Well, one thing is special about it, but it takes a while to explain. 


**Please let me know, whether this number occurs in other special occasions as well.**


The explanation: Let $p$ be a complex polynomial of degree $n$ with simple zeros $z\_j$ and simple critical points $z'\_k$ and consider the matrix


\begin{equation}
A\_p \ := \ 
\begin{pmatrix} 
\frac{1}{n} & \ldots & \frac{1}{n} \\
\frac{1}{n} \, \frac{H\_c(z'\_1)}{(z\_1 - z'\_1)^2} & \ldots & 
\frac{1}{n} \, \frac{H\_c(z'\_1)}{(z\_n - z'\_1)^2} \\
\vdots & & \vdots \\
\frac{1}{n} \, \frac{H\_c(z'\_{n-1})}{(z\_1 - z'\_{n-1})^2} & \ldots & 
\frac{1}{n} \, \frac{H\_c(z'\_{n-1})}{(z\_n - z'\_{n-1})^2} \\
\end{pmatrix} \, ,
\end{equation} 
where $H\_c(z'\_k)$ is defined as follows:
$$
H\_c(z'\_k) \ := \ \frac{n}{\sum^{n}\_{j=1} \frac{1}{(z\_j - z'\_k)^2}} \, .
$$


It is a complex relative gain array, so its row and column sums equal 1 and it has eigenvalues 0 and 1, shared with $A^t\_pA\_p$, $A\_pA^t\_p$, $A^\*\_pA\_p$, and $A\_pA^\*\_p$, the former complex symmetric, the latter hermitian. 


Some years ago I proved $A^t\_pA\_p$ also has the eigenvalue $n-2/n$ and the corresponding eigenvector $(z\_1, \ldots, z\_n)^t$, if we assume the centroid of the zeros to be zero itself. $A\_pA^t\_p$ has the corresponding eigenvector $(0, z'\_1, \ldots, z'\_{n-1})^t$, whose centroid is zero too in that case.


If this eigenvalue is simple, then the matrix governs the geometry of the zeros of $p$, as there will be a euclidean transformation/scaling transforming any set of zeros generating $A\_p$ to any other set generating the same $A\_p$, $A\_p$ is invariant concerning these transformations.


Unfortunately, this eigenvalue is not always simple, for example consider $A\_q$ of $q(z) \, = \, z^4 - z$. 


That was my knowledge of this particular situation until some pupils at my school considered rhombi and rectangles, whose vertices were taken as zeros of the considered polynomials, under my guidance. 


We found the following:


Let $r(z) \, = \, (z^2 - r^2)(z^2 + 1)$, where $1 \, < \, r \, < \, \infty$. Then $A\_rA^t\_r \, = \, A\_qA^t\_q$ if and only if $r \, = \, 2 + \sqrt{3}$, corresponding to a double eigenvalue 1/2 indeed.


So there are some different geometric configurations leading to the same matrices but they are very rare. 


Ever since then I have asked myself the question given as headline. I have no idea where to look and would be glad for any suggestions. It might well be just an isolated curiosity, but this is the place to find out, I assume.



What is special about 2 + $\sqrt{3}$?
The fun fact in a comment of @LucGuyot about the fundamental unit of $\mathbb{Q}[\sqrt{3}]$ is on target, and not only for $\mathbb{Q}[\sqrt{3}]$ or units. I hope the empirical facts I give here will be of use to people interested in quadratic number fields. The following relation between the geometric configuration and one of the eigenvalues of $A\_p^\*A\_p$ was discovered empirically in an equivalent form for rectangles by Linda Heinrich, one of my pupils: Let $a/b$ be the ratio of the length of different sides of a rectangle and $a \, > \, b$. Then one eigenvalue $x$ of $A\_p^\*A\_p$ is given by $$ x \ = \ \frac{1}{2} \, \frac{a^2 + b^2}{a^2 - b^2} \, . $$ Based on work of another of my pupils, Jakob Taube, who investigated the rhombi of the question, I found the same formular to apply in that case as well, where the ratio $a/b \, > \, 1$ is between the length of the diagonals then. The double eigenvalue in $A\_p^tA\_p$ only occurs for rhombi: The fourth eigenvalue $y$ of $A\_p^\*A\_p$ is given by $$ y \ = \ \frac{3}{8} \, \left( \frac{2ab}{a^2 - b^2} \right)^2 $$ in the case of rectangles and by $$ y \ = \ \frac{3}{8} \, \left( \frac{a^2 + b^2}{a^2 - b^2} \right)^2 $$ in the case of rhombi. It is the absolute value of the fourth eigenvalue of $A\_p^tA\_p$. For rhombi this eigenvalue is positive as well, but it is negative for rectangles, so $1/2$ always is a simple eigenvalue of $A\_p^tA\_p$ for rectangles. The connection to quadratic number fields is due to the following: Let $e \, = \, c \, + \, d\sqrt{m}$ be an element of the real quadratic number field $\mathbb{Q}[\sqrt{m}]$ such that $|e| \, > \, \sqrt{|N(e)|}$, where $N(e) \, = \, c^2 - md^2$ is the norm of $e$ and put $a \, = \, |e|$ and $b \, = \, \sqrt{|N(e)|}$. Then $c$ and $d$ have the same sign and an elementary computation shows $$ x \ = \ \frac{1}{2} \left(\frac{c}{d \sqrt{m}}\right)^{N(e)/|N(e)|} \, .$$ So I will begin to look into quadratic number fields and the purpose of the question is fulfilled.


