# Professor Halfbrain and the tilted cube

When I ran into professor Halfbrain this morning, he told me that he has constructed a cube that can be tilted and balanced on a plane table such that

• exactly one of the eight corners touches the table (and lies in the table plane),
• while the distances of the other seven corners to the table plane are respectively 10, 20, 30, 40, 50, 60, and 70 centimeters.

Question: Does such a cube really exist, or has the professor once again comitted one of his phenomenal mathematical blunders?

Such a cube

does exist

The first thing I noticed that this essential the same question as

Can you rotate a unit cube in such a way that the height to the table of the corners form an arithmetic sequence for some order of the corners?

Because then you can just multiply it's size by something to the required distances.

Now let's take this cube:

for convenience let's define $f(AB)$ as the height difference between corner A and B.
The lines AB, DC, EF and HG are all in parralel so $f(AB)$, $f(DC)$, $f(EF)$ and $f(HG)$ are all the same no matter how you rotate the cube. The same holds true for AD, BC, EH and FG.

Let us then first tilt the cube only a little bit by rotating it around the AD line.

Now we rotate the cube around the line AB in such a way that D goes in between A nd B in terms of height. The situation we now have is that $f(AD)=f(DB)=f(BC)=f(EH)=f(HF)=f(FG)$. Now all we need to get is that this also equals $f(CE)$

$f(CE)$ is determined by the initial tilt we did. We only did a little tilt to show the corner E will end up a lot heigher than corner C. Now let's say our initial tilt was 45 degrees for example. We then see E will be lower then corner C*. By the Intermediate value theorem we can say that at some initial angle $f(CE)$ will hold our desired value so professor Halfbrain is right.

*: sorry no proof but it can be seen very easy when you try it out. I have an actual cube here in front of me that gave a visual confirmation

As a supplement to Ivo Beckers's answer, I provide an explicit construction of Professor Halfbrain's cube:

The cube has edge length $10\sqrt{21}$. The angles between the three bottom edges and the table plane are (colored red, green, and blue, respectively):

$$\color{red}{\sin^{-1}\left(\frac{1}{\sqrt{21}}\right)} \\ \color{lime}{\sin^{-1}\left(\frac{2}{\sqrt{21}}\right)} \\ \color{blue}{\sin^{-1}\left(\frac{4}{\sqrt{21}}\right)}$$

As Ivo conjectures, this arrangement is unique (up to reflection and rotation around the vertical axis). We can prove this by using Euler angles to describe the rotation of the cube. (I will use the z-y-z convention.) We can ignore the first rotation about the z-axis, since it does not affect the angles between the cube and the table surface, only spinning the cube about its vertex. Calling the remaining two angles $\alpha$ and $\beta$, we can write the transformation matrix:

$$\left(\begin{matrix} \cos\alpha\cos\beta & -\cos\alpha\sin\beta &\sin\alpha\\ \sin\beta &\cos\beta & 0\\ -\sin\alpha\cos\beta &\sin\alpha\sin\beta &\cos\alpha\\\end{matrix}\right)$$

and its application to the three unit vectors:

$$\left( \begin{matrix} \cos\alpha\cos\beta\\ \sin\beta\\ -\sin\alpha\cos\beta\\ \end{matrix} \right),\left( \begin{matrix} -\cos\alpha\sin\beta\\ \cos\beta\\ \sin\alpha\sin\beta\\ \end{matrix} \right),\left( \begin{matrix} \sin\alpha\\ 0\\ \cos\alpha\\ \end{matrix} \right)$$

With this we have the height of three vertices of the cube, as well as the heights of the three edges from those vertices to the bottommost vertex. The position of each vertex is a sum of some subset of those edges.

The height of the shortest edge must be $10$; otherwise—as the height of each edge is positive—no vertex could have height $10$. Next, since each of the three different heights may be included at most once, the next edge must have height $20$, otherwise no vertex could have height $20$. Similarly, the third edge must have a height of $40$. This gives us enough to reach all the way to $10+20+40=70$; and, since our choice of heights were fixed at each step, we know the solution must be unique.

