My Math Forum Polygons in a cube

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 June 16th, 2017, 05:51 PM #1 Newbie   Joined: Jun 2017 From: Moon Posts: 2 Thanks: 0 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?
 June 16th, 2017, 07:30 PM #2 Senior Member   Joined: May 2016 From: USA Posts: 1,310 Thanks: 552 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.
 June 17th, 2017, 04:41 AM #3 Newbie   Joined: Jun 2017 From: Moon Posts: 2 Thanks: 0 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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