Correction: icosahedral tensegrity structure not icosahedral

In my previous post  Building a tensegrity structure I implied that when the thread was pulled taut an icosahedron would form.  I didn't directly use the word "icosahedron" but I did claim that you needed to allow enough thread for a stick length to triangle edge ratio of $(1+\sqrt 5)/2$, which amounts to the same thing.

I have since been wondering how to prove this is the case, and I have discovered it is not in fact true, which is disappointing.

Let's model the energy as

$$
E = \sum_{(i,j) \in \text{triangle edges}} \frac{1}{2} \lvert\mathbf{x_i} - \mathbf{x_j}\rvert^2
$$

I.e. we imagine that the string is elastic and needs to be stretched to reach from vertex to vertex.

 

In the above diagram I've shown just the connections in which the left vertical stick is involved.  If we take partial derivatives relative to $x_1, y_1, z_1, x_2, y_2, z_2$ we get

$$
\begin{align}
\partial_{x_{1}}E &= 4 x_{1} - (x_{5} + x_{6} + x_{10} + x_{12}) \\
\partial_{x_{2}}E &= 4 x_{2} - (x_{7} + x_{8} + x_{10} + x_{12}) \\
\partial_{y_{1}}E &= 4 y_{1} - (y_{5} + y_{6} + y_{10} + y_{12}) \\
\partial_{y_{2}}E &= 4 y_{2} - (y_{7} + y_{8} + y_{10} + y_{12}) \\
\partial_{z_{1}}E &= 4 z_{1} - (z_{5} + z_{6} + z_{10} + z_{12}) \\
\partial_{z_{2}}E &= 4 z_{2} - (z_{7} + z_{8} + z_{10} + z_{12})
\end{align}
$$

Now, if Energy is stationary in the entire structure it must be the case that a translation of the entire stick by $\epsilon$ should not change the energy, to $\mathcal{O}(\epsilon)$.  Likewise, moving $\mathbf{x_1}$ or $\mathbf{x_2}$ by $\epsilon$ in the x-y plane should have zero effect, to order $\epsilon$.  But it works the other way round too: if we can show that these moves have zero effect on energy (to $\mathcal{O}(\epsilon)$) then energy is stationary (since every change to the $\mathbf{x_i}$, subject to the constraints that the rods are equal length and rigid, can be acheived via these moves).

This leaves us just having to find any $\mathbf{x_i}$ satisfying 

$$
\begin{align}
\partial_{x_{1}}E + \partial_{x_{2}}E &= 0 \\
\partial_{y_{1}}E + \partial_{y_{2}}E &= 0 \\
\partial_{z_{1}}E + \partial_{z_{2}}E &= 0 \\
\partial_{x_{1}}E & = 0 \\
\partial_{y_{1}}E & = 0 \\
\partial_{x_{2}}E & = 0 \\
\partial_{y_{2}}E & = 0 \\
\end{align}
$$

This is a 7x24 matrix equation which will have a lot of solutions as it is degenerate.  But we only need to find one.  And we can cheat by just inserting a guess (this is where maths lecturers would say "by inspection")

$$
\begin{align}
\mathbf{x_{1}} &= (1,0,-2) \\
\mathbf{x_{2}} &= (1,0,2) \\
\mathbf{x_{5}} &= (0,2,-1) \\
\mathbf{x_{6}} &= (0,-2,-1) \\
\mathbf{x_{7}} &= (0,2,1) \\
\mathbf{x_{8}} &= (0,-2,1) \\
\mathbf{x_{10}} &= (2,1,0) \\
\mathbf{x_{12}} &= (2,-1,0)
\end{align}
$$

This is a solution where all the rods are 4 units in length and the parallel pairs are two units apart.  Applying Pythagoras then gives us a triangle side of $\sqrt 6$.  This gives us a ratio of rod length to triangle side of around 1.633, which is a bit over the golden ratio of 1.618 - the shape you'd get if the structure was icosahedral.  This explains why my original construction had a bit more give in it than I'd hoped for after measuring out the string so carefully.