Using $s$ as the edge length, we can now set up a system of equations to solve for the Euler angles and size of the cube:

\begin{align} -s\sin\alpha\cos\beta &= 10 \\ s\sin\alpha\sin\beta &= 20 \\ s\cos\alpha &= 40 \end{align}

Immediately we can write $s=40\sec\alpha$ and remove $s$ from the first two equations:

\begin{align} -4 \cos \beta \tan \alpha &= 1 \\ 2\sin \beta \tan \alpha &= 1 \end{align}

Then divide and write:

$$-\frac{1}{2}\tan\beta = 1 \\ \beta = -\tan^{-1} 2$$

Then we can back-substitute and solve for $\alpha$:

$$-\frac{4}{\sqrt{5}}\tan\alpha = 1 \\ \alpha = -\tan^{-1}\left(\frac{\sqrt{5}}{4}\right)=\cos^{-1}\left(\frac{4}{\sqrt{21}}\right)$$

Finally, using $s=40\sec\alpha$ we then obtain:

$$s = 10\sqrt{21}$$

For completeness, here is the resulting rotation matrix:

$$\left( \begin{matrix} \frac{4}{\sqrt{105}} & \frac{8}{\sqrt{105}} & -\sqrt{\frac{5}{21}} \\ -\frac{2}{\sqrt{5}} & \frac{1}{\sqrt{5}} & 0 \\ \frac{1}{\sqrt{21}} & \frac{2}{\sqrt{21}} & \frac{4}{\sqrt{21}} \\ \end{matrix} \right)$$

Yes, it's possible.

Placing the base vertex at the origin, the $8$ vertices are the subset-sums of the $3$ edge vectors leaving the base vertex. If their respective $z$-components can be made to be $1$, $2$, and $4$, then the vertices will have $z$-components $0$ through $7$ by binary encoding.

Take the vector $(1,2,4)$, and complete it to an orthogonal basis with two vectors of the same length as it. Thinking of these as rows of a $3 \times 3$ scaled-orthogonal matrix with $(1,2,4)$ the last row, the columns are also equal-length and orthogonal, and have $z$-coordinates $1,2,4$. These satisfy all the conditions we needed of the base vectors.

An addition to the other answers: I was hoping for a construction where every vertex of the cube had integer, three-dimensional coordinates, but I don't think this is possible. However, if you try to answer the problem in 4-dimensions, not only does such a hypercube exist but you can construct an example with integer coordinates. Think of all possible sums of the vectors:

$$\mathbf{v_1} = (-8, 4,-2, 1) \\ \mathbf{v_2} = (-4, -8, 1, 2) \\ \mathbf{v_4} = (2, -1, -8, 4) \\ \mathbf{v_8} = (1, 2, 4, 8) \\$$ One can check these are all the same length and mutually orthogonal. If you include the empty sum $(0, 0, 0, 0)$, these form the vertices of a hypercube whose last coordinate is all possible values from $0$ to $15$.

Now if you only take the subset-sums of $\{\mathbf{v_1}, \mathbf{v_2},\mathbf{v_4}\}$, you get a cube embedded in four-space. Does this cube satisfy the requirements? Well, not exactly, since trying to project this cube back into three-space results in something that isn't a cube. However, if you go the other way and project the line $L = t (0, 0, 0, 1)$ into the three-dimensional space $S$ that contains this cube to create a line $L'$, then the points of the cube must be evenly spaced if projected onto $L'$. Hence if you take the plane orthogonal to $L'$ in $S$ going through the original and call this your table", it all sort of works up to a scaling factor.

In other words, there is an example with integer coordinates. It just requires four-space, and you lose a nice coordinate representation of the table (and you are also off by a scaling factor).