of rotation, where d is a parameter to go up and down.
Then my plane E is (a|a|d) + r*(0|1|tan(alpha)) + s*(1|0|tan(beta)) where alpha and beta are elements of (0,Pi/2) and r,s the parameters.
Do you think this is correct?

If yes, how can I go on from here?
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