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Watching the logo of windows made me think of the following: If I have four points on a plane, how can I tell whether those four points could be the projection of a square from some point onto the plane. I guess this is two different questions because a rectangle could only be a projection of a square if the point is at infinity, so I guess, one question without points at infinity and one with.Naraht (talk) 14:51, 23 August 2014 (UTC)[reply]
Isn't the rectangle formed by the ordered triplets (1,1,1), (1,2,3), (2,1,1), (2,2,3) the projection of the square formed by the ordered pairs (1,1), (1,2), (2,1), (2,2) onto the plane containing that rectangle? To be clear, the rectangle has the square as its regular projection onto the x-y plane (i.e. a surface integral over the rectangular surface is (most easily) evaluated by a double integral over the square); through a rotation of coordinates, one could make a similar argument for the rectangle being a projection of a square.
Clearly, though, your definition of a projection is obviously different from mine. If you mean projection in the optical sense, then the above example is basically a projection viewed from infinitely far away (at least in Euclidean geometry). In the case of a finite distance, it is sufficient to show that the four points do form a square if you already know them to be coplanar (but not necessary - if you view from an angle relative to the plane, I'm not entirely sure how to handle it).--Jasper Deng(talk)17:41, 23 August 2014 (UTC)[reply]
This question seems to fit into projective geometry. Since the cross-ratio is essentially the only invariant under projective transformations, I expect that you'd find a similar cross-ratio between four of the six lengths of the segments defined by the four points that would determine whether the original figure could have been a square. —Quondum19:06, 23 August 2014 (UTC)[reply]
Any four random points on a plane could be the projection of a square from some point onto the plane. The image of every other point is determine by the four points though. Dmcq (talk) 20:22, 23 August 2014 (UTC)[reply]
There appear to be four conditions to meet and at least five degrees of freedom, so it is likely that there is always a solution (irrespective of whether the given points are coplanar). You can choose the size of the square too. --catslash (talk) 21:18, 23 August 2014 (UTC)[reply]
Wouldn't any projection of a square from a point to a plane at least be convex? That would imply that not every four points can be such a projection. AndrewWTaylor (talk) 21:58, 23 August 2014 (UTC)[reply]
For any projection plane that intersects the square, you can draw parallel planes that do not intersect it, on which the projected image ought to be the same except for scale. —Tamfang (talk) 00:32, 24 August 2014 (UTC)[reply]
The four points of the square need not all be projected to the same side of the projection point. Consider a square with the projection point above it, then we'll project onto another plane going through one of the corners and almost perpendicular to the first plane but leaning away a little facing a little above the projection point. Then the opposite corner will project to the opposite side of the projection point high up and the two sides ones down below. The original corner point will be within the three other points. Dmcq (talk) 12:01, 24 August 2014 (UTC)[reply]
I don't quite understand the construction here, but it cannot produce a non-convex quadrilateral. The projection map will send line segments to line segments and, vice versa, the inverse projection will map line segments to line segments. So if a line segment in the projected quadrilateral is drawn between two of its interior points, the inverse image of that line segment must be interior to the square (because the square is convex), and so its projection back down into the quadrilateral must also be interior to that quadrilateral. Since the line segment between any two points of the quadrilateral lies inside the quadrilateral, it is convex. Sławomir Biały (talk) 13:42, 24 August 2014 (UTC)[reply]
concave projection of a squareOkay I've set up and uploaded a slightly better construction. ABCD is the original square and O is the point to project througgh and is above the centre of the square. the new plane is given by the points B and D and a point A' which is on the opposite side of O. B and D project to themselves. Then the point C is projected to C' which is on intersection of the line OC, and the line between A' and the mid-point M of BD - this is because the line is given by A'BD and the plane AA'C. As you can see this new point C' is inside the triangle formed by A'BD. Dmcq (talk) 16:09, 24 August 2014 (UTC)[reply]
Yes, it looks like that exchanges the interior of the square with the exterior. So although straight lines are sent to straight lines, the lines may pass through a point at infinity (so the quadrilateral in the image has a "convex" exterior rather than interior). Clever. Sławomir Biały (talk) 23:31, 24 August 2014 (UTC)[reply]
A convex outside, that's a good one ;-) Anyway I think a simple modification of that construction can give a way of projecting a square ABCD to a square ABD'C' which is also a bit surprising. Dmcq (talk) 00:30, 25 August 2014 (UTC)[reply]
Convex exteriorAnyway I've drawn a picture of a convex exterior with a circle showing the line at infinity. Thanks for that idea. As a puzzle a person could try doing one for the square ABD'C' I described just above. Dmcq (talk) 10:06, 25 August 2014 (UTC)[reply]
