Why do sp³ orbitals have tetrahedral symmetry when they are linear combinations of orbitals with octahedral symmetry?

I'm trying to understand how orbitals hybridize. What I understand—or at least think I understand—is the following:


1. The "standard" $\mathrm{s}, \mathrm{p}, \mathrm{d}, …$ orbitals are orthonormal and can therefore be freely linearly combined to form any other solution to the Schrödinger equation.
2. The squares of the coefficients in the linear combination of the wave functions must sum to $1$. (Not the coefficients themselves.)
3. Let's look at $\ce{CH4}$: When the $2\mathrm{s}$ and the three $2\mathrm{p}$ orbitals of the carbon atom hybridize, they might form four $\mathrm{sp}^3$ orbitals. Let $\mathrm{s}, \mathrm{p}\_x, \mathrm{p}\_y, \mathrm{p}\_z$ denote the actual wave functions of the orbitals we started with. Then, one of the resulting hybrid orbitals might have the wave function
$$\frac{1}{2}\mathrm{s} + \frac{1}{2}\mathrm{p}\_x + \frac{1}{2}\mathrm{p}\_y + \frac{1}{2}\mathrm{p}\_z$$
and another one might have
$$\frac{1}{2}\mathrm{s} + \frac{1}{2}\mathrm{p}\_x - \frac{1}{2}\mathrm{p}\_y - \frac{1}{2}\mathrm{p}\_z.$$
There are a total of eight combinations of how the signs of the $\mathrm{p}$ orbitals can be distributed. But two of them are always the negative of each other and form the different halves of basically the same orbital.  

(Sloppy description, I know, but I hope you know what I mean.)


Now, here's my problem:


The four $\mathrm{sp}^3$ orbitals form a tetrahedron. If you told me, there are four equally shaped orbitals and nothing else, I would immediately accept why they would form a tetrahedron. But since we started with the $p$ orbitals—which form a octahedron—I don't see how we arrive there via these linear combinations:


Let's choose the symmetry axes of the $\mathrm{p}$ orbitals as our actual coordinate system. Then, the sum $\mathrm{p}\_x + \mathrm{p}\_y + \mathrm{p}\_z$ should have its symmetry axis pointing in the $\begin{pmatrix}1\\1\\1\end{pmatrix}$ direction. Scaling with $\frac{1}{2}$ doesn't change that, of course, and adding the $\mathrm{s}$ wave function also shouldn't change this symmetry direction, as $\mathrm{s}$ itself is fully rotational symmetric. But since the linear combinations of the $\mathrm{sp}^3$ orbitals only differ by the signs of the $\mathrm{p}$ summands, the symmetry axis of any $\mathrm{sp}^3$ orbital should always points towards a vector of the form $\begin{pmatrix}\pm 1\\\pm 1\\\pm 1\end{pmatrix}$. None of which form a tetrahedron. Can anyone explain to me, where my mistake is?


TL;DR: If linear combinations of $\mathrm{s}$ and $\mathrm{p}$ orbitals work the way I think they work, they should always have octahedral/cubical symmetry. But $\mathrm{sp}^3$ orbitals form a tetrahedron and I don't see how to get there.




Since I understood it myself in the end, here a bit more about my thought process:


1. Of course, a cube contains a tetrahedron! You can find an interactive visualization of it here: <https://cindyjs.org/gallery/main/NestedPolytopes/>  

Concretely, you can choose the corners $$\begin{pmatrix}1\\1\\1\end{pmatrix}, \begin{pmatrix}-1\\-1\\1\end{pmatrix}, \begin{pmatrix}1\\-1\\-1\end{pmatrix}, \begin{pmatrix}-1\\1\\-1\end{pmatrix}.$$  

As set up there, these describe the sign of the $\mathrm{p}$-summands in the wave functions of the $\mathrm{sp}^3$ hybrid orbitals. I.e., the actual wave functions are
\begin{alignat}{3}
\frac{1}{2}\mathrm{s} \enspace &+\enspace \frac{1}{2}\mathrm{p}\_x \enspace &&+\enspace \frac{1}{2}\mathrm{p}\_y \enspace &&+\enspace \frac{1}{2}\mathrm{p}\_z,\\
\frac{1}{2}\mathrm{s} \enspace &-\enspace \frac{1}{2}\mathrm{p}\_x \enspace &&-\enspace \frac{1}{2}\mathrm{p}\_y \enspace &&+\enspace \frac{1}{2}\mathrm{p}\_z,\\
\frac{1}{2}\mathrm{s} \enspace &+\enspace \frac{1}{2}\mathrm{p}\_x \enspace &&-\enspace \frac{1}{2}\mathrm{p}\_y \enspace &&-\enspace \frac{1}{2}\mathrm{p}\_z \qquad\text{and}\\
\frac{1}{2}\mathrm{s} \enspace &-\enspace \frac{1}{2}\mathrm{p}\_x \enspace &&+\enspace \frac{1}{2}\mathrm{p}\_y \enspace &&-\enspace \frac{1}{2}\mathrm{p}\_z,
\end{alignat}
respectively. So, the symmetry axes of the standard orbitals $\mathrm{p}\_x,\mathrm{p}\_y,\mathrm{p}\_z$ have octahedral symmetry; and also an implicit cubical symmetry (since a cube is a dual of an octahedron). This cubical structure becomes visible if you look at all eight possible linear combinations of the $\mathrm{p}$-orbitals, which only differ in the signs of the coefficients.  

And if you choose four of them in the right way, a tetrahedron emerges.
2. My mistake was to not think about it geometrically. I was looking solely at the linear combinations, and my brain wanted to find four of them which look symmetric/cyclic. Somehow I was working under the assumption that if we start with a symmetric configuration and end up with another symmetric configuration, the computations leading from one to the other must also be somehow symmetric. Silly me...


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