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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)

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Consider a three-dimensional Cartesian coordinate system, with your trapezium lying on the $x-y$ coordinate plane. It looks to me that you can always find a rotation about the $x$ or $y$-axis such that the shadow of this rotated trapezium is a square... –  J. M. Dec 9 '11 at 14:13
    
en.wikipedia.org/wiki/Trapezium "The word trapezium has several meanings" Please specify. –  Phira Dec 9 '11 at 14:54
    
@Phira A quadrilateral with one pair of parallel sides, or "trapezoid" in American English it seems. I wasn't aware that there was more than one definition, my apologies. –  Sp3000 Dec 9 '11 at 15:05

2 Answers 2

up vote 2 down vote accepted

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).

enter image description here

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I appreciate the inclusion of a diagram but I'm still a little confused about how the positioning of the viewpoint in the diagram is related to the source plane and the virtual/normal projections. What confuses me is that the viewpoint is between the virtual projection and the source plan. Can you please clarify this? –  Sp3000 Jan 10 '12 at 3:17
    
@Sp3000: Looking at the projection plane, you don't see the virtually projected points because, as you say, they are behind the viewpoint. However, the virtual projections are on the line containing the viewpoint and the source point, so mathematically they get mapped onto the projection plane. –  robjohn Jan 10 '12 at 7:40

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.

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In fact, this means that it is true for paralleogram. –  FiniteA Dec 9 '11 at 15:44
    
Does $PO = PA = PB = PC$ have to hold in order for it to look like a square? For example, if you pick any point on the ray PA (P being the end point, direction is towards A), wouldn't it appear to be in the same position as point A itself? –  Sp3000 Dec 10 '11 at 6:18
    
Right, then what it would make something looks like a square? –  FiniteA Dec 10 '11 at 13:54

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