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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.

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Consider a tetrahedron with a face parallel to the complex plane, and its 4th vertex above the origin. Say projections $a$, $b$, $c$ form an equilateral triangle, and $d=0$. The tetrahedral result then follows from the triangular one (noting $a+b+c=0$). Now, convince yourself the tetrahedral formula is invariant under motions of the tetrahedron: for instance, rotations about the altitude through the 4th vertex clearly preserve the result, as do reflections in planes through that altitude, and all translations. What about reflections in arbitrary planes and/or rotations about arbitrary axes? –  Blue Jul 25 '11 at 23:49
    
By the way: Years ago, I attended a talk concerning projections of any Platonic solid into the complex plane. Sadly, I have forgotten any particularly-elegant relationships that may have arisen among projected vertices, though I recall thinking that what I had seen would be helpful in my then-current work (so there had to be something cool there, right?). I don't recall who gave the talk. –  Blue Jul 25 '11 at 23:58
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Forgive me if this is a stupid question, but I'm too lazy to think right now: What kind of projection are we talking about? Stereographic? $(x,y,z) \mapsto x+iy$? –  Hans Lundmark Jul 26 '11 at 8:11
    
@Hans If it doesn't mension a "stereographic" word it is clear to me that it's the latter, usual one. Also the tetrahedron is arbitrally placed in euclidean 3-space, to answer doubts. –  old Jul 26 '11 at 8:24
    
I see. Thank you. –  Hans Lundmark Jul 26 '11 at 14:22

3 Answers 3

up vote 4 down vote accepted

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 $$4 \sum_k z_k^2 = 4z_4^2 =\bigl(\sum_k z_k\bigr)^2\ .$$

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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.

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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

$$\left( \sum p_i \right)^2 = n \sum p_i^2$$

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@Don higher dimensional figures have to make through 3-space anyway being one of those allowed solids don't you think? –  old Jul 28 '11 at 2:38
    
@user13763: I'm afraid I don't understand your question. –  Blue Jul 28 '11 at 23:53

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