Perspective problem - trapezium turned square True or false: If you draw a trapezium on the ground, there always exists a point above (but not necessarily directly above) the trapezium such that the trapezium looks like a square from that point.
Intuitively this seems true to me, but I'm not sure how you would go about proving/disproving this. Is there an easy-to-understand method of proving/disproving the above statement?
(By "looks like a square" I mean if you take a photograph of the trapezium from that point, the four corners of the trapezium form a square on the 2D photograph)
 A: This is possible for any strictly-convex quadrilateral. In another answer, I compute a $2{-}\mathrm{D}$ to $2{-}\mathrm{D}$ perspective map that maps any $4$ points to any other $4$ points. This perspective map essentially projects $4$ points in one plane to another plane by projecting from a given viewpoint.
Strict-convexity insures that no points of the square need to be virtually projected from behind the viewpoint onto the plane of the quadrilateral. For example, if $1$ or $3$ points of the square are virtually projected, the quadrilateral will be concave; if $2$ points of the square are virtually projected, two sides of the quadrilateral will cross.
Edit: In the case of virtual projection,  I said that the quadrilateral will be convex or that two of its sides will cross. That would be if the corners were mapped and connected in order. If the sides were projected, they would actually shoot off to infinity since the sides of the square cross the source horizon (the plane parallel to the projection plane, but passing through the viewpoint).

A: I think it is false, at least not for every trapezium. Without loss of generality, consider the case of right-angled trapezium, with one angle at the origin. Let OABC be a trapezium, where A is on the y-axis and C is on the x-axis. Suppose O=(0,0,0), A=(0,a,0), B=(b,a,0), C=(c,0,0). Let P(x,y,z) be the point of observer. Basically we want, PO=PA=PB=PC so it looks like a square. That is, $x^2+y^2+z^2=x^2+(y-a)^2+z^2=(x-b)^2+(y-a)^2+z^2=(x-c)^2+y^2+z^2$
which gives us

$y^2=(y-a)^2$
$x^2=(x-b)^2$
$x^2=(x-c)^2$

and

$x=\frac{b}{2}=\frac{c}{2}$

which may not be consistent.
