How generate n equidistant points in a n-1 dimensional space As said, i want to build a program to generate n equidistant points in an euclidian space.
From what i know


1d : all couple of points
2d : all equilateral triangles
3d : all equilateral tetrahedra
up to 3d : i suppose it's called an equilateral hypertriangle


So my problem is as follow, in a n-1 euclidian space, giving a defined point build the n-1 other in order to have an equilateral hypertriangle with a distant d between each points.

I suppose that we can start as follow with for example a 3d space.


p1 = (x1,y1,z1) fixed
p2 = (x2,y2,z2)
p3 = (x3,y3,z3)
p4 = (x4,y4,z4)
d


We start to fix p2 knowing d and p1 


$d²=(x_1-x_2)^2 + (y_1-y_2)^2 + (z_1-z_2)^2$


We have 3 variables x2,y2,z2.
We can fix randomly two of them and determine the third without problem.

Then for the second point we have now 2 equations to defined it :


$d²=(x_1-x_3)^2 + (y_1-y_3)^2 + (z_1-z_3)^2$
$d²=(x_2-x_3)^2 + (y_2-y_3)^2 + (z_2-z_3)^2$


As previous, i presume that we can fix 2 variables to determine the third.

For the last point we now have 3 equations which defined it.

So for a n-1 dimensional space we have n-1 equation to defined the last point.

I don't know how to solve this kind of system composed of quadratic equation with one variable and if the process which consist to fix n-1 dimension to determine the last one lead to an equidistant hypertriangle. Moreover it exist maybe others methods with an smaller complexity and easier to implement.

I hope i was clear enough and i thank you for your help.
I assume we are working in $\mathbb{R}^n$.

First of all, observe that one regular $n-$simplex effectively determines all the others. In fact, if $S_1, S_2$ are two sets of points in $\mathbb{R}^n$ that satisfy the regularity condition, then they can be obtained from each other by composing at most an isometry and an homothetic transformation of the affine space (the converse is also true).

Therefore, it is sufficient to construct an unitary $n-$simplex centered in the origin. We visualize each vertex of the simplex $v_1, v_2, \dots v_{n+1}$ as an element of the real $n$-dimensional vector space.

Consider two vertices $v_1, v_2$ of the simplex, let $O$ be the origin and $\pi$ be the plane passing through $v_1, v_2, O$. The angle $\vartheta =\angle v_1 O v_2$ is exactly $\arccos (-1/n)$. To prove that, we observe that:

$$
0 = \left \| \sum_{i} v_i \right \|^2 = (n+1) + 2 \cos\vartheta \binom{n+1}{2} 
$$

We deduce that the following hold:


$\forall i \quad \| v_i \| = 1$
$\forall i \neq j \quad \left \langle v_i, v_j\right \rangle = -1/n$


We may assume without loss of generality that $v_1, v_2$ lie on the same line, that $v_3$ lies on the same plane as the others and so on (in general, that for each $k$, $v_1, v_2, \dots v_k$ lie on the same subspace of $\mathbb{R}^n$ with dimension $k$). Therefore, we may write the vectors by coordinates as follows:

$$
\begin{array}{r c l}
\begin{array}{r}
v_1 \\ v_2 \\ \cdots \\ v_{n+1}
\end{array} &amp;
\begin{array}{r}
= \\ = \\ \\ =
\end{array} &amp;
\begin{array}{c c c c c c}
(x_{1,1} &amp; 0 &amp; 0 &amp; 0 &amp; \dots &amp; 0)^T \\ 
(x_{2,1} &amp; x_{2,2} &amp; 0 &amp; 0 &amp; \dots &amp; 0)^T \\
\\
(x_{n+1, 1} &amp; x_{n+1, 2} &amp; x_{n+1, 3} &amp; x_{n+1, 4} &amp; \dots &amp; x_{n+1, n+1})^T \\ 
\end{array}
\end{array}
$$

The first equation uniquely determines $x_{1,1}$ and the second all the $x_{m,1}$. Now we use again the first to compute $x_{2,2}$ and with the second we determine all the remaining $x_{2, m}$.

Continuing the procedure in a similar fashion computes the coordinate values of all the vertices.