# 2D different integer valued vertices coordinates of cube projection

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?

• It is really a hexagonal parallelogon. Commented Jul 28, 2023 at 1:13
• @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)? Commented Jul 28, 2023 at 1:19
• 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. Commented Jul 28, 2023 at 1:29
• @EthanBolker : very true ! thanks, but what about 8 different vertex projections, I will adjust question and title Commented Jul 28, 2023 at 1:30
• @EthanBolker I added an example below with 8 different integer 2D coordinates. Commented Jul 28, 2023 at 4:13

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.

• To prove impossibility of the orignal example I would start with the method outlined here: math.stackexchange.com/questions/4667386/… Commented Jul 28, 2023 at 21:06
• @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. Commented Jul 28, 2023 at 21:14
• @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? Commented Jul 28, 2023 at 23:26
• @FirstNameLastName Right, the projection would be parallel to the $z$-direction, i.e. to drop the $z$-coordinate and only keep $x,y$ unchanged. Commented Jul 29, 2023 at 0:08
• @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... Commented Jul 29, 2023 at 0:13

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 ...

• 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. Commented Jul 29, 2023 at 1:15
• @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. Commented Jul 29, 2023 at 12:21
• 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. Commented Jul 31, 2023 at 21:30