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 (n2)*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: 785 Thanks: 312 
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 xyplane and intersects the cube will form a rectangle. Now alter the orientation of the intersecting plane with respect to the xyplane without making it parallel to the yzplane or or xzplane. 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 (aad) as the anchor of rotation, where d is a parameter to go up and down. Then my plane E is (aad) + r*(01tan(alpha)) + s*(10tan(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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cube, plane, polygons 
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