<s>[INST] Assume that there are $n+1$ vectors of dimension $n$ and all angles of any two vectors equal, is the cosine of the angle $-\frac{1}{n}$? When $n=2$, the $3$ vectors construct a regular triangle, and the consine of the angle is $-\frac{1}{2}$
When $n=3$, the $4$ vectors construct a regular tetrahedon, and the consine of the angle is $-\frac{1}{3}$
Is it true for any $n$? I guess so, but how to prove?
 [/INST] You can do this without needing any specific assignments of coordinates except for having one node at the origin. This is how you do it.
Let one node be at the origin (obviously). Now, you have a set of $n$ vectors pointing to each of the other nodes. We call these $\mathbf{v}_i$. As all distances between nodes are of the same length, we have that $|\mathbf{v}_i|=|\mathbf{v}_j-\mathbf{v}_i|=C$ for $j\neq i$. Let the centre be $\mathbf{c}=\frac1{n+1}\sum_{i=1}^n \mathbf{v}_i$.
Now, we consider the centre, the origin node, and one other node - without loss of generality, we use $\mathbf{v}_1$. These three points form a triangle, and from the Law of Cosines, we have
$$
\mathbf{c}\cdot(\mathbf{c}-\mathbf{v}_1) = |\mathbf{c}||\mathbf{c}-\mathbf{v}_1|\cos \theta
$$
Now, noting the symmetry of the system, we have
$$
\mathbf{c}\cdot\mathbf{v}_1 = \frac{C^2+(n-1)\mathbf{v_2}\cdot\mathbf{v_1}}{n+1}
$$
Each face forms an equilateral triangle, so $\mathbf{v}_2\cdot\mathbf{v}_1=\frac{C^2}2$. This allows us to write
$$
\mathbf{c}\cdot\mathbf{v}_1 = \frac{C^2}2
$$
Similarly, we have
$$
|\mathbf{c}|^2 = n\frac{C^2+(n-1)\mathbf{v}_2\cdot\mathbf{v}_1}{(n+1)^2}=\frac{nC^2}{2(n+1)}
$$
Substituting these into our equation, and noting that $|\mathbf{c}-\mathbf{v}_1|=|\mathbf{c}|=\frac{C}{\sqrt{2}}$, we have
$$
\cos\theta = \frac{\frac{nC^2}{2(n+1)}-\frac{C^2}2}{\frac{nC^2}{2(n+1)}}=\frac{\frac{n}{n+1}-1}{\frac{n}{n+1}} = \frac{\ \frac{-1}{n+1}\ }{\frac{n}{n+1}} = -\frac1n
$$
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