Angle between $n + 1$ equidistant unit vectors in $\mathbb{R}^n$ My question is basically as posed in the title. Suppose we are given $n + 1$ unit vectors in $\mathbb{R}^n$ so that the angle between any pair of them is the same. What is that angle? I have (unfounded) reason to believe that the angle is arccos$(\frac{-1}{n})$, but I don't know how to prove it. For example, we can find three unit vectors $\begin{pmatrix} 1 \\ 0\end{pmatrix}$, $\begin{pmatrix} \frac{-1}{2} \\ \frac{\sqrt{3}}{2}\end{pmatrix}$, $\begin{pmatrix} \frac{-1}{2} \\ \frac{-\sqrt{3}}{2}\end{pmatrix}$, and the angle between any pair of them is arccos$(\frac{-1}{2})$, since the dot product of any pair of them is $\frac{-1}{2}$. Can anyone supply the reasoning for general $\mathbb{R}^n$, or correct me if I'm incorrect?
 A: Since they're spaced symmetrically,
$$u_1+u_2+u_3+\cdots+u_n+u_{n+1}=0.$$
Now take the dot product with $u_1$:
$$1+u_1\cdot u_2+u_1\cdot u_3+\cdots+u_1\cdot u_n+u_1\cdot u_{n+1}=0.$$
Again by symmetry, these last $n$ dot products should be the same:
$$1+u_1\cdot u_2+u_1\cdot u_2+\cdots+u_1\cdot u_2+u_1\cdot u_2=0$$
$$1+nu_1\cdot u_2=0$$
$$u_1\cdot u_2=-1/n.$$
A: Build an $n\times(n+1)$ matrix $A$ by letting the $k$-th column be your $k$-th vector.
Consider the matrix $B=A^tA$ where $A^t$ is the transpose of $A$. Then $B$ has
size $(n+1)\times(n+1)$ and must be singular, as its rank is at most the rank
of $A$, which is $\le n$.
The $(j,k)$ entry of $B$ is the dot-product of your $j$-th and $k$-th vectors. This
is $1$ when $j=k$ and $x$ when $j\ne k$ where $x$ is the cosine of the angle you
seek. Thus
$$0=\det B=\det\pmatrix{1&x&x&\cdots &x\\
x&1&x&\cdots &x\\
x&x&1&\cdots &x\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
x&x&x&\cdots&1}.$$
It's a routine calcutaion to get
$$\det(B)=(nx+1)(1-x)^n$$
so $x=1$ or $x=-1/n$, and we can rule out $x=1$.
A: To fill a gap in mr_e_man's beautiful argument, we prove the following claim . . .

Claim:$\;$If $v_1,...,v_{n+1}$ are distinct unit vectors in $\mathbb{R}^n$ such that all the dot products $v_i{\,\cdot\,}v_j$ with $i\ne j$ are equal, then 
$$\;v_1+\cdots +v_{n+1}=0$$
Proof:

Let $c$ be the common value of those dot products.

From $v_1\ne v_2$, it follows that $c\ne1$.

Since $v_1,...,v_{n+1}\in\mathbb{R}^n$, it follows that $v_1,...,v_{n+1}$ are linearly dependent.

Thus$\;a_1v_1+\cdots a_{n+1}v_{n+1}\!=0\;$for some $a_1,...,a_{n+1}\in\mathbb{R}$, not all zero.

Without loss of generality, assume $a_1\ne 0$.

Let $A=a_1+\cdots+ a_{n+1}$.

Fix $k\in\{1,...,n+1\}$.
\begin{align*}
\text{Then}\;\;&
a_1v_1+\cdots a_{n+1}v_{n+1}=0\\[4pt]
\implies\;&
v_k{\,\cdot\,}(a_1v_1+\cdots a_{n+1}v_{n+1})=0\\[4pt]
\implies\;&
a_k+(A-a_k)c=0\\[4pt]
\implies\;&
a_k=\frac{Ac}{c-1}\\[4pt]
\implies\;&
a_1=\cdots=a_{n+1}\\[12pt]
\text{hence}\;\;&
a_1v_1+\cdots a_{n+1}v_{n+1}=0\\[4pt]
\implies\;&
a_1(v_1+\cdots +v_{n+1})=0\\[4pt]
\implies\;&
v_1+\cdots +v_{n+1}=0\\[4pt]
\end{align*}
