radius of the inscribed sphere of the pyramid in $\mathbb{R^5}$

Question

How to find the radius of the inscribed sphere of the pyramid in $\mathbb{R^5}$ with the vertex in $(1,0,0,0,0)$, which base is a regular $4$-dimensional simplex, luying in the hyperplane $x_1=0$ with inscribed sphere of radius 1 with center at $(0,0,0,0,0)$?

Answer 1

4-dimensional symplex $\Delta_4$ has five face $\Delta_3$ which is 3-dimensional symplex Hence $$ {\rm vol}\ \Delta_4=5 \cdot{\rm vol}\ \Delta_3\cdot 1 \cdot \frac{1}{4} $$

When $R$-sphere is in pyramid, then faces are 5 faces $f={\rm conv}\ \{v,\Delta_3\}$ and $\Delta_4$ : Here since $ \Delta_3$ has a distance 1 from origin so that ${\rm vol}\ f={\rm vol}\ \Delta_3\cdot \sqrt{2}\cdot \frac{1}{4} $

$$ {\rm vol}\ \Delta_4\cdot 1\cdot \frac{1}{5}= {\rm vol}\ \Delta_5= {\rm vol}\ \Delta_4\cdot R\cdot\frac{1}{5} + {\rm vol}\ f\cdot 5\cdot R \cdot \frac{1}{5} $$

Hence $R =\sqrt{2}-1$