# What quadrilaterals are the image of projective transformations of squares?

What quadrilaterals in the real projective plane can be obtained by a projective transformation of the real projective plane from a square?

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"Real" in my question does not refer to "non-projective", it refers to the projective plane over the reals as opposed to the projective plane over the complex numbers etc. – Phira Jan 16 '14 at 18:44
@rschwieb Done, sorry for the confusion. This question comes from reflecting on the method of trivializing a certain class of contest geometry problems, and there, the interesting decision is between real projective plane and complex projective line. – Phira Jan 16 '14 at 19:20
OK: looks good! – rschwieb Jan 16 '14 at 19:21

With a (unique) projective transformation of the projective plane you can take any four points, no three of which are collinear, to any other four points, no three of which are collinear. (You can arrange for the points all to be in your real plane inside the projective plane, but some points of the plane may conceivably get sent to infinity by the projective transformation.) You can prove this just by writing down matrices, although there are somewhat more elegant proofs, too.

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In the meantime, I found a way myself to do it by sending a line to infinity. I will write it as an answer for later reference, but I would appreciate a sketch of an alternate elegant approach. – Phira Jan 16 '14 at 19:10
@Pete: I stipulated that the 4 points be in general position (no three collinear). So there is no cross ratio :) – Ted Shifrin Jan 16 '14 at 19:49
@Ted: Thanks. I agree: I was thinking of the case of the projective line instead. – Pete L. Clark Jan 16 '14 at 20:08

One way to prove that all quadrilaterals $ABCD$ without three collinear vertices can be mapped by a projective transformation to a square is as follows:

Call the intersection points of opposite sides $X$ and $Y$. Map the line $XY$ to the line at infinity. (This is easy to do, just project to a plane that is parallel to the plane formed by the line $XY$ and the center of the projection.)

On the line at infinity, we can map any two points on any other two points (this is actually true for three, but obvious for two points). Therefore, we can choose to map $X$ and $Y$ to two points at infinity that correspond to orthogonal directions.

This forces the quadrilateral to have two pairs of parallel sides that are orthogonal to each other, i.e. the quadrilateral became a rectangle. Now, it is clear that you can tilt your plane and project it to make a square out of the rectangle.

Second (similar) approach:

On the two diagonals, take the point that is harmonic with respect to the two vertices of the quadrilateral and the intersection point of the diagonals. Map these two points again to two points at infinity that correspond to orthogonal directions.

The result is a quadrilateral whose diagonals halve each other (i.e. a parallelogram) and whose diagonals are orthognal to each other. Therefore, it is a square.

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