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On paper we, orthogonal, project a cube as in provided image. But is it always really a cube? And, in this particular example, the 8 different 2D coordinates of the vertices have integer values. But is that even possible? I believe it is not possible (integer valued coordinates for distinct projections), but, is there a, perhaps simple, proof of this general impossibility, or, is there an example of a possibility?

enter image description here

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    $\begingroup$ It is really a hexagonal parallelogon. $\endgroup$
    – peterwhy
    Commented Jul 28, 2023 at 1:13
  • $\begingroup$ @peterwhy : the example is visually not very convincing as example of a valid possibility, but how to prove it is not possible to have integer valued orthogonal projection coordinates (if so)? $\endgroup$ Commented Jul 28, 2023 at 1:19
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    $\begingroup$ If you project parallel to an edge of the cube the image is a square and integer coordinates are possible. I suspect it's impossible otherwise. I hope someone posts a solution. $\endgroup$ Commented Jul 28, 2023 at 1:29
  • $\begingroup$ @EthanBolker : very true ! thanks, but what about 8 different vertex projections, I will adjust question and title $\endgroup$ Commented Jul 28, 2023 at 1:30
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    $\begingroup$ @EthanBolker I added an example below with 8 different integer 2D coordinates. $\endgroup$
    – peterwhy
    Commented Jul 28, 2023 at 4:13

2 Answers 2

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Possible with all $8$ vertices having different integer $(x,y)$ coordinates, for example:

$$\require{cancel} \begin{array}{cccc} & (-9,20,\cancel{12}) & ----- & (11,20,\cancel{27})\\ (-21,5,\cancel{28}) & ----- & (-1,5,\cancel{43}) & \mid\\ \mid & \vdots & \mid & \mid\\ \mid & \vdots & \mid & \mid\\ \mid & \vdots & \mid & \mid\\ \mid & \vdots & \mid & \mid\\ \mid &(0,0,\cancel0) & \cdots\cdots\mid \cdots\cdots& (20,0,\cancel{15})\\ (-12,-15,\cancel{16}) & ----- & (8,-15,\cancel{31})\\ \end{array}$$

This example is based on the $3-4-5$ right-angled triangle, by rotating a cube twice along two axes by $\tan^{-1}(3/4)$. The cube side length is $25$.


To create this type of example, first start with a unit cube with vertices $\mathbf v\in\{0,1\}^3$.

Pick a Pythagorean triple where $a^2+b^2=c^2$, then scale each vertex by $c$ and rotate about the $x$-axis:

$$\mathbf v \mapsto c\begin{pmatrix} 1 & 0 & 0\\ 0 & b/c & -a/c\\ 0 & a/c & b/c \end{pmatrix}\mathbf v = \begin{pmatrix} c & 0 & 0\\ 0 & b & -a\\ 0 & a & b \end{pmatrix}\mathbf v$$

Then pick another (or the same) Pythagorean triple where $d^2+e^2=f^2$, then scale the above result by $f$ and rotate about the $y$-axis:

$$\begin{align*} T(\mathbf v) &= f\begin{pmatrix} e/f & 0 & -d/f\\ 0 & 1 & 0\\ d/f & 0 & e/f \end{pmatrix}\begin{pmatrix} c & 0 & 0\\ 0 & b & -a\\ 0 & a & b \end{pmatrix}\mathbf v\\ &= \begin{pmatrix} e & 0 & -d\\ 0 & f & 0\\ d & 0 & e \end{pmatrix}\begin{pmatrix} c & 0 & 0\\ 0 & b & -a\\ 0 & a & b \end{pmatrix}\mathbf v\\ &= \begin{pmatrix} ce & -ad & -bd\\ 0 & bf & -af\\ cd & ae & be \end{pmatrix}\mathbf v\\ \end{align*}$$

Lastly, project the above result parallel to the $z$-axis, and only keep the $(x,y)$-coordinates unchanged:

$$\begin{align*} P(T(\mathbf v)) &= \begin{pmatrix} 1 & 0&0\\ 0 & 1 & 0 \end{pmatrix}T(\mathbf v)\\ &= \begin{pmatrix} ce & -ad & -bd\\ 0 & bf & -af \end{pmatrix}\mathbf v\\ \end{align*}$$

The example above is from picking $3^2+4^2=5^2$ twice.


For your initial projection, by hand calculation, for now I can only fit a $2\sqrt{10}\times 2\sqrt{10}\times 4\sqrt{3}$ cuboid, i.e. a $\sqrt5 :\sqrt 5 :\sqrt 6$ cuboid.

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  • $\begingroup$ To prove impossibility of the orignal example I would start with the method outlined here: math.stackexchange.com/questions/4667386/… $\endgroup$
    – MathFont
    Commented Jul 28, 2023 at 21:06
  • $\begingroup$ @MathSensei : thanks, and, sure, original example is not possible ... which is why I asked for generalization or possible example. I am sorry if the way I expressed my question was not very clear. $\endgroup$ Commented Jul 28, 2023 at 21:14
  • $\begingroup$ @peterwhy : just to be sure I understand the visual point of view in the construction, one rotates over x and y so one looks orthogonal direction z, right? $\endgroup$ Commented Jul 28, 2023 at 23:26
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    $\begingroup$ @FirstNameLastName Right, the projection would be parallel to the $z$-direction, i.e. to drop the $z$-coordinate and only keep $x,y$ unchanged. $\endgroup$
    – peterwhy
    Commented Jul 29, 2023 at 0:08
  • $\begingroup$ @MathSensei Thanks for the link. There was a deleted answer here that tried to find a cube for the original example, and the answer looks reasonable, but I am not sure why it was deleted by the author... $\endgroup$
    – peterwhy
    Commented Jul 29, 2023 at 0:13
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This is not an answer but just a compliment to @peterwhy's excellent solution and a visual illustration of that solution: a projection of a cube with side length 25.

fwiw: it 'looks' much more like a cube than my example ...

enter image description here

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    $\begingroup$ It may look like a cube, "But is it always really a cube?" has to be false. Say if I scale my example cube non-uniformly along the $z$-direction alone, then I get a parallelepiped with the same projection. $\endgroup$
    – peterwhy
    Commented Jul 29, 2023 at 1:15
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    $\begingroup$ @peterwhy : Right, I will adjust text to not harm the spirit of your answer. So correct property of these drawings is if they can or cannot be a projection of a cube. Your example is extra elegant in that the projection is along one of the cube's directions but such was not even required. $\endgroup$ Commented Jul 29, 2023 at 12:21
  • $\begingroup$ fwiw: I spotted a somewhat loosely related result: sum of the squares of the lengths of the orthogonal projections of the edges of a cube with edge length a to a plane equals 8a^2. $\endgroup$ Commented Jul 31, 2023 at 21:30

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