intersection of 2 simplices

Question

the permuted of the point $(1,2,3)$ is :

$(1,2,3)$, $(3,1,2)$, $(2,3,1)$

now suppose I have 2 simplices in $R^{10}$. the vertices of the first simplex are permuted $(1,1,1,1,1,0,0,0,0,0)$ and thee vertices of the other one is permuted $(3,2,0,0,0,0,0,0,0,0)$. as it is obvious, both of them have the same equation: $$x1+x2+x3+x4+x5+x6+x7+x8+x9+x10=5$$

they have intersection because they have the same barycentric which is: $$d=(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2})$$

my question is that how I can find intersection points of these simplices.

Answer 1

Your mentioned sets aren't simplices at all!
A simplex within 10 dimension would have 11 vertices.
The set of all permutions of $(1,1,1,1,1,0,0,0,0,0)$ has $\frac{10!}{5!\cdot5!}=252$ members.
The set of permutions of $(3,2,0,0,0,0,0,0,0,0)$ has $\frac{10!}{1!\cdot1!\cdot8!}=90$ members.

--- rk

Answer 2

You also can consider the circumradius of the hull polytopes of either permutation set.
$p_1=(1,1,1,1,1,0,0,0,0,0)$ resp. $p_2=(3,2,0,0,0,0,0,0,0,0)$, then
$r_1=|p_1-d|=\frac{\sqrt{10}}{2}=1.581...$
$r_2=|p_2-d|=\frac{\sqrt{42}}{2}=3.240...$

--- rk