Projection of tetrahedron to complex plane It is widely known that:
distinct points $a,b,c$ in the complex plane form equilateral triangle iff $ (a+b+c)^{2}=3(a^{2}+b^{2}+c^{2}). $
New to me is this fact:
let $a,b,c,d$ be the images of vertices of regular tetrahedron projected to complex plane, then $(a+b+c+d)^{2}=4(a^{2}+b^{2}+c^{2}+d^{2}).$
I wonder if somebody would came up with intresting proof, maybe involving previous statement.
What I try is some analytic geometry but things get messy enough for me to quit. 
 A: Actually, every harmonic polynomial $P$ in $\mathbb R^3$ of degree not exceeding $2$ has this "mean value property", which remains true for other platonic solids as well. All we need to check for that is that the linear functional $A\mapsto \sum_v(Av,v)$ where $A$ is a symmetric matrix with real entries and the sum is taken over all vertices of a platonic solid centered at the origin is just a multiple of the trace functional. This would immediately follow if we show that the expression $\langle x,y\rangle=\sum_v (x,v)(y,v)$ is a multiple of the usual scalar product $(x,y)$. Now, $\langle x,y\rangle$ is a scalar product and it is invariant under any orthogonal operator $U$ that preserves the platonic solid. Thus, the set of $x$ that maximize $\langle x,x\rangle$ under the condition $|x|=1$ is also invariant under such $U$. This set is just the eigenspace corresponding to the largest eigenvalue of the symmetric matrix $B$ defined by $\langle x,y\rangle=(Bx,y)$. But the group of rotations preserving the platonic solid is rich enough in the sense that it has no non-trivial invariant subspaces (that's all that we need from the automorphism group of a solid to get the MVP for the harmonic polynomials of degree 2, so many semiregular solids will work too), so the eigenspace of $B$ is the full space, which is possible only if $B$ is a multiple of the identity matrix.    
A: Let $z_k$ be the projected vertices. Note that for arbitrary $w\in{\mathbb C}$ one has
$$4\sum_k(z_k+w)^2-\bigl(\sum_k(z_k+w)\bigr)^2 = 4 \sum_k z_k^2 -\bigl(\sum_k z_k\bigr)^2\ ,$$
whence the statement is translation invariant. 
We assume the four vertices $a_k$ of the tetrahedron being four vertices of a unit cube with one vertex at $0$. To be exact:   $a_1$, $a_2$, $a_3$ are the endpoints of (and can be identified with) an orthonormal basis of ${\mathbb R}^3$, and $a_4:=a_1+a_2+a_3$.
For any vector $x\in{\mathbb R}^3$ we get the complex number $z$ belonging to the orthogonal projection of $x$ onto the $(e_1,e_2)$-plane by means of the formula
$$z=  \langle e_1,x\rangle + i\langle e_2,x\rangle\ .$$
As this map ${\mathbb R}^3\to{\mathbb C}$ is linear we have
$$\sum_k z_k = \sum_k' z_k \ + z_4=2 z_4\ ,\qquad(*)$$
where  $'$ denotes summation over $1\leq k\leq 3$ only.
Now comes the decisive point of this setup: We have
$$\eqalign{\sum_k' z_k^2 &=\sum_k' \langle e_1,a_k\rangle^2 - \sum_k'\langle e_2,a_k\rangle^2 + 2i \sum_k' \langle e_1,a_k\rangle  \langle e_2,a_k\rangle \cr &= |e_1|^2 -|e_2|^2 + 2i\langle e_1,e_2\rangle =0\ ,\cr}$$
because the $a_k$ $\ (1\leq k\leq 3)$ form an orthonormal basis of ${\mathbb R}^3$, as do the $e_i$. Using $(*)$ we conclude that
$$4 \sum_k z_k^2 = 4z_4^2 =\bigl(\sum_k z_k\bigr)^2\ .$$
A: As I mentioned in my comment, the tetrahedral formula is invariant under translations, so let's focus on regular tetrahedra conveniently centered at the origin.
Let $T$ be the coordinate matrix such a tetrahedron; that is, the matrix whose columns are coordinates in $\mathbb{R}^3$ of the tetrahedron's vertices. The columns of the matrix obviously sum to zero, but there's something less-obvious that we can say about the rows:
Fact: The rows of $T$ form an orthogonal set of vectors of equal magnitude, $m$.
For example (and proof-of-fact), take the tetrahedron that shares vertices with the double-unit cube, for which $m=2$:
$$T = \begin{bmatrix}1&1&-1&-1\\1&-1&1&-1\\1&-1&-1&1\end{bmatrix} \hspace{0.25in}\text{so that}\hspace{0.25in} T T^\top=\begin{bmatrix}4&0&0\\0&4&0\\0&0&4\end{bmatrix}=m^2 I$$
Any other origin-centered regular tetrahedron is similar to this one, so its coordinate matrix has the form $S = k Q T$ for some orthogonal matrix $Q$ and some scale factor $k$. Then 
$$SS^\top = (kQT)(kQT)^\top = k^2 Q T T^\top Q^\top = k^2 Q (m^2 I) Q^\top = k^2 m^2 (Q Q^\top) = k^2 m^2 I$$
demonstrating that the rows of $S$ are also orthogonal and of equal magnitude. (Fact proven.)
For the general case, take $T$ as follows
$$T=\begin{bmatrix}a_x&b_x&c_x&d_x\\a_y&b_y&c_y&d_y\\a_z&b_z&c_z&d_z\end{bmatrix}$$
Now, consider the matrix $J := \left[1,i,0\right]$. Left-multiplying $T$ by $J$ gives $P$, the coordinate matrix (in $\mathbb{C}$) of the projection of the tetrahedron into the coordinate plane:
$$P := J T = \left[a_x+i a_y, b_x+ib_y, c_x+i c_y, d_x + i d_y\right] = \left[a, b, c, d\right]$$
where $a+b+c+d=0$. Observe that 
$$P P^\top = a^2 + b^2 + c^2 + d^2$$
On the other hand,
$$PP^\top = (JT)(JT)^\top = J T T^\top J^\top = m^2 J J^\top = m^2 (1 + i^2) = 0$$
Therefore, 
$$(a+b+c+d)^2=0=4(a^2 + b^2 + c^2 + d^2)$$

Note: It turns out that the Fact applies to all the Platonic solids ... and most Archimedeans ... and a great many other uniforms, including wildly self-intersecting realizations (even in many-dimensional space). The ones for which the Fact fails have slightly-deformed variants for which the Fact succeeds. (The key is that the coordinate matrices of these figures are (right-)eigenmatrices of the vertex adjacency matrix. That is, $TA=\lambda T$. For the regular tetrahedron, $\lambda=-1$; for the cube, $\lambda = 1$; for the great stellated dodecahedron, $\lambda=-\sqrt{5}$; for the small retrosnub icosicosidodecahedron, $\lambda\approx-2.980$ for a pseudo-classical variant whose pentagrammic faces have non-equilateral triangular neighbors.)
The argument of my answer works for all "Fact-compliant" origin-centered polyhedra, so that $(\sum p_i)^2 = 0 = \sum p_i^2$ for projected vertices $p_i$. Throwing in a coefficient --namely $n$, the number of vertices-- that guarantees translation-invariance, and we have
$$\left( \sum p_i \right)^2 = n \sum p_i^2$$
