The question is pretty much what the title says, with the additional clarification that I'm not interested in the trivial solution where one rectangle is twice as big and twice as far as the first.

Here is how I have formulated the problem so far:

Say I have two rectangles, $Q_1$ and $Q_2$, whose corners are the columns of the matrices

$$Q_i = \begin{pmatrix} w_i & -w_i & -w_i & w_i\\ h_i & h_i & -h_i & -h_i\\ 0 & 0 & 0 & 0\\ 1 & 1 & 1 & 1 \end{pmatrix}$$

(The rectangles are axis-aligned in the $xy$ plane, expressed in homogeneous coordinates.)

Each rectangle is allowed to move somewhere arbitrary, eg transformed by

$$ T_i = \begin{pmatrix} R_i &t_i \\ \mathbf{0}_{1\times 3} & 1 \end{pmatrix}, \qquad R_i(R_i)^T = I, \quad t_i = (x_i, y_i, z_i)^T$$

The rectangles are then projected onto the plane $z=d$ via $$ P=\begin{pmatrix} d &0 &0 &0\\ 0 &d &0 &0 \\ 0 &0 &1 &0 \end{pmatrix}$$ so that the projected points, in homogeneous coordinates in the 2D projective plane, equal

$$ q_i = P T_i Q_i $$

Finally, my question can be expressed formally as under what circumstances does $q_1^j \sim q_2^j \, \forall j \in \{1..4\}$, where $q_i^j$ is the $j^{th}$ column of matrix $q_i$? In particular, is it possible in this scenario to have $R_1 \ne R_2$, given an appropriate choice of $h_i, w_i, t_i$? (The real problem I'm trying to solve is if I can uniquely recover $R_i$ given $q_i$.)

  • $\begingroup$ I have read that the projection of any quadrilateral can be of exactly two squares and infinitely many rectangles. $\endgroup$ Aug 24, 2017 at 3:40

1 Answer 1


Yes, it's possible for two 3D rectangles to have the same projection.

Suppose the center of projection is $(0,0,0)$, and we're projecting onto the plane $z=1$ using the projection mapping is $p(x,y,z) =(x/z, y/z, 1)$.

For the first rectangle, $R_1$, use the corners $$ A_1 = (1,1,1), \; B_1 = (-1,1,1), \; C_1 = (-1,-1,1), \; D_1 = (1,-1,1) $$

For the second rectangle, $R_2$, use the corners $$ A_2 = (1,1,1), \; B_2 = (-1,1,1), \; C_2 = (-2,-2,2), \; D_2 = (2,-2,2) $$ Then $p(A_2) = p(A_1)$, $p(B_2) = p(B_1)$, $p(C_2) = p(C_1)$, and $p(D_2) = p(D_1)$ -- so the two rectangles have the same projection.

Geometrically, imagine a symmetrical vertical pyramid whose centerline is along the $z$-axis. If you slice this pyramid with any two planes that are parallel to the $x$-axis but not parallel to each other, you'll get two rectangles that have different aspect ratios but the same projection. Here's a picture. The $z$-axis is vertical, and we're looking along the $x$-axis.


You can not recover the 3D rectangle from its projection.

  • $\begingroup$ In your example, $R_2$ is a parallelogram, not a rectangle. $\endgroup$ Sep 23, 2013 at 14:47
  • $\begingroup$ Well, it's an isosceles (symmetric) trapezoid, actually. But, you're right, it's not a rectangle. Sorry about that. I'll rethink. $\endgroup$
    – bubba
    Sep 23, 2013 at 14:55
  • $\begingroup$ Wow, of course it's trapezoid not parallelogram, wtf were my fingers thinking as I typed that... $\endgroup$ Sep 23, 2013 at 14:58
  • $\begingroup$ I would be great to see a corrected version of this answer! $\endgroup$ Aug 24, 2017 at 4:04

